business growthBusiness Management Study Manuals Certificate in Business Management INTRODUCTION TO QUANTITATIVE METHODS The Association of Business Executives A B E S T U D Y M A N U A L I N T R O D U C T I O N T O Q U A N T I T A T I V E M E T H O D S The Association of Business Executives 5th Floor, CI Tower • St Georges Square • High Street • New Malden Surrey KT3 4TE • United Kingdom Tel: + 44(0)20 8329 2930 • Fax: + 44(0)20 8329 2945 E-mail:
[email protected] • www.abeuk.com i Certificate in Business Management INTRODUCTION TO QUANTITATIVE METHODS Contents Unit Title Page Introduction to the Manual iii Syllabus v Formulae, rules and tables provided with examination paper vii 1 Basic Numerical Concepts 1 Introduction 3 Number Systems 4 Numbers – Approximation and Integers 5 Arithmetic 7 Dealing with Negative Numbers 11 Fractions 13 Decimals 22 Percentages 26 Ratios 29 Further Key Concepts 31 2 Algebra, Equations and Formulae 39 Introduction 40 Some Introductory Definitions 40 Algebraic Notation 43 Solving Equations 50 Formulae 55 3 Basic Business Applications 63 Introduction 64 Currency and Rates of Exchange 64 Discounts and Commission 68 Simple and Compound Interest 70 Depreciation 76 Pay and Taxation 79 4 Simultaneous, Quadratic, Exponential and Logarithmic Equations 89 Introduction 90 Simultaneous Equations 91 Formulating Problems as Simultaneous Equations 96 Quadratic Equations 98 Logarithmic and Exponential Functions 106 ii Unit Title Page 5 Introduction to Graphs 119 Introduction 120 Basic Principles of Graphs 121 Graphing the Functions of a Variable 124 Graphs of Equations 126 Gradients 141 Using Graphs in Business Problems 147 6 Introduction to Statistics and Data Analysis 171 Introduction 172 Data for Statistics 173 Methods of Collecting Data 175 Classification and Tabulation of Data 177 Frequency Distributions 185 7 Graphical Representation of Information 193 Introduction 194 Graphical Representation of Frequency Distributions 194 Pictograms 204 Pie Charts 206 Bar Charts 207 Scatter Diagrams 209 General Rules for Graphical Presentation 210 8 Measures of Location 221 Introduction 222 The Arithmetic Mean 223 The Mode 232 The Median 236 9 Measures of Dispersion 249 Introduction 250 Range 251 The Quartile Deviation, Deciles and Percentiles 251 Standard Deviation 255 The Coefficient of Variation 262 Skewness 263 10 Probability 275 Introduction 276 Estimating Probabilities 277 Types of Event 279 The Two Laws of Probability 280 Conditional Probability 286 Tree Diagrams 288 Sample Space 294 Expected Value 295 The Normal Distribution 298 iii Introduction to the Study Manual Welcome to this study manual for Introduction to Quantitative Methods. The manual has been specially written to assist you in your studies for the ABE Certificate in Business Management and is designed to meet the learning outcomes specified for this module in the syllabus. As such, it provides a thorough introduction to each subject area and guides you through the various topics which you will need to understand. However, it is not intended to "stand alone" as the only source of information in studying the module, and we set out below some guidance on additional resources which you should use to help in preparing for the examination. The syllabus for the module is set out on the following pages and you should read this carefully so that you understand the scope of the module and what you will be required to know for the examination. Also included in the syllabus are details of the method of assessment – the examination – and the books recommended as additional reading. Following the syllabus, we also set out the additional material which will be supplied with the examination paper for this module. This is provided here for reference only, to help you understand the scope of the syllabus, and you will find the various formulae and rules given there fully explained later in the manual. The main study material then follows in the form of a number of study units as shown in the contents. Each of these units is concerned with one topic area and takes you through all the key elements of that area, step by step. You should work carefully through each study unit in turn, tackling any questions or activities as they occur, and ensuring that you fully understand everything that has been covered before moving on to the next unit. You will also find it very helpful to use the additional reading to develop your understanding of each topic area when you have completed the study unit. Additional resources ABE website – www.abeuk.com. You should ensure that you refer to the Members Area of the website from time to time for advice and guidance on studying and preparing for the examination. We shall be publishing articles which provide general guidance to all students and, where appropriate, also give specific information about particular modules, including updates to the recommended reading and to the study units themselves. Additional reading – It is important you do not rely solely on this manual to gain the information needed for the examination on this module. You should, therefore, study some other books to help develop your understanding of the topics under consideration. The main books recommended to support this manual are included in the syllabus which follows, but you should also refer to the ABE website for further details of additional reading which may be published from time to time. Newspapers – You should get into the habit of reading a good quality newspaper on a regular basis to ensure that you keep up to date with any developments which may be relevant to the subjects in this module. Your college tutor – If you are studying through a college, you should use your tutors to help with any areas of the syllabus with which you are having difficulty. That is what they are there for! Do not be afraid to approach your tutor for this module to seek clarification on any issue, as they will want you to succeed as much as you want to. Your own personal experience – The ABE examinations are not just about learning lots of facts, concepts and ideas from the study manual and other books. They are also about how these are applied in the real world and you should always think how the topics under consideration relate to your own work and to the situation at your own workplace and others with which you are familiar. Using your own experience in this iv way should help to develop your understanding by appreciating the practical application and significance of what you read, and make your studies relevant to your personal development at work. It should also provide you with examples which can be used in your examination answers. And finally … We hope you enjoy your studies and find them useful not just for preparing for the examination, but also in understanding the modern world of business and in developing in your own job. We wish you every success in your studies and in the examination for this module. The Association of Business Executives September 2008 Unit Title: Introduction to Quantitative Methods Unit Code: IQM Level: 3 Learning Hours: 100 Learning Outcomes and Indicative Content: Candidates will be able to: 1. Demonstrate the rules of numeracy 1.1 Apply the four rules to whole numbers, fractions and decimals 1.2 Express numbers in standard form 1.3 Multiply and divide negative numbers 2. Apply calculations 2.1 Compare numbers using ratios, proportions and percentages; 2.2 Obtain values for simple financial transactions involving purchases, wages, taxation, discounts 2 p g purchases, wages, taxation, discounts 2.3 Determine values for simple and compound interest depreciation 2.4 Convert foreign currency 2.5 Make calculations using a scientific calculator including roots and powers; logarithms and exponential values 2.6 Evaluate terms involving a sequence of operations and use of brackets 2.7 Interpret, transpose and evaluate formulae 2.8 Approximate data using rounding, significant figures 3. Use algebraic methods 3.1 Solve linear and simultaneous equations 3.2 Solve quadratic equations using factorisation and formulae 3.3 Solve equations using roots or logarithms 3.4 Determine the equation of a straight line through two points and also when given one point and its gradient 3.5 Determine the gradient and intercepts on the x or y axes for a straight line 4. Construct and use: graphs, charts and diagrams 4.1 Draw charts and diagrams derived from tabular data: eg bar charts, pie charts, scatter diagrams 4.2 Plot graphs, applying general rules and principles of graphical construction including axes, choice of scale and zero 4.3 Plot and interpret mathematical graphs for simple linear, quadratic, exponential and logarithmic equations 4.4 Identify points of importance on graphs eg maximum, minimum and where they cut co-ordinate axes Determine values for simple and compound interest, and for depreciation of an asset using the straight line method and the reducing balance method 2.3 v vi INTRODUCTION TO QUANTITATIVE METHODS FORMULAE FOR BUSINESS MATHEMATICS AND STATISTICS INTEREST The formula for calculating compound interest: where: A = Accrued amount P = Original principal r = Rate of interest (for a particular time period, usually annual) n = Number of time periods. DEPRECIATION ● Straight-line method: Cost of asset Annual depreciation = –––––––––––– Useful life (Cost of asset) – (Value at end of useful life) or Annual depreciation = –––––––––––––––––––––––––––––––––––– Useful life ● Reducing balance method: D = B(1 – i ) n where: D = Depreciated value at the end of the nth time period B = Original value at beginning of time period i = Depreciation rate (as a proportion) n = Number of time periods (normally years) STRAIGHT LINE A linear function is one for which, when the relationship is plotted on a graph, a straight line is obtained. The expression of a linear function, and hence the formula of a straight line, takes the following form: y = mx + c Note that: c = the y intercept (the point where the line crosses the y axis) m = the gradient (or slope) of the line A P r n = + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 100 vii QUADRATIC EQUATION A quadratic equation of the form ax 2 + bx + c = 0 can be solved using the following formula: RULES FOR LOGARITHMS 1. log(p × q) = log p + log q 2. 3. log p n = n log p 4. If y = ax n then n = (log y – log a) ÷ log x PROBABILITY ● Probability rules: Probability limits: 0 ≤ P(A) ≤ 1 Total probability rule: ΣP = 1 (for all outcomes) For complementary events: P(A) + P(A – ) = 1 For two mutually exclusive events: P(A and B) = 0 For independent events: P(A) = P(A| B) and/or P(B) = P(B| A) ● Multiplication rules: For independent events: P(A and B) = P(A)P(B) For dependent events: P(A and B) = P(A)P(B| A) ● Additional rules: For mutually exclusive events: P(A or B) = P(A) + P(B) For non-mutually exclusive events: P(A or B) = P(A) + P(B) – P(A and B) ● Conditional rules: P(A and B) P(A and B) P(A| B) = –––––––––– and P(B| A) = ––––––––– P(B) P(A) Expected value of variables x with associated probabilities P(x) is E(x) = ΣxP(x) log log log p q p q ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = – x b b ac a = ± – – 2 4 2 viii STATISTICAL MEASURES ● Mean for ungrouped data: and where N and x – are the population and sample means respectively. ● Mean for grouped data: μ = Σmf/N and x – = Σmf/n where m is the midpoint and f is the frequency of a class. ● Median for ungrouped data: = Value of the th observation in a ranked data set, where the number of = observations is odd and where n is the number of observations. ● Range = Largest value – Smallest value. ● Standard deviation for ungrouped data: and where σ and s are the population and sample standard deviations respectively. ● Standard deviation for grouped data: and ● Pearson’s measure of skewness: Mean – Mode 3(Mean – Median) Psk = ––––––––––––––– or –––––––––––––––– Standard deviation Standard deviation μ = ∑ x N = ∑ x x n n + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 σ = ∑ ( ) ∑x x N N 2 2 – = ∑ ( ) ∑x x n n 2 2 1 – – s σ = ∑ ( ) ∑m f mf N N 2 2 – = ∑ ( ) ∑m f mf n n 2 2 1 – – s ix STANDARD NORMAL DISTRIBUTION The table of values of the standard normal distribution set out below provides a means of determining the probability of an observation (x ) lying within specified standard deviations (σ) of the mean of the distribution (μ). 0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641 0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247 0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3874 .3936 .3897 .3859 0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483 0.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121 0.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776 0.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451 0.7 .2420 .2389 .2358 .2327 .2296 .2266 .2236 .2206 .2177 .2148 0.8 .2119 .2090 .2061 .2033 .2005 .1977 .1949 .1922 .1894 .1867 0.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .1660 .1635 .1611 1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379 1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170 1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985 1.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823 1.4 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .0708 .0694 .0681 1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559 1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455 1.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367 1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 2.0 .02275 .02222 .02169 .02118 .02068 .02018 .01970 .01923 .01876 .01831 2.1 .01786 .01743 .01700 .01659 .01618 .01578 .01539 .01500 .01463 .01426 2.2 .01390 .01355 .01321 .01287 .01255 .01222 .01191 .01160 .01130 .01101 2.3 .01072 .01044 .01017 .00990 .00964 .00939 .00914 .00889 .00866 .00842 2.4 .00820 .00798 .00776 .00755 .00734 .00714 .00695 .00676 .00657 .00639 2.5 .00621 .00604 .00587 .00570 .00554 .00539 .00523 .00508 .00494 .00480 2.6 .00466 .00453 .00440 .00427 .00415 .00402 .00391 .00379 .00368 .00357 2.7 .00347 .00336 .00326 .00317 .00307 .00298 .00289 .00280 .00272 .00264 2.8 .00256 .00248 .00240 .00233 .00226 .00219 .00212 .00205 .00199 .00193 2.9 .00187 .00181 .00175 .00169 .00164 .00159 .00154 .00149 .00144 .00139 3.0 .00135 .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 AREAS IN TAIL OF THE STANDARD NORMAL DISTRIBUTION x μ (x – μ) –––––- σ x 1 © ABE and RRC Study Unit 1 Basic Numerical Concepts Contents Page Introduction 3 A. Number Systems 4 Denary Number System 4 Roman Number System 4 Binary System 4 B. Numbers – Approximation and Integers 5 Approximation 5 Integers 6 C. Arithmetic 7 Addition (symbol +) 7 Subtraction (symbol ÷) 8 Multiplication (symbol ×) 8 Division (symbol ÷) 9 The Rule of Priority 9 Brackets 10 D. Dealing with Negative Numbers 11 Addition and Subtraction 11 Multiplication and Division 13 E. Fractions 13 Basic Rules for Fractions 14 Adding and Subtracting with Fractions 17 Multiplying and Dividing Fractions 20 F. Decimals 22 Arithmetic with Decimals 23 Limiting the Number of Decimal Places 25 (Continued over) 2 Basic Numerical Concepts © ABE and RRC G. Percentages 26 Arithmetic with Percentages 27 Percentages and Fractions 27 Percentages and Decimals 27 Calculating Percentages 28 H. Ratios 29 I. Further Key Concepts 31 Indices 31 Expressing Numbers in Standard Form 32 Factorials 33 Answers to Questions for Practice 35 Basic Numerical Concepts 3 © ABE and RRC INTRODUCTION The skill of being able to handle numbers well is very important in many jobs. You may often be called upon to handle numbers and express yourself clearly in numerical form – and this is part of effective communication, as well as being essential to understanding and solving many business problems. Sometimes we can express numerical information more clearly in pictures – for example, in charts and diagrams. We shall consider this later in the course. Firstly though, we need to examine the basic processes of manipulating numbers themselves. In this study unit, we shall start by reviewing a number of basic numerical concepts and operations. It is likely that you will be familiar with at least some of these, but it is essential to ensure that you fully understand the basics before moving on to some of the more advanced applications in later units. Throughout the course, as this is a practical subject, there will be plenty of practice questions for you to work through. To get the most out of the material, we recommend strongly that you do attempt all these questions. One initial word about calculators. It is perfectly acceptable to use calculators to perform virtually all arithmetic operations – indeed, it is accepted practice in examinations. You should therefore get to know how to use one quickly and accurately. However, you should also ensure that you know exactly how to perform all the same operations manually. Understanding how the principles work is essential if you are to use the calculator correctly. Objectives When you have completed this study unit you will be able to: identify some different number systems round-up numbers and correct them to significant figures carry out calculations involving the processes of addition, subtraction, multiplication and division manipulate negative numbers cancel fractions down to their simplest terms change improper fractions to mixed numbers or integers, and vice versa add, subtract, multiply and divide fractions and mixed numbers carry out calculations using decimals calculate ratios and percentages, carry out calculations using percentages and divide a given quantity according to a ratio explain the terms index, power, root, reciprocal and factorial. 4 Basic Numerical Concepts © ABE and RRC A. NUMBER SYSTEMS Denary Number System We are going to begin by looking at the number system we use every day. Because ten figures (i.e. symbols 0 to 9) are used, we call this number system the base ten system or denary system. This is illustrated in the following table. Figure 1.1: The denary system Seventh column Sixth column Fifth column Fourth column Third column Second column First column Millions Hundreds of thousands Tens of thousands Thousands Hundreds Tens Units The number of columns indicates the actual amounts involved. Consider the number 289. This means that, in this number, there are: two hundreds (289) eight tens (289) nine units (289). Before we look at other number systems, we can note that it is possible to have negative numbers – numbers which are less than zero. Numbers which are more than zero are positive. We indicate a negative number by a minus sign (÷). A common example of the use of negative numbers is in the measurement of temperature – for example, ÷20°C (i.e. 20°C below zero). We could indicate a positive number by a plus sign (+), but in practice this is not necessary and we adopt the convention that, say, 73 means +73. Roman Number System We have seen that position is very important in the number system that we normally use. This is equally true of the Roman numerical system, but this system uses letters instead of figures. Although we invariably use the normal denary system for calculations, Roman numerals are still in occasional use in, for example, the dates on film productions, tabulation, house numbers, etc. Here are some Roman numbers and their denary equivalent: XIV 14 XXXVI 36 IX 9 LVlll 58 Binary System As we have said, our numbering system is a base ten system. By contrast, computers use base two, or the binary system. This uses just two symbols (figures), 0 and 1. In this system, counting from 0 to 1 uses up all the symbols, so a new column has to be started when counting to 2, and similarly the third column at 4. Basic Numerical Concepts 5 © ABE and RRC Figure 1.2: The binary system Denary Binary 0 0 1 1 2 10 3 11 4 100 5 101 6 110 7 111 8 1000 9 1001 This system is necessary because computers can only easily recognise two figures – zero or one, corresponding to "components" of the computer being either switched off or on. When a number is transferred to the computer, it is translated into the binary system. B. NUMBERS – APPROXIMATION AND INTEGERS Approximation Every figure used in a particular number has a meaning. However, on some occasions, precision is not needed and may even hinder communication. For example, if you are calculating the cost of the journey, you need only express the distance to the nearest mile. It is important to understand the methods of approximation used in number language. (a) Rounding off Suppose you are given the figure 27,836 and asked to quote it to the nearest thousand. Clearly, the figure lies between 27,000 and 28,000. The halfway point between the two is 27,500. The number 27,836 is more than this and so, to the nearest thousand, the answer would be 28,000. There are strict mathematical rules for deciding whether to round up or down to the nearest figure: Look at the digit after the one in the last required column. Then: for figures of four or less, round down for figures of five or more, round up. Thus, when rounding off 236 to the nearest hundred, the digit after the one in the last required column is 3 (the last required column being the hundreds column) and the answer would be to round down to 200. When rounding off 3,455 to the nearest ten, the last required column is the tens column and the digit after that is 5, so we would round up to 3,460. 6 Basic Numerical Concepts © ABE and RRC (However, you should note that different rules may apply in certain situations – for example, in VAT calculations, the Customs and Excise rule is always to round down to the nearest one pence.) (b) Significant figures By significant figures we mean all figures other than zeros at the beginning or end of a number. Thus: 632,000 has three significant figures as the zeros are not significant 000,632 also has only three significant figures 630,002 has six significant figures – the zeros here are significant as they occur in the middle of the number. We can round off a number to a specified number of significant figures. This is likely to be done to aid the process of communication – i.e. to make the number language easier to understand – and is particularly important when we consider decimals later in the unit. Consider the figure 2,017. This has four significant figures. To write it correct to three significant figures, we need to drop the fourth figure – i.e. the 7. We do this by using the rounding off rule – round down from 4 or round up from 5 – so 2,017 correct to three significant figures becomes 2,020. To write 2,017 correct to one significant figure, we consider the second significant figure, the zero, and apply the rule for rounding off, giving 2,000. Do not forget to include the insignificant zeros, so that the digits keep their correct place values. Integers Integers are simply whole numbers. Thus, 0, 1, 2, 3, 4, etc, are integers, as are all the negative whole numbers. The following are not integers: ¾, 1½, 7¼, 0.45, 1.8, 20.25 By adjusting any number to the nearest whole number, we obtain an integer. Thus, a number which is not an integer can be approximated to become one by rounding off. Questions for Practice 1 1. Round off the following the figures to the nearest thousand: (a) 3,746 (b) 8,231 (c) 6,501 2. Round off the following figures to the nearest million: (a) 6,570,820 (b) 8,480,290 (c) 5,236,000 Basic Numerical Concepts 7 © ABE and RRC 3. Correct the following to three significant figures: (a) 2,836 (b) 3,894 (c) 7,886 (d) 9,726 4. How many significant figures are in the following? (a) 1,736 (b) 22,300 (c) 1,901 (d) 26,000 Now check your answers with those given at the end of the unit. C. ARITHMETIC We will now revise some basic arithmetic processes. These may well be familiar to you, and although in practice it is likely you will perform many arithmetic operations using a calculator, it is important that you know how to do the calculations manually and understand all the rules that apply and the terminology used. When carrying out arithmetic operations, remember that the position of each digit in a number is important, so layout and neatness matter. Addition (symbol +) When adding numbers, it is important to add each column correctly. If the numbers are written underneath each other in the correct position, there is less chance of error. So, to add 246, 5,322, 16, 3,100 and 43, the first task is to set the figures out in columns: 246 5,322 16 3,100 43 Total 8,727 The figures in the right hand column (the units column) are added first, then those in the next column to the left and so on. Where the sum of the numbers in a column amount to a figure greater than nine (for example, in the above calculation, the sum of the units column comes to 17), only the last number is inserted into the total and the first number is carried forward to be added into the next column. Here, it is 1 or, more precisely, one ten which is carried forward into the "tens" column. Where you have several columns of numbers to add (as may be the case with certain statistical or financial information), the columns can be cross-checked to produce check totals which verify the accuracy of the addition: 8 Basic Numerical Concepts © ABE and RRC Columns A B C D Totals 246 210 270 23 749 211 203 109 400 923 29 107 13 16 165 300 99 7 3 409 47 102 199 91 439 Rows Totals 833 721 598 533 2,685 Note that each row is added up as well as each column, thus providing a full cross-check. Subtraction (symbol ÷) Again, it is important to set out the figures underneath each other correctly in columns and to work through the calculation, column by column, starting with the units column. So, to subtract 21 from 36, the first task is to set the sum out: 36 ÷21 Total 15 As with addition, you must work column by column, starting from the right – the units column. Now consider this calculation: 34 ÷18 Total 16 Where it is not possible to subtract one number from another (as in the units column where you cannot take 8 from 4), the procedure is to "borrow" one from the next column to the left and add it to the first number (the 4). The one that you borrow is, in fact, one ten – so adding that to the first number makes 14, and subtracting 8 from that allows you to put 6 into the total. Moving on to the next column, it is essential to put back the one borrowed and this is added to the number to be subtracted. Thus, the second column now becomes 3 ÷ (1 + 1), or 3 ÷ 2, which leaves 1. Multiplication (symbol ×) The multiplication of single numbers is relatively easy – for example, 2 × 4 = 8. When larger numbers are involved, it is again very helpful to set out the calculation in columns: 246 × 23 In practice, it is likely that you would use a calculator for this type of "long multiplication", but we shall briefly review the principles of doing it manually. Before doing so though, we shall note the names given to the various elements involved in a multiplication operation: Basic Numerical Concepts 9 © ABE and RRC the multiplicand is the number to be multiplied – in this example, it is 246 the multiplier is the number by which the multiplicand is to be multiplied – here it is 23 and the answer to a multiplication operation is called the product – so to find the product of two or more numbers, you multiply them together. Returning to our example of long multiplication, the first operation is to multiply the multiplicand by the last number of the multiplier (the number of units, which is 3) and enter the product as a subtotal – this shown here as 738. Then we multiply the multiplicand by the next number of the multiplier (the number of tens, which is 2). However, here we have to remember that we are multiplying by tens and so we need to add a zero into the units column of this subtotal before entering the product for this operation (492). Thus, the actual product of 246 × 20 is 4,920. Finally, the subtotals are added to give the product for the whole operation: 246 × 23 Subtotal (units) 738 Subtotal (tens) 4,920 Total product 5,658 Division (symbol ÷) In division, the calculation may be set out in a number of ways: 360 ÷ 24, or 360/24, or most commonly 24 360 Just as there were three names for the various elements involved in a multiplication operation, so there are three names or terms used in division: the dividend is the number to be divided – in this example, it is 360 the divisor is the number by which the dividend is to be divided – here it is 24 and the answer to a division operation is called the quotient – here the quotient for our example is 15. Division involving a single number divisor is relatively easy – for example, 16 ÷ 2 = 8. For larger numbers, the task becomes more complicated and you should use a calculator for all long division. The Rule of Priority Calculations are often more complicated than a simple list of numbers to be added and/or subtracted. They may involve a number of different operations – for example: 8 × 6 ÷ 3 + 2 8 When faced with such a calculation, the rule of priority in arithmetical operations requires that: division and multiplication are carried out before addition and subtraction. 10 Basic Numerical Concepts © ABE and RRC Thus, in the above example, the procedure would be: (a) first the multiplication and division: 8 × 6 = 48; 2 8 = 4 (b) then, putting the results of the first step into the calculation, we can do the addition and subtraction: 48 ÷ 3 + 4 = 49. Brackets Brackets are used extensively in arithmetic, and we shall meet them again when we explore algebra. Look at the previous calculation again. It is not really clear whether we should: (a) multiply the eight by six; or (b) multiply the eight by "six minus three" (i.e. by three); or even (c) multiply the eight by "six minus three plus four" (i.e. by seven). We use brackets to separate particular elements in the calculation so that we know exactly which operations to perform on which numbers. Thus, we could have written the above calculation as follows: (8 × 6) ÷ 3 + 2 8 This makes it perfectly clear what numbers are to be multiplied. Consider another example: 12 ÷ 3 + 3 × 6 + 1 Again there are a number of ways in which this calculation may be interpreted, for example: (12 ÷ 3) + (3 × 6) + 1 12 ÷ (3 + 3) × (6 + 1) (12 ÷ 3) + 3 × (6 + 1) (12 ÷ [3 + 3]) × 6 + 1 Note that we can put brackets within brackets to further separate elements within the calculation and clarify how it is to be performed. The rule of priority explained above now needs to be slightly modified. It must be strictly followed except when brackets are included. In that case, the contents of the brackets are evaluated first. If there are brackets within brackets, then the innermost brackets are evaluated first. Example: 2 + 3 × (6 ÷ 8 ÷ 2) = 2 + 3 × (6 ÷ 4) (within the brackets, the division first) = 2 + 3 × 2 = 2 + 6 (multiplication before addition) = 8 Thus we can see that brackets are important in signifying priority as well as showing how to treat particular operations. Basic Numerical Concepts 11 © ABE and RRC Questions for Practice 2 1. Work out the answers to the following, using a calculator if necessary. (a) 937 × 61 (b) 1,460 × 159 (c) 59,976 ÷ 63 (d) 19,944 ÷ 36 2. Add each of the four columns in the following table and cross-check your answer. A B C D Totals 322 418 42 23 219 16 600 46 169 368 191 328 210 191 100 169 Totals 3. Work out the answers to the following questions. (a) (6 + 3) × 6 + 4 (b) (6 + 3) × (6 + 4) (c) 10 × (5 + 2) ÷ (10 ÷ 5) (d) 24 + 24 ÷ (2 × 2) (e) 6 × (24 ÷ [4 × 2]) (f) (15 ÷ [3 + 2] + 6 + [4 × 3] × 2) ÷ 11 Now check your answers with those given at the end of the unit. D. DEALING WITH NEGATIVE NUMBERS Addition and Subtraction We can use a number scale, like the markings on a ruler, as a simple picture of addition and subtraction. Thus, you can visualise the addition of, say, three and four by finding the position of three on the scale and counting off a further four divisions to the right, as follows: 0 1 2 3 4 5 6 7 8 This brings us to the expected answer of seven (3 + 4). Similarly we can view subtraction as moving to the left on the scale. Thus, 6 minus 4 is represented as follows: 12 Basic Numerical Concepts © ABE and RRC 0 1 2 3 4 5 6 7 8 This illustrates the sum 6 ÷ 4 = 2. Negative numbers can also be represented on the number scale. All we do is extend the scale to the left of zero: ÷6 ÷5 ÷4 ÷3 ÷2 ÷1 0 1 2 3 4 5 6 We can now visualise the addition and subtraction of negative numbers in the same way as above. Consider the sum (÷3) + 4. Start at ÷3 on the scale and count four divisions to the right: ÷4 ÷3 ÷2 ÷1 0 1 2 3 4 Therefore, (÷3) + 4 = 1. (Note that, here, we have shown the figure of "minus 3" in brackets to avoid confusion with the addition and subtraction signs.) The order of the figures of these calculations does not matter, so we can also write: (÷3) + 4 = 4 + (÷3) = 1 We can illustrate the second sum as follows: ÷3 ÷2 ÷1 0 1 2 3 4 5 This is the same as 4 ÷ 3. In effect, therefore, a plus and a minus together make a minus. With subtraction, the process is slightly different. To subtract a minus number, we change the subtraction sign to a plus and treat the number as positive. For example: 4 ÷ (÷3) = 4 + 3 = 7 To sum up: a plus and minus make a minus two minuses make a plus. Consider the following examples. (Draw your own number scales to help visualise the sums if you find it helpful.) 7 ÷ (÷3) = 7 + 3 = 10 2 + (÷6) = 2 ÷ 6 = ÷4 Basic Numerical Concepts 13 © ABE and RRC Multiplication and Division The rules for multiplication and division are as follows: Type of calculation Product/Quotient Example Minus times plus minus ÷3 × 7 = ÷21 Minus times minus plus ÷7 × (÷3) = 21 Minus divide by plus minus ÷21 ÷ 3 = ÷7 Minus divide by minus plus ÷21 ÷ (÷3) = 7 Plus divide by minus minus 21 ÷ (÷3) = ÷7 Questions for Practice 3 Solve the following sums, using a number scale if necessary: 1. 7 + (÷4) 2. (÷4) + 7 3. 8 ÷ (÷6) 4. 6 + (÷8) 5. (÷9) + 3 6. (÷8) ÷ (÷2) 7. (÷5) + (÷4) 8. (÷3) × (÷5) 9. (÷7) ÷ (÷7) 10. 10 × (÷10) Now check your answers with those given at the end of the unit. E. FRACTIONS A fraction is a part of a whole number. When two or more whole numbers are multiplied together, the product is always another whole number. In contrast, the division of one whole number by another does not always result in a whole number. Consider the following two examples: 12 ÷ 5 = 2, with a remainder of 2 10 ÷ 3 = 3, with a remainder of 1 The remainder is not a whole number, but a part of a whole number – so, in each case, the result of the division is a whole number and a fraction of a whole number. The fraction is expressed as the remainder divided by the original divisor and the way in which it is shown, then, is in the same form as a division term: 12 ÷ 5 = 2, with a remainder of 2 parts of 5, or two divided by five, or 5 2 14 Basic Numerical Concepts © ABE and RRC 10 ÷ 3 = 3, with a remainder of 1 part of 3, or one divided by three, or 3 1 Each part of the fraction has a specific terminology: the top number is the numerator and the bottom number is the denominator. Thus, for the above fractions, the numerators are 2 and 1, and the denominators are 5 and 3. When two numbers are divided, there are three possible outcomes. (a) The numerator is larger than the denominator In this case, the fraction is known as an improper fraction. The result of dividing out an improper fraction is a whole number plus a part of one whole, as in the two examples above. Taking two more examples: 9 42 = 4 whole parts and a remainder of 6 parts of 9, or 4 9 6 24 30 = 1 whole part and a remainder of 6 parts of 24, or 1 24 6 (b) The denominator is larger than the numerator In this case, the fraction is known as a proper fraction. The result of dividing out a proper fraction is only a part of one whole and therefore a proper fraction has a value of less than one. Examples of proper fractions are: 5 1 , 3 2 , 7 4 , 5 4 , etc. (c) The numerator is the same as the denominator In this case, the expression is not a true fraction. The result of dividing out is one whole part and no remainder. The answer must be 1. For example: 5 5 = 5 ÷ 5 = 1 Fractions are often thought of as being quite difficult. However, they are not really hard as long as you learn the basic rules about how they can be manipulated. We shall start by looking at these rules before going on to examine their application in performing arithmetic operations on fractions. Basic Rules for Fractions Cancelling down Both the numerator and the denominator in a fraction can be any number. The following are all valid fractions: 5 2 , 50 20 , 93 47 , 9 6 The rule in expressing fractions, though, is always to reduce the fraction to the smallest possible whole numbers for both its numerator and denominator. The resulting fraction is then in its lowest possible terms. Basic Numerical Concepts 15 © ABE and RRC The process of reducing a fraction to its lowest possible terms is called cancelling down. To do this, we follow another simple rule. Divide both the numerator and denominator by the largest whole number which will divide into them exactly. Consider the following fraction from those listed previously: 50 20 Which whole number will divide into both 20 and 50? There are several: 2, 5, and 10. However the rule states that we use the largest whole number, so we should use 10. Dividing both parts of the fraction by 10 reduces it to: 5 2 Note that we have not changed the value of the fraction, just the way in which the parts of the whole are expressed. Thus: 50 20 = 5 2 A fraction which has not been reduced to its lowest possible terms is called a vulgar fraction. Cancelling down can be a tricky process since it is not always clear what the largest whole number to use might be. For example, consider the following fraction: 231 165 In fact, both terms can be divided by 33, but you are very unlikely to see that straight away. However, you would probably quickly see that both are divisible by 3, so you could start the process of cancelling down by dividing by three: 3 231 3 165 ÷ ÷ = 77 55 This gives us a new form of the same fraction, and we can now see that both terms can be further divided – this time by 11: 11 77 11 55 ÷ ÷ = 7 5 So, large vulgar fractions can often be reduced in stages to their lowest possible terms. Changing the denominator to required number The previous process works in reverse. If we wanted to express a given fraction in a different way, we can do that by multiplying both the numerator and denominator by the same number. So, for example, 5 2 could be expressed as 20 8 by multiplying both terms by 4. If we wanted to change the denominator to a specific number – for example, to express ½ in eighths – the rule is to divide the required denominator by the existing denominator (8 ÷ 2 = 4) and then multiply both terms of the original fraction by this result: 4 2 4 1 × × = 8 4 16 Basic Numerical Concepts © ABE and RRC Note again that we have not changed the value of the fraction, just the way in which the parts of the whole are expressed. Changing an improper fraction into a whole or mixed number Suppose we are given the following improper fraction: 8 35 The rule for converting this is to divide the numerator by the denominator and place any remainder over the denominator. Thus we find out how many whole parts there are and how many parts of the whole (in this case, eighths) are left over: 8 35 = 35 ÷ 8 = 4 and a remainder of three = 4 8 3 This result is known as a mixed number – one comprising a whole number and a fraction. Changing a mixed number into an improper fraction This is the exact reverse of the above operation. The rule is to multiply the whole number by the denominator in the fraction, then add the numerator of the fraction to this product and place the sum over the denominator. For example, to write 3 3 2 as an improper fraction: (a) first multiply the whole number (3) by the denominator (3) = 3 × 3 = 9, changing the three whole ones into nine thirds (b) then add the numerator (2) to this = 2 + 9 = 11, thus adding the two thirds to the nine thirds from the first operation (c) finally place the sum over the denominator = 3 11 Questions for Practice 4 1. Reduce the following fractions to their lowest possible terms: (a) 36 9 (b) 64 8 (c) 70 21 (d) 54 18 (e) 49 14 (f) 150 25 (g) 81 36 (h) 135 45 (i) 192 24 (j) 88 33 Basic Numerical Concepts 17 © ABE and RRC 2. Change the following fractions as indicated: (a) 5 1 to 10ths (b) 5 3 to 20ths (c) 9 4 to 81sts (d) 11 9 to 88ths 3. Change the following improper fractions to mixed numbers: (a) 5 12 (b) 8 17 (c) 6 25 (d) 3 31 4. Change the following mixed numbers to improper fractions: (a) 2 3 2 (b) 3 8 3 (c) 4 8 1 (d) 9 4 3 Now check your answers with those given at the end of the unit. Adding and Subtracting with Fractions (a) Proper fractions Adding or subtracting fractions depends upon them having the same denominator. It is not possible to do these operations if the denominators are different. Where the denominators are the same, adding or subtracting is just a simple case of applying the operation to numerators. For example: 7 3 + 7 2 = 7 5 8 5 ÷ 8 1 = 8 4 = 2 1 Where the denominators are not the same, we need to change the form of expression so that they become the same. This process is called finding the common denominator. Consider the following calculation: 5 2 + 10 3 We cannot add these two fractions together as they stand – we have to find the common denominator. In this case, we can change the expression of the first fraction by multiplying both the numerator and denominator by 2, thus making its denominator 10 – the same as the second fraction: 10 4 + 10 3 = 10 7 18 Basic Numerical Concepts © ABE and RRC Not all sums are as simple as this. It is often the case that we have to change the form of expression of all the fractions in a sum in order to give them all the same denominator. Consider the following calculation: 5 3 + 4 1 ÷ 10 1 The lowest common denominator for these three fractions is 20. We must then convert all three to the new denominator by applying the rule explained above for changing a fraction to a required denominator: 5 3 + 4 1 ÷ 10 1 = 20 12 + 20 5 ÷ 20 2 = 20 15 = 4 3 The rule is simple: find the lowest common denominator – and where this is not readily apparent, a common denominator may be found by multiplying the denominators together change all the fractions in the calculation to this common denominator then add (or subtract) the numerators and place the result over the common denominator. (b) Mixed numbers When adding mixed numbers, we deal with the integers and the fractions separately. The procedure is as follows: first add all the integers together then add the fractions together as explained above, and then add together the sum of the integers and the fractions. Consider the following calculation: 3 + 4 4 1 + 5 2 1 + 2 4 3 = (3 + 4 + 5 + 2) + | . | \ | + + 4 3 2 1 4 1 = 14 + | . | \ | + + 4 3 4 2 4 1 = 14 + 4 6 = 14 + 1 2 1 = 15 2 1 In general, the same principles are applied to subtracting mixed numbers. For example: 4 6 5 ÷ 1 2 1 = (4 ÷ 1) + | . | \ | ÷ 2 1 6 5 = 3 + | . | \ | ÷ 6 3 6 5 = 3 + 6 2 = 3 3 1 Basic Numerical Concepts 19 © ABE and RRC However, the procedure is slightly different where the second fraction is larger than the first. Consider this calculation: 4 2 1 ÷ 1 6 5 Here, it is not possible to subtract the fractions even after changing them to the common denominator: 6 3 ÷ 6 5 To complete the calculation we need to take one whole number from the integer in the first mixed number, convert it to a fraction with the same common denominator and add it to the fraction in the mixed number. Thus: 4 2 1 ÷ 3 + | . | \ | + 2 1 1 = 3 + | . | \ | + 6 3 6 6 = 3 6 9 We can now carry out the subtraction as above: 4 2 1 ÷ 1 6 5 = 3 6 9 ÷ 1 6 5 = (3 ÷ 1) + | . | \ | ÷ 6 5 6 9 = 2 + 6 4 = 2 3 2 Questions for Practice 5 Carry out the following calculations: 1. 3 1 + 4 1 2. 6 1 + 10 1 3. 8 1 + 3 2 4. 5 1 + 6 1 5. 5 2 + 4 3 6. 3 2 ÷ 4 1 7. 6 5 ÷ 3 1 8. 8 7 ÷ 5 1 9. 7 5 ÷ 3 1 10. 4 3 ÷ 8 3 11. 2 2 1 + 1 4 3 12. 4 8 3 + 2 4 3 + 1 2 1 13. 3 2 + 3 + 1 2 1 + 2 5 1 14. 1 4 3 ÷ 3 2 15. 4 9 5 ÷ 2 3 1 16. 6 8 1 ÷ 2 3 1 17. 2 8 1 + 2 3 1 ÷ 2 6 1 18. 4 4 3 + 1 3 2 ÷ 2 2 1 Now check your answers with those given at the end of the unit. 20 Basic Numerical Concepts © ABE and RRC Multiplying and Dividing Fractions (a) Multiplying a fraction by a whole number In this case, we simply multiply the numerator by the whole number and place the product (which is now the new numerator) over the old denominator. For example: 3 2 × 5 = 3 5 2× = 3 10 = 3 3 1 4 3 × 7 = 4 7 3 × = 4 21 = 5 4 1 We often see the word "of" in a calculation involving fractions – for example, ¼ of £12,000. This is just another way of expressing multiplication. (b) Dividing a fraction by a whole number In this case, we simply multiply the denominator by the whole number and place the old numerator over the product (which is now the new denominator). For example: 4 3 ÷ 5 = 5 4 3 × = 20 3 3 2 ÷ 5 = 5 3 2 × = 15 2 (c) Multiplying one fraction by another In this case, we multiply the numerators by each other and the denominators by each other, and then reduce the result to the lowest possible terms. For example: 4 3 × 3 2 = 3 4 2 3 × × = 12 6 = 2 1 8 3 × 5 3 = 5 8 3 3 × × = 40 9 (d) Dividing by a fraction In this case, we turn the divisor upside down and multiply. For example: 4 ÷ 4 3 = 4 × 3 4 = 3 16 = 5 3 1 6 1 ÷ 3 1 = 6 1 × 1 3 = 6 3 = 2 1 (e) Dealing with mixed numbers All that we have said so far in this section applies to the multiplication and division of proper and improper fractions. Before we can apply the same methods to mixed numbers, we must complete one further step – convert the mixed number to an improper fraction. For example: 5 2 1 × 11 4 = 2 11 × 11 4 = 22 44 = 2 4 4 3 ÷ 2 1 = 4 19 × 1 2 = 4 38 = 9 2 1 Basic Numerical Concepts 21 © ABE and RRC (f) Cancelling during multiplication We have seen how it is possible to reduce fractions to their lowest possible terms by cancelling down. This involves dividing both the numerator and the denominator by the same number, for example: 8 6 cancels down to 4 3 by dividing both terms by 2. We often show the process of cancelling down during a calculation by crossing out the cancelled terms and inserting the reduced terms as follows: 3 12 2 8 = 3 2 (It is always helpful to set out all the workings throughout a calculation. Some calculations can be very long, involving many steps. If you make a mistake somewhere, it is much easier to identify the error if all the workings are shown. It is also the case that, in an examination, examiners will give "marks for workings", even if you make a mistake in the arithmetic. However, if you do not show all the workings, they will not be able to follow the process and see that the error is purely mathematical, rather than in the steps you have gone through.) Cancellation can also be used in the multiplication of fractions. In this case it can be applied across the multiplication sign – dividing the numerator of one fraction and the denominator of the other by the same number. For example: 4 1 3 × 4 12 1 = 16 1 Remember that every cancelling step must involve a numerator and a denominator. Thus, the following operation would be wrong: 1 4 3 × 3 12 1 = 3 3 Note, too, that you cannot cancel across addition and subtraction signs. However, you can use cancellation during a division calculation at the step when, having turned the divisor upside down, you are multiplying the two fractions. So, repeating the calculation involving mixed numbers from above, we could show it as follows: 4 4 3 ÷ 2 1 = 2 4 19 × 1 1 2 = 2 19 = 9 2 1 22 Basic Numerical Concepts © ABE and RRC Questions for Practice 6 Carry out the following calculations: 1. 7 × 4 3 2. 8 × 3 2 3. 5 × 6 3 4. 6 × 7 3 5. 3 2 ÷ 6 6. 4 3 ÷ 6 7. 8 7 ÷ 3 8. 7 5 ÷ 4 Now check your answers with those given at the end of the unit. F. DECIMALS Decimals are an alternative way of expressing a particular part of a whole. The term "decimal" means "in relation to ten", and decimals are effectively fractions expressed in tenths, hundredths, thousandths, etc. Unlike fractions though, decimals are written completely differently and this makes carrying out arithmetic operations on them much easier. Decimals do not have a numerator or a denominator. Rather, the part of the whole is shown by a number following a decimal point: the first number following the decimal point represents the number of tenths, so 0.1 is one-tenth part of a whole and 0.3 is three-tenths of a whole the next number to the right, if there is one, represents the number of hundredths, so 0.02 is two-hundredths of a whole and 0.67 is six-tenths and seven-hundredths (i.e. 67 hundredths) of a whole the next number to the right, again if there is one, represents the number of thousandths, so 0.005 is five-thousandths of a whole and 0.134 is 134 thousandths of a whole and so on up to as many figures as are necessary. We can see the relationship between decimals and fractions as follows: 0.1 = 10 1 , 0.3 = 10 3 , 1.6 = 1 10 6 0.01 = 100 1 , 0.67 = 100 67 , 3.81 = 3 100 81 0.001 = 1000 1 , 0.054 = 1000 54 , 5.382 = 5 1000 382 We can see that any fraction with a denominator of ten, one hundred, one thousand, ten thousand, etc. can be easily converted into a decimal. However, fractions with a different denominator are also relatively easy to convert by simple division. Basic Numerical Concepts 23 © ABE and RRC Arithmetic with Decimals Performing addition, subtraction, multiplication and division with decimals is exactly the same as carrying out those operations on whole numbers. The one key point to bear in mind is to get the position of the decimal point in the right place. Layout is all important in achieving this. (a) Adding and subtracting To add or subtract decimals, arrange the numbers in columns with the decimal points all being in the same vertical line. Then you can carry out the arithmetic in the same way as for whole numbers. For example, add the following: 3.6 + 7.84 + 9.172 The layout for this is as follows: 3.6 7.84 9.172 Total 20.612 Similarly, subtraction is easy provided that you follow the same rule of ensuring that the decimal points are aligned vertically. For example, subtract 18.857 from 63.7: 63.7 – 18.857 Total 44.843 (b) Multiplying decimals It is easier to use a calculator to multiply decimals, although you should ensure that you take care to get the decimal point in the correct place when both inputting the figures and when reading the answer. The principles of multiplying decimals manually are the same as those considered earlier in the study unit for whole numbers. Again, it is important to get the decimal point in the right place. There is a simple rule for this – the answer should have the same number of decimal places as there were in the question. Consider the calculation of 40 × 0.5. The simplest way of carrying this out is to ignore the decimal points in the first instance, multiply the numbers out as if the figures were whole numbers, and then put back the decimal point in the answer. The trick is to get the position of the decimal point right! So, ignoring the decimal points, we can multiply the numbers as follows: 40 × 5 = 200 Now we need to put the decimal point into the answer. To get it in the right place, we count the number of figures after a decimal point in the original sum – there is just one, in the multiplier (0.5) – and place the decimal point that number of places from the right in the answer: 20.0 or just 20 If the sum had been 0.4 × 0.5, what would the answer be? 24 Basic Numerical Concepts © ABE and RRC The multiplication would be the same, but this time there are 2 figures after decimal points in the original sum. So, we insert the decimal point two places to the right in the answer: 0.2 One final point to note about multiplying decimals is that, to multiply a decimal by ten, you simply move the decimal point one place to the right. For example: 3.62 × 10 = 36.2 Likewise, to multiply by 100, you move the decimal point two places to the right: 3.62 × 100 = 362 This can be a very useful technique, enabling quick manual calculations to be done with ease. (c) Division with decimals Again, it is far easier to use a calculator when dividing with decimals. However, when dividing by ten or by one hundred (or even by a thousand or a million), we can use the technique noted above in respect of multiplication to find the answer quickly without using a calculator: to divide by ten, simply move the decimal point one place to the left – for example, 68.32 ÷ 10 = 6.832 to divide by one hundred, simply move the decimal point two places to the left – for example, 68.32 ÷ 100 = 0.6832. (d) Converting fractions to decimals, and vice versa Fractions other than tenths, hundredths, thousandths, etc. can also be converted into decimals. It is simply a process of dividing the numerator by the denominator and expressing the answer in decimal form. You can do this manually, but in most instances it is easier to use a calculator. For example, to change 5 3 into decimal form, you divide 3 by 5 (i.e. 3 ÷ 5). Using a calculator the answer is easily found: 0.6 However, not all fractions will convert so easily into decimals in this way. Consider the fraction 3 1 . Dividing this out (or using a calculator) gives the answer 0.3333333 or however many 3s for which there is room on the paper or the calculator screen. This type of decimal, where one or more numbers repeat infinitely, is known as a recurring decimal. In most business applications, we do not need such a degree of precision, so we abbreviate the decimal to a certain size – as we shall discuss in the next section. To change a decimal into a fraction, we simply place the decimal over ten or one hundred etc, depending upon the size of the decimal. Thus, the decimal becomes the numerator and the denominator is 10 or 100 etc. For example: 0.36 = 100 36 = 25 9 3.736 = 3 1000 736 = 3 125 92 Basic Numerical Concepts 25 © ABE and RRC Limiting the Number of Decimal Places We noted that certain fractions do not convert into exact decimals. The most common examples are one-third and two-thirds, which convert into 0.3333333... and 0.6666666... respectively. (When referring to these types of recurring decimals, we would normally just quote the decimal as, say, "0.3 recurring".) However, there are also plenty of other fractions which do convert into exact decimals, but only at the level of thousandths or greater. For example: 7 2 = 0.2857142, or 17 11 = 0.6470568 This gives us a problem: how much detail do we want to know? It is very rare that, in business applications, we would want to know an answer precise to the millionth part of a whole, although in engineering, that level of precision may be critical. Such large decimals also present problems in carrying out arithmetic procedures – for example, it is not easy to add the above two decimals together and multiplying them does not bear thinking about. In most business circumstances then, we limit the number of decimal places to a manageable number. This is usually set at two or three decimal places, so we are only interested in the level of detail down to hundredths or thousandths. We then speak of a number or an answer as being "correct to two (or three) decimal places". For decimals with a larger number of decimal places than the required number, we use the principles of rounding discussed earlier in the study unit to reduce them to the correct size. Thus, we look at the digit in the decimal after the one in the last required place and then: for figures of five or more, round up the figure in the last required place for figures of four or less, round down by leaving the figure in the last required place unchanged. So, if we want to express the decimal form of two-sevenths (0.2857142) correct to three decimal places, we look at the digit in the fourth decimal place: 7. Following the rule for rounding, we would then round up the number in the third decimal place to give 0.286. Now consider the decimal forms of one-third and two-thirds. If we were to express these correct to two decimal places, these would be 0.33 and 0.67 respectively. Working to only two or three decimal places greatly simplifies decimals, but you should note that the level of accuracy suffers. For example, multiplying 0.33 by two gives us 0.66, but we have shown two-thirds as being 0.67. These types of small error can increase when adding, subtracting, multiplying or dividing several decimals which have been rounded – something which we shall see later in the course in respect of a set of figures which should add up to 100, but do not. In such cases, we need to draw attention to the problem by stating that "errors are due to rounding". Questions for Practice 7 1. Carry out the following calculations: (a) 3.67 + 4.81 + 3.2 (b) 8.36 + 4.79 + 6.32 (c) 7.07 + 8.08 + 11.263 (d) 11.14 + 8.16 + 3.792 (e) 8.84 ÷ 3.42 (f) 7.36 ÷ 2.182 (g) 10.8 ÷ 4.209 (h) 12.6 ÷ 3.43 26 Basic Numerical Concepts © ABE and RRC (i) 3.62 × 2.8 (j) 4.8 × 3.3 (k) 5.8 × 4.1 (l) 0.63 × 5.2 (m) 154 ÷ 2.2 (n) 2.72 ÷ 1.7 (o) 11.592 ÷ 2.07 (p) 2.6 ÷ 0.052 2. Change the following fractions into decimal form, correct to the number of decimal places (dp) specified: (a) 5 2 to 2 dp (b) 8 3 to 3 dp (c) 4 3 to 2 dp (d) 8 7 to 3 dp (e) 8 5 to 3 dp (f) 6 5 to 3 dp (g) 7 4 to 3 dp (h) 16 5 to 3 dp 3. Change the following decimals into fractions and reduce them to their lowest possible terms: (a) 0.125 (b) 0.60 (c) 1.75 (d) 0.750 Now check your answers with those given at the end of the unit. G. PERCENTAGES A percentage is a means of expressing a fraction in parts of a hundred – the words "per cent" simply mean "per hundred", so when we say "percentage", we mean "out of a hundred". The expression "20 per cent" means 20 out of 100. The per cent sign is %, so 20 per cent is written as 20%. Some examples: 4% means four-hundredths of a whole, or four parts of a hundred = 100 4 28% means twenty-eight-hundredths of a whole, or twenty-eight parts of a hundred = 100 28 100% means one hundred-hundredths of a whole, or a hundred parts of a hundred = 100 100 = 1 Percentages are used extensively in many aspects of business since they are generally more convenient and straightforward to use than fractions and decimals. It is therefore important that you are fully conversant with working with them. Basic Numerical Concepts 27 © ABE and RRC Arithmetic with Percentages Multiplication and division have no meaning when applied to percentages, so we need only be concerned with their addition and subtraction. These processes are quite straightforward, but you need to remember that we can only add or subtract percentages if they are parts of the same whole. Take an example: if Peter spends 40% of his income on rent and 25% on household expenses, what percentage of his income remains? Here both percentages are parts of the same whole – Peter's income – so we can calculate the percentage remaining as follows: 100% ÷ (40% + 25%) = 35% However, consider this example: if Alan eats 20% of his cake and 10% of his pie, what percentage remains? Here the two percentages are not parts of the same whole – one refers to cake and the other to pie. Therefore, it is not possible to add the two together. Percentages and Fractions Since a percentage is a part of one hundred, to change a percentage to a fraction you simply divide it by 100. In effect all this means is that you place the percentage over 100 and then reduce it to its lowest possible terms. For example: 60% = 100 60 = 5 2 24% = 100 24 = 25 6 Changing a fraction into a percentage is the reverse of this process – simply multiply by 100. For example: 8 3 × 100 = 8 300 % = 37½% or 37.5% 3 2 × 100 = 3 200 % = 66.67% As you become familiar with percentages you will get to know certain equivalent fractions well. Here are some of them: 12½% = 8 1 20% = 5 1 25% = 4 1 33.3% = 3 1 50% = 2 1 75% = 4 3 Percentages and Decimals To change a percentage to a decimal, again you divide it by 100. As we saw in the last section, this is easy – you simply move the decimal point two places to the left. Thus: 11.5% = 0.115 Working the conversion the other way is equally easy. To change a decimal to a percentage, move the decimal point two places to the right: 0.36% = 36% 28 Basic Numerical Concepts © ABE and RRC Calculating Percentages To find a particular percentage of a given number, change the percentage rate into a fraction or a decimal and then multiply the given number by that fraction/decimal. For example, to find 15% of £260: £260 × 100 15 = 100 3900 £ = £39 or £260 × 0.15 = £39 To express one number as a percentage of another, we convert the numbers into a fraction with the first number as the numerator and the second the denominator, and then change the fraction into a percentage by multiplying by 100. For example, what percentage of cars have been sold if 27 have gone from a total stock of 300? 300 27 × 100 = 300 2700 = 9% Do not forget that percentages refer to parts of the same whole and therefore can only be calculated where the units of measurement of the quantities concerned are the same. For example, to show 23 pence as a percentage of £2, we need to express both quantities in the same units before working out the percentage. This could be both as pence or both as pounds: 200 23 × 100 = 11.5% or 2 23 . 0 × 100 = 11.5% There are occasions when we deal with percentages greater than 100%. For example, if a woman sells a car for £1,200 and says that she made 20% profit, how much did she buy it for? The cost plus the profit = 120% of cost = £1,200. The cost, therefore, is 100% To work out the cost, we need to multiply the selling price by 120 100 : Cost = 1,200 × 120 100 = £1,000 To prove that this is correct, we work the calculation through the other way. The cost was £1,000 and she sold it at 20% profit, so the selling price is: £1,000 + 20% of £1,000 = £(1,000 + 0.2 × 1,000) = £1,200, as given above. Basic Numerical Concepts 29 © ABE and RRC Questions for Practice 8 1. Change the following fractions into percentages: (a) 100 3 (b) 8 3 2. Change the following percentages into fractions and reduce them to their lowest possible terms (a) 2% (b) 8% (c) 7½% (d) 18% 3. Calculate the following: (a) 15% of £200 (b) 25% of 360° (c) 12½% of 280 bottles (d) 27% of £1,790 (e) 17.5% of £138 4. Calculate the following: (a) 60 as a percentage of 300 (b) 25 as a percentage of 75 (c) £25 as a percentage of £1,000 (d) 30 as a percentage of 20 Now check your answers with those given at the end of the unit. H. RATIOS A ratio is a way of expressing the relationship between two quantities. It is essential that the two quantities are expressed in the same units of measurement – pence, number of people, etc. – or the comparison will not be valid. For example, if we wanted to give the ratio of the width of a table to its depth where the width is 2 m and the depth is 87 cm, we would have to convert the two quantities to the same units before giving the ratio. Expressing both in centimetres would give the ratio of 200 to 87. Note that the ratio itself is not in any particular unit – it just shows the relationship between quantities of the same unit. We use a special symbol (the colon symbol : ) in expressing ratios, so the correct form of showing the above ratio would be: 200: 87 (pronounced "200 to 87") There are two general rules in respect of ratios: They should always be reduced to their lowest possible terms (as with vulgar fractions). To do this, express the ratio as a fraction by putting the first figure over the second and cancel the resulting fraction. Then re-express the ratio in the form of numerator : denominator. 30 Basic Numerical Concepts © ABE and RRC For example, consider the ratio of girls to boys in a group of 85 girls and 17 boys. The ratio would be 85 : 17. This can be reduced as follows: 17 85 = 5 1 The ratio of 85 to 17 is, therefore, 5 : 1. Ratios should always be expressed in whole numbers. There should not be any fractions or decimals in them. Consider the ratio of average miles per gallon for cars with petrol engines to that for cars with diesel engines, where the respective figures are 33½ mpg and 47 mpg. We would not express this as 33½ : 47, but convert the figures to whole numbers by, here, multiplying by 2 to give: 67 : 94 Ratios are used extensively in business to provide information about the way in which one element relates to another. For example, a common way of analysing business performance is to compare profits with sales. If we look at the ratio of profits to sales across two years, we may be able to see if this aspect of business performance is improving, staying the same or deteriorating. (This will be examined elsewhere in your studies.) Another application is to work out quantities according to a particular ratio. For example, ratios are often used to express the relationship between partners in a partnership – sharing profits on a basis of, say, 50 : 50 or 80 : 20. Consider the case of three partners – Ansell, Boddington and Devenish – who share the profits of their partnership in the ratio of 3 : 1 : 5. If the profit for a year is £18,000, how much does each partner receive? To divide a quantity according to a given ratio: first add the terms of the ratio to find the total number of parts then find what fraction each term of the ratio is to the whole, and finally divide the total quantity into parts according to the fractions. For the Ansell, Boddington and Devenish partnership, this would be calculated as follows: The profit is divided into 3 + 1 + 5 = 9 parts. Therefore: Ansell receives 9 3 of the profits Boddington receives 9 1 of the profits and Devenish receives 9 5 of the profits. Therefore: Ansell gets 9 3 × £18,000 = £6,000 Boddington gets 9 1 × £18,000 = £2,000 Devenish gets 9 5 × £18,000 = £10,000 Basic Numerical Concepts 31 © ABE and RRC Questions for Practice 9 1. Express the following sales and profits figures as profit to sales ratios for each firm: Firm Sales Profits £ £ A 100,000 25,000 B 63,000 7,000 C 30,000 10,000 D 75,000 15,000 2. Calculate the distribution of profits for the following partnerships: (a) X, Y and Z share profits in the ratio of 3 : 4: 5. Total profit is £60,000. (b) F, G and H share profits in the ratio of 2 : 3 : 3. Total profit is £7,200. Now check your answers with those given at the end of the unit. I. FURTHER KEY CONCEPTS In this last section we shall briefly introduce a number of concepts which will be used extensively later in the course. Here we concentrate mostly on what these concepts mean, and you will get the opportunity to practice their use in later study units. However, we do include practice in relation to the expression of numbers in standard form. Indices Indices are found when we multiply a number by itself one or more times. The number of times that the multiplication is repeated is indicated by a superscript number to the right of the number being multiplied. Thus: 2 × 2 is written as 2 2 (the little "2" raised up is the superscript number) 2 × 2 × 2 is written as 2 3 2 × 2 × 2 × 2 × 2 × 2 × 2 is written as 2 7 etc. The number being multiplied (here, 2) is termed the base and the superscript number is called the exponent or index of its power. Thus in the case of 2 6 we refer to the term as "two raised to the sixth power" or "two to the power of six". (Strictly, 2 2 should be referred to as "two to the power of two", but we normally call it "2 squared"; similarly, 2 3 is referred to as "2 cubed".) The power of a number is the result of multiplying that number a specified number of times. Thus, 8 is the third power of 2 (2 3 ) and 81 is the fourth power of 3 (3 4 ). 32 Basic Numerical Concepts © ABE and RRC It is quite possible to have a negative exponent, as in the case of 2 ÷3 . A negative exponent indicates a reciprocal. The reciprocal of any number is one divided by that number. For example: the reciprocal 5 is 5 1 and 2 ÷3 represents 3 2 1 ÷ = 8 1 = 0.125 Indices also do not have to be whole numbers – they can be fractional, as in the case of 16 ¼ . This indicates that the base is raised to the power of a quarter. It is more usual though to refer to this as the fourth root of the base and to write it as follows: 4 16 The symbol \ indicates a root. A root is the number that produces a given number when raised to a specified power. Thus, the fourth root of 16 is 2 (2 × 2 × 2 × 2 = 16). A square root of any number is the number which, when multiplied by itself, is equal to the first number. Thus 2 9 = 3 (3 × 3 = 9) The root index (the little 4 or little 2 in the above examples) can be any number. It is, though, customary to omit the 2 from the root sign when referring to the square root. Thus 36 refers to the square root of 36, which is 6. Expressing Numbers in Standard Form The exploration of indices leads us on to this very useful way of expressing numbers. Working with very large (or very small) numbers can be difficult, and it is helpful if we can find a different way of expressing them which will make them easier to deal with. Standard form provides such a way. Consider the number 93,825,000,000,000. It is a very big number and you would probably have great difficulty stating it, let alone multiplying it by 843,605,000,000,000. In fact, can you even tell which is the bigger? Probably not. It is made a little easier by the two numbers being very close to each other on the page, but if one was on a different page, it would be very difficult indeed. However, help is at hand. The method of expressing numbers in standard form reduces the number to a value between 1 and 10, followed by the number of 10s you need to multiply it by in order to make it up to the correct size. The above two numbers, under this system are: 9.3825 × 10 13 and 8.43605 × 10 14 . The 13 and 14 after the 10 are index numbers, as discussed above. Written out in full, the first number would be: 9.3825 × (10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10). Basic Numerical Concepts 33 © ABE and RRC This is not particularly useful, so an easier way of dealing with a number expressed in standard form is to remember that the index number after the 10 tells you how many places the decimal point has to be moved to the right in order to create the full number (remembering also to put in the 0s). So, the second number from above, written out in full would be: 8 . 4 3 6 0 5 0 0 0 0 0 0 0 0 0 = 843,605,000,000,000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Converting numbers in this way makes arithmetic operations much easier – something which we shall return to later in your studies – and also allows us to compare numbers at a glance. Thus, we can see immediately that 8.43605 × 10 14 is larger than 9.3825 × 10 13 . Just as with indices we can also have numbers expressed in standard form using a negative index – for example, 2.43 × 10 ÷2 . Before we consider this in more detail, it would be helpful to recap exactly what the index number means. We know that 10 × 10 = 100, so we can say that 10 2 = 100. We can go on multiplying 10 by itself to get: 10 3 = 1,000 10 4 = 10,000, etc. We also know from the above that, when there is a minus number in the index, it indicates a reciprocal – i.e. one divided by the number as many times as the index specifies. Thus: 10 ÷1 = 10 1 = 0.1. We could go on to say that 10 ÷2 = 0.01 and 10 ÷3 = 0.001, etc. Again, you will note that a minus index number after the 10 tells you how many places the decimal point has to be moved to the left in order to create the full number (remembering also to put in the 0s as well). Now you should be able to work out that 2.43 × 10 ÷2 = 0.0243. Factorials A factorial is the product of all the whole numbers from a given number down to one. For example, 4 factorial is written as 4! (and is sometimes read as "four bang") and is equal to 4 × 3 × 2 × 1. You may encounter factorials in certain types of statistical operations. 34 Basic Numerical Concepts © ABE and RRC Questions for Practice 10 1. Express the following numbers in standard form: (a) 234 (b) 0.045 (c) 6,420 (d) 0.0005032 (e) 845,300,000 (f) 0.00000000000000003792 2. Write out in full the following numbers which are expressed in standard form. (a) 4.68 × 10 2 (b) 9.35 × 10 ÷2 (c) 1.853 × 10 6 (d) 6.02 × 10 ÷5 (e) 7.653 × 10 12 (f) 8.16 × 10 ÷10 Now check your answers with those given at the end of the unit. Basic Numerical Concepts 35 © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. (a) 4,000 (b) 8,000 (c) 7,000 2. (a) 7,000,000 (b) 8,000,000 (c) 5,000,000 3. (a) 2,840 (b) 3,880 (c) 7,890 (d) 9,730 4. (a) Four (b) Three (c) Four (d) Two Questions for Practice 2 1. (a) 57,157 (b) 232,140 (c) 952 (d) 554 2. The check on your answers is achieved by adding a column into which you put the sum of each row. The total of the check column should equal the sum of the four column totals. A B C D Check 322 418 42 23 805 219 16 600 46 881 169 368 191 328 1,056 210 191 100 169 670 Totals 920 993 933 566 3,412 3. (a) (6 + 3) × 6 + 4 = 9 × 6 + 4 = 54 + 4 = 58 (b) (6 + 3) × (6 + 4) = 9 × 10 = 90 (c) 10 × (5 + 2) ÷ (10 ÷ 5) = 10 × 7 ÷ 2 = 70 ÷ 2 = 68 (d) 24 + 24 ÷ (2 × 2) = 24 + 24 ÷ 4 = 24 + 6 = 30 (e) 6 × (24 ÷ [4 × 2]) = 6 × (24 ÷ 8) = 6 × 3 = 18 (f) (15 ÷ [3 + 2] + 6 + [4 × 3] × 2) ÷ 11 = (15 ÷ 5 + 6 + 12 × 2) ÷ 11 = (3 + 6 + 24) ÷ 11 = 33 ÷ 11 = 3 Questions for Practice 3 1. 3 2. 3 3. 14 4. ÷2 5. ÷6 6. ÷6 7. ÷9 8. 15 9. 1 10. ÷100 36 Basic Numerical Concepts © ABE and RRC Questions for Practice 4 1. (a) 4 1 (b) 8 1 (c) 10 3 (d) 3 1 (e) 7 2 (f) 6 1 (g) 9 4 (h) 3 1 (i) 8 1 (j) 8 3 2. (a) 10 2 (b) 20 12 (c) 81 36 (d) 88 72 3. (a) 2 5 2 (b) 2 8 1 (c) 4 6 1 (d) 10 3 1 4. (a) 3 8 (b) 8 27 (c) 8 33 (d) 4 39 Questions for Practice 5 1. 12 7 2. 15 4 3. 24 19 4. 30 11 5. 1 20 3 6. 12 5 7. 2 1 8. 40 27 9. 21 8 10. 8 3 11. 4 4 1 12. 8 8 5 13. 7 30 11 14. 1 12 1 Basic Numerical Concepts 37 © ABE and RRC 15. 2 9 2 16. 3 24 19 17. 2 24 7 18. 3 12 11 Questions for Practice 6 1. 4 21 = 5 4 1 2. 3 16 = 5 3 1 3. 6 15 = 2 2 1 4. 7 18 = 2 7 4 5. 18 2 = 9 1 6. 24 3 = 8 1 7. 24 7 8. 28 5 Questions for Practice 7 1. (a) 11.68 (b) 19.47 (c) 26.413 (d) 23.092 (e) 5.42 (f) 5.178 (g) 6.591 (h) 9.17 (i) 10.136 (j) 15.84 (k) 23.78 (l) 3.276 (m) 70 (n) 1.6 (o) 5.6 (p) 50 2. (a) 0.40 (b) 0.375 (c) 0.75 (d) 0.875 (e) 0.625 (f) 0.833 (g) 0.571 (h) 0.313 3. (a) 1000 125 = 8 1 (b) 100 60 = 5 3 (c) 100 175 = 1 4 3 (d) 1000 750 = 4 3 38 Basic Numerical Concepts © ABE and RRC Questions for Practice 8 1. (a) 3% (b) 37.5% 2. (a) 50 1 (b) 25 2 (c) 40 3 (d) 50 9 3. (a) £30 (b) 90° (c) 35 bottles (d) £483.30 (e) £24.15 4. (a) 20% (b) 33.33% (c) 2.5% (d) 150% Questions for Practice 9 1. Firm A: 1 : 4 Firm B: 1 : 9 Firm C: 1 : 3 Firm D: 1 : 5 2. (a) X: £15,000 Y: £20,000 Z: £25,000 (b) F: £1,800 G: £2,700 H: £2,700 Questions for Practice 10 1. (a) 2.34 × 10 2 (b) 4.5 × 10 ÷2 (c) 6.42 × 10 3 (d) 5.032 × 10 ÷4 (e) 8.453 × 10 8 (f) 3.792 × 10 ÷15 2. (a) 468 (b) 0.0935 (c) 1,853,000 (d) 0.0000602 (e) 7,653,000,000,000 (f) 0.000000000816 39 © ABE and RRC Study Unit 2 Algebra, Equations and Formulae Contents Page Introduction 40 A. Some Introductory Definitions 40 Algebra 40 Equations 42 Formulae 42 B. Algebraic Notation 43 Addition and Subtraction 43 Multiplication 44 Division 45 Indices, Powers and Roots 45 Brackets in Algebra 48 C. Solving Equations 50 Equality of Treatment to Both Sides of the Equation 50 Transposition 52 Equations with the Unknown Quantity on Both Sides 54 D. Formulae 55 Formulating a Problem 55 Constants and Variables 56 Substitution 56 Rearranging Formulae 57 Answers to Questions 59 40 Algebra, Equations and Formulae © ABE and RRC INTRODUCTION In this study unit we continue our review of basic concepts by looking at algebra. Algebra essentially involves the substitution of letters for numbers in calculations, so that we can establish rules and procedures for carrying out mathematical operations which can be applied whatever the actual numbers involved. In this sense then, algebra enables general expressions to be developed and general results to be obtained. We shall then go on to introduce the subject of equations. An equation is simply a statement of equality between two mathematical expressions, and is generally used to enable one or more unknown quantities to be worked out. There are various rules and procedures which can be applied to equations to help this process, and these will be examined in full. Here we shall be concerned with "simple" equations, but later in the course we shall consider more complex forms. However, the same basic rules are applied, so it is important to understand these at the outset. Finally, we look briefly at formulae. A formula is one particular form of generalised mathematical statement, usually expressed in the form of an equation, that sets out a rule which may be applied in particular situations. It enables us to work out an unknown quantity or value provided we know certain specific quantities or values. There are formulae which describe a great many physical situations or mathematical problems and you will encounter these throughout your studies, both in this course and many others. Objectives When you have completed this study unit you will be able to: outline the basic principles of algebra apply the basic arithmetic operations of addition, subtraction, multiplication and division to algebraic notation, and simplify algebraic expressions by the process of collecting like terms define equations and formulae and outline their uses find unknown quantities and values from simple equations by using transposition outline the principles of formulae and how they are constructed rearrange the terms in a formula to isolate different unknowns. A. SOME INTRODUCTORY DEFINITIONS Algebra Algebra is a branch of mathematics in which, instead of using numbers, we use letters to represent numbers. We all know that 2 + 3 = 5. Suppose, though, that we substitute letters for the first two numbers, so that: 2 = a 3 = b We can then write: a + b = 5 All that has happened is that we have replaced the numbers with letters. However, a number is a specific quantity – for example, 5 is more than 4, but less than 6 – whereas a letter can be used to represent any number. Thus, in the previous expression, "a" could be 4 and "b" Algebra, Equations and Formulae 41 © ABE and RRC could be 1. We only know that they are 2 and 3 respectively because we defined them as such before. The main consequence of this is that algebra uses general expressions and gives general results, whereas arithmetic (using numbers) uses definite numbers and gives definite results. Arithmetic is specific, whereas algebra is general. Let us consider some examples to illustrate this further. Example 1: Suppose you have a piece of wood which is 7 metres long and from it you wish to cut a piece 4 metres long. The length of the remaining piece is 3 m, calculated as follows: 7 ÷ 4 = 3 This is a specific arithmetic statement relating to cutting a specific amount from this particular piece of wood. We could translate this into an algebraic expression by substituting letters for the specific lengths: let the original length of the piece of wood = x metres and the length of the piece cut off = y metres. The calculation can now be shown as: x ÷ y = 3 This is now a general statement for cutting one length of wood from another to leave a piece 3 metres in length. Example 2: To find the area of a floor measuring 10 m long and 9 m wide, we multiply one dimension by the other: area = 10 × 9 square metres = 90 sq metres. Substituting the letters "l" and "w" to represent the actual length and width, we can reformulate the expression as: area = l × w Again, this is a general expression which can be made specific by putting in particular values for "l" and "w". Example 3: The distance travelled by a train in 3 hours at a speed of 60 miles per hour is easily calculated as: 3 × 30 = 180 miles. Here again, letters may be substituted to give us a general expression: s × t = d where: speed = s mph time = t hours distance travelled = d miles. 42 Algebra, Equations and Formulae © ABE and RRC Thus, using algebra – working with letters instead of numbers – allows us to construct general mathematical expressions. This is not particularly helpful in itself, but is very important when we come to consider equations. Equations An equation is simply a mathematical statement that one expression is equal to another. So, for example, the statement that "2 + 2 = 4" is an equation. Note that an equation has two sides. Here, they are "2 + 2" and "4". The two sides must always be in equality for the statement to be an equation. If we now introduce the concept of algebra into this, we have the makings of an extremely useful mathematical tool. For example, given that a certain number multiplied by 3 is 6, we can write this as: 3 × the required number = 6. To save writing "the required number" (or the " unknown" number), it is more convenient to call it "x", so that we can then write: 3 × x = 6 Here are a few simple statements which you can easily write as equations for practice: (a) A certain number is added to 4 and the result is 20. (b) A certain number is multiplied by 4 and the result is 20. (c) If 4 is taken from a certain number the result is 5. (d) If a certain number is divided by 3 the result is 1. The correct equations are: (a) x + 4 = 20 (b) x × 4 = 20 (c) x ÷ 4 = 5 (d) x ÷ 3 = 1 The point of constructing equations in this way is that it provides a means by which you can work out the unknown value (here, "x"). These examples are very simple and you can probably solve them (i.e. work out the unknown) very easily. However, equations can get very complicated. Don't worry, though – there are a number of simple rules which can be used to help solve them. Formulae A mathematical formula (plural "formulae") is a special type of equation which can be used for solving a particular problem. In fact, we have already introduced two formulae in the preceding section on algebra: area = length × width distance travelled = speed × time. The essence of these two statements is that they are always true, whatever the actual values of length, width, speed etc. are. Thus, a formula is an equation which always applies to a particular mathematical problem, whatever the actual values. It provides a set of rules which can be used in a particular situation in order to solve the problem. Algebra, Equations and Formulae 43 © ABE and RRC If we know the actual values, we could put them into the equation, but if not we would have to treat them as unknown. Thus, we can formulate a general statement using algebra as follows: A = l × w where: A = area l = length w = width. d = s × t where: speed = s mph time = t hours distance travelled = d miles. Having a formula for a particular problem set out as a general equation using algebra enables us to work out the answer by substituting actual values back into it for the unknowns. For example, if we know we have been driving for 2 hours at an average speed of 60 mph, we can easily work out the distance travelled by substituting 60 for "s" and 2 for "t" in the above formula: d = s × t = 60 × 2 = 120 miles. There are many formulae used in mathematics and other physical sciences such as chemistry or physics. You will meet a good many in your studies – principally in this subject and in financial subjects. Again, do not worry too much about them. They are usually quite straightforward and, by using the general rules governing equations, can easily be used to solve the problem faced. B. ALGEBRAIC NOTATION As algebraic letters simply represent numbers, the operations of addition, subtraction, multiplication and division are still applicable in the same way. However, in algebra it is not always necessary to write the multiplication sign. So, instead of "a × b", we would write simply "ab" (or sometimes "a.b", using the full stop to represent multiplication). Addition and Subtraction Addition and subtraction follow the same rules as in arithmetic. Thus, just as: 3 + 3 + 3 + 3 + 3 + 3 = 3 × 6 so y + y + y + y + y + y = y × 6, or 6y You should note that this holds good only for like terms. Like terms are those of the same symbol or value – for example, all the terms to be added above were "y". 44 Algebra, Equations and Formulae © ABE and RRC When unlike terms are to be added, we can simplify an arithmetical statement, but not an algebraic statement. For example: 3 + 6 + 7 = 16 y + x + z = y + x + z The three numbers, when added, can be reduced to a simpler form, i.e. 16. However, there is not a simpler means of expressing "x + y + 3". We can, though, simplify expressions when they involve like terms. Just as the result of adding two numbers does not depend upon their order, so the result of addition in algebra is not affected by order. Thus: 2 + 7 = 7 + 2 = 9 and x + y + 3 = 3 + y + x = y + 3 + x, etc. In simplifying an algebraic expression, we can collect like terms together whenever possible. For example: 5x + 2y ÷ x + 3y + 2 = (5x ÷ x) + (2y + 3y) + 2 = 4x + 5y + 2 We have collected together all the "x"s (5x ÷ x) and all the "y"s (2y + 3y), but cannot do anything with the "2" which is unlike the other terms. This distinction between like and unlike terms, and the way in which they can be treated, is an important point. Multiplication As we have seen, the product of a number and a letter, for example 3 × a, may be written as 3a. Always place the number before the letter, thus 3a, not a3. Since multiplication by 1 does not alter the multiplicand, the expression 1a is not used; you just write a. To multiply unlike terms, we simply write them together with no sign. Thus: a × b × c × d = abcd If numerical factors are involved then they are multiplied together and placed in front of the simplified term for the letters: 3a × 2b × 4c × d = 24abcd That this is correct can be shown by writing it in full and changing the order of multiplication: 3 × a × 2 × b × 4 × c × d = 3 × 2 × 4 × a × b × c × d = 24abcd Multiplying like terms (for example, x × x) gives a result of the term being raised to the power of the multiplier (here, x 2 ). We shall consider this in more detail later. Algebra, Equations and Formulae 45 © ABE and RRC Division Division should present little difficulty if we follow the rules used in arithmetic. Consequently, just as we write: 2 4 for 4 ÷ 2 so we can write: y x for x ÷ y Remember that any number goes into itself once. This will prevent you from saying that 6 ÷ 6 = 0 instead of 6 ÷ 6 = 1. In the same way x ÷ x = 1. It is then easy to cancel algebraic fractions. For example: 4x 2xy can be cancelled by dividing both the numerator and denominator by x to give 4 2y 4 2y can be further cancelled by dividing both the numerator and denominator by 2 to give 2 y or we could simply say we can cancel 4x 2xy by dividing both parts by 2x. We can always cancel like terms in the numerator and denominator. Indices, Powers and Roots You will remember that indices are found when we multiply the same number by itself several times. The same rules apply in algebra, but we can investigate certain aspects further by using algebraic notation. To recap: x × x is written as x 2 (referred to as "x squared") x × x × x is written as x 3 (referred to as "x cubed") x × x × x × x is written as x 4 (referred to as "x to the power four" or "x raised to the fourth power") and so on. In, for example, x 5 , x is termed the base and 5 is called the exponent or index of the power. Note two particular points before we go on: x = x 1 x 0 = 1 (a) Multiplication of indices You will note that, for example: x 2 × x 4 = (x × x) × (x × x × x × x) = x × x × x × x × x × x = x 6 = x 2 + 4 46 Algebra, Equations and Formulae © ABE and RRC and x 3 × x 5 = (x × x × x) × (x × x × x × x × x) = x × x × x × x × x × x × x × x = x 8 = x 3 + 5 There is in fact a general rule that, when we have two expressions with the same base, multiplication is achieved by adding the indices. Thus: x m × x n = x m + n (Rule 1) (b) Division of indices When we divide expressions with the same base, we find we can achieve this by subtracting the indices. For example: 3 6 x x = x x x x x x x x x × × × × × × × = x × x × x = x 3 = x 6 ÷ 3 and x x 2 = x x x × = x = x 1 = x 2 ÷ 1 The general rule is: n m x x = x m ÷ n (Rule 2) (c) Showing reciprocals In Rule 2 above, let us put m = 0, so that the expression becomes : n 0 x x = x 0 ÷ n We know that x 0 = 1, so n x 1 = x 0 ÷ n = x ÷ n Thus, negatives denote reciprocals. (d) Raising one power by another To work out an expression such as (x 2 ) 3 – i.e. x squared all cubed – we have x 2 × x 2 × x 2 as it is the cube of x 2 . Algebra, Equations and Formulae 47 © ABE and RRC Thus (x 2 ) 3 = x 2 × x 2 × x 2 = x × x × x × x × x × x = x 6 = x 2 × 3 In general, we have (x m ) n = x mn (Rule 3) (e) Roots Indices do not have to be whole numbers. They can be fractional. Let us consider x ½ (i.e. x raised to the power half). We know from Rule (3) that: (x ½ ) 2 = x ½ × 2 = x Thus, x ½ when multiplied by itself gives x. However, this is exactly the property that defines the square root of x. Therefore, x ½ is the square root of x which you will sometimes see denoted as: 2 x or, more commonly, as . x Therefore, x ½ ÷ . x (The symbol "÷" means "identical with".) Similarly, ( 3 1 x ) 3 = x. Thus, 3 1 x is the cube root of x which is denoted by the symbol 3 x . Similarly, x ¼ is the fourth root of x denoted by 4 x . Thus, in general, we can say that: n 1 x is the n th root of x or n x . Note that in all these expressions the number to the left of the root sign – for example the 3 in 3 x – is written so that it is level with the top of the root sign. A number written level with the foot of the root sign – for example, x 3 – would imply a multiple of the square root of x – in this example, 3 times the square root of x. We shall often need to use square roots in later work, particularly in statistics and in stock control. (f) Roots of powers There is one more type of index you may meet – an expression such as: or 3 5 x , or in general n m x . To interpret these, remember that the numerator of a fractional index denotes a power and that the denominator denotes a root. Thus, 2 3 x is (x 3 ) ½ , i.e. the square root of (x cubed) or 2 3 x . Alternatively, it is (x ½ ) 3 , i.e. the cube of (the square root of x) or ( x ) 3 . Let us consider its value when x = 4: 2 3 4 = 2 3 4 = 4 4 4 × × = 64 = 8 (since 8 × 8 = 64) or 2 3 4 = ( ) 4 3 = 2 3 = 2 × 2 × 2 = 8 Therefore it does not matter in which order you perform the two operations, and we have: n m x = n m x = ( n x ) m 48 Algebra, Equations and Formulae © ABE and RRC (g) Collecting like terms We saw earlier how we can collect like items. Let us look at a couple of examples where indices are involved. It is no more complicated, even though it may look that way. a 2 + 2a 2 = a 2 + a 2 + a 2 = 3a 2 These are like terms and can be added. b 2 + 2b 2 + 2b 3 = b 2 + b 2 + b 2 + b 3 + b 3 = 3b 2 + 2b 3 Note that b 2 and b 3 are not like terms, so they cannot be combined. Brackets in Algebra The rules for brackets are exactly the same in algebra as in arithmetic. However, if we have unlike terms inside the brackets, it is not possible to collect them together before removing the brackets. In such cases, we need to be very careful when removing them and always obey the following three rules If the bracket has a term (a number or a letter) in front of it, everything within the bracket must be multiplied by that term. If a bracket has a + sign before it, or before the term by which the terms inside the brackets are to be multiplied, the signs + or ÷ within the brackets remain unchanged. If a bracket has a ÷ sign before it, or before the term by which the terms inside the brackets are to be multiplied, the signs + or ÷ within the brackets are changed so that + becomes ÷ and vice versa. These rules need a little explanation, so let us consider some examples. Example 1: 4x + 2(x ÷ y) The terms inside the brackets are unlike and cannot be further collected. All the terms within the brackets must be multiplied by the term outside before removing the brackets, so this gives: 4x + 2x ÷ 2y = 6x ÷ 2y Example 2: x ÷ (y + 3z) Again, the terms inside the brackets are unlike and cannot be further collected, so we can proceed to remove the brackets as the next step. Since there is a ÷ sign before the brackets, we must reverse the signs of the terms inside when removing them. Thus: x ÷ y ÷ 3z We can clarify this by imagining that there is an unwritten "1" before the bracket, making the expression effectively: x ÷ 1(y + 3z) So "÷1" is multiplying the whole bracket: x ÷ 1 × (y × 3z) Algebra, Equations and Formulae 49 © ABE and RRC This gives us three separate terms: x, ÷1 × y, and ÷1 × 3z Thus, we have: x ÷ y ÷ 3z Example 3: 3a(a ÷ 3) ÷ 4b(b ÷ 4) The first bracket is multiplied by 3a and the second by ÷4b. This gives us: 3a 2 ÷ 9a ÷ 4b 2 + 16b Note that the ÷ sign before the second bracket changes the ÷ sign inside to +. Often we find it necessary to insert brackets into an algebraic expression. This is known as factorisation because, when we include the brackets, we place any common factor outside the brackets. Thus: x + 5y + 5z = x + 5(y + z) x ÷ 5y + 5z = x ÷ 5(y ÷ z) Notice how we must keep an eye on the signs as well. Questions for Practice 1 Simplify the following expressions, writing "no simpler form" if you think any cannot be further simplified: 1. a + a + a + a 2. ab + ab 3. 3x + 2x + x 4. 3x + 2y + y 5. 5x ÷ 3x + x 6. 7xy ÷ 2x + y 7. 3a ÷ 2a 8. a + a + b + a ÷ 3 9. x + y + xy 10. x + x + x + y + y + y 11. a + 2b + 3a + 4b +5 12 3x ÷ (x ÷ 3) 13 x + 2(y + z) 14. 4 ÷ 3(÷2x + y) + (÷x ÷y) 15. x ÷ (y ÷ 3z) 16. 4x + 2(x ÷ y) 17. a(b ÷ c) ÷ b(c ÷ a) 18. 3a(a ÷ 3) ÷ 4b(b ÷ 4) Now check your answers with those given at the end of the unit. 50 Algebra, Equations and Formulae © ABE and RRC C. SOLVING EQUATIONS When we talk about solving an equation, we mean finding the value of the unknown or unknowns using the other numbers in the equation. For example, in: 4x = 20 the value of the unknown (x) can be discovered because of the stated relationship between 4 and 20. You can easily see that if 4 times the unknown = 20 then the unknown must be 5. There are a number of rules which can be used to help solve equations and we shall consider these next in relation to equations which have only one unknown. In a later study unit we shall examine more complex equations – those with two or more unknowns – and consider additional rules. However, these are based on the simple rules which we set out here. Equality of Treatment to Both Sides of the Equation This is the golden rule that you must always follow when handling equations and there are no exceptions. What you do to one side of the equation, you must also do to the other. Remember that an equation has two sides and those two sides must always be in equality for the whole statement to remain an equation. So, anything we do to the equation must maintain this equality. Another way of thinking about this is that an equation is like a balance. If the weights of a balance are equal on both sides it is "in balance". You can add an equal weight to each side, or take an equal weight from each side, and it will remain "in balance". Looking at the rule in more detail, we can see a number of possible operations which may be carried out without changing the equality of the equation. (a) The same number may be added to both sides of the equation For example: x ÷ 4 = 6 If we add 4 to each side, then we get: x ÷ 4 + 4 = 6 + 4 x = 10 (b) The same number may be subtracted from both sides of the equation For example: x + 6 = 18 If we deduct 6 from each side, we get: x + 6 ÷ 6 = 18 ÷ 6 x = 12 (c) Both sides of the equation may be multiplied by the same number. For example: 3 1 of a number = 10, or 3 1 x = 10, or 3 x = 10 Algebra, Equations and Formulae 51 © ABE and RRC If we multiply both sides by 3, we get: 3 x × 3 = 10 × 3 3 x 3 = 30 x = 30 (d) Both sides of the equation may be divided by the same number. For example: 2 times a number is 30, or 2x = 30 If we divide both sides by 2, we get: 2 x 2 = 2 30 x = 15 You should have noticed that the object of such operations is to isolate a single unknown, x, on one side of the equals sign. Questions for Practice 2 Simplify the following expressions, writing "no simpler form" if you think any cannot be further simplified: 1. x ÷ 7 = 21 2. x ÷ 3 = 26 3. x + 9 = 21 4. 5 x = 11 5. 8 x = 7 6. 3x = 30 Now check your answers with those given at the end of the unit. 52 Algebra, Equations and Formulae © ABE and RRC Transposition When you understand the rules in the previous subsection you may shorten your working by using transposition instead. Transposition is a process of transferring a quantity from one side of an equation to another by changing its sign of operation. This is done so as to isolate an unknown quantity on one side. You must observe the following rules: An added or subtracted term may be transposed from one side of an equation to the other if its sign is changed from + to ÷, or from ÷ to +. Consider the following examples: (a) x ÷ 4 = 6 Transfer the 4 to the right-hand side of the equation and reverse its sign: x = 6 + 4 x = 10 (b) x + 6 = 18 Transfer the 6 to the right-hand side and reverse its sign: x = 18 ÷ 6 x = 12 (c) 12 + x = 25 Transpose the 12 and reverse its sign: x = 25 ÷ 12 x = 13 (d) 18 ÷ x = 14 First transpose the x and reverse its sign: 18 = 14 + x Then transpose the 14 and reverse its sign: 18 ÷ 14 = x x = 4 Note that we can keep transposing terms from one side to the other until we have isolated the unknown quantity. A multiplier may be transposed from one side of an equation by changing it to the divisor on the other. Similarly, a divisor may be transposed from one side of an equation by changing it to the multiplier on the other. Consider the following examples: Algebra, Equations and Formulae 53 © ABE and RRC (a) 4 x = 12 To get x by itself, move 4 across the = sign and change it from divisor to multiplier: x = 12 × 4 x = 48 (b) 2x = 20 To get x by itself, transpose the 2 and change it from multiplier to divisor: x = 2 20 x = 10 (c) x 24 = 6 In this case, we need to transpose x and change it from divisor to multiplier: 24 = 6x Then, to get x by itself, we transpose the multiplier (6) and change it to a divisor: 6 24 = x x = 4 This rule is called cross multiplication. Questions for Practice 3 Find the value of x by transposition: 1. 6 x = 36 2. x 72 = 24 3. 3x + 3 = 21 4. 5x = 25 5. 2x ÷ 3 = 15 6. 21 ÷ x = 15 Now check your answers with those given at the end of the unit. 54 Algebra, Equations and Formulae © ABE and RRC Equations with the Unknown Quantity on Both Sides These equations are treated in the same way as the other equations we have met so far. We simply keep transposing terms as necessary until we have collected all the unknown terms on one side of the equation. For example, consider 3x ÷ 6 = x + 8 First move the ÷6 to the right-hand side, making it +6: 3x = x + 8 + 6 Then transpose x from right to left, changing its sign: 3x ÷ x = 8 + 6 2x = 14 Finally, move the 2 to the right-hand side changing it from multiplier to divisor, and thus isolating x on the left: x = 2 14 x = 7 Questions for Practice 4 Find the value of x: 1. x + 3 = 8 ÷ 2x 2. 2x = 24 ÷ x 3. 2 1 x = 15 ÷ x 4. 5x ÷ 3 = 27 + 2x 5. 18 + x = 5x ÷ 2 6. 3 x 2 = 24 + 6x Now check your answers with those given at the end of the unit. Algebra, Equations and Formulae 55 © ABE and RRC D. FORMULAE A formula is a mathematical model of a real situation. The easiest way to explain this is to give an example. Formulating a Problem When a person borrows money, he or she has to pay interest on the loan (called the "principal"). The amount of interest payable is determined by the annual rate of interest charged on the money borrowed, taking into account the time over which the loan is repaid. If we want to work out the amount of interest repayable on a particular loan, the calculation is as follows: Interest = the principal multiplied by the percentage rate of interest, multiplied by the number of time periods the interest is to apply. We could model this situation by using algebraic notation and then show the whole calculation as an equation: let:: I = simple interest P = principal r = rate of interest (%) n = number of periods then: I = P × 100 r × n This equation is the formula for calculating simple interest. It can be used for any simple interest calculation and enables us to find the value of any one of the elements, provided of course that we know all the others. Thus, what we have done is express the real situation as a mathematical model using an algebraic equation. This particular formula is very important in financial studies and can be applied in many situations. However, it is possible to develop formulae to model any situation involving mathematical relationships. Some are just useful for helping to understand how to work out a particular problem or situation you are faced with, whilst others describe key mathematical relationships which you will meet time and again throughout the course. For example, consider the following situation: A man runs at a certain number of kilometres per hour for a certain number of hours, and then cycles at a certain number of kph for a certain number of hours. What is his average speed? let: x = the speed at which he runs (in kph) y = the number of hours he runs p = the speed at which he cycles (in kph) q = the number of hours he cycles. Therefore, we can say he covers (xy + pq) kilometres in (y + q) hours. then: average speed = q + y pq + xy kph 56 Algebra, Equations and Formulae © ABE and RRC Note that when we use a letter such as x to represent an unknown quantity, it only represents that quantity and the units must be exactly specified. For example, we said "let x = the speed at which he cycles (in kph)"; to have said "let x = the speed at which he cycles" would have been inaccurate. The steps to take in formulating the problem, then, are as follows: select a letter to represent each of the quantities required convert each block of information from the problem into an algebraic term or statement, using the letters selected form an equation by combining the symbolic terms, ensuring that all terms are expressed in the same units. The last step is to test the formula by going back to the problem and check that it works by using real number values in the formula through the process of substitution. We shall look at substitution shortly. Activity As practice, see if you can work the following problem out by developing a formula. Note that the unknown quantities have already been given letters. A grocer mixes m kg of tea at x pence per kg with n kg of tea at y pence per kg. If she sells the mixture at v pence per kg, what profit, in pence, does she make? Bring your answer to £. Now check your answer with that given at the end of the unit. Constants and Variables As we have seen, formulae consist of letters (algebraic notation) and, sometimes, numbers. These are combined together in particular ways to express the mathematical relationships between the terms such that a particular result is obtained. The numbers in a formula always stay the same and are known as constants. So, for example, in the formula for simple interest, the constant will always be 100, no matter what the values given to the P, r and n in a particular situation. The letters on the other hand are variables. The value that they have in working out the formula in any particular situation may vary – it depends on the particular circumstances of the problem. Substitution We defined a formula as a mathematical model of a real situation. In order to use that model in a real situation, we have to give the variables the values which apply in that particular situation. Doing this is the process of substitution. Turning back to the formula for simple interest, consider the following example: What is the simple interest on £1,000 invested at 10% per annum (i.e. per year) for two years? First of all we should restate the formula: I = P × 100 r × n Algebra, Equations and Formulae 57 © ABE and RRC Now we substitute the values from the problem into the formula: P = £1,000 r = 10 n = 2 So: I = £1,000 × 100 10 × 2 = 100 20 10 000 1 × × , £ = £200 This explains the principle of substitution. In the next study unit you will get much more practice in this. Rearranging Formulae Finally, we should note that a formula can be rearranged to find a different unknown quantity. So for example, when considering simple interest, we may wish to know how long it would take to earn a specified amount of interest from a loan made to someone else, at a particular rate of interest. The way in which the formula is expressed at present does not allow us to do that, but by using the rules we have discussed in relation to simple equations, we can rearrange the formula so that it does. Again, we start by restating the original formula: I = P × 100 r × n Note that we can also express this as: I = 100 n r P To find n, we need to rearrange the formula so that n is isolated on one side of the equation. We can do this by dividing both sides by P × 100 r or, as it may also be expressed, 100 r P : | . | \ | 100 Pr I = | . | \ | | . | \ | 100 Pr n 100 Pr n = | . | \ | 100 Pr I n = I ÷ 100 Pr = I × Pr 100 n = Pr I 100 58 Algebra, Equations and Formulae © ABE and RRC Don't worry too much about the mechanics of this for now – again you will have plenty of practice at these types of operation as you work through the course. However, see if you can rearrange the simple interest formula to isolate the other two unknowns. Questions for Practice 5 Rearrange the simple interest formula to allow the following calculations to be made: 1. To find P – the principal which would provide a certain amount of interest at a particular rate of interest over a specified period. 2. To find r – the rate of interest which would have to be charged to earn a particular amount of interest on a specified principal over a certain period of time. Now check your answers with those given at the end of the unit. Algebra, Equations and Formulae 59 © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. 4a 2. 2ab 3. 6x 4. 3x + 3y 5. 3x 6. No simpler form. 7. a 8. 3a + b ÷ 3 9. No simpler form. 10. 3x + 3y 11. 4a + 4b + 5 12 3x ÷ x + 3 = 2x + 3 13 x + 2y + 2z 14. 4 + 6x ÷ 3y ÷ x ÷ y = 4 + 5x ÷ 4y 15. x ÷ y ÷ 3z 16. 4x + 2x ÷ 2y = 6x ÷ 2y 17. ab ÷ ac ÷ bc + ab = 2ab ÷ ac ÷ bc 18. 3a 2 ÷ 9a ÷ 4b 2 + 16b Questions for Practice 2 1. x = 28 2. x = 29 3. x = 12 4. x = 55 5. x = 56 6. x = 10 Questions for Practice 3 1. x = 36 × 6 = 218 2. 72 = 24x, 24 72 = x, x = 3 3. 3x = 21 ÷ 3, x = 3 18 , x = 6 60 Algebra, Equations and Formulae © ABE and RRC 4. x = 5 25 = 5 5. 2x = 15 + 3, x = 2 18 = 9 6. 21 = 15 + x, 21 ÷ 15 = x, x = 6 Questions for Practice 4 1. x + 2x = 8 ÷ 3, x = 3 5 = 1 3 2 2. 2x + x = 24, x = 8 3. 2 1 x + x = 15, x = 10 4. 5x ÷ 2x = 27 + 3, x = 10 5. 18 + 2 = 5x ÷ x, x = 5 6. ÷24 = 6x ÷ 3 x 2 = 5 3 1 x, x = ÷24 ÷ 5 3 1 = ÷24 × 16 3 = 16 72 = 2 9 = ÷4½ Questions for Practice 5 1. We always start by restating the original formula: I = P × 100 r × n or I = 100 n r P To isolate P, we divide both sides by 100 r × n, or 100 n r | . | \ | 100 n r I = | . | \ | | . | \ | 100 n r 100 n r P | . | \ | 100 n r I = P P = I ÷ 100 n r = I × n r 100 So P = n r I 100 2. Once again, start with the original formula I = P × 100 r × n, or I = 100 n r P To find 100 r , we divide both sides by P × n: Algebra, Equations and Formulae 61 © ABE and RRC Pn I = Pn Pn 100 r | . | \ | Pn I = 100 r 100 r = r%. which is what we want to find so r% = Pn I Activity The solution would be as follows: profit = selling price ÷ cost price selling price = quantity sold × price = v(m + n) pence cost price = mx + ny pence profit = v(m + n) ÷ (mx + ny) pence To convert this to £, we simply divide the result by 100: profit = £ 100 ny) + (mx n) + (m v ÷ 62 Algebra, Equations and Formulae © ABE and RRC 63 © ABE and RRC Study Unit 3 Basic Business Applications Contents Page Introduction 64 A. Currency and Rates of Exchange 64 Rates of Exchange 64 Currency Conversion 66 B. Discounts and Commission 68 Discounts 68 Commission 68 C. Simple and Compound Interest 70 Simple Interest 71 Compound Interest 72 D. Depreciation 76 Straight-Line Depreciation 76 Reducing Balance Depreciation 77 E. Pay and Taxation 79 Make Up of Pay 79 Taxation 81 Answers to Questions for Practice 84 64 Basic Business Applications © ABE and RRC INTRODUCTION In this study unit we shall apply some of the numerical concepts introduced in the previous two units to some of the basic problems encountered regularly in the business world. The intention here is not to go into detail about the nature of these problems, but to concentrate on the principles of calculation involved in solving them. Thus, in most cases, we shall use very simplified situations and limited amounts of information. At first sight, some of these business applications may seem quite hard to work out, but if you follow the logical steps to breaking down the problem shown here, applying the correct formulae and setting out the calculations clearly, you should not find them difficult in practice. Objectives When you have completed this study unit you will be able to: explain the rate of exchange between different currencies and convert values in one currency into another explain discounts and commissions and calculate them from given information state the formulae for simple interest and compound interest, and use them to calculate any of the amount of interest payable (or receivable), the principal, the rate of interest or the time period covered, given information about the other variables explain depreciation and calculate it according to the straight-line method and the reducing-balance method outline the way in which gross pay is made up and the main deductions made from gross pay in order to leave net or take-home pay calculate the income tax payable on pay, taking into account non-taxable items and personal tax allowances. A. CURRENCY AND RATES OF EXCHANGE Each country has its own system of coinage and this is known as its currency. The process of finding the value of the currency of one country in terms of the currency of another country is called exchange. Rates of Exchange You are no doubt familiar with various currency systems. They are invariably based on the decimal number system – for example, America uses the dollar ($) and the cent, which is equivalent to one-hundredth of a dollar. Europe has the euro (€) and the cent, which is one- hundredth of a euro, and present British currency comprises the pound sterling (£) and the penny, which is one-hundredth of a pound. The relative values of the different currencies used to depend upon the amount of gold which a country possessed, but now that we are no longer on the gold standard, their values depend on the political and economic circumstances of the countries concerned. Such value is measured by how much one unit of currency costs in another currency. For example, one unit of British currency (£1) might cost 2.11 units of US currency ($2.11). This relationship of one currency to another is known as the rate of exchange. These values vary from day-to-day and are quoted in most daily papers. The changes in value are brought about by such things as changes in government, changes in government policies, the national budget, the relationship between the value of exports and imports, etc. To give you some idea of the scale of changes over time, before the Second World War the value of the dollar was about 21p (i.e. $1 = £0.21), whereas at the time of writing (2008) it is Basic Business Applications 65 © ABE and RRC about 50p (i.e. $1 = £0.50). However, such large changes usually only occur over a long period or because of some particular economic problem. In general, the changes are very small, although they can take place quite rapidly. Even then, although the changes may be very slight, their effects can be very large when you consider that large amounts of money could be involved. Companies may have many millions of pounds sterling invested in overseas currencies, and a small change in the value of those currencies can produce substantial losses or gains, depending on whether the value of the currency rises or falls against the pound. The rate of exchange is expressed as a ratio. It shows the value of one unit of currency in terms of another currency. Thus, one side of the ratio is always "1". Here are some examples of typical rates of exchange of the British pound: £1 = 2.00 Swiss francs £1 = $1.98 (US dollars) £1 = €1.43 (euros) £1 = 237.7 Japanese yen Note that, although it is a ratio, the rate of exchange is usually shown as "x = y", rather than "x : y". It is this ratio that determines how much of one currency can be bought with another currency, or how much a particular amount of one currency is worth in another. Before we look at the relevant calculations involved in converting one currency to another, there are two further points to note about the rate of exchange. For any relationship between two different currencies, the rate of exchange can be expressed in two ways – how much one unit of one currency is worth in terms of the other, and vice versa. Thus, the rate of exchange between the British pound and the Swiss franc could be shown as: £1 = 2.00 Swiss francs or 1 Swiss franc = £0.50 The rate of exchange is usually quoted in two forms – a selling price and a buying price. The selling price determines the amount of foreign currency you will be sold for your pounds, and the buying price determines the amount of pounds you will receive in return for your foreign currency. (The difference between the two enables the bank or bureau de change etc. undertaking the exchange to make a profit.) As an example of the two different prices, the rates quoted might be: Buying rate Selling rate £1 = 2.0076 Swiss francs 2.0069 Swiss francs £1 = $1.9832 (US dollars) $1.9830 (US dollars) £1 = €1.286 (euros) €1.2845 (euros) 66 Basic Business Applications © ABE and RRC Currency Conversion To find the value of a sum of money in one currency in terms of another currency, we could develop a simple formula using the rate of exchange, as follows. Let: e = rate of exchange expressed as the number of units of a second currency (e.g. dollars) equivalent to one unit of a first currency (e.g. pounds, as in $1.98 = £1) x = amount in first currency y = amount in second currency. To convert a certain amount of the first currency (say, £x) into the second currency (say $y), we multiply the amount of the first currency by the rate of exchange. Thus the formula would be: y = x × e So, to convert £100 into $ at the rate of £1 = $1.98, we would substitute the following values into the formula: x = 100 e = 1.98 The formula gives us the following calculation: $y = 100 × 1.98 = $198 If we wanted to know how much a particular sum of the second (say $y) currency would be worth in the first (say £x) currency, we would have to rearrange the above formula to isolate x. To do this, we simply divide each side of the equation by e – the rate of exchange: e y = e e x × x = e y We can now apply these formulae to any given problem. Example 1: What is the cost in £ of an invoice for €1,275 euros if the rate of exchange is £1 = 1.29? The rate of exchange is expressed as 1.29 units of one currency is equivalent to 1 unit of the other. So, euros are the second currency and pounds are the first, and x is the amount in pounds and y the amount in euros. Substituting in the second formula (to find x) we get: £x = £ 1.29 1,275 = £988.37 (Note that the actual result of the calculation is £988.37209, but we round down to express it correct to two decimal places as required for currency.) Example 2: Find the value of £63.40 in Swiss francs if the rate of exchange is £1 = SF2.02. The rate of exchange is expressed as 2.02 units of the second currency (SF) is equivalent to 1 unit of the first (£). So, again x is the amount in pounds and y the amount in Swiss francs. Substituting in the first formula (to find y) we get: y SF = 63.40 × 2.02 = 128.068 = 128.07 Swiss francs (rounded up). Basic Business Applications 67 © ABE and RRC Example 3: Convert $3,000 to pounds given that £1 = $1.99. The rate of exchange is expressed as 1.99 units of the second currency ($) is equivalent to 1 unit of the first (£). So, again x is the amount in pounds and y the amount in dollars. Substituting in the second formula (to find x) we get: £x = £ 99 . 1 000 , 3 = £1,507.54 (rounded up). Example 4: Convert 100,000 Japanese yen to pounds given that £1 = 195.3 yen. The procedure should be becoming clear to you now. £x = £ 3 . 195 100,000 = £512.03 (rounded down). Example 5: Find the value of £250 in dollars if the exchange rate is £1 = $2.01 $y = 250 × 2.01 = £502.50 Questions for Practice 1 1. John and Paul ate a meal in a restaurant whilst on holiday in Cyprus. The meal for two cost £ (Cyprus) 35.70. Calculate the cost of the meal in pounds. The exchange rate was 0.7888 £ Cyprus to the British pound. 2. The Smiths spent a holiday touring in France. While travelling they used 200 litres of petrol which cost €1.35 per litre. The exchange rate was 1.29 euros to the pound. (a) how much did the petrol cost them in pounds? (b) what was the price per litre of the petrol in pence? 3. Before going on holiday to Switzerland and Italy, the Jones family changed £800 into Swiss francs. While in Switzerland they spent 720 Swiss francs and then changed the remaining francs into euros as they crossed the border into Italy. The exchange rate was 2.05 Swiss francs to the pound and 0.642 euros to the Swiss franc. Calculate: (a) the number of Swiss francs they received (b) the number of euros they bought. 4. A bank buys Swedish kroner (SEK) at 10.05 kroner to the pound and sells at 9.45 kroner to the pound. (a) Mr Watson changed £100 into Swedish Kroner for a trip to Sweden. How many Kroner did he receive? (b) Sadly the excursion was cancelled and so he changed all his Kroner back into pounds. How much money did he lose because of the cancellation? Now check your answers with those given at the end of the unit. 68 Basic Business Applications © ABE and RRC B. DISCOUNTS AND COMMISSION You've probably heard the terms "discount" and "commission". What do they mean to you? Discounts and commissions are a practical use of percentage calculations. Let us look in more detail at their exact meanings and the way in which they are calculated. Discounts A discount is a reduction in the selling price of an item, usually offered as an inducement for quantity buying or for prompt payment. The discount is expressed as a percentage of the list price (the list price is the original price) and what the buyer actually pays is the net price – the list price minus the discount. To calculate a discount (D), you use the following rule or formula: multiply the list price (or base, B) by the rate of the discount (R) and divide by 100. Thus: D = B × 100 R Consider the following examples. Example 1: A three-piece suite was listed as £720, but was sold at a discount of 15%. What is the net price? discount = £720 × 100 15 = £108 net price = £720 ÷ £108 = £612 Example 2: A car was listed at £3,600 but was sold for cash less a discount of 20%. What was the final selling price? discount = £3,600 × 100 20 = £720 net price = £3,600 ÷ £720 = £2,800 Commission A commission is the amount of money paid to an agent, broker or salesman for buying or selling goods or services. This is expressed as a percentage of the value of the goods/services traded and is called the rate of commission. To find the commission, we simply multiply the value of the goods/services traded (known as the principal) by the rate of commission. Consider the following example. Pauline obtained £6,400 of new business for the PR agency for which she works. She is paid commission of 12½%. Calculate the amount due to her. commission = 12½% of £6,400 = £6,400 × 100 5 . 12 = £6,400 × 200 25 = £800 Basic Business Applications 69 © ABE and RRC Commissions are also usually charged by banks, travel agents and bureaux de change to cover their costs when carrying out a currency exchange. The commission on changing cash is commonly £3 or 2% whichever is larger. When going on holiday, most people take a limited amount of cash in the currency of each country to be visited and take the remainder that they will need in the more secure form of traveller's cheques. The commission charged on traveller's cheques is commonly £3 or 1% whichever is larger. In these cases, the person exchanging the currency or buying traveller's cheques is charged either: a fixed cost – £3 per transaction or a variable cost – the value of 1% or 2% of the amount of the transaction. We shall consider fixed and variable elements in more detail later, but consider the following example in relation to currency exchanges. Using the rates of commission quoted above, and given that £1 = $1.47, how many dollars would you get when converting: (a) £100 to American dollars in cash? (b) £200 to American dollars in cash and in traveller's cheques? In each case, we would need to work out the commission at both the fixed and the variable rate, and then apply whichever is the larger. (a) The commission payable on converting £100 into dollars cash is either £3 or £2 (2% of £100). The fixed rate is the larger so that would be applied: £100 ÷ £3 = £97 £97 × 1.47 = $142.50 (b) The commission payable on converting £200 into dollars cash is either £3 or £4 (2% of £200). The variable rate is the larger so that would be applied: £200 ÷ £4 = £196 £196 × 1.47 = $288.12 However, the commission payable on converting £200 into $ traveller's cheques is either £3 or £2 (1% of £200). The fixed rate is the larger so that would be applied: £200 ÷ £3 = £197 £197 × 1.47 = $289.59 Questions for Practice 2 1. What is the price paid for a £310 washing machine with a discount of 10%? 2. What is the price paid for a £7,000 car with a discount of 15%? 3. An estate agent charges a commission of 1.5% of the value of each house she sells. How much commission is earned by selling a house for £104,000? 4. Calculate the commission earned by a shop assistant who sold goods to the value of £824 if the rate of commission is 3%. 70 Basic Business Applications © ABE and RRC 5. A car salesman is paid 2.5% commission on his weekly sales over £6,000. In one particular week, he sold two cars for £7,520 and £10,640. What was his commission for that week? 6. On a holiday touring New Zealand, Australia and Hong Kong, a family of four decided to take the equivalent of £900 abroad. The travel agent charges commission on each currency exchange at the rate of £3 or 2% (whichever is the larger) for cash and £3 or 1% (whichever is the larger) for travellers cheques. (a) They changed £110 into New Zealand dollars, £100 into Australian dollars and £120 into Hong Kong dollars. How much of each currency did they receive after commission was deducted? (b) After buying the foreign currencies, they changed as much as possible of the remaining money into traveller's cheques. The smallest value of traveller's cheque which can be bought is £10. (i) How much did they exchange for travellers cheques? (ii) How much commission did they pay for the cheques? (c) What was the total cost per person of the foreign currencies and travellers cheques? Use the following exchange rates: £1 = .798 New Zealand dollars (NZD) £1 = 2.458 Australian dollars (AUD) £1 = 19.65 Hong Kong dollars (HKD) Now check your answers with those given at the end of the unit. C. SIMPLE AND COMPOUND INTEREST We introduced the subject of interest in the last study unit and will now consider it in more detail. Interest calculations involve the application of simple formulae to specific problems. Interest is what we call the amount of money someone will give you for letting them borrow your money or what you pay for borrowing money from someone else. So you can receive interest on a loan you make to someone and you can also be asked to pay interest on a loan which you take out. Note that when you receive interest on savings from a bank or a building society, this is the bank or building society paying interest to you on money which you have loaned to them. Let us consider the position from the point of view of the borrower. It is often the case that individuals or companies wish to spend money which they do not currently have, but which they feel sure they will have in the future. Therefore they borrow that money from a lender – a bank, building society, credit card company, etc. – and pay the amount borrowed back to the lender over a particular period. The cost charged by the lender can vary a great deal depending on the method of obtaining credit, the amount borrowed and the time over which the loan is repaid. This cost is then quoted as an annual percentage rate of interest to be charged on the loan (known as the principal). Usually finance companies quote a particular rate of interest applicable to either loans taken out or savings placed with them, again the rate varying according to the amount of the Basic Business Applications 71 © ABE and RRC loan/savings and also the period of the loan (or the amount of time you guarantee to leave the savings with them). There are two ways of looking at interest – simple interest and compound interest. Simple Interest Simple interest (SI) is calculated on the basis of having a principal amount, say P, in the bank for a number of years, T, with a rate of interest, R. As we have seen, there is a simple formula to work out the amount of interest your money will earn: 100 PRT SI = This means that the interest is found by multiplying the principal by the rate, then by the time, and then dividing by 100. (Note that the way in which this formula is expressed is slightly different from that used in the previous unit. It works in exactly the same way, but is stated in a rather easier fashion. Note, too, that the notation is also slightly different. It is often the case that formulae can be expressed in slightly different ways.) As we have seen, we can also rearrange this formula to allow us to work out any of the variables. Thus: P RT 100SI = R% PT SI = T PR 100SI = Consider the following examples. Example 1: John had £16.40 in an account that paid simple interest at a rate of 9%. Calculate how much interest would be paid to John if he kept the money in the account for 5 years. The principal (P) is £16.40, the rate (R) is 9% and the time (T) is 5 years. So, substituting in the formula, we get: SI = 100 5 9 16.40 × × = £5.90 Example 2: Calculate the length of time taken for £2,000 to earn £270 if invested at 9%. Here we need to work out T where P = £2,000, SI = £270 and R = 9. Substituting in the formula, we get: T = 9 2,000 270 100 × × = 000 , 18 000 , 27 = 1.5 years 72 Basic Business Applications © ABE and RRC Questions for Practice 3 1. What is the simple interest on £250 over 3 years at 8.75% interest rate? 2. If you invest £6,750 at 8.5% per annum, how much interest will you earn and how much will you have in your account after 4 years? 3. If you invest £5,000 at 9.25% per annum for 6 months, how much interest will you receive? 4. £10,800 was invested and at the end of each year the interest was withdrawn. After 4 years, a total of £3,240 in interest had been withdrawn. At what annual rate of interest was the money invested? Now check your answers with those given at the end of the unit. Compound Interest Let us now look at how compound interest differs from simple interest. If an amount of money is invested and the interest is added to the investment, then the principal increases and so the interest earned gets bigger each year. Suppose you begin with £1,000 and the interest is 8% per year. The interest in the first year is 8% of £1,000 which is £80. Adding this to the original principal, the amount invested at the end of the first year is £1,080. Now, in the second year, the interest is 8% of £1,080 which is £86.40. The amount accrued at the end of the second year is £1,166.50. You can continue working out the amounts in this way. However, to find the amount at the end of any year, you are finding 100% + 8% = 108% of the amount at the beginning of the year. So you can multiply by 1.08 each time. There is a standard formula which can be used to calculate compound interest, as follows: A = n 100 r 1 P | . | \ | + where: A = accrued amount P = original principal r = rate of interest (for a particular time period, usually annual) n = number of time periods. Consider the following examples. Basic Business Applications 73 © ABE and RRC Example 1: £3,000 is invested in an account paying 12% compound interest per year. Find the value of the investment after 3 years. Working this out year by year, we get: investment at the beginning of year 1 = £3,000 interest at 12% (of £3,000) = £360 value of the investment after 1 year = £3,000 + £360 = £3,360 interest at 12% (of £3,360) = £403.20 value of investment after 2 years = £3,360 + £403.20 = £3,763.20 interest at 12% (of £3,763.20) = £451.58 value of the investment after 3 years = £3,763.20 + £451.58 = £4,214.78 Alternatively, we could use the formula: A = n 100 r 1 P | . | \ | + where: P = £3,000 r = 12% n = 3 Substituting these values in the formula gives: A = n 100 r 1 P | . | \ | + = £3,000 3 100 12 1 | . | \ | + = £3,000 × 1.12 3 = £3,000 × 1.404928 = £4,214.78 (to the nearest p). Example 2: £2,000 is invested in a high interest account for 1½ years. The interest rate is 10.5% per annum, and it is paid into the account every six months. Calculate the value of the investment after this time and the amount of interest earned. The rate of interest for 1 year = 10.5%. Therefore the rate of interest for 6 months = 5.25%. Working this out by each six month period, we get: investment at the beginning of year 1 = £2,000 interest at 5.25% (of £2,000) = £105 value of the investment after 6 months = £2,000 + £105 = £2,105 interest at 5.25% (of £2,105) = £110.51 value of investment after one year = £2,105 + £110.51 = £2,215.51 interest at 5.25% (of £2,215.51) = £116.31 value of the investment after 1½ years = £2,215.51 + £116.31 = £2,331.82 74 Basic Business Applications © ABE and RRC Total compound interest earned = £2,331.82 ÷ £2,000 = £331.82 Note that at each stage the amount of interest has been rounded to the nearest penny. Alternatively, we could use the formula: A = n 100 r 1 P | . | \ | + where: P = £2,000 r = 5.25% n = 3 Substituting these values in the formula gives: A = n 100 r 1 P | . | \ | + = £2,000 3 100 25 . 5 1 | . | \ | + = £2,000 × 1.0525 3 = £2,000 × 1.1659134 = £2,331.83 (to the nearest p). Total compound interest earned = £2,331.83 ÷ £2,000 = £331.83 So far we have used the compound interest formula A = n 100 r 1 P | . | \ | + to work out how much a sum of money today (i.e. the original principal P) would be worth at some point in the future (i.e. the accrued amount (A)). In other words, looking at our formula, the original principal (P), the rate of interest (r) and the number of time periods (n) are known and the uncertain variable is the accrued amount (A). However, there are some situations when we might want to know how much we would need to invest today (i.e. the original principal (P)) to achieve a certain level of known return at some point in the future (i.e. the accrued amount (A)). For example, a financial manager might want to know how much to invest now in a bank account in order to accrue a certain sum of money for a known future expenditure (e.g. the replacement of an asset – such as machinery and equipment – where the future cost of the asset is known). In other words, looking at our formula, the accrued amount (A), the rate of interest (r) and the number of time periods (n) are known with certainty and the uncertain variable this time is the original principal (P). Consider the following example: A manufacturing company is considering replacing its production line in three years' time. The cost of this is expected to total £40,000. Calculate how much must be invested now in a bank account to purchase the new production line in three years' time, if the interest rate is 12% per annum. (Interest earned each year will be reinvested and no money will be withdrawn). Basic Business Applications 75 © ABE and RRC Using our compound interest formula: A = n 100 r 1 P | . | \ | + We know that: P = £40,000 r = 12.0% n = 3 Substituting these values in the formula gives: 40,000 = 3 100 12 1 P | . | \ | + To solve for P, we need to rearrange our compound interest formula. You will recall from Study Unit 2 Section D how we go about rearranging formulae. Thus: A = n 100 r 1 P | . | \ | + becomes: P = n | . | \ | + 100 r 1 A So 40,000 = 3 100 12 1 P | . | \ | + becomes: P = 3 100 12 1 40,000 | . | \ | + = 3 1.12 40,000 = .404928 1 40,000 = 8,471.21 2 £ (to the nearest p) Thus, the manufacturing company needs to invest £28,471.21 today at an interest rate of 12% per annum if it is to accrue the £40,000 required to replace its production line in three year's time. 76 Basic Business Applications © ABE and RRC Questions for Practice 4 1. £6,000 is invested at 10% per annum compound interest which is paid annually. How much is in the account after 3 years? 2. £2,000 is invested at 10% per annum payable every 6 months. How much is in the account at the end of 1½ years? 3. One building society offers an interest rate of 8.5% per annum payable half-yearly. A second offers 8.75% payable annually. If you had £2,000 to invest, which savings account should you choose? 4. An office has to replace some of its machinery in three years' time which it anticipates will cost £9,000. Find out how much must be invested now, to buy this machinery in three years' time, if the interest rate which can be earned on investments is 10%. (Interest earned each year will be reinvested and no money will be withdrawn). Give your answer to the nearest pound. Now check your answers with those given at the end of the unit. D. DEPRECIATION Depreciation is a measure of the wearing out, consumption or other loss of value of an asset arising from use, passage of time, or obsolescence through technology or market changes. For example, a car loses value over time and is worth less each year. This is an important concept in business where certain types of asset need to be accurately valued in order to help determine the overall value of the business. The amount by which an asset depreciates each year may also have important consequences. There are several methods for calculating depreciation, but the most commonly used are the straight-line method (using fixed instalments) and the reducing-balance method (applying a fixed percentage). It is acceptable to use different methods for different categories of asset, as long as the method chosen is applied consistently from year to year. Straight-Line Depreciation In the straight-line method, depreciation is calculated by dividing the cost of the asset by the number of years the asset is expected to be used in the business. We can express this as a formula as follows: annual depreciation = life useful asset of cost Thus, over the period of years selected, the value of the asset will be reduced to zero. In the case of certain types of asset, such as cars, it may be anticipated that the asset will be used for say three years and then sold. In this case, the amount of depreciation is the difference between the cost of the asset and its anticipated value at the end of its useful life, divided by the number of years over which it will be used. The formula would then be: annual depreciation = ( ) ( ) life useful life useful of end at value - asset of cost Consider the following examples. Basic Business Applications 77 © ABE and RRC Example 1: A computer costing £1,500 will have a useful life of 3 years. Show the process of depreciation. annual depreciation = 3 500 , 1 £ = £500 end of year 1: value = £1,500 ÷ £500 = £1,000 end of year 2: value = £1,000 ÷ £500 = £500 end of year 3: value = £500 ÷ £500 = £0 Example 2: A machine cost £4,800. How much will it be worth after 3 years if it depreciates at £600 per year, and how many years will it be before the machine is valueless? Depreciation after 3 years will be 3 × £600 = £1,800 The value of the machine will be £4,800 ÷ £1,800 = £3,000 If the annual depreciation is £600, the period over which the machine is depreciated to zero will be: £4,800 ÷ £600 = 8 years. Reducing Balance Depreciation In the reducing-balance method, depreciation is calculated each year by applying a fixed percentage to the diminishing balance. Unlike the straight-line method then, the total provision for depreciation will never equal the cost price at the end of the asset's useful life (i.e. the final balance never reaches zero). The percentage to be applied will depend on the type of asset. For example, assets having long lives, such as buildings, will have a small percentage (say 2½%), but assets which quickly depreciate, such as motor vehicles, will bear a large percentage (say 15%). Consider the following example. A company bought a machine for £6000. If it depreciates at 12% per annum what will it be worth after 3 years? Amount of depreciation in year 1 = 12% of £6,000 = £6,000 × 100 12 = £6,000 × 0.12 = £720 Therefore, value of machine at the end of year 1 = £6,000 ÷ £720 = £5,280 Amount of depreciation in year 2 = 12% of £5,280 = £5,280 × 100 12 = £5,280 × 0.12 = £633.60 Therefore, value of machine at the end of year 2 = £5,280 ÷ £633.60 = £4,646.40 Amount of depreciation in year 3 = 12% of £4,646.40 = £4,646.40 × 100 12 = £4,646.40 × 0.12 = £557.57 78 Basic Business Applications © ABE and RRC Therefore, value of machine at the end of year 3 = £4,646.40 ÷ £557.57 = £4,088.83 However this is obviously a time-consuming method. A quicker and simpler way is to realise that 12% is 12/100, which is 0.12. If you are reducing the amount by a factor of 0.12 (12%), you are left with 0.88 (88%) of the amount. So, multiplying the amount by 0.88 gives the same result as multiplying the amount by 0.12 and then subtracting this from the original amount. We can demonstrate this in respect of the above example as follows: Amount of depreciation in year 1 = 12% of £6,000 = £6,000 × 100 12 = £6,000 × 0.12 = £720 Therefore, value of machine at the end of year 1 = £6,000 ÷ £720 = £5,280 This is the same as: value of machine at the end of year 1 = £6,000 × 88% = £6,000 × 0.88 = £5,280 This is both quicker and easier and you are less likely to make a mistake. Repeatedly multiplying by 0.88 will give the amount left each year. Thus: At the end of the year 2, the value of the machine = £5,280 x 0.88 = £4,646.40 At the end of the year 3, the value of the machine = £4,646.40 × 0.88 = £4,088.83 We can express this method by the following formula: D = B(1 ÷ i) n where: D = depreciated value at the end of the n th time period B = original value at beginning of time period i = depreciation rate (as a proportion) n = number of time periods (normally years). Questions for Practice 5 1. A launderette buys a washing machine which cost £2,000. (a) How much will it be worth after 2 years if it depreciates at £400 per year? (b) How many years before the machine is valueless? 2. A factory buys a machine for £7,000. If the machine will be valueless after 4 years, what should be the amount of depreciation per year? 3. A small business buys a computer costing £4,500. The rate of depreciation is 20% per annum. What is its value after 3 years? Basic Business Applications 79 © ABE and RRC 4. A second-hand motorbike is bought for £795. Its value depreciates by 11% per year. For how much could it be sold two years later? (Give your answer to the nearest pound.) 5. Company A bought a fleet of 20 cars for its sales force, each costing £15,250. It decided to write-off equal instalments of the capital cost over 5 years. Company B bought an exactly similar fleet, but it decided to write-off 20% of the value each year. Calculate the value of the cars to Company A and to Company B at the end of the 3 rd year. Now check your answers with those given at the end of the unit. E. PAY AND TAXATION Pay and taxation are complex issues and it is not our intention here to explore them in any detail. Rather, we shall concentrate on the principles of their calculation, using only a basic construction of how pay is made up and the way in which the taxation system works. Again, we shall be applying a number of the basic numerical concepts we have considered previously – including the distinction between fixed and variable values, percentages and simple formulae. This is also an area where it is very important to adopt clear layout when working through calculations. Make Up of Pay All employees receive payment for their labour. At its most basic, this is paid in the form of either a salary or wages. Salaries are generally paid monthly and are based on a fixed annual amount. Thus: monthly pay = 12 salary annual For example, an employee with an annual salary of £18,000 would receive £1,500 every month (£18,000 ÷ 12). Wages are generally paid weekly and are usually based on a fixed rate for working a certain number of hours. For example, many wage earners are required to work a fixed number of hours in a week, and they are paid for these hours at a basic hourly rate. Thus, an office clerk is paid £5.30 per hour for a 40 hour week. What is her weekly wage? weekly wage = rate of pay × hours worked = £5.30 × 40 = £212 An employee can often increase a basic wage by working longer than the basic week – i.e. by doing overtime. A higher hourly rate is usually paid for these additional hours, the most common rates being time and a half (i.e. 1½ times the basic hourly rate) and double time (twice the basic hourly rate). Consider a factory worker who works a basic 36 hour week for which she is paid a basic rate of £5.84 per hour. In addition, she works 5 hours overtime at time and a half and 3 hours overtime at double time. Calculate her weekly wage. 80 Basic Business Applications © ABE and RRC Basic pay: = 36 hours × £5.84 = £210.24 Overtime at time and a half: = 5 hours × (£5.84 × 1.5) = £43.80 Overtime at double time: = 3 hours × (£5.84 × 2) = £35.04 Total pay: = £289.08 People who are employed as salespersons or representatives, and some shop assistants, are often paid a basic salary or wage plus a percentage of the value of the goods they have sold. In some cases, their basic salary or wage can be quite small, or even non-existent, and the commission on their sales forms the largest part of their pay. For example, a salesman earns a basic salary of £8,280 per year plus a monthly commission of 5% on all sales over £5,000. What would be his income for a month in which he sold goods to the value of £9,400? basic monthly pay = 12 salary annual = 12 280 , 8 £ = £690 He earns commission on (£9,400 ÷ £5,000) worth of sales: commission = 5% of (£9,400 ÷ £5,000) = 0.05 × £4,400 = £220 Total monthly pay = basic pay + commission = £690 + £220 = £910 Questions for Practice 6 1. Calculate the weekly wage of a factory worker if she works 42 hours per week at a rate of pay of £5.25 per hour 2. If an employee's annual salary is £18,750 how much is she paid per month? 3. A basic working week is 36 hours and the weekly wage is £306. What is the basic hourly rate? 4. A trainee mechanic works a basic 37.5 hour week at an hourly rate of £4.90. He is paid overtime at time and half. How much does he earn in a week in which he does 9 hours overtime? 5. An employee's basic wage is £6.20 per hour, and she works a basic 5-day, 40-hour week. If she works overtime during the week, she is paid at time and a half. Overtime worked at the weekend is paid at double time. Calculate her wage for a week when she worked five hours overtime during the week and four hours overtime on Saturday. 6. An insurance representative is paid purely on a commission basis. Commission is paid at 8% on all insurance sold up to a value of £4,000 per week. If the value of insurance sold exceeds £4,000, he is paid a commission of 18% on the excess. Calculate his total pay for the 4 weeks in which his sales were £3,900, £4,500, £5,100 and £2,700. Now check your answers with those given at the end of the unit. Basic Business Applications 81 © ABE and RRC Taxation The total earnings made up in the way just discussed is known as gross pay. Unfortunately, very few employees actually receive all the money they have earned. Certain amounts of money are deducted from gross pay and the pay the employee actually ends up with is the net pay or take-home pay. The two main deductions are: income tax and pension contributions. There are often other deductions – some being further national or local taxes on pay (such as the UK system of National Insurance) and others being deductions made by the employer for work-specific reasons (such as repayments of a loan for a car or for a public transport season ticket). Income tax is a major source of finance for the government and is used to fund all types of public expenditure. The tax is a percentage of gross pay, based on the amount a person earns in a tax year that begins on 6 th April and ends on 5 th April the following year. The percentage rate at which the tax is applied varies according to the level of income. The amount of gross income on which tax is payable (known as taxable pay) may be affected by the provision of certain allowances within the tax system and also by certain elements of pay or deductions being exempted from tax (non-taxable items). The rates of tax applied to particular bands of income, and the levels of allowances, are decided each year by the government as part of the Budget. Most people have income tax deducted from their pay by their employer before they receive it; the employer then pays the tax to the government. This method of paying income tax is called PAYE (Pay As You Earn). FromApril 2008, the UK system divides taxable pay into two bands of income, with the following rates of tax applied to each: Table 3.1: Taxation bands Annual taxable income Tax rate 0 – £36,000 20% £36,001 and above 40% This means that, for a person with a total taxable pay of £20,000 in a year, the total tax liability would be: £20,000 × 20% = £4,000 For a person with a total taxable pay of £45,000 in a year, the total tax liability would be: £36,000 × 20% £7,200 (£45,000 ÷ £36,000) × 40% = £3,600 £10,800 These annual tax bands can be converted into monthly or weekly amounts for calculating tax on a PAYE basis and deducting the correct amount from an employees gross pay: 82 Basic Business Applications © ABE and RRC Table 3.2: Annual, monthly and weekly taxable income Tax rate Annual taxable income Monthly taxable income Weekly taxable income 20% 0 – £36,000 0 – £3,000 0 – £692 40% £36,001 and above £3,001 and above £693 and above There are many possible allowances and non-taxable items which affect the level of taxable pay. Here we shall consider just two. Personal allowances – the amount of money an individual is allowed to earn each year before he or she starts to pay tax. This amount (sometimes called free-pay) depends on the individual's circumstances – for example, whether they are single or married, number of children, etc. If your personal allowances are greater than your actual pay, then you would pay no income tax at all. The level of personal allowances are generally set as an annual amount, but this can be converted into monthly or weekly amounts of "free-pay". Pension contributions – the amounts paid by an individual into a pension scheme, such as an employer's occupational pension scheme, are not taxable. Therefore, pension contributions, where they are deducted from pay by the employer, may also be deducted from gross pay before calculating tax. (Note that this does not necessarily apply to any other types of deductions from pay. Where pension contributions are not paid through the employer, an adjustment may be made to a person's personal allowances to allow for them.) Now we can develop a framework for calculating the amount of tax payable annually, using the tax bands set out here. gross taxable pay = gross pay ÷ non-taxable items net taxable pay = gross taxable pay ÷ personal allowances tax liability = first £36,000 of net taxable pay × 20% plus remainder of net taxable pay × 40% Consider the following examples, using the bands and tax rates set out here. Example 1: How much annual tax will be paid by someone who earns £18,600 per annum and has personal allowances of £5,800? What is the monthly tax deduction? There are no non-taxable items specified, so: gross taxable pay = gross pay ÷ non-taxable items = £18,600 net taxable pay = gross taxable pay ÷ personal allowances = £18,600 ÷ £5,800 = £12,800 tax liability = £12,800 × 20% = £2,560 The monthly income tax deducted will be £2,560 divided by 12, i.e. £213 rounded down. Basic Business Applications 83 © ABE and RRC Example 2: How much annual tax will be paid by someone who earns £43,000 per annum, has personal allowances of £4,400 and pays pension contributions through his employer at a rate of 5% of his annual salary. Here, the pension contributions are non-taxable, so: gross taxable pay = £43,000 ÷ (£43,000 × 5%) = £40,850 net taxable pay = £40,850 ÷ £4,400 = £36,450 tax liability = £36,000 × 20% + (£36,450 ÷ £36,000) × 40% = £7,200 + £180 = £7,380 Note that taxation questions are often set without reference to income bands with different taxation rates applying, or without reference to deductions from gross pay or personal allowances. Always answer the question using the information supplied. For example: How much tax will be paid each month by someone who earns £24,000 per annum if the tax rate is 25%? Net taxable pay per month, based on the information supplied = 12 000 , 24 £ = £2,000 Tax liability = £2,000 × 25% = £500 per month. Questions for Practice 7 1. An employee is paid a salary of £24,492 per annum which he receives in 12 monthly payments. He has personal allowances amounting to £4,800. If tax is payable at the rate of 24%, calculate his monthly pay. 2. If, in the next Budget, the Chancellor changes the rate of income tax from 24% to 23%, but wages remain the same, how much less tax will be paid by the employee in Question 1 above. 3. Someone whose personal tax allowance is £4,800 is paid a weekly wage of £80. Would this person have to pay tax? 4. An employee has a gross income of £23,000 and personal allowances of £5,000. She pays £3,600 tax. What is the rate of income tax? 5. A weekly paid worker receives a basic wage of £5.50 per hour for a 37.5 hour week. Overtime is paid at a rate of time and a half for the first five hours, and at double time for any additional hours worked. She also belongs to an occupational pension scheme, making contributions at a rate of 6% of basic earnings only. As a single mother, she has a personal taxation allowance of £5,850. Calculate her net pay in a week when she works 48 hours if the tax bands and rates are as set out above (i.e. as per 2008 in the UK). Now check your answers with those given at the end of the unit. 84 Basic Business Applications © ABE and RRC ANSWERS TO QUESTIONS Questions for Practice 1 1. £45.26 2. (a) £209.30 (b) 105p (rounded). 3. (a) 1640 Swiss francs (b) 590.64 euros. 4. (a) 945 Swedish kroner (b) £5.97 Questions for Practice 2 1. £31 2. £5,950 3. £1,560 4. £24.72 5. £304 6. There are three parts to this question. (a) They would be charged £3 for each cash transaction. Therefore, they would be changing: £107 into NZD = 299.39 NZD £97 into AUD = 238.43 AUD £117 into HKD = 2299.05 HKD. (b) (i) £570 (ii) £5.70 (c) Total cost is the £900 worth of foreign currency plus the commission of £9 for the cash transactions and £5.70 for the travellers cheques = £914.70. Spread between the four members of the family, cost per person is £228.68. Questions for Practice 3 1. £65.63 2. £2,295 interest and £9,045 in the account. 3. £231.25 4. 7.5% Questions for Practice 4 1. £7,986 2. £2,315.25 Basic Business Applications 85 © ABE and RRC 3. To compare the effects of the two rates, work out the accrued amount after one year in each case: For the first building society, the rate of 8.5% payable six monthly equates to 4.25% as a six-monthly rate. Thus, using the formula for calculating the accrued amount over two six month periods is: A = n 100 r 1 P | . | \ | + = £2,000 2 100 25 . 4 1 | . | \ | + = £2,000 × 1.0425 2 = £2,000 × 1.1659134 = £2,173.61 (to the nearest p) The calculation for the second building society rate of 8.75% paid annually is: A = £2,000 1 100 75 . 8 1 | . | \ | + = £2,000 × 1.0875 = £2,175.00 Therefore, the second building society, with the rate of 8.75% paid annually, is the better investment. 4. Using the formula: P = n | . | \ | + 100 r 1 A = 3 100 10 1 9,000 | . | \ | + = ( ) 3 .10 1 9,000 = 1.331 9,000 = £6,762 Therefore, £6,762 would need to be invested today at an interest rate of 10% per annum, if the office is to accrue the £9,000 required to replace some of its machinery in three year's time. Questions for Practice 5 1. (a) £1,200 (b) 5 years 2. £1,750 3. The value at the end of each year is 80% of the value at the beginning of the year. Thus: value at the end of year 1 = £4,500 × 80% = £3,600 value at the end of year 2 = £3,600 × 80% = £2,880 value at the end of year 3 = £2,880 × 80% = £2,304. 86 Basic Business Applications © ABE and RRC 4. The value at the end of each year is 89% of the value at the beginning of the year. Thus: value at the end of year 1 = £795 × 89% = £707.55 value at the end of year 2 = £707.55 × 89% = £629.72 = £630 to the nearest pound. 5. Total cost of the fleets to both companies = £15,250 × 20 = £305,000. Company A: annual amount of depreciation = 5 000 , 305 £ = £61,000 after 3 years, total depreciation = £61,000 × 3 = £183,000 value of the fleet after 3 years = £305,000 ÷ £183,000 = £122,000 Company B: The value at the end of each year is 80% of the value at the beginning of the year. Thus: value at the end of year 1 = £305,000 × 80% = £244,000 value at the end of year 2 = £244,000 × 80% = £195,200 value at the end of year 3 = £195,200 × 80% = £156,160. Questions for Practice 6 1. £220.50 2. £1,562.50 3. £8.50 4. £249.90 5. £344.10 6. £1,456 Questions for Practice 7 1. £1,247.16 per month 2. £16.41 per month 3. No 4. 20% 5. Gross pay is made up as follows: 37.5 hours at the basic rate of £5.50 per hour £206.25 5 hours at (£5.50 × 1.5) per hour £41.25 5.5 hours at (£5.50 × 2) per hour £60.50 gross pay £308.00 Non-taxable items are pension contributions amounting to: £206.25 × 6% = £12.38 Basic Business Applications 87 © ABE and RRC Gross taxable pay = gross pay ÷ non-taxable items = £308.00 ÷ £12.38 = £295.62 Net taxable pay = gross taxable pay ÷ weekly personal allowances = £295.62 ÷ 52 850 , 5 £ = £295.62 ÷ £54.81 = £240.81 Tax liability is based on the weekly income tax bands, and this level of weekly net taxable income falls within the 20% liability band (0 – £692). tax liability = £240.81 × 20% = £48.16 Net pay = gross pay ÷ pension contribution ÷ tax = £308.00 ÷ £12.38 ÷ £48.16 = £247.46 88 Basic Business Applications © ABE and RRC 89 © ABE and RRC Study Unit 4 Simultaneous, Quadratic, Exponential and Logarithmic Equations Contents Page Introduction 90 A. Simultaneous Equations 91 Solution by Elimination 91 B. Formulating Problems as Simultaneous Equations 96 C. Quadratic Equations 98 Solving a Quadratic Equation by Factorisation 99 Solving Quadratic Equations with no Terms in x 102 Solving Quadratic Equations by Formula 103 D. Logarithmic and Exponential Functions 106 The Exponential Number e 106 Logarithms 106 Relationship between Logarithms and the Exponential Number 108 Laws of Logs and Exponents 108 Solving Equations using Logarithms and the Exponential Number 110 Answers to Questions for Practice 111 90 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC INTRODUCTION In Study Unit 2 we introduced the concept of equations and looked at the basic principles of solving them – i.e. finding out the value of the unknown quantity. In that study unit, we considered simple equations where there was only one unknown quantity. Now we shall take our study of equations further by examining equations where there is more than one unknown quantity – simultaneous equations. We shall also consider the distinction between linear and quadratic equations, a difference which will be developed in the next study unit when we look at drawing graphs to represent equations. Finally, we will examine the principles and rules of logarithms and exponential values, and learn how to solve equations using them. Before we start though, we should recap the key principles already examined in relation to equations. It is essential that you fully understand these before studying this unit, so if you are unsure of any points, go back to Study Unit 2 and revise the section on equations there. An equation is a statement of equality between two expressions. It has two sides and they must always be equal. Solving an equation means finding the value of the unknown quantity (or quantities) in the equation – the variable – by using the other numbers, the constants. To do this, the expressions which form the equation may be manipulated with the objective of isolating the unknown on one side of the equation. The golden rule in manipulating equations is equality of treatment to both sides of the equation. What you do to one side of the equation, you must also do to the other. Thus, the same number may be added or subtracted from both sides of the equation without changing its equality, and both sides may be multiplied or divided by the same number without changing its equality. Quantities may be transposed – transferred from one side of the equation to the other – by changing their signs. Thus, an added or subtracted term may be transposed if its sign is changed from + to ÷, or from ÷ to +. Similarly, a multiplier may be transferred from one side of an equation by changing it to the divisor on the other, and a divisor may be transferred from one side by changing it to the multiplier on the other. Unknowns may also be transposed, using the same rules, in order to collect all the unknown terms on the same side of the equation. Objectives When you have completed this study unit you will be able to: use the elimination method to solve simultaneous equations including two unknowns formulate problems as simultaneous equations define a quadratic equation and distinguish it from a linear equation solve quadratic equations by the factorisation method simplify and solve equations using both base 10 and natural logarithms. Simultaneous, Quadratic, Exponential and Logarithmic Equations 91 © ABE and RRC A. SIMULTANEOUS EQUATIONS As we said in the introduction, up until now, when solving equations, we had only one unknown quantity to find. However, in practice, many situations will arise when there is more than one unknown. If an equation involves two unknown quantities, we may find any number of pairs of values to satisfy it. Consider the following example: x + y = 7 This may be satisfied by: x = 3 and y = 4 x = 2½ and y = 4½ x = ÷1.3 and y = +8.3, etc. To determine a particular pair of unknown values then, we need more information. One way of getting more information is to have another equation – for example: x + y = 7 and 2x + y = 12 Now there is only one solution pair: x = 5 and y = 2. Similarly, if x + y = 7 and x ÷ y = 2, there is only one solution: x = 4½ and y = 2½. Therefore to find two unknowns we need two equations; to find three unknowns we need three equations, etc. These groups of equations are known as simultaneous equations. Note though, that for equations to be simultaneous they must satisfy two conditions – they must be consistent and independent. For example: the equations x + y = 7 and 2x + 2y = 5 are inconsistent – they cannot be true simultaneously, and no pair of values which satisfies one will satisfy the other the equations 3x + 3y = 21 and 2x + 2y = 14 are not independent – they are really the same equations, and all pairs of values which satisfy the one will satisfy the other. Solution by Elimination If two equations are true simultaneously, any other equation obtained from them will also be true. The method of solution by elimination depends on this fact. By adding or subtracting suitable multiples of the given equations, we can eliminate one of the unknowns and obtain an equation containing only one unknown. Once we have done that, the resulting equation can be solved using the methods we already know, and then the other unknown may be discovered by substitution in the original equation. The following examples explain the method. Study them carefully. Example 1: Consider the following pair of equations: 3x + y = 11 (1) x ÷ y = ÷3 (2) We can add the two equations together to derive a third equation which will eliminate y: 92 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC 3x + y = 11 x ÷ y = ÷3 4x = 8 Thus, by adding the two equations, we are left with 4x = 8. Therefore, x = 2. If we now substitute that value for x in (1), so that 3x becomes 6: 6 + y = 11 y = 5 (the symbol means "therefore"). We can check that the values we have determined for x and y are correct by substituting both in equation (2): x ÷ y = 2 ÷ 5 = ÷3 Note that eliminating one of the unknowns (in this case, y) was possible because both the "y"s in the original two equations had a coefficient of 1 – i.e. the coefficients were equal. (The coefficient is simply the factor or number applied to an algebraic term, so the coefficient of the term 3x is 3, for example.) If we want to eliminate one of the unknowns by adding or subtracting the two equations, the coefficients of the unknowns must be equal. Example 2: Consider the following pair of equations: 6x + 5y = ÷6 (1) 18x + 7y = 6 (2) We want one of the unknowns to have equal coefficients, but unlike in Example 1, this is not the case here. However, we can see that the coefficient of x in (2) is a multiple of the coefficient of x in (1). To make the coefficients equal, we derive a changed version of the same equation by multiplying (1) by 3 and then subtract (2) from this equation: Multiplying (6x + 5y = ÷6) by 3 gives: 18x + 15y = ÷18 Note that, providing we apply the same proportionate change to each term in the equation (by multiplying or dividing each by the same number), the equation has not been changed. It is not a new, independent equation, but simply a different way of expressing the same equation. We can now subtract equation (2) from this: 18x + 15y = ÷18 18x ÷ 7y = 6 8y = ÷24 y = ÷3 To find x, we substitute the value of y = ÷3 in (1) so that 5y becomes ÷15: 6x + (÷15) = ÷6 By transposition, we get: 6x = ÷6 + 15 = 9 x = 1.5 Simultaneous, Quadratic, Exponential and Logarithmic Equations 93 © ABE and RRC It is always good practice to check the values by substitution in the other equation, but we will skip that step from now on. Example 3: Consider the following pair of equations: 13x ÷ 7y = 33 (1) 9x + 2y = 16 (2) This time, there are no equal or multiple coefficients, so there is no obvious unknown term which can be easily eliminated. We can then choose which unknown to eliminate and, to avoid large numerical terms, we shall decide to eliminate y. Where there are no equal or multiple coefficients, we need to change the way in which both equations are expressed so that we can make the coefficients of one of the unknowns equal. To do this here, we multiply (1) by 2 and (2) by 7, so that the coefficient of y in both equations is 14. We can then eliminate y by adding the two equations: 26x ÷ 14y = 66 63x + 14y = 112 89x = 178 x = 2 To find y, we substitute the value of x = 2 in (2) so that 9x becomes 18: 18 + 2y = 16 By transposition, we get: 2y = 16 ÷ 18 = ÷2 y = ÷1 Important note: When working through the processes of transposition and substitution, it is not necessary to write ÷ (÷7) and + (÷15), etc., but be sure to pay special attention to all calculations involving negative numbers. Questions for Practice 1 Solve the following simultaneous equations by finding values for x and y. 1. 3x + 4y = 1 9x + 5y = 10 2. 5x ÷ 3y = 14 7x + 5y = 38 3. 3x ÷ y = 8 x + 5y = 4 4. 2x ÷ 11y = 23 7x ÷ 8y = 50 94 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC 5. 9x + 10y = 12 2x + 15y = ÷5 6. x + 3 2 y = 16 2 1 x + 2y = 14 7. 2 1 1 x ÷ 5 3 y = 12 x + 15 7 y = 21 Now check your answers with those given at the end of the unit. Now we shall look at some harder examples! (a) Where the unknowns are denominators Consider the following equations: x 4 ÷ y 3 = 4 (1) x 3 – y 7 = 2 (2) Here we can treat x 1 and y 1 as the unknowns and then solve the problem in the usual way. To eliminate y 1 , multiply (1) by 7 and (2) by 3 and then subtract: x 28 ÷ y 21 = 28 (1) x 9 ÷ y 21 = 6 (2) x 19 = 22 x 1 = 19 22 x = 22 19 Substituting for x 1 in (1) gives: 19 22 4× ÷ y 3 = 4 19 88 ÷ 4 = y 3 Simultaneous, Quadratic, Exponential and Logarithmic Equations 95 © ABE and RRC y 3 = 19 76 88 ÷ = 19 12 y 1 = 19 4 y = 4 19 = 4 3 4 Alternatively, you may find it easier to put x 1 = m and y 1 = n, and rewrite the equations as: 4m ÷ 3n = 4 (1) 3m ÷ 7n = 2 (2) Solve these in the usual way, but remember to write your answer as x = ? and y = ?, not as m and n. (b) Forming simultaneous equations from single equations Equations are sometimes given in the form: x + 5y = 2x ÷ 4y + 3 = 3x + 2y ÷ 4 In order to solve this, we need to reformulate the equation as two simultaneous equations. Taking the first pair of expressions, we can rearrange the terms to form one equation in the usual form: x + 5y = 2x ÷ 4y + 3 ÷x + 9y = 3 Now take the second pair of expressions and rearrange them: 2x ÷ 4y + 3 = 3x + 2y ÷ 4 2x ÷ 3y = 4 We now have two simultaneous equations as follows: ÷x + 9y = 3 (1) 2x ÷ 3y = 4 (2) These can be solved in the usual way. To eliminate y, multiply (2) by 3 and add to (1): 5x = 15 x = 3 y = 3 2 96 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC Questions for Practice 2 Solve the following equations for x and y: x 3 + y 20 = 13 x 7 + y 10 = 2 Now check your answers with those given at the end of the unit. B. FORMULATING PROBLEMS AS SIMULTANEOUS EQUATIONS As we have seen previously, it is very useful to develop mathematical models of real situations to help analyse and solve problems. The use of algebraic notation to represent variables in a situation and their combination with numerical constants to form equations are the building blocks of such mathematical models. We have seen that simple equations can be developed to model a situation and enable us to find the value of one unknown variable, where we know the values of all the other variables. In the same way, we can develop models which will enable us to find two or more unknown variables by using simultaneous equations. The key points about formulating problems as simultaneous equations are the same as doing the same with simple equations: Always state clearly what your unknowns represent and give the units you are using. Having found the number for which the unknowns stand, give the answer to the question in words. Check your answer in the question given, not in your own equation. Consider the following examples and try to form the equations for yourself before looking at the solutions given. Note too, that unless you are told to find both or all of the unknowns, it is not essential to do so, but it is most advisable, as otherwise you have no means of checking your answer. Example 1: The expenditure of 10 men and 8 boys amounts to £160. If 4 men together spend £18 more than 6 boys, how much does each man and boy spend? (Assume each man spends the same amount and each boy spends the same amount.) Let: x = expenditure of 1 man in £ y = expenditure of 1 boy in £ Then we can express the information given as equations: 10x + 8y = 160 (all terms must be expressed in the same unit, i.e. here pounds) and 4x ÷ 6y = 18 Solving these two equations will answer the problem. Simultaneous, Quadratic, Exponential and Logarithmic Equations 97 © ABE and RRC Example 2: In a bag containing black and white balls, half the number of white is equal to a third the number of black, and twice the total number of balls exceeds three times the number of black balls by four. How many balls did the bag contain? Let: x = number of white balls y = number of black balls. Expressing the information given as equations, we get: 2 1 x = 3 1 y and 2(x + y) ÷ 3y = 4 Example 3: Pay special attention to the following problem. Note how the equations are built up. A certain number, of two digits, is such that it is diminished by 9 if the digits are reversed. The number itself is six times the sum of its digits. Find the number. Let: x = the tens figure y = the units figure Then the number is 10x + y, and the reversed number is 10y + x. Expressing the information given as equations, we get: 10x + y ÷ 9 = 10y + x (1) and 10x + y = 6(x + y) (2) Rearranging these equations to isolate the constant in (1) and simplify both, we get: for (1): 9x ÷ 9y = 9 or x ÷ y = 1 (A) for (2): 4x = 5y (B) If we now multiply (A) by 4, we get: 4x ÷ 4y = 4 We can then substitute the value of 4x from (B) to get: 5y ÷ 4y = 4 y = 4 Substituting for y in (A), we get: x = 5 The required number is 54. 98 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC Questions for Practice 3 1. Half the sum of two numbers is 20 and three times their difference is 18. Find the numbers. 2. 6 kg of tea and 11 kg of sugar cost £13.30. 11 kg of tea and 6 kg of sugar cost £17.30. Find the cost of tea and sugar per kilo. 3. The income from advertisements and sales for a college magazine amounted, in a year, to £670. In the following year the income from advertisements was increased by 12½% and the income from sales decreased by 16 3 2 %. The total income decreased by £12.50. Find the original income from advertisements and sales using a calculation method. 4. A train from London to Birmingham takes 2 hours for the journey. If the average speed of the train were decreased by 5 miles per hour, the train would take 12 minutes longer. Find the distance from London to Birmingham and the average speed of the train, using simultaneous equations. Now check your answers with those given at the end of the unit. C. QUADRATIC EQUATIONS All the equations we have studied so far have been of a form where all the variables are to the power of 1 and there is no term where x and y are multiplied together – i.e. xy does not feature in the equation. (The types of equation studied so far are known as linear equations since, as we shall see in the next study unit, if we draw a graph of the outcome of the equation for all the possible values of the unknowns, the result is a straight line.) By contrast, a quadratic equation is one that contains the square of the unknown number, but no higher power. For example, the following are quadratic expressions in x: 4x 2 + 7x 2x 2 ÷ x + 1 2 1 x 2 ÷ 2 We can also have quadratic expressions in x 2 . For example, 3x 4 + 2x 2 + 5 is a quadratic in x 2 , since it may be also be written as: 3(x 2 ) 2 + 2(x 2 ) + 5, or 3y 2 + 2y + 5, where y stands for x 2 Note that these are examples of quadratic expressions. A quadratic equation is obtained by making such an expression equal to a similar expression or to a number. One of the features of quadratic equations is that, usually, they cannot be solved definitively. There are generally two possible values for the unknown. Simultaneous, Quadratic, Exponential and Logarithmic Equations 99 © ABE and RRC A second feature, reflecting this, is that when we draw a graph of the outcome of a quadratic equation for all the possible values of the unknown, the result is a curve. Again, we shall examine this later. Solving a Quadratic Equation by Factorisation If an equation contains two unknown numbers, we cannot determine their values unless we are told something more. For example, if x and y are two quantities and all we are told about them is that their product is 12, we cannot tell either of their values – x may be 3, in which case y must be 4; or x may be ÷½, in which case y must be ÷24, etc. This is true whatever the value of the product, with the one exception that if xy = 0, then either x = 0 or y = 0. If x = 0, y may have any value, and if y = 0, x may have any value. They need not both be 0, but one of them must be, since the product of two numbers is zero only when one of them is itself zero. This important fact provides the easiest method of solving a quadratic equation, as is shown in the following examples. Example 1: Consider the following equation: x(x ÷ 3) = 0 Here, the product of the two factors x and (x ÷ 3) is zero. Therefore, one of the factors must be zero: either: x = 0 or: (x ÷ 3) = 0 further, if (x ÷ 3) = 0, then x = 3 so the equation is satisfied if x = 0, or if x = 3. Having understood the principle of this procedure, we do not need to give it in full every time. The following example simplifies the steps. Example 2: Consider the following equation: x(x ÷ 5) = 0 Either, x = 0, or (x ÷ 5) = 0 x = 0, or x = 5 The important thing to remember is that this applies only to a product the value of which is zero. Therefore, the above method of solving an equation applies only when the RHS (right- hand side) of the equation is zero, while the LHS can be expressed in factors. Example 3: Consider the following equation: (x ÷ 5)(x ÷ 2) = 4 We must rearrange this so that the RHS is zero. First of all, we have to expand the product of the factors: x 2 ÷ 7x + 10 = 4 100 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC Then we rearrange the equation so that the RHS is zero: x 2 ÷ 7x + 6 = 0 Next we have to re-express the LHS in factors: (x ÷ 6)(x ÷ 1) = 0 Now we can say that either (x ÷ 6) = 0, or (x ÷ 1) = 0. x = 6, or x = 1 Check this in the original equation just to justify to ourselves that the method works: (x ÷ 5)(x ÷ 2) = 4 if x = 6: (6 ÷ 5)(6 ÷ 2) = 1 × 4 = 4 if x = 1: (1 ÷ 5)(1 ÷ 2) = (÷4)(÷1) = +4 Questions for Practice 4 Solve the following equations: 1. x(x ÷ 1) = 0 2. (x ÷ 1)(x + 2) = 0 3. 3x(x + 1) = 0 4. 4x 2 = 0 5. (2x ÷ 1)(x + 3) = 0 6. (3x ÷ 2)(2x + 3) = 0 Now check your answers with those given at the end of the unit. You should have found these questions very simple and you may even have been able to work them out in your head. However, no matter how simple an equation is, always copy it down first exactly as it is set. Then, if you are sure of the working, you may give the answer at once, but do not just guess. On the other hand, you may prefer to omit the "either ... or ..." step. If you do this, remember to change the sign of the number as you transfer it to the RHS of the equation. For example: (4x ÷ 3)(2x + 5) = 0 4x = 3, or 2x = ÷5 x = ¾, or x = ÷2½ One of the reasons that the questions just given were so easy is that the LHS of the equations were already expressed in factor terms. Commonly, though, equations are not so simple – the factors may not already have been found, or the terms are not already collected on one side of the equation. You will, therefore, have to work as follows: Simultaneous, Quadratic, Exponential and Logarithmic Equations 101 © ABE and RRC Collect all the terms on the LHS, so that the RHS is zero. Factorise the LHS expression – the product of the factors is then zero. Equate each factor in turn to zero, where possible. Finally, give your answers as alternative values for x. Work through the following two examples carefully to make sure that you understand the process thoroughly. Example 4: Consider the following equation: 7x 2 + 23x = 60 First, collect all the terms on the LHS, so that the RHS is zero: 7x 2 + 23x ÷ 60 = 0 Next factorise the LHS expression: (7x ÷ 12)(x + 5) = 0 Then equate each factor in turn to zero: either (7x ÷ 12) = 0, or (x + 5) = 0 which may be expressed as: 7x = 12, or x = ÷5 Finally, give your answers as alternative values for x: x = 7 5 1 or ÷5 Example 5: Solve x(5x ÷ 2) = 3(2x ÷ 1) Expand the product of the factors: 5x 2 ÷ 2x = 6x ÷ 3 Collect the terms: 5x 2 ÷ 8x + 3 = 0 Factorise the LHS: (5x ÷ 3)(x ÷ 1) = 0 5x = 3, or x = 1 x = 5 3 or 1 102 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC Questions for Practice 5 Solve the following equations: 1. 7 ÷ x 2 = 5 ÷ x 2. x 2 + 4x + 4 = 0 3. x 2 ÷ 10x = 24 4. x(x ÷ 2) = 2(2 ÷ x) 5. x 2 ÷ 11x + 30 = 0 6. 2x 2 + 5 = 11x 7. 10x 2 + 11x + 3 = 0 8. (x + 3)(x ÷ 3) = 8x 9. (x ÷ 1) 2 + (x + 1) 2 = 34 10. (x + 1) 2 ÷ (x ÷ 1) 2 = x 2 11. 9x 2 = 25 12. 12x 2 + 12x = 9 Now check your answers with those given at the end of the unit. Solving Quadratic Equations with no Terms in x In particular circumstances, there is a quicker and easier method than the factor method of solving a quadratic equation. Consider the following equation: 25x 2 ÷ 36 = 0 Using the factor method, this would be solved as follows: (5x ÷ 6)(5x + 6) = 0 5x = 6, or 5x = ÷6 x = 1.2 or ÷1.2. There is, however, a simpler method of obtaining this answer and you should use it whenever there is no term in x in the given equation: 25x 2 ÷ 36 = 0 25x 2 = 36 x 2 = 25 36 Simultaneous, Quadratic, Exponential and Logarithmic Equations 103 © ABE and RRC To solve this, take the square root of each side, remembering that there are two square roots numerically equal, but opposite in sign: 2 x = 25 36 x = ± 5 6 or ±1.2 Questions for Practice 6 Solve the following equations: 1. x 2 = 16 2. 4x 2 = 9 3. 3x 2 ÷ 27 = 0 4. x 2 ÷ a 2 = 0 5. 9x 2 = 1 Now check your answers with those given at the end of the unit. Solving Quadratic Equations by Formula Since the method just used for solving a quadratic equation applies to all cases, even where no real roots can be found, it can be used to establish a formula. Consider the equation ax 2 + bx + c = 0, where a, b and c may have any numerical value. Rearranging the equation we get: ax 2 + bx + c = 0 ax 2 + bx = ÷c x 2 + a b x = ÷ a c At this point, we need to "complete the square", which means adding or subtracting the expression that is needed in order to be able to factorise the LHS. In this case, it is: 2 a 2 b | . | \ | Remember that this needs to be added to both sides of the equation to keep it balanced. x 2 + a b x + 2 a 2 b | . | \ | = a c ÷ + 2 2 a 4 b 104 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC Factorising the LHS we get: 2 2a b + x | . | \ | = 2 2 4a 4ac - b x + 2a b = ± 2 2 4a ac 4 b ÷ x = ÷ 2a b ± 2a ac 4 b 2 ÷ x = a 2 ac 4 b b 2 ÷ ± ÷ This is an important formula. When using it, always write it down first, then write the values of a, b, c in the particular case, paying great attention to the signs. It can be used to solve any quadratic equation, although you should use the factor method whenever possible. Example 1: Consider the following equation: 5x 2 ÷ 15x + 11 = 0 x = a 2 ac 4 b b 2 ÷ ± ÷ where: a = 5, b = ÷15, c = 11 Therefore: x = 10 220 225 15 ÷ ± + 10 236 . 2 15± = = 1.72 or 1.28 (to two decimal places) Example 2: Consider the following equation: 3x 2 + 4x ÷ 5 = 0 x = a 2 ac 4 b b 2 ÷ ± ÷ where: a = 3, b = 4, c = ÷5 Therefore: x = 6 60 16 4 + ± ÷ = 6 718 . 8 4 ± ÷ = 6 718 . 4 or 6 718 . 12 ÷ = 0.79 or ÷2.12 (to two decimal places) Simultaneous, Quadratic, Exponential and Logarithmic Equations 105 © ABE and RRC Example 3: Consider the following equation: 3 ÷ 2x 2 = 4x First arrange equation with x 2 term positive: 2x 2 + 4x ÷ 3 = 0 Then apply the formula as usual. Questions for Practice 7 Solve the following by means of the formula: 1. 2x 2 + 13x + 4 = 0 2. 5x 2 ÷ 3x ÷ 4 = 0 3. x 2 ÷ 5x + 2 = 0 4. 4x 2 + 7x + 1 = 0 5. x 2 ÷ x ÷ 1 = 0 Now check your answers with those given at the end of the unit. Remember that the square of a number is always positive, and that no real square root can be found for a negative number. Therefore, if the term b 2 ÷ 4ac is negative, no real roots of the equation can be found. For example: 2x 2 + 4x + 5 = 0 Applying the formula: x = a 2 ac 4 b b 2 ÷ ± ÷ where: a = 2, b = 4, c = 5 Therefore: x 4 40 16 4 ÷ ± ÷ = = 4 24 4 ÷ ± ÷ There are no real roots to this expression. If b 2 ÷ 4ac is a perfect square, then the square root will be a whole number and you should have solved the equation by the factor method. 106 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC D. LOGARITHMIC AND EXPONENTIAL FUNCTIONS So far in this study unit we have considered simultaneous and quadratic equations. In this section we will now move on to consider logarithmic and exponential functions. The Exponential Number e There is a particular constant used in many mathematical calculations whose value (to seven decimal places) is 2.7182818. This value occurs so frequently when mathematics is used to model physical and economic phenomena that it is convenient to write it simply as e; it is known as the exponential constant or exponential number. Have a go on your scientific calculator now to obtain the value of e. To do so, simply locate and press the e x button on your calculator and enter 1. (Calculators differ, so you may have to experiment with your own calculator to find out how to obtain the value of e. Some scientific calculators may require you to enter 1 before pressing the e x button.) By performing this calculation, you have just demonstrated that: e 1 = 2.7182818 Questions for Practice 8 Using your scientific calculator, find the following values: 1. e 0 2. e 3 3. e -2 4. e 1/2 5. e -3 Now check your answers with those given at the end of the unit. Logarithms A logarithm (also known simply as a log) is used to write an expression involving powers. Having mastered the application of powers in Study Unit 2, you should have no problem handling logarithms! In essence, a logarithm is the power to which a fixed number (base) must be raised in order to produce a given number. Let us work through some examples to aid our understanding of the application of logarithms. It is common to work with logarithms using the number 10 as the base, referred to as common logarithms. Let us first consider the expression 10 2 . From our understanding of powers, 10 2 = 100 (i.e. 10 × 10 = 100). In other words, 10 is the base and 2 is the power or index. Using common logarithms we can rewrite this expression as: log 10 100 = 2 Simultaneous, Quadratic, Exponential and Logarithmic Equations 107 © ABE and RRC This statement can be read as the "log to the base 10 of 100 is 2". In other words, 2 is the power to which the base number 10 must be raised in order to produce the number 100. Thus, a logarithm is basically an alternative way of writing an expression involving powers. Using this logic, then: log 10 1000 = 3 (i.e. 3 is the power to which the base number 10 must be raised in order to produce the number 1000). log 10 10 = 1 (i.e. 1 is the power to which the base number 10 must be raised in order to produce the number 10). Using your scientific calculator, have a go at confirming the above results. To do so, simply locate and press the log button on your calculator and enter the number you want to find the common logarithm of. (Note – calculators differ, so you may have to experiment with your own calculator to find out how to obtain the common log of a number. Some calculators may require you to enter the number you want to find the logarithm of before pressing the log button.) Questions for Practice 9 Using your scientific calculator, find the logs of the following values: 1. log 10 50 2. log 10 45 3. log 10 2 4. log 10 0.4 5. log 10 4.4 Now check your answers with those given at the end of the unit. Another base that we frequently use when working with logarithms (in addition to base 10) is the exponential number e (i.e. the value of 2.7182818). Logarithms to the base e are commonly referred to as natural logarithms and are denoted by ln or loge. For example: ln(2) = 0.693 This statement can be read as the "log to the base e of 2 is 0.693" or "the natural log of 2 is 0.693". In other words, 0.693 is the power to which the base value of e (i.e. 2.7182818) must be raised in order to produce the number 2. Using your scientific calculator, have a go at confirming this result. To do so, simply locate and press the ln button on your calculator and enter the number you want to find the natural logarithm of. (Note – calculators differ, so you may have to experiment with your own to find out how to obtain the natural log of a number. Some calculators may require you to enter the number you want to find the natural logarithm of before pressing the ln button.) 108 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC Questions for Practice 10 Using your scientific calculator, find the natural logarithms of the following values: 1. ln(1.76) 2. ln(0.3) 3. ln(2002) 4. ln(10) 5. ln(38) Now check your answers with those given at the end of the unit. Relationship between Logarithms and the Exponential Number From this discussion of natural logarithms and the exponential number, you may have identified that a relationship exists between the two. The natural logarithm function f(x) = e x is the inverse of the exponential function f(y) = ln(x). Notice that the variables x and y have been reversed. This is not a mistake! When we take the natural logarithm of an exponential function the result is the exponent, so if y = e x , then ln(y) = x. Use your calculator to confirm this natural logarithmic and exponential relationship. First, think of a number for y and then calculate the natural logarithm of this number to find x (i.e. using the formula y = e x ). Then, using this value for x, calculate the value of y using the exponential function y = e x . The resulting value of y should be the same as the initial number you chose for y, thus demonstrating that an inverse relationship exists between the natural logarithm function f(x) = e x and the exponential function f(y) = ln(x). Thus, we can conclude that: ln(e x ) = x Laws of Logs and Exponents Numbers expressed in logarithmic or exponential form obey the same rules applicable to powers and indices, which have already been described in Study Unit 2. Here we give the various rules of logarithms (which can be equally applied to common logarithms (i.e. to the base 10) and natural logarithms (to the base e)). Since logarithms are basically powers, there should be some logic to the following rules. Thus, for any base: - log(p × q) = log p + log q - log | . | \ | q p = log p ÷ log q - log p n = n log p - if y = ax n then n = (log y ÷ log a) ÷ log x - log 1 = 0 and finally, do not forget that: - ln(e x ) = x Simultaneous, Quadratic, Exponential and Logarithmic Equations 109 © ABE and RRC Using these laws, we can simplify some expressions involving logarithms to a single log term. Example 1: Simplify the following logarithm expression into a single log term: log(x) + 2 log(y) Using the rule [log p n = n log p] we can rearrange the second part of the expression: log(x) + log(y 2 ) Using the rule [log(p × q) = log p + log q] we can rearrange the expression: log(xy 2 ) Example 2: Simplify the following logarithm expression into a single log term: 3ln(A) ÷ 4ln(B) Using the rule [log p n = n log p] we can rearrange both parts of the expression: ln(A 3 ) – ln(B 4 ) Using the rule [log | . | \ | q p = log p ÷ log q] we can rearrange the expression: ln | . | \ | 4 3 B A Questions for Practice 11 Simplify the following logarithmic expressions to a single log term: 1. log(x) + log(x ÷ 4) 2. 4 log(2) + 2 log(3) ÷ log(12) 3. ln(x2y) ÷ ln(y) + 3 ln(x) Now check your answers with those given at the end of the unit. 110 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC Solving Equations using Logarithms and the Exponential Number Before completing this section on logarithms and exponents, let us bring together everything we have learnt about them, as well as how to solve equations in general, by answering the following questions. Questions for Practice 12 Solve the following equations, giving your answer correct to 2 decimal places (you will need a scientific calculator to evaluate logs and natural logs): 1. 4x = 100 2. e2x = 10 3. log 3x = 3 Now check your answers with those given at the end of the unit. Simultaneous, Quadratic, Exponential and Logarithmic Equations 111 © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. 3x + 4y = 1 (i) 9x + 5y = 10 (ii) To eliminate x, multiply (i) by 3 so that the coefficients are equal and then subtract (ii): 9x + 12y = 3 9x + 5y = 10 7y = ÷7 y = ÷1 Substituting y = ÷1 in (i) we get: 3x ÷ 4 = 1 3x = 5 x = 3 2 1 (Check in (ii): 15 ÷ 5 = 10) 2. 5x ÷ 3y = 14 (i) 7x + 5y = 38 (ii) To eliminate y, multiply (i) by 5 and (ii) by 3 so that the coefficients are equal and then add: 25x ÷ 15y = 70 21x + 15y = 114 46x = 184 x = 4 Substituting x = 4 in (i) we get: 20 ÷ 3y = 14 3y = 6 y = 2 (Check in (ii): 28 + 10 = 38) The full workings for the next three answers are not given, but the method is indicated. 3. 3x ÷ y = 8 (i) x + 5y = 4 (ii) To eliminate y, multiply (i) by 5 and add the equations: 16x = 44, x = 4 3 2 or to eliminate x, multiply (ii) by 3 and subtract (i): 6y = 4, y = 4 1 112 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC 4. 2x ÷ 11y = 23 (i) 7x ÷ 8y = 50 (ii) To eliminate y, subtract 8 times (i) from 11 times (ii): x = 6 or to eliminate x, subtract 7 times (i) from 2 times (ii): y = ÷1 5. 9x + 10y = 12 (i) 2x + 15y = ÷5 (ii) To eliminate y, subtract 2 times (ii) from 3 times (i): x = 2 or to eliminate x, subtract 2 times (i) from 9 times (ii): y = ÷ 5 3 6. x + 3 2 y = 16 (i) 2 1 x + 2y = 14 (ii) To eliminate x, subtract 2 times (ii) from (i): 3 3 1 y = 12, 10y = 36, y = 10 6 3 substituting for y in (ii): 2 1 x + 5 6 6 = 14 2 1 x = 14 ÷ 5 1 7 2 1 x = 5 4 6 x = 5 3 13 7. 2 1 1 x ÷ 5 3 y = 12 (i) x + 15 7 y = 21 (ii) Notice that (i) has a factor 3, so divide by this to get a simpler equation: 2 1 x ÷ 5 1 y = 4 (i) To eliminate x, subtract 2 times (i) from (ii): ( 5 2 15 7 + )y = 13, 15 13 y = 13, y = 15 Substituting in (i) for y gives: 2 1 x ÷ 3 = 4 2 1 x = 7 x = 14 Simultaneous, Quadratic, Exponential and Logarithmic Equations 113 © ABE and RRC Questions for Practice 2 Let a x 1 = , and b y 1 = . Then express the equations using these terms: 3a + 20b = 13 7a ÷ 10b = 2 Solving these in the normal way gives: a = 1, and b = 2 1 Therefore: x = 1, and y = 2 Questions for Practice 3 1. Let: x = first number y = second number Then, if half the sum of two numbers is 20: 2 1 (x + y) = 20 or x + y = 40 (i) and, if three times their difference is 18: 3(x ÷ y) = 18 or x ÷ y = 6 (ii) Eliminating x by subtraction, we get: 2y = 34 y = 17 Substituting for y in (i), we get: x + 17 = 40 x = 23 Therefore, the numbers are 17 and 23. 2. As the units must all be the same, change each amount of money to pence. Let: x = cost of tea per kg in pence y = cost of sugar per kg in pence. Then 6x + 11y = 1,330 (i) 11x + 6y = 1,730 (ii) 114 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC To eliminate x, subtract 6 times (ii) from 11 times (i) to get: 85y = 4,250 y = 50 Substituting for y in (i) we get: 6x + 550 = 1,330 x = 6 550 1,330÷ x = 130 Therefore, the cost of tea = £1.30 per kilo, and the cost of sugar = £0.50 per kilo. 3. Let: x = the original income from advertisements (in £) y = the original income from sales (in £). Thus, in the first year: x + y = 670 (i) In the second year: (x + 12 2 1 %x) + (y ÷ 16 3 2 %y) = 670 ÷ 12.50 (x + 8 1 x) + (y ÷ 6 1 y) = 657 2 1 8 9 x + 6 5 y = 657 2 1 Multiplying this by 24 to eliminate the fractions, we get: 27x + 20y = 15,780 (ii) To eliminate y from the pair of equations, multiply (i) by 20: 20x + 20y = 13,400 and subtract this from (ii) to get: 7x = 2,380 x = 340 Substituting for x in (i): 340 + y = 670 y = 330 Thus, the original income from: advertisements = £340 sales = £330. 4. Let: x = the average speed (in miles per hour) y = the distance (in miles) We can use a formula for the time taken: speed distance = time taken Simultaneous, Quadratic, Exponential and Logarithmic Equations 115 © ABE and RRC So at average speed: x y = 2 hours and at the slower speed: ) 5 (x y ÷ = 2 hours 12 minutes = 5 1 2 hours Rearranging these equations into standard linear form, we get: y = 2x (i) y = 5 1 2 x ÷ 11 (ii) Subtracting (ii) from (i) gives: 0 = 5 1 x ÷ 11 11 = 5 1 x 55 = x x = 55 Substituting for x in (i), we get: y = 110 Thus: average speed for the journey = 55 mph distance from London to Birmingham = 110 miles. Questions for Practice 4 1. x = 0 or 1 2. x = 1 or ÷2 3. x = 0 or ÷1 4. x = 0 5. x = 2 1 or ÷3 6. x = 3 2 or 1 2 1 Questions for Practice 5 1. x = 2 or ÷1 2. x = ÷2 3. x = 12 or ÷2 4. x = 2 or ÷2 5. x = 5 or 6 6. x = 2 1 or 5 7. x = ÷ 5 3 or ÷ 2 1 116 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC 8. x = 9 or ÷1 9. x = 4 or ÷4 (written as x = ±4) 10. x = 0 or 4 11. x = ± 3 2 1 12. x = ÷1 2 1 or 2 1 Questions for Practice 6 1. x = ±4 2. x = ±1 2 1 3. x = ±3 4. x = ±a 5. x = ± 3 1 Questions for Practice 7 1. ÷0.32, ÷6.18 2. 1.24, ÷0.64 3. 4.56, 0.44 4. ÷0.16, ÷1.59 5. 1.62, ÷0.62 Questions for Practice 8 1. e 0 = 1 2. e 3 = 20.086 3. e -2 = 0.135 4. e 1/2 = 1.649 5. e -3 = 0.050 Questions for Practice 9 1. log 10 50 = 1.699 2. log 10 45 = 1.653 3. log 10 2 = 0.301 4. log 10 0.4 = -0.398 5. log 10 4.4 = 0.643 Simultaneous, Quadratic, Exponential and Logarithmic Equations 117 © ABE and RRC Questions for Practice 10 1. ln(1.76) = 0.5653 2. ln(0.3) = ÷1.204 3. ln(2002) = 7.602 4. ln(10) = 2.303 5. ln(38) = 3.638 Questions for Practice 11 1. Using the rule [log(p × q) = log p + log q]: log(x) + log(x ÷ 4) = log(x (x ÷ 4)) = log(x 2 ÷ 4x) 2. Using the rule [log p n = n log p]: 4 log(2) = log(2 4 ) = log(16), and 2 log(3) = log(3 2 ) = log(9) Using the rules [log(p × q) = log p + log q] and [log | . | \ | q p = log p ÷ log q]: 4 log(2) + 2 log(3) ÷ log(12) = log(16) + log(9) ÷ log(12) = log | . | \ | × 12 9 16 = log(12) 3. Using the rules [log(p × q) = log p + log q] and [log p n = n log p]: ln(x 2 y) ÷ ln(y) + 3ln(x) = ln(x 2 ) + ln(y) ÷ ln(y) + ln(x 3 ) = ln(x 5 ) Questions for Practice 12 1. 4 x = 10, 4 log 100 log x ÷ = , 32 . 3 x = 2. e 2x = 10, ) 10 ( In x 2 = , 2 10 In x ÷ = 15 1 x · = 3. log 3x = 3, 3x = 10 3 , x = 10 3 ÷3 x = 333.33 118 Simultaneous, Quadratic, Exponential and Logarithmic Equations © ABE and RRC 119 © ABE and RRC Study Unit 5 Introduction to Graphs Contents Page Introduction 120 A. Basic Principles of Graphs 121 Framework of a Graph 121 Drawing a Graph 122 B. Graphing the Functions of a Variable 124 Representation 125 Ordered Pairs 126 C. Graphs of Equations 126 Linear Equations 127 Quadratic Equations 132 Equations with x as the Denominator 136 Exponential and Logarithmic Equations 139 D. Gradients 141 Gradient of a Straight Line 142 Equation of a Straight Line and Gradient 145 E. Using Graphs in Business Problems 147 Business Costs 147 Break-even Analysis 151 Economic Ordering Quantities 152 Plotting Time Series Data 155 Answers to Questions for Practice 160 120 Introduction to Graphs © ABE and RRC INTRODUCTION For many problems and issues which fall within the remit of quantitative methods, it is possible to analyse and solve them both mathematically and graphically. This study unit introduces the principles and role of graphs in such problems. There are many different types of graph, but here we will only be concerned with linear (straight line), quadratic (curved line), exponential and logarithmic graphs. Line graphs illustrate the relationships between two variables, showing graphically what happens to one variable when the other changes. For example, many graphs involve time as one of the variables, and thus provide a measure of something that changes with time. Whenever one measurable quantity changes as a result of another quantity changing, it is possible to draw a graph. You will immediately realise that since equations are a mathematical expression of the relationship between two variables, it is possible to represent equations by means of graphs. We shall spend some time on this in the unit. The importance of graphs is that they give a visual picture of the behaviour of the two variables. From this, by knowing how to interpret it, information can very quickly be found. The picture may, for example, indicate a trend in the data from which we might be able to make further predictions. The graph might also be used to approximate a solution to a particular problem. Graphical approaches to problem solving are often easier to carry out than mathematical ones. However, they are less accurate, since the answers must be read from the graph and hence cannot be more accurate than the thickness of the pencil. For this reason, it is essential that you are very careful when drawing graphs, and we cover some of the key design aspects here. One element of this is the need to use graph paper. You will find it very useful to have a supply to hand as you work through this unit. Objectives When you have completed this study unit you will be able to: explain the construction of graphs calculate functions of x (that is, f(x)) calculate and tabulate the values of y from an equation in x draw graphs from linear, quadratic, exponential and logarithmic equations identify the characteristics of a straight line in terms of its equation, and calculate its gradient from its graph or its equation decide, from an equation, which way its graph will slope identify, from looking at their equations, whether two lines are parallel describe a shaded region on a graph in terms of the lines bounding it use graphs to solve a variety of problems. Introduction to Graphs 121 © ABE and RRC A. BASIC PRINCIPLES OF GRAPHS A graph is a diagram illustrating one or more mathematical relationships between two variables. The following figures show the relationship between the height of a plant and the time since it was planted. We shall use this to explain some of the basic features of graphs and their construction. We shall start by considering the framework of the graph. Framework of a Graph Figure 5.1: Sample graph framework A graph has two axes along which the values of the variables are shown: The y-axis is the vertical axis and usually displays the dependent variable – i.e. the variable which changes according to changes in the other variable. Here this is the height of the plant (which changes in accordance with the time since planting). The x-axis is the horizontal axis and usually displays the independent variable – i.e. the variable which causes change in the other variable. Here this is the time since planting (which causes the height of the plant to increase). The values of the variables are marked along each axis according to a scale (here, 0–9 on the x-axis and 0–6 on the y-axis). Not all values need to be marked, but the dividing marks do need to be equally spaced along the axis and the scale used must be clear to the reader. The scale need not be the same on both axes. Each axis should be titled to show clearly both what it represents and the units of measurement. The area within the axes (known as the plot area) is divided into a grid by reference to the scale on the axes. This grid may actually be marked on the paper (as is the case with graph paper, which is why it is much easier to draw graphs using graph paper) or may be invisible. We have shown the main gridlines here as grey, dotted lines. Particular points within the graph are identified by a pair of numbers called co-ordinates which fix a point by reference to the scales along the axes. The first co-ordinate number always refers to a value on the scale along the x-axis and the second to the scale on the y- 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 Time since planting (days) Height of plant (cm) y-axis origin co-ordinate (3,3) co-ordinate (6,2) x-axis 122 Introduction to Graphs © ABE and RRC axis. We have shown two points on the framework in Figure 5.1 – make sure you understand their referencing. The point at which the two axes meet is called the origin. This has the co-ordinate reference (0,0). Finally, note that the x and y-axes may extend to include negative numbers. In this case, the framework of the graph will look like that shown in Figure 5.2. Figure 5.2: Graph framework including negative scales Drawing a Graph When drawing a graph, follow this procedure for constructing the framework. Determine the range of values which will need to be shown on each of the axes. You should start by considering the x-axis, since the range of values for the independent variable will determine the range of values needed for the dependent variable on the y- axis. Draw the axes and mark off the scales along them. The aim is usually to draw as large a graph as possible since this will make it easier to plot the co-ordinates accurately, and also make it easier to read. Remember that the scales must be consistent along each axis, but do not need to be the same on both. If you are using graph paper, use the thick lines as your main divisions. If you are not using graph paper, use a ruler to mark the scale and select easy-to-identify measurements for the divisions (such as one cm for each main unit). Give each axis a title, number the main divisions along each scale and give the whole graph a title. Now you are ready to start plotting the graph itself. Points on the graph are identified by their co-ordinates. These are the values given (or determined) for firstly, the independent variable and, secondly, for the dependent variable. Thus, you may be given the following data about the growth of a plant: 0 y-axis x-axis co-ordinate (3,1) 3 2 1 3 2 1 1 2 3 1 2 3 co-ordinate (2,2) co-ordinate (1,2) Introduction to Graphs 123 © ABE and RRC 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 Time since planting (days) Height of pl ant (cm) Height of plant (cm) Day 1 1 Day 3 2 Day 5 3 Day 7 4 The points should be plotted with a small x using a sharp pencil. To plot the co-ordinate (1,1) you count one unit across (along the x-axis) and then count one unit up (along the y-axis). Be careful not to get these in the wrong order or when you come to plot the co-ordinate (7,4) you will have plotted the point (4,7), which is somewhere completely different! (Remember that if the x co-ordinate is negative you count to the left and if the y co-ordinate is negative you count downwards.) Once you have plotted all the given points, you should join them all up. If your graph is a linear graph, all the points will be in a straight line and you can join them using a ruler. (In fact, to draw a linear graph, you only need to plot two points which, when joined and extended across the whole scale of the x-axis, give the complete graph for all values of the independent variable.) However, if your graph is a quadratic graph, then the plot line will form a curve. The points should be joined with a smooth curve. The easiest way of doing this is to trace through the points with a pencil, but without touching the paper, to find the shape of the curve. When you are satisfied that the path of the curve is smooth, draw the curve through the points in one movement. The full graph derived from the previous data on the height of a plant and extended over nine days is given in Figure 5.3. Figure 5.3: Graph of height of plant against time Now that we have reviewed the basic principles, we shall move on to consider in detail the way in which variables and equations are represented on graphs. 124 Introduction to Graphs © ABE and RRC B. GRAPHING THE FUNCTIONS OF A VARIABLE The statement that something is a "function of x" simply means that the value of that something depends upon the value of a variable quantity, x. It is denoted by f(x). If y f(x), y will have a different value for each value of x, so that we can obtain a succession of values for y by giving successive values to x. For example, if y x 3, then when x is 0, y will be equal to 3; when x 1, y 4; when x 2, y 5; when x 3, y 6, and so on. For convenience, we usually tabulate our values for x and y, as follows: x 0 1 2 3 4 5 y x 3 3 4 5 6 7 8 Let's look at another example. Suppose you are given the equation y x 2 3 and are asked to find the values of y from x 2 to x 4. You could start by making the following table: x 2 1 0 1 2 3 4 x 2 4 1 0 1 4 9 16 3 3 3 3 3 3 3 3 y x 2 3 7 4 3 4 7 12 19 Alternatively, as this is a simple problem, you could tabulate just the first and the last lines, working the rest in your head as you go along. We can also find a single function of x by substituting that value directly into the given equation, for example: f(x) x 2 2x Replace x by 3 and find the value: f(3) 3 2 2 3 15 Replace x by 2 and find the value f(2) (2) 2 2(2) 0 Introduction to Graphs 125 © ABE and RRC Questions for Practice 1 1. Complete the table for y 3x 2. x 0 0.5 1 1.5 2 2.5 y 3x 2 2. Which one of the following functions could be represented by the following table? y x 4 y x 2 6x 16 y 2x 1 y x 3 1 x 3 5 7 9 11 y 7 11 15 19 23 3. If f(x) 5x 2 find: (a) f(2) (b) f(3). 4. If f(x) x 2 2 find: (a) f(2) (b) f(3) Now check your answers with those given at the end of the unit. Representation From the tables we see that as the value of x changes, so does the value of y. One way of showing how these values change is by means of a graph. The values in the table can be expressed as a pair of co-ordinates which may be plotted on a graph. The first figure is the value of x along the scale of the x (horizontal) axis, and the second figure is the value of y along the scale of the y (vertical) axis. For example, in Figure 5.4, A is the point (2,1), B is (4,2), C is (3,1), D is (5, 2 1 ), E is (1,0) and F is (0,1 2 1 ). 126 Introduction to Graphs © ABE and RRC Figure 5.4: Co-ordinates for y f(x) Ordered Pairs Another way of describing the varying values of x and y is by means of a set of ordered pairs. This is a very useful and helpful way of noting the points we required in the graphical representation of a function. The table in Question 2 of Questions for Practice 1 could be written as a set of ordered pairs as follows: (3,7) (5,11) (7,15) (9,19) (11,23) These are then the points which we would plot if we were going to draw the graph of the function in question. When dealing with ordered pairs, you will normally be told the values of x with which you are concerned. Thus, if we had to write down the ordered pairs of the function 2x 1 for integral (whole number) values of x from 1 to 2 inclusive, we would write (1,1), (0,1), (1,3), (2,5); and if the function was x 2 3 for the same values of x it would be (1,4), (0,3), (1,4), (2,7). C. GRAPHS OF EQUATIONS In this section we shall consider graphs of linear, quadratic, exponential, logarithmic and hyperbolic equations. Linear equations – equations of the first degree in x and y (i.e. containing no higher power of x than the first, such as y 2x 3, y 5x 1). As the name implies, the graph of expressions like these will always be a straight line. Quadratic equations – those of the second degree in x and y (i.e. containing the square of the independent variable, but no higher power, such as y 2x 2 x 1, y ½x 2 3). Such equations always give a curved graph with a characteristic "U" shape, called a parabola. Hyperbolic equations – we shall look at graphs from the expression y x a where a is a constant. Such graphs have a different curved shape – a hyperbola. 2 4 6 2 0 3 1 2 2 1 4 6 A B C D E F Introduction to Graphs 127 © ABE and RRC Exponential and logarithmic equations – an exponential or logarithmic function occurs when the independent variable appears as the exponent or as a logarithm. When plotted, equations of both exponentials and logarithms produce non-linear lines. Linear Equations A linear equation is of the form y mx c, where m and c are constants. This will always give a straight line. (Note that such an equation can be written in different forms, e.g. y b ax.) To draw a line we must know two points on it. It is even better if we know three points, using the third as a check. So, the minimum number of points you ought to plot for a linear graph is three. If we choose any three values of x we can find the corresponding values of y and this will give us the co-ordinates of three points which lie on the line. Example 1: Draw the graph of y 2x 3. Suppose we choose x 0, x 2 and x 2. Taking the given equation, y 2x 3: when x 0: y 0 3 3 when x 2: y (4 3) 1 when x 2: y (4 3) 7 This is better tabulated as: x 0 2 2 y 2x 3 3 1 7 We now have three points which we can plot and, on joining them, we have the graph of y 2x 3. Figure 5.5: Graph of y 2x – 3 -8 -6 -4 -2 0 2 4 -3 -2 -1 0 1 2 3 x y 128 Introduction to Graphs © ABE and RRC Example 2: Try this one yourself. Draw the graph of y 3x 2, for values of x from 3 to 2. First choose three values of x, say 3, 0 and 2. Then draw up a table of values – fill in the spaces for yourself to find the co-ordinates of the points to be plotted. x 3 0 2 y 3x 2 You should have found the points: (3, 7), (0, 2), (2, 8). Now draw your two axes to cover the range of values for x specified (although it is good practice to extend them beyond this – leaving some spare room at each end of each axis), mark the scale, plot your points and join them up. Then compare your result with Figure 5.6. Figure 5.6: Graph of y 2x + 2 If you look carefully at Figure 5.6 you will notice that the line y 2x 2 meets the y-axis at the point where y 2. This is because the constant c in the general equation y mx c is always the point where the line passes through the y-axis (i.e. where x 0). A further fact which emerges is that when x goes from 0 to 2 (i.e. increases by two), y rises from 2 to 8 (i.e. increases by four). The change in the value of y for a change of one in the value of x is given in the equation y mx c by the constant m. We have thus found that in the equation y mx c: the constant c the point where the line crosses the y-axis the constant m the slope of the line. The point at which the line crosses an axis is called the intercept. The y intercept is where the line crosses the y-axis and the x intercept is where the line crosses the x-axis. We shall return to these points in some detail later. -8 -6 -4 -2 0 2 4 6 8 -4 -2 0 2 4 x y Introduction to Graphs 129 © ABE and RRC Example 3: Suppose we are asked to draw the graph of 3y 2x 1, from x 4 to 2. Before we can work out the table of values in this case, we must change the equation to the form y mx c. So we rewrite the equation, having divided both sides of it by 3: 3 1 2x y Now choose some values of x – let's say 4, 1, 2. Working these out gives us the following table of values: x 4 1 2 y 3 1 1 Now plot these points and join them. Note that, as in the title to Figure 5.7, this is still a graph of the original equation. Figure 5.7: Graph of 3y 2x + 1 -3 -1 1 3 -2 -1 0 1 2 3 4 x y 130 Introduction to Graphs © ABE and RRC Example 4: We are not limited to having just one line plotted on a graph. We can draw two or more graphs using the same axes – a very useful method of comparing different equations of the same variables. For example, we could draw the graphs of y 3 1 x and y 8 x on the same axes. Taking values of x between 4 and 10, we get the following tables of values: For y 3 1 x x 3 0 3 y 1 0 1 For y 8 x x 2 4 10 y 10 4 2 The graphs are shown in Figure 5.8. Figure 5.8: Graphs of y ⅓ x and y 8 – x You will notice that the lines intersect (i.e. cross) at A (6,2) -2 0 2 4 6 8 10 -4 -2 0 2 4 6 8 10 x y A Introduction to Graphs 131 © ABE and RRC Questions for Practice 2 1. Draw the graph of y 5x 4, from x 1 to x 2. 2. Draw the graph of y 2x 3, from x 4 to x 2. 3. Draw the graph of y 2 x, for values of x from 4 to 6. 4. On the same axes, plot the following pairs of points on squared paper, and draw the line through them: (i) (3,0) (0,6) (ii) (1,0) (0,5) (iii) (3,2) (2,3) (iv) (4,3) (1,2) Write down the co-ordinates of the points where (i) and (ii) each cut the lines (iii) and (iv). 5. For values of x from 3 to 6, draw on the same axes the graphs of: (i) y 5 (ii) x 5 (iii) y 5x (iv) y 1 5 x Describe briefly the resultant lines. 6. For values of x from 3 to 6, draw on the same axes the graphs of: (i) y 2x (ii) y 2x 3 (iii) y 2x 4 Describe briefly the resultant lines and their intercepts. 7. For values of x from 3 to 6, draw on the same axes the graphs of: (i) y x 2 (ii) 3y 6 x Identify the point of intersection. Now check your answers with those given at the end of the unit. Characteristics of Straight-Line Graphs If you consider the graphs we have drawn as examples, and those you drew in the Questions for Practice, we can start to identify a number of characteristics of straight-line graphs. The general equation for a straight line is y mx c where m and c are constants which may have any numerical value, including 0. The value of m is known as the coefficient of x, and plays a very important part in calculations concerning a straight-line graph. Lines parallel to an axis Consider the case where m 0. This means y c. Whatever value x has, y is always equal to c. The graph of y c must, then, be a line parallel to the x-axis at a distance c from it. Note that y 0 is the x-axis itself. In the same way, x c is a line parallel to the y-axis at a distance c from it. x 0 is the y-axis. 132 Introduction to Graphs © ABE and RRC Lines through the origin If c 0, then the equation of the line is y mx. When x 0 it is clear that y 0, so the graph of y mx is a line through the origin (0,0). Parallel lines When the coefficient of x is the same for several lines, the lines are parallel. Thus, if m is kept the same and c has different values, we get a set of parallel straight lines. Direction of slope If the coefficient of x is negative (for example, y 4 3x) the line will slope down from left to right instead of up, as it does when the coefficient of x is positive. Thus, in y mx c, if m is positive the line slopes up from left to right; if m is negative the line slopes down. Intercepts The distances from the origin to the point where the line cuts the axes are called the intercepts on the axes. They are found by putting x 0 and y 0 in the equation of the graph. In y mx c, the intercept on the y-axis is always at c. Quadratic Equations A quadratic equation is of the form y ax 2 bx c, where a, b and c are constants. This will always give you a curved graph, and the part that is usually asked for is the part that does a "U" turn. The general shape of a quadratic equation will be as shown in Figure 5.9. Figure 5.9: General shape of a quadratic equation In contrast to linear graphs, in order to draw a quadratic graph accurately, you need as many points as possible – especially around the dip. The maximum and minimum values of the quadratic equation can be found by careful inspection of the graph. or where a is negative y x where a is negative y x Introduction to Graphs 133 © ABE and RRC Example 1: Draw the graph of y x 2 taking whole number values of x from x 3 to x 3 inclusive. What are the co-ordinates at which the function y x 2 is minimised? As ever, the first step is to draw up a table of values for the given range of x. x 3 2 1 0 1 2 3 y x 2 9 4 1 0 1 4 9 Plot these seven points on the graph and then join the points with a smooth curve, as shown in Figure 5.10. From the graph you will see that function y x 2 is minimised where y 0 and x 0 (i.e. the co-ordinates where y x 2 is minimised are (0,0)). Figure 5.10: Graph of y x 2 -2 0 2 4 6 8 10 -3 -2 -1 0 1 2 3 x y y x 2 134 Introduction to Graphs © ABE and RRC Example 2: Draw the graph of y 2x 2 from x 3 to 3. Then from the graph find: (a) the values of x when y 10 (b) the value of y when x 1.25 (c) the values of x and y where the function y 2x 2 is minimised. Table of values: x 3 2 1 0 1 2 3 x 2 9 4 1 0 1 4 9 y 2x 2 18 8 2 0 2 8 18 Plot the points and draw the graph as in Figure 5.11. Figure 5.11: Graph of y 2x 2 Reading off from this: (a) when y 10, there are two values of x, at points A and B: x 2.25 and x 2.25 (b) when x 1.25, the value of y is at the point C: y 3.3 (c) when the function y 2x 2 is minimised, x 0 and y 0. 0 2 4 6 8 10 12 14 16 18 -3 -2 -1 0 1 2 3 x y y 2x 2 y 10 y 1.25 A B C Introduction to Graphs 135 © ABE and RRC Example 3: Draw the graph of y x 2 2x 1 from x 2 to 3. At what value of x does the minimum value of y occur? Table of values: x 2 1 0 1 2 3 x 2 4 1 0 1 4 9 2x 4 2 0 2 4 6 1 1 1 1 1 1 1 y x 2 2x 1 9 4 1 0 1 4 Plot the points and draw the graph as in Figure 5.12. Figure 5.12: Graph of y x 2 – 2x + 1 Reading from the graph, we can see that the minimum value of y occurs when x 1. 0 2 4 6 8 10 -2 -1 0 1 2 3 x y 136 Introduction to Graphs © ABE and RRC Questions for Practice 3 1. Draw the graph of y 2 1 x 2 from x 4 to 4. From the graph find: (a) the value of y when x 3.2 (b) the values of x when y 7. 2. Draw the graph of y x 2 from x 3 to 3. 3. Draw the graph of y 4x 2 from x 2 to 2. Find the values of x when y 9. 4. Draw the graph of the profit (£000s) function x 2 8x 1 from x 0 to 8, where x ('000) represents output. From the graph find: (a) the output at which profit is maximised (b) the output range at which profits generated amount to at least £8,000. Now check your answers with those given at the end of the unit. Equations with x as the Denominator These equations (hyperbolic equations) take the form of y x a where a is a constant. They result in a characteristic curve, the direction of which may vary. Example 1: Let's look at the case where a 1: y x 1 We'll take x to have values from x 1 to x 6, which gives the table following and the graph in Figure 5.13. x 1 2 3 4 5 6 y x 1 1 0.5 0.33 0.25 0.2 0.17 Note that we have avoided taking x 0, since the expression 0 1 is not defined (i.e. it has no meaning). Introduction to Graphs 137 © ABE and RRC 0 0.2 0.4 0.6 0.8 1 1.2 0 2 4 6 0 1 2 3 4 0 2 4 6 8 Figure 5.13: Graph of y x 1 This graph illustrates the general shape of graphs representing y x a . In each case, as x gets bigger, y gets smaller. There is also no point corresponding to x 0. Example 2: Draw the graph of y x 4 for values of x from x 1 to x 8. From the graph, find the value of x when y 2.5. Table of values: x 1 2 3 4 5 6 7 8 y x 4 4 2 1.33 1 0.8 0.67 0.57 0.5 Figure 5.14: Graph of y x 4 Reading from the graph, when y 2.5, x 1.7. 138 Introduction to Graphs © ABE and RRC Questions for Practice 4 1. Draw the graph of y x 2 from x 1 to x 6. From your graph, find the value of x when y 1.5. Now check your answers with those given at the end of the unit. Just as we saw that the shape of the quadratic graphs changed according to whether the coefficient of x 2 is positive or negative, so the shape of graphs developed from equations where x is the denominator change: x a (i.e. where both the a and x are positive) gives a graph which slopes down from left to right x a gives a graph which slopes down from left to right x a gives a graph which slopes up from right to left x a gives a graph which slopes up from right to left. These are illustrated in Figure 5.15. Figure 5.15: Shape of graphs from the equation x a x 0 y x a x a x a x a Introduction to Graphs 139 © ABE and RRC 0 20 40 60 80 100 120 140 160 -2 -1 0 1 2 3 4 5 Exponential and Logarithmic Equations So far in this unit we plotted and interpreted graphs for linear and quadratic equations (also known as polynomial functions). In the final part of this section we will now plot graphs for exponential and logarithmic equations. The mathematics behind exponentials and logarithms was discussed in Study Unit 4. Example 1: Let us begin by plotting the graph of an exponential function. Using a scientific calculator, or otherwise, calculate the value of y e x for the values of x from x 2 to x 5. Plot a graph of this exponential function and from the graph find the value of x when y 10. Table of values: x -2 -1 0 1 2 3 4 5 y e x 0.14 0.37 1.00 2.72 7.39 20.1 54.60 148.41 Figure 5.16: Graph of equation y e x Reading from the graph, when y 10, x 2.5 (approximately). 140 Introduction to Graphs © ABE and RRC -1 -0.5 0 0.5 1 1.5 0 1 2 3 4 Example 2: In this example we will plot a graph of the natural logarithmic function. Using a scientific calculator, calculate the values of y ln(x) for the values of x from x 0.5 to x 4.0. Plot a graph of this logarithmic function and from the graph find the value of x when y 0. Table of values: x 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 ln (x) 0.693 0.00 0.405 0.693 0.916 1.099 1.253 1.386 Figure 5.17: Graph of equation ln(x) Reading from the graph, when y 0, x 1. Questions for Practice 5 1. Use a scientific calculator to construct a table of values for y log 10 (x) when x 0,5, 10, 15, 20 and 25 (give your answers to 2 decimal places). Plot these values on a graph. 2. The sales S (in thousands of units) of a product after it has been on the market for t months is given by the equation: S 100(1 e 0.1625t ). (a) Construct a table of values for S for t 0,5, 10, 15, 20 and 25 (give S values to the nearest whole number). (b) Use the table to draw a graph of S against t and from your graph estimate the value of S when t is very large. Now check your answers with those given at the end of the unit. Introduction to Graphs 141 © ABE and RRC D. GRADIENTS The gradient of a line is a measure of how much it slopes. A horizontal line has a gradient of zero: A vertical line has an infinitely large gradient: In between, the gradient increases: or decreases Now, consider the five slopes illustrated in Figure 5.18. In the five shapes, the slope of the line(s) increases as we go from (i) through to (v). Figure 5.18: Gradient of a slope In (i), the gradient (s) is zero (no slope). In the other cases, the gradient is defined as the height of the right-angled triangle divided by the base. Therefore in (ii), the gradient of s is 1/2 0.5 in (iii), the gradient of s is 2/2 1 in (iv), the gradient of s is 4/2 2 in (v), the gradient of s is 8/2 4 2 (ii) s 2 (i) s 1 2 (iii) s 2 2 (iv) s 4 2 (v) s 8 142 Introduction to Graphs © ABE and RRC If a line slopes down from left to right (instead of up), then the gradient is negative. This is denoted by introducing a negative sign, as illustrated by Figure 5.19. Figure 5.19: Negative gradient The gradient of the line s is 2 2 1. Gradient of a Straight Line We can easily calculate the gradient of a straight line joining two points from the co-ordinates of those points. Firstly, we shall examine the method by reference to plotting the line on a graph. Example 1: Plot the points (1,3) and (4,8) and find the gradient of the line joining them. The procedure would be as follows: (a) Plot the two points on a graph and join them to give a sloping line, as shown in Figure 5.20. Figure 5.20: Gradient of the line connecting points (1, 3) and (4, 8) (b) Make this line the hypotenuse of a right-angled triangle by drawing in a horizontal "base" and a vertical "height". 2 s 2 x y 5 3 1 2 3 4 5 9 4 6 5 3 2 7 1 8 Introduction to Graphs 143 © ABE and RRC (c) Work out the lengths of the base and the height from the graph: base 3 height 5. As we have seen, the gradient is given by the equation base height 3 5 3 2 1 Note that the slope is positive. Therefore, the gradient of the straight line joining (1, 3) and 4, 8) is 3 2 1 Note that, at step (c), we could have worked out the lengths of the base and the height from the co-ordinates of the points: length of the base difference between x co-ordinates 4 1 3 length of the height difference between y co-ordinates 8 3 5 Example 2: Plot the points (1,4) and (2,3) and find the gradient of the line joining them. Plot the points, join them and construct a right-angled triangle with the line forming the hypotenuse: Figure 5.21: Gradient of the line connecting points (–1, 4) and (2, –3) This time we shall work out the lengths of the base and the height from the co-ordinates of the points. base 2 (1) 3 height 4 (3) 7 We can check that this is correct by reference to the plotted triangle on the graph. Note that the slope is negative. Therefore the gradient base height 3 7 2 3 1 2 1 2 3 1 x y 2 1 1 2 3 3 4 7 3 144 Introduction to Graphs © ABE and RRC General formula We can now develop a general formula for the gradient of a straight line by reference to its co-ordinates. To find the gradient of a line, take any two points A and B on the line. DrawAC parallel to the x-axis, and CB parallel to the y-axis. The gradient is then given by the ratio CB : AC. Figure 5.22: Gradient of a straight line Note that these lengths are directed lengths: AC is positive if the movement fromA to C is in the same direction as O to x, and negative if the movement fromA to C is in the opposite direction to O to x. In the same way, CB is positive if the movement from point C to B is in the same direction as O to y. (The point at O is the origin, i.e. (0,0).) Taking the lines in Figure 5.22, we find the following: In line (i), AC and CB are both positive, hence the gradient AC CB is positive. In line (ii), AC is negative, CB is positive, and the gradient is negative. We can further see that AC is the difference between the co-ordinates along the x-axis and CB is the difference between the co-ordinates along the y-axis. Therefore, we could calculate the gradient where the points are, say, (5,1) and (7,2) as follows: 5 7 1 2 2 1 For points (1,2) and (2,3), the gradient would be: ) 1 ( 2 ) 2 ( 3 1 2 2 3 3 1 B x y A C C A B (i) (ii) O Introduction to Graphs 145 © ABE and RRC Equation of a Straight Line and Gradient We can now go back to consider aspects of the equation of a straight line which we have met before. Consider a general straight line drawn on a graph. D is the point on that line where it crosses the y-axis, and P is another point along the line. This is shown in Figure 5.23. Figure 5.23: Proving the equation of a straight line OAPB is a rectangle and OAGD is a rectangle. Let: OA x, OB y, the intercept OD c and the gradient of DP m. Then: m DG PG OA PG OA BD OA DO BO x c y Rearranging the equation m x c y we get: mx y c y mx c This is the general equation of any straight line. Note that c is the intercept that the line makes with the y-axis. If the line cuts the y-axis below the origin, its intercept is negative, c. Using this method, given any straight-line graph, its equation can be found immediately. Also, given any linear equation, we can find the gradient of the line (m) immediately. For example: Find the gradient of a line with the equation 6x 3y 19. We need to put the equation into the general form y mx c: y 3 19 x 6 P x y A G B O D 146 Introduction to Graphs © ABE and RRC 0 2 4 6 8 10 0 2 4 6 8 Therefore, the gradient is 3 6 2 Finally, consider the graph in Figure 5.24. Figure 5.24: Deriving the equation of a straight line The gradient of the line, m 6 5 10 6 5 The intercept, c 5. Therefore, the equation of this line is: y 5 x 6 5 or 6y 5x 30 Questions for Practice 6 1. Find the gradients of the straight lines joining the following points. (a) (1,3) and (3,7) (b) (2,1) and (5,6) (c) (2,6) and (1,1) (d) (4,5) and (2,0) (e) (7,0) and (1,4) (f) (0,2) and (4,7) 2. Find the gradients of the lines with the following equations: (a) 2y 3x 6 0 (b) 7y 4x (c) 7y 1 2x Introduction to Graphs 147 © ABE and RRC The next three questions bring together aspects of the whole unit so far. They are actual examination questions. 3. Draw an x-axis from 3 to 5 and a y-axis from 4 to 4. On them draw the graphs of y x 1 and 2y 4 x. Write down the co-ordinates of the point where the two lines intersect. 4. Draw the graph of y 2 2 x 3 for values of x from 3 to 3. On the same axes draw the line y 5. Write down the co-ordinates of the points where the line and the curve intersect. 5. Plot the points (6,5) and (3,3). Find (a) the gradient and (b) the equation of the straight line joining them. Now check your answers with those given at the end of the unit. E. USING GRAPHS IN BUSINESS PROBLEMS In this final part of the unit, we shall consider how we can use the properties of graphs we have been examining to analyse and solve business problems. Business Costs All businesses incur costs in the production and supply of the goods and/or services which they provide. These are an important consideration in many aspects of business finance, not least in calculating profit, which at heart is simply: revenue from sales less costs of production and supply (or cost of sales) The analysis of costs is a subject of its own and, within that, there are many ways of looking at costs. One of these is to divide them into fixed and variable costs. Fixed costs are those which the business must always meet, regardless of the number of units produced – such as rent on buildings. Variable costs are those which vary with output – such as the cost of raw materials used as part of the production process. Thus, if we take the case of a printing company, its fixed costs would include the rent on its offices, printing room and warehouse, and the variable costs would include paper and ink. Whether they print six or six hundred books a week, the fixed costs remain the same, but the variable costs rise with the number of books produced. (Note that fixed costs are only fixed in the short term – over a long period, they may change, for example by acquiring different premises which may be cheaper.) If we consider variable costs, we can say that they are a function of the level of production. This means that, as the level of production changes, so do the variable costs. We can turn this observation into an algebraic expression: let: x number of units produced (level of production) b variable costs 148 Introduction to Graphs © ABE and RRC Then: b f(x) This is where we started the unit! We know that we can draw a graph of this relationship if we know some of the values for x and y. Let us suppose that we do have such values and can produce such a graph (Figure 5.25): Figure 5.25: Graph of relationship between variable costs and level of production We can work out the gradient of the line on this graph, based on two co-ordinates. One of those co-ordinates must be (0, 0), since when there is no production, no variable costs will be incurred. If we know that one other point on the graph is (9, 3), then the gradient (which is m) is: m 0 9 0 3 9 3 3 1 We also know that the intercept (c) of the line is 0, so we can express the relationship of variable costs to units of production by the following equation in the form of y mx c: a 3 1 x 0 or 3a x Now consider fixed costs. These do not vary with the level of production. So, if we know the level of fixed costs, we can show this on the same graph as in Figure 5.26. Variable costs 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 Costs (£000s) Production (thousands of units) Introduction to Graphs 149 © ABE and RRC Figure 5.26: Graph of fixed costs and level of production We can see that this line is parallel to the x-axis, with an intercept at 2. This is represented by the equation: b 2 (where b is fixed costs) Businesses are not only interested in costs as either fixed or variable, but in their total costs. We can say that: total costs variable costs fixed costs If total costs y, and using the notation from above, we get the following equation: y a b Substituting from the equations established for a and b, we get: y 3 1 x 2 This is a standard linear equation. We could work out a table of values for it (which we shall skip here) and draw a graph. This has been done, again on the same graph as in Figure 5.26, in Figure 5.27. Note that the line for total costs is in fact a combination of the lines for variable and fixed costs. Its intercept is at the level of fixed costs and its gradient is that of variable costs. Variable costs 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 Costs (£000s) Production (thousands of units) Fixed costs 150 Introduction to Graphs © ABE and RRC Figure 5.27: Graph of total costs and level of production Note that the chart is only a model, which is why we can show the variable and total costs lines as linear. In real life, variable costs do not necessarily rise at a constant rate with increases in production. Despite this, it is a useful way of looking at costs. Consider the following situation. A business wishes to use a courier company on a regular basis. Company A charge a call out charge of £5.00 plus £1.00 per mile, whilst company B charge £2.00 per mile but no call out charge. Which one is more economical? We can develop equations for these charges and then use a graph to show which company would be more economical at different journey lengths: Let: x number of miles y total cost (£) Then, for company A, total cost may be defined as : y x 5 For company B, the costs is: y 2x The graph is shown in Figure 5.28. The point at which the two lines intersect means that the values of x and y are the same for each line. Thus at that point, the cost of using each company is the same. This takes place at 5 miles. For any distance above this, company A is cheaper, whereas for distances less than five miles, it is cheaper to use company B. For example: at 7 miles, A charges £12 and B charges £14 at 2 miles, A charges £7 and B charges £4. Some of these situations are shown by dotted lines on the graph. Variable costs 0 1 2 3 4 5 6 0 1 2 3 4 5 6 7 8 9 Costs (£000s) Production (thousands of units) Fixed costs Total costs Introduction to Graphs 151 © ABE and RRC Figure 5.28: Comparison of courier company charges Break-even Analysis We can further develop the costs graph by including revenue on it. Revenue is the total income from sales of the units produced. If we assume that each unit sells for £1, we can say that y x where y is the income from sales (in £) and x is the number of units sold. Adding the graph of this to the costs graph shown in Figure 5.27, we get Figure 5.29. (Note that we are assuming that all units produced are sold.) Figure 5.29: Costs and revenue graph The point at which the revenue and cost lines intersect is known as the break-even point. At that point, revenue exactly equals costs. Here, this takes place at sales of 3,000 units, with both costs and revenue being £3,000. 0 5 10 15 20 0 2 4 6 8 10 Costs (£) Journey length (miles) Company A Company B 0 2 4 6 8 0 1 2 3 4 5 6 7 8 9 Costs (£000s) Sales(thousands of units) Total costs Revenue Break-even point 152 Introduction to Graphs © ABE and RRC Any sales above the break-even point will mean that revenue exceeds costs, and there will be a profit. At sales levels below the break-even point though, costs exceed revenue and there will be a loss. For example: at sales of 2,000 units, costs are £2,667 whilst revenue is only £2,000 – there will a loss of £667 at sales of 6,000 units, costs are £4,000 whilst revenue is £6,000 – there will be a profit of £2,000. The break-even point represents the minimum position for the long term survival of a business. Questions for Practice 7 1. Teapots R U Ltd makes and sells a small range of teapots. It has worked out that the following costs for its best selling teapot "Standard" are: fixed costs £20,150 variable costs £5.50 per Standard teapot. If the price of a Standard teapot is £12.00, how many must Teapots R U Ltd. sell each week to break even? Use a fixed and variable costs chart (break-even chart) to find the answer. 2. A company makes bicycles and sells them for £80.00. The fixed cost of producing them is £80,000 with a variable cost per bicycle of £36.00. Using a fixed and variable costs chart (break-even chart), find the number of bicycles needed to be sold for the company to break even. Give your answer to the nearest 100 bicycles. Now check your answers with those given at the end of the unit. Economic Ordering Quantities We shall now look at the role of graphs in analysing and helping to solve a purchasing problem. We shall not work through the problem in detail, but just review the general approach and the way in which graphs can be used. One of the central issues in purchasing materials to be used as part of the production process is how much to order and when. Many producers and most retailers must purchase the materials they need for their business in larger quantities than those needed immediately. However, there are costs in the ordering of materials – for example, each order and delivery might incur large administrative and handling costs, including transportation costs, so buying- in stock on a daily basis might not be appropriate. On the other hand, there are also costs involved in holding a stock of unused materials – very large stocks can incur large warehousing costs. Set against this, buying in larger quantities has the effect of spreading the fixed costs of purchase over many items, and can also mean getting cheaper prices for buying in bulk. The purchaser must determine the optimum point. By "economic order quantity" we mean the quantity of materials that should be ordered in order to minimise the total cost. In order to do this, we need to look at the relationships between the usage, supply and costs involved. Introduction to Graphs 153 © ABE and RRC Suppose we have the following problem: We want to order 3,600 units over one year and these units are used at a constant rate. As soon as we order, we receive the whole order instantaneously – unrealistic but let us keep the problem simple. We order as soon as existing stock runs out. The cost of placing the order is £30. The cost of keeping one unit in stock for a year is £2. We have several options as to how we could tackle this: (a) Place a single order for the whole year. (b) Make 12 monthly orders. (c) Make three orders at four-monthly intervals. (d) Make six orders bimonthly. (e) Make 36 orders at 10-day intervals. There are many other possibilities but these will serve our purpose. The relevant costs for each option above are: (a) £30 for one order storage costs of £2 1,800 £3,630. (As we place an order for 3,600 units and use them evenly through the year, the average stock held in storage will be 1,800 units.) (b) (12 £30) for 12 orders (£2 150) storage £660 (In this case, the orders are for 300 each, so the average holding is 150.) (c) (3 £30) (£2 600) £1,290. (600 is the average holding on orders of 1,200 each.) (d) (6 £30) (£2 300) £780. (Six orders of 600 each, giving the average holding of 300.) (e) (36 £30) (£2 50) £1,180 (36 orders of 100 each, giving the average holding of 50.) We can see that the cheapest option is (b), i.e. monthly orders of 300. But is it really the cheapest? We would need to try many other options before we could be sure. Let us consider another problem. A company called Oily Engineers Ltd uses 1,000 sprockets every day. It could purchase daily, but considers this to be unacceptable as the cost of raising an order and sending a van to a cash-and-carry to collect the goods has been estimated at £100. If it buys in lots of 1,000Q sprockets, then the stock will last for Q days. The graph of stock level is a saw-tooth. There will be 1,000Q sprockets in stock immediately after a purchase has been made. Q days later this will be down to zero, when another purchase will be made. The average stock level is 500Q (see Figure 5.30). 154 Introduction to Graphs © ABE and RRC Figure 5.30: Stock levels over time The warehousing cost is linear – if the amount purchased in each order is doubled, then the warehousing costs are doubled as there are, on average, twice as many items in the warehouse. This shown in Figure 5.31. Figure 5.31: Warehousing costs The cost of purchasing is not linear. The cost per annum depends on the number of orders placed, as it is a fixed amount per order. For example, assume that there are 250 working days each year: (a) Sprockets purchased in batches of 1,000, i.e. a batch every day: cost £100 250 £25,000 (b) Sprockets purchased in batches of 2,000, i.e. a batch every other day: cost £100 125 £12,500 (c) Sprockets purchased in batches of 4,000, i.e. a batch every fourth day: cost £100 62.5 £6,250. Each time the amount purchased is doubled, the cost per annum halves. The graph is shown in Figure 5.32. Size of purchased lot Cost Stock level 0 2Q Days 3Q Q 1,000Q Introduction to Graphs 155 © ABE and RRC Figure 5.32: Cost of purchasing (per annum) These costs can be added to the warehouse costs to give a total cost per annum. The graph of this is shown in Figure 5.33. Figure 5.33: Total costs of purchasing Thus, we can identify the economic order quantity from the minimum cost position on the graph. Plotting Time Series Data Businesses and governments statistically analyse information and data collected at regular intervals over extensive periods of time, to plan future strategies and policies. For example, sales values or unemployment levels recorded at yearly, quarterly or monthly intervals could be examined in an attempt to predict their future behaviour. Such sets of values observed at regular intervals over a period of time are called time series. The analysis of this data is a complex problem as many variable factors may influence the changes which take place. The first step in analysing the data is to plot the observations on a scattergram, with time evenly spaced across the x-axis and measures of the dependent variable forming the y-axis. This can give us a good visual guide of the trend in the time series data. Cost Size of purchased lot Cost Size of purchased lot Total cost Purchasing cost Storage or warehousing cost Minimum cost 156 Introduction to Graphs © ABE and RRC Let us consider a factory employing a number of people in producing a particular commodity, say thermometers. Naturally, at such a factory during the course of a year some employees will be absent for various reasons. The following table shows the number of days lost through sickness over the last five years. Each year has been broken down into four quarters of three months. We have assumed that the number of employees at the factory remained constant over the five years. Year Quarter Days Lost 2003 1 30 2 20 3 15 4 35 2004 1 40 2 25 3 18 4 45 2005 1 45 2 30 3 22 4 55 2006 1 50 2 32 3 28 4 60 2007 1 60 2 35 3 30 4 70 Table 5.1: Days lost (per quarter) through sickness at a thermometer factory We can plot this information using a particular type of scattergram, as shown in Figure 5.34. The scattergram of a time series is often called a historigram. (Do not confuse this with a histogram, which is a type of bar chart.) Note the following characteristics of a historigram: It is usual to join the points by straight lines. The only function of these lines is to help your eyes to see the pattern formed by the points. Intermediate values of the variables cannot be read from the historigram. A historigram is simpler than other scattergrams since no time value can have more than one corresponding value of the dependent variable. It is possible to change the scale of the y-axis to highlight smaller changes. Every historigram will look similar to the one in Figure 5.34, depending on the scale of the y- axis. Introduction to Graphs 157 © ABE and RRC Figure 5.34: Historigram (scattergram) of days lost (per quarter) through sickness (2003–2007) There are four factors that influence the changes in a time series over time. These are the trend, seasonal variations, cyclical fluctuations, and irregular or random fluctuations. However, you only need to concern yourself with the trend. The trend is the change in general level over the whole time period and is often referred to as the secular trend. You can see in Figure 5.34 that the trend is definitely upwards, in spite of the obvious fluctuations from one quarter to the next. A trend can thus be defined as a clear tendency for the time series data to travel in a particular direction in spite of inter-period fluctuations (be they large or small). If we were to draw a simple linear trend line through the time series data shown in Figure 5.34, then we would get a trend line similar to that shown in Figure 5.35. Based on this trend line, we can clearly see that lost days from work (per quarter) have increased over the period, from approximately 25 days lost per quarter in 2003 to approximately 50 days lost per quarter in 2007. Lost days 2003 2004 2005 2006 2007 0 10 20 30 40 50 60 70 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 158 Introduction to Graphs © ABE and RRC Figure 5.35: Historigram (scattergram) of days lost (per quarter) through sickness (2003–2007), with linear trend line Lost days 2003 2004 2005 2006 2007 0 10 20 30 40 50 60 70 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Introduction to Graphs 159 © ABE and RRC Questions for Practice 8 The following table shows sales of premium bonds at a large post office over five years: Year and Quarter Sales (£000) 2004 1 320 2 200 3 150 4 400 2005 1 400 2 210 3 180 4 450 2006 1 800 2 400 3 340 4 900 2007 1 950 2 500 3 400 4 980 1. Plot the time series on a historigram. 2. Using a ruler, draw a linear trend line through the time series plotted in Question 1 and comment on the trend. Now check your answers with those given at the end of the unit. 160 Introduction to Graphs © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. x 0 0.5 1 1.5 2 2.5 y 3x 2 2 0.5 1 2.5 4 5.5 2. y 2x 1 is the only function which satisfies all values in the table. 3. (a) f(2) 10 2 8 (b) f(3) 15 2 17 4. (a) f(2) 4 2 6 (b) f(3) 9 2 11 Questions for Practice 2 1. Table of values: x 1 0 2 y 5x 4 9 4 6 Graph of y 5x – 4 -10 -8 -6 -4 -2 0 2 4 6 -2 -1 0 1 2 3 x y Introduction to Graphs 161 © ABE and RRC 2. Table of values: x 4 0 2 y 2x 3 5 3 7 Graph of y 2x + 3 3. Table of values: x 4 1 6 y 2 x 6 1 4 Graph of y 2 – x -6 -4 -2 0 2 4 6 8 -4 -3 -2 -1 0 1 2 x y -4 -2 0 2 4 6 -4 -2 0 2 4 6 x y 162 Introduction to Graphs © ABE and RRC Note that for the next four questions, we do not show the graphs themselves. 4. (i) cuts (iii) at ( 3 2 1 , 3 2 2 ) or (1.7, 2.7) to 1 decimal place. (i) cuts (iv) at (2.6, 0.8). (ii) cuts (iii) at (1, 0). (ii) cuts (iv) at (1.4, 2.1) to 1 decimal place. 5. (i) y 5 is a line parallel to the x-axis and 5 units above it. (ii) x 5 is a line parallel to the y-axis and 5 units to the left of it. (iii) y 5x is a line through the origin, sloping up from left to right with a gradient 5 in 1 (i.e. if you go along 1 unit you go up 5 units). (iv) y 5 1 x is a line through the origin, sloping down from left to right with a gradient of 1 in 5 (i.e. along 5 and down 1). 6. These three graphs are parallel lines: (i) passes through the origin. (ii) makes an intercept of 3 on the y-axis and 1½ on the x-axis (i.e. crosses the y- axis at 3 and the x-axis at 1½). (iii) makes an intercept of 4 on the y-axis and 2 on the x-axis. 7. The lines intersect at (3, 1). Questions for Practice 3 1. Table of values: x 4 3 2 1 0 1 2 3 4 x 2 16 9 4 1 0 1 4 9 16 y 2 1 x 2 8 4½ 2 ½ 0 ½ 2 4½ 8 The graph is shown on the next page. Reading off from it, we get: (a) when x 3.2, y 5.1 (b) when y 7, x 3.7 and 3.7 Introduction to Graphs 163 © ABE and RRC 0 2 4 6 8 -4 -2 0 2 4 Graph of y 2 1 x 2 2. Table of values: x 3 2 1 0 1 2 3 y x 2 9 4 1 0 1 4 9 Graph of y –x 2 -10 -8 -6 -4 -2 0 -3 -2 -1 0 1 2 3 164 Introduction to Graphs © ABE and RRC 3. Table of values: x 2 1 0 1 2 x 2 4 1 0 1 4 4x 2 16 4 0 4 16 y 4x 2 16 4 0 4 16 Graph of y –4x 2 Reading from the graph, when y 9, x 1.5 and 1.5. 4. Table of values: x 0 1 2 3 4 5 6 7 8 -x 2 0 -1 -4 -9 -16 -25 -36 -49 -64 8x 0 8 16 24 32 40 48 56 64 1 1 1 1 1 1 1 1 1 1 y x 2 8x 1 1 8 13 16 17 16 13 8 1 The graph is shown on the next page. Reading off from it, we get: (a) Profit is maximised (£17,000) at 4,000 units of output, i.e. when x 4. (b) At an output of between 1,000 and 7,000 units (i.e. between x 1and x 7), profits generated are in excess of £8,000 (i.e. y 8 or above). -16 -12 -8 -4 0 -2 -1 0 1 2 Introduction to Graphs 165 © ABE and RRC Graph of y x 2 + 8x + 1 Questions for Practice 4 1. Table of values: x 1 2 3 4 5 6 y x 2 2 1 0.67 0.5 0.4 0.33 Graph of y x 2 When y 1.5, x 1.3. 0 0.5 1 1.5 2 0 1 2 3 4 5 6 0 2 4 6 8 10 12 14 16 18 0 1 2 3 4 5 6 7 8 Profit (£000s) Output (£000s) 166 Introduction to Graphs © ABE and RRC 0 0.2 0.4 0.6 0.8 1 1.2 1.4 0 5 10 15 20 25 0 20 40 60 80 100 120 0 5 10 15 20 25 Questions for Practice 5 1. Table of values for y log 10 (x): x 0 5 10 15 20 25 y log 10 (x) 0.00 0.70 1.00 1.18 1.30 1.40 Graph of y log 10 (x) 2. Table of values for S 100(1 e 0.1625t ): t 0 5 10 15 20 25 S 0 56 80 91 96 98 Graph of S 100(1 e 0.1625t ) From the graph, when t is very large S will be approximately 100,000. Introduction to Graphs 167 © ABE and RRC Questions for Practice 6 1. (a) 1 3 3 7 2 4 2 (b) 3 5 ) 1 ( 6 3 7 3 1 2 (c) 1 ) 2 ( ) 1 ( 6 3 7 3 1 2 (d) 2 ) 4 ( 0 5 6 5 6 5 (e) ) 1 ( 7 ) 4 ( 0 8 4 2 1 (f) 0 4 ) 2 ( 7 4 9 2 1 2 2. (a) y 2 6 x 3 gradient 2 3 (b) y 7 4 x gradient 7 4 (c) y 7 1 x 2 gradient 7 2 3. Tables of values: x 3 1 5 y x 1 4 0 4 2y 4 x y 2 2 1 x x 2 0 4 2 2 2 2 2 1 x 1 0 2 y 2 2 1 x 3 2 0 The graph is shown on the following page. 168 Introduction to Graphs © ABE and RRC Graph of y x – 1 and 2y 4 – x Reading from this, the lines intersect at point (2, 1) 4. Table of values: x 3 2 1 0 1 2 3 y 2 x 3 2 13.5 6 1.5 0 1.5 6 13.5 Graph of y 2 3x 2 and y 5 From the graph, you will see that the line and curve intersect at (1.8, 5) and (1.8, 5). -4 -3 -2 -1 0 1 2 3 4 -3 -2 -1 0 1 2 3 4 5 y x 1 2y 4 x 0 2 4 6 8 10 12 14 -3 -2 -1 0 1 2 3 y 5 y 2 x 3 2 Introduction to Graphs 169 © ABE and RRC 0 1 2 3 4 5 0 1 2 3 4 5 6 5. (a) Gradient 3 6 3 5 3 2 (b) The line cuts the y-axis at (0,1). Therefore, the equation is: y 3 2 x 1 or 3y 2x 3 Questions for Practice 7 1. Break-even chart: The break-even point is at sales of 3,100 Standard teapots. 0 10 20 30 40 50 0 1 2 3 4 Revenue/ Costs (£000s) Sales (000s) 170 Introduction to Graphs © ABE and RRC 2. Break-even chart The break-even point is at sales of 1,800 bicycles (to the nearest 100). Questions for Practice 8 1. Time series historigram: 2. Linear trend is shown on the historigram above. The historigram clearly shows that sales of premium bonds have increased over the period, from approximately £250,000 per quarter in 2003 to approximately £700,000 per quarter in 2007. 0 50 100 150 200 250 0 1 2 3 Revenue/ Costs (£000s) Sales (000s) Sales (£000s) 2004 2005 2006 2007 0 200 400 600 800 1000 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 171 © ABE and RRC Study Unit 6 Introduction to Statistics and Data Analysis Contents Page Introduction 172 A. Data for Statistics 173 Preliminary Considerations 173 Types of Data 174 Requirements of Statistical Data 174 B. Methods of Collecting Data 175 Published Statistics 175 Personal Investigation/Interview 176 Delegated Personal Investigation/Interview 176 Questionnaire 176 C. Classification and Tabulation of Data 177 Forms of Tabulation 179 Secondary Statistical Tabulation 181 Rules for Tabulation 183 D. Frequency Distributions 185 Simple Frequency Distributions 186 Grouped Frequency Distributions 186 Choice of Class Interval 188 Cumulative Frequency Distributions 189 Relative Frequency Distributions 189 Answers to Questions for Practice 191 172 Introduction to Statistics and Data Analysis © ABE and RRC INTRODUCTION This unit is the first of five concerned specifically with statistics. "Statistics" is a word which is used in a variety of ways and with a variety of meanings, but, in whatever way it is used, it is always concerned with numerical information. There are two particular meanings of the word which concern us, namely: The numerical facts themselves – for example, we talk of the "statistics" of steel production. The methods of analysing the facts – in this sense, "statistics" is the title of a subject like "arithmetic" or "chemistry" or "physics". Sometimes the subject is called "statistical method". Thus, statistics is concerned with numerical facts about a problem or issue – the facts themselves, and with methods of analysing these facts in order to solve or shed light on the problem or issue. The numerical facts with which statistics are concerned are commonly called data. We therefore talk about the collection and analysis of data in respect of a particular problem. The methods used for analysing data can be divided into two categories – descriptive and inferential. Descriptive methods simply manipulate the available data in order to convey information about it, often in the form of summaries or graphical displays, and usually for the purpose of simplifying comprehension or appreciation of the facts. These methods are concerned only with the available data, even if that data is only a part of a (possibly much larger) whole. Inferential methods use the available data in order to make general statements about the whole from which that data is only a part. In other words, they allow us to infer something about the whole mass of data from a study of just a part of it. Predictions and estimates are typical of the general conclusions which may be made by using inferential methods. For example, a statistical analysis may be made of the ages of the eleven players in the England football team. The ages of the individual players form the data. If we determine the average age of the team from that data, we are applying descriptive methods to tell us something about the eleven players as a whole. However, the players are also a part of a much larger group – for example, all professional English football players – and clearly their average age will not necessarily be an accurate estimate of the average age of this larger group. If we wanted to draw conclusions about this larger group from the study of the eleven players in the England team, we would need to use inferential methods. In this course we shall only be concerned with descriptive statistical methods. In this study unit we shall review the basis of the statistical method – the collection of facts (data) about a problem or issue – and then consider initial ways of analysing this data so that it has meaning in relation to the problem. Objectives When you have completed this study unit you will be able to: explain the concept of data in statistics distinguish between primary and secondary data explain the differences between quantitative and qualitative variables and between continuous and discrete variables Introduction to Statistics and Data Analysis 173 © ABE and RRC explain the reasons for classifying and tabulating data, and prepare tables from given data explain the concept of a frequency distribution and define the terms class frequency, class boundary, class limits and class interval prepare simple, grouped, cumulative and relative frequency distributions from given data. A. DATA FOR STATISTICS A statistical investigation involves a number of stages: definition of the problem or issue collection of relevant data classification and analysis of the collected data presentation of the results. So even before the collection of data starts, there are some important points to consider when planning a statistical investigation. Preliminary Considerations It is important to be aware of these issues as they impact on the data which is to be collected and analysed. Exact definition of the problem This is necessary in order to ensure that nothing important is omitted from the enquiry, and that effort is not wasted by collecting irrelevant data. The problem as originally put to the statistician is often of a very general type and it needs to be specified precisely before work can begin. Definition of the units The results must appear in comparable units for any analysis to be valid. If the analysis is going to involve comparisons, then the data must all be in the same units. It is no use just asking for "output" from several factories – some may give their answers in numbers of items, some in weight of items, some in number of inspected batches and so on. Scope of the enquiry No investigation should be commenced without defining the field to be covered. Are we interested in all departments of our business, or only some? Are we to concern ourselves with our own business only, or with others of the same kind? Accuracy of the data To what degree of accuracy is data to be recorded? For example, are ages of individuals to be given to the nearest year or to the nearest month or as the number of completed years? If some of the data is to come from measurements, then the accuracy of the measuring instrument will determine the accuracy of the results. The degree of precision required in an estimate might affect the amount of data we need to collect. In general, the more precisely we wish to estimate a value, the more readings we need to take. 174 Introduction to Statistics and Data Analysis © ABE and RRC Types of Data There are a number of different ways in which data may be categorised. Primary and secondary data In its strictest sense, primary data is data which is both original and has been obtained in order to solve the specific problem in hand. Primary data is therefore raw data, and has to be classified and processed using appropriate statistical methods in order to reach a solution to the problem. Secondary data is any data other than primary data. Thus it includes any data which has been subject to the processes of classification or tabulation or which has resulted from the application of statistical methods to primary data; it includes all published statistics. Quantitative and qualitative data Variables may be either quantitative or qualitative. Quantitative variables, to which we shall restrict discussion here, are those for which observations are numerical in nature (for example a person's height or the number of children in a school). Qualitative variables relate to non-numeric observations, such as colour of hair. However each possible non-numeric value (hair colour say) may be associated with a numeric frequency of occurrence. Continuous and discrete data Quantitative variables may be either continuous or discrete. A continuous variable may take any value between two stated limits (which may possibly be minus and plus infinity). Height, for example, is a continuous variable. This is because a person's height may (with appropriately accurate equipment) be measured to any minute fraction of a millimetre. A discrete variable however can take only certain values occurring at intervals between stated limits. For most (but not all) discrete variables, these interval values are the set of integers (whole numbers). For example, if the variable is the number of children per family, then the only possible values are 0, 1, 2, etc. because it is impossible to have other than a whole number of children. However, in Britain, shoe sizes are stated in half-units, and so here we have an example of a discrete variable which can take the values 1, 1½, 2, 2½, etc. Requirements of Statistical Data Having decided upon the preliminary matters about the investigation, the statistician must look in more detail at the actual data to be collected. The desirable qualities of statistical data are as follows. Homogeneity The data must be in properly comparable units. "Five houses" means little since five slum hovels are very different from five ancestral mansions. Houses cannot be compared unless they are of a similar size or value. If the data is found not to be homogeneous, there are two methods of adjustment possible. (a) Break down a non-homogeneous group into smaller component groups which are homogeneous, and study them separately. (b) Standardise the data. Use units such as "output per person-hour" to compare the output of two factories of very different size. Alternatively, determine a relationship between the different units so that all may be expressed in terms of one – in food consumption surveys, for example, a child may be considered equal to half an adult. Introduction to Statistics and Data Analysis 175 © ABE and RRC Completeness Great care must be taken to ensure that no important aspect is omitted from the enquiry. Accurate definition Each term used in an investigation must be carefully defined; it is so easy to be slack about this and to run into trouble. For example, the term "accident" may mean quite different things to the injured party, the police and the insurance company! Watch out also, when using other people's statistics, for changes in definition. Acts of Parliament may, for example, alter the definition of an "indictable offence" or of an "unemployed person". Uniformity The circumstances of the data must remain the same throughout the whole investigation. It is no use comparing the average age of employees in an industry at two different times if the age structure has changed markedly. Likewise, it is not much use comparing a firm's profits at two different times if the working capital has changed. B. METHODS OF COLLECTING DATA When all the foregoing matters have been considered, we come to the question of how to collect the data we require. The methods usually available are as follows: use of published statistics personal investigation/interview delegated personal investigation/interview questionnaire. Published Statistics Sometimes we may be attempting to solve a problem that does not require us to collect new information, but only to reassemble and reanalyse data which has already been collected by someone else for some other purpose. We can often make good use of the great amount of statistical data published by governments, the United Nations, nationalised industries, chambers of trade and commerce and so on. When using this method, it is particularly important to be clear on the definition of terms and units and on the accuracy of the data. The source must be reliable and the information up-to-date. This type of data is sometimes referred to as secondary data (as we have mentioned), in that the investigator has not been personally responsible for collecting it, and it thus came to the investigator "second-hand". By contrast, data which has been collected by the investigator for the particular survey in hand is called primary data. The information you require may not be found in its entirety in one source, but parts of it may appear in several different sources. Although the search through these may be time- consuming, it can lead to data being obtained relatively cheaply; this is one of the advantages of this type of data collection. Of course, the disadvantage is that you could spend a considerable amount of time looking for information which may not be available. Another disadvantage of using data from published sources is that the definitions used for variables and units may not be the same as those you wish to use. It is sometimes difficult to establish the definitions from published information, but, before using the data, you must establish what it represents. 176 Introduction to Statistics and Data Analysis © ABE and RRC Personal Investigation/Interview In this method the investigator collects the data personally. This method has the advantage that the data will be collected in a uniform manner and with the subsequent analysis in mind. However the field that may be covered by personal investigation may be limited. Additionally, there is sometimes a danger that the investigator may be tempted to select data that accords with preconceived notions. The personal investigation method is also useful if a pilot survey is carried out prior to the main survey, as personal investigation will reveal the problems that are likely to occur. Delegated Personal Investigation/Interview When the field to be covered is extensive, the task of collecting information may be too great for one person. Then a team of selected and trained investigators or interviewers may be used. The people employed should be properly trained and informed of the purposes of the investigation. Their instructions must be very carefully prepared to ensure that the results are in accordance with the requirements described in the previous section of this study unit. If there are many investigators, personal biases may tend to cancel out. Care in allocating the duties to the investigators can reduce the risks of bias. For example, suppose you were investigating the public attitude to a new drug in two towns. Do not put investigator A to explore town X and investigator B to explore town Y. If you do, any difference that is revealed might be due to the towns being different, or it might be due to different personal biases on the part of the two investigators. In such a case, you would try to get both people to do part of each town. Questionnaire In some enquiries the data consists of information which must be supplied by a large number of people. The most convenient way to collect such data is to issue questionnaires to the people concerned, asking them to fill in the answers to a set of printed questions. This method is usually cheaper than delegated personal investigation and can cover a wider field. (A carefully thought out questionnaire is also often also used in personal or delegated interview in order to reduce the effect of personal bias.) The distribution and collection of questionnaires by post suffers from two main drawbacks: The forms are completed by people who may be unaware of some of the requirements and who may place different interpretations on the questions – even the most carefully worded ones! There may be a large number of forms not returned, and these may be mainly by people who are not interested in the subject or who are hostile to the enquiry. The result is that we end up with completed forms only from a certain kind of person and thus have a biased sample. For this reason, it is essential to include a reply-paid envelope to encourage people to respond. If the forms are distributed and collected by interviewers, a greater response is likely and queries can be answered. (This is the method used, for example, in the UK Population Census.) However care must be taken that the interviewers do not lead respondents in any way. Introduction to Statistics and Data Analysis 177 © ABE and RRC C. CLASSIFICATION AND TABULATION OF DATA Having completed the survey and collected the data, we need to organise it so that we can extract useful information and then present our results. Very often the data will consist of a mass of figures in no special order. For example, we may have a card index of the 3,000 employees in a large factory. The cards are probably kept in alphabetical order of names, but they will contain a large amount of other data such as wage rates, age, sex, type of work, technical qualifications and so on. If we are required to present to the factory management a statement about the age structure of the workforce (both male and female), then the alphabetical arrangement does not help us, and no one could possibly gain any idea about the topic from merely looking through the cards as they are. What is needed is to classify the cards according to the age and sex of the employee and then present the results of the classification as a tabulation. The data in its original form, before classification, is usually known as raw data (i.e. it is unprocessed). Example: Of course we cannot give an example involving 3,000 cards, so we will look at a shortened version, involving only a small number. Let us assume that our raw data about employees consists of 15 cards in alphabetical order, as follows. The age of each employee is stated in years. Archer, L Mr 39 Black, W Mrs 20 Brown, A Mr 22 Brown, W Miss 22 Carter, S Miss 32 Dawson, T Mr 30 Dee, C Mrs 37 Edwards, D. Mr 33 Edwards, R Mr 45 Gray, J Mrs 42 Green, F Miss 24 Gunn, W Mr 27 Jackson, J Miss 28 Johnson, J Mr 44 Jones, L Mr 39 As it stands, this data does not allow us to draw any useful conclusions about the group of employees. In order to do this, we need to classify the data in some way. One way would be to classify the individual cases by sex: 178 Introduction to Statistics and Data Analysis © ABE and RRC Male Female Archer, L Mr 39 Black, W Mrs 20 Brown, A Mr 22 Brown, W Miss 22 Dawson, T Mr 30 Carter, S Miss 32 Edwards, D Mr 33 Dee, C Mrs 37 Edwards, R Mr 45 Gray, J Mrs 42 Gunn, W Mr 27 Green, F Miss 24 Johnson, J Mr 44 Jackson, J Miss 28 Jones, L Mr 39 An alternative way of classifying the group would be by age, splitting them into three groups as follows: Age between 20 and 29 Age between 30 and 39 Black, W Mrs 20 Archer, L Mr 39 Brown, A Mr 22 Carter, S Miss 32 Brown, W Miss 22 Dawson, T Mr 30 Green, F Miss 24 Dee, C Mrs 37 Gunn, W Mr 27 Edwards, D Mr 33 Jackson, J Miss 28 Jones, L Mr 39 Age over 40 Edwards, R Mr 45 Gray, J Mrs 42 Johnson, J Mr 44 Again, although the data has now acquired some of the characteristics of useful information, it is still a collection of data about individuals. In statistics though, we are concerned with things in groups rather than with individuals. In comparing the height of Frenchmen with the height of Englishmen, we are concerned with Frenchmen in general and Englishmen in general, but not with Marcel and John as individuals. An insurance company, to give another example, is interested in the proportion of men or women who die at certain ages, but it is not concerned with the age at which John Smith or Mary Brown, as individuals, will die. We need to transform the data into information about the group, and we can do this by tabulation. The number of cards in each group, after classification, is counted and the results presented in a table. Introduction to Statistics and Data Analysis 179 © ABE and RRC Table 6.1: Employees by age and sex Sex Age group Male Female Total 20 – 29 2 4 6 30 – 39 4 2 6 40 – 49 2 1 3 Total 8 7 15 You should look through this example again to make quite sure that you understand what has been done. Of particular importance to note is the presentation of the original age data, which is given to the nearest year (and not broken down further into months, days, etc.). This has allowed the use of the age groups 20 – 29, 30 – 39 and 40 – 49. As we learnt in section A of this unit, age is a continuous variable, which means it is possible to measure a person’s age broken down into months, days, minutes and seconds. If the age of employees had been presented more precisely in the original data, then it is possible that Mr Archer and Mr Jones, for example, may be slightly older than 39 years. Let us assume that Mr Archer is 39 years, 4 months and 3 days old and Mr Jones is 39 years, 2 months and 20 days old. In which age group in Table 6.1 would Mr Archer and Mr Jones now be classified? Clearly they are both older than (exactly) 39 years, therefore they should not be included in the age group 30 – 39, but they are not as old as (exactly) 40 years, therefore they should not be included in the age group 40 – 49! Thus, when tabulating continuous variables such as a person’s age, we need to ensure that the data groups chosen are appropriate. In most cases (i.e. where a more precise breakdown of a person’s age is given), we could use the age groups 20 to < 30 (i.e. 20 to less than 30), 30 to < 40 and 40 to < 50. Mr Archer and Mr Jones would now clearly fall into the 30 to less than 40 years age group (i.e. 30 to < 40). You should note that we did not use the age groups 20 – 30, 30 – 40 and 40 – 50 years. This is because an employee that is, for example, exactly 40 years old, could not be classified into a single group, as this person would fall into both the 30 – 40 and 40 – 50 age groups! So choice of date grouping (often referred to as "class intervals") is an important aspect of the classification and tabulation of data. We shall look in more detail at the use of class intervals in section D of this unit when we look "frequency distribution". Having understood this, you are now in a position to appreciate the purpose behind classification and tabulation –- it is to condense an unwieldy mass of raw data to manageable proportions and then to present the results in a readily understandable form. Forms of Tabulation We can classify the process of tabulation into simple tabulation and complex or matrix tabulation. (a) Simple tabulation This covers only one aspect of a set of figures. The idea is best conveyed by an example. Consider the card index of 3,000 employees mentioned earlier. Assume that each card carries the name of the workshop in which the person works. A question as to how the workforce is distributed can be answered by sorting the cards and preparing a simple table as in Table 6.2. 180 Introduction to Statistics and Data Analysis © ABE and RRC Table 6.2: Distribution of employees by workshop Workshop Number employed A 600 B 360 C 660 D 840 E 540 Total 3,000 Another question might have been "What is the wage distribution of the employees?" The answer can be given in another simple table, Table 6.3. Table 6.3: Wage distribution Wages group Number of employees £40 but less than £60 105 £60 but less than £80 510 £80 but less than £100 920 £100 but less than £120 1,015 £120 but less than £140 300 £140 but less than £160 150 Total 3,000 Note that such simple tables do not tell us very much – although it may be enough to answer the initial question at the moment. (b) Complex tabulation This deals with two or more aspects of a problem at the same time. In the problem just studied, it is very likely that the two questions presented in Tables 6.2 and 6.3 would be asked at the same time. We could therefore present the answers in a single, although more complex, table or matrix (as shown in Table 6.4). Introduction to Statistics and Data Analysis 181 © ABE and RRC Table 6.4: Distribution of employees and wages by workshop Wages group (£ per week) Workshop 40 – 59.99* 60 – 79.99 80 – 99.99 100 – 119.99 120 – 139.99 140 – 159.99 Total no. employed A 20 101 202 219 29 29 600 B 11 52 90 120 29 58 360 C 19 103 210 200 88 40 660 D 34 167 303 317 18 1 840 E 21 87 115 159 136 22 540 Total 105 510 920 1,015 300 150 3,000 * Note: 40 – 59.99 is the same as "40 but less than 60" and similarly for the other columns. This table is much more informative than are the two previous simple tables, although it is more complicated. We could have complicated it further by dividing the groups into male and female employees, or into age groups. Later in this unit, we will look at a list of the rules you should try to follow in compiling statistical tables. At the end of that list you will find a table relating to our 3,000 employees, which you should study as you read the rules. Secondary Statistical Tabulation So far, our tables have merely classified the already available figures – the primary statistics. However, we can go further than this and do some simple calculations to produce other figures – secondary statistics. As an example consider again Table 6.2, and this time calculate how many employees there are on average per workshop. This is obtained by dividing the total (3,000) by the number of workshops (5), and the table appears as follows: Table 6.5: Distribution of employees by workshop Workshop Number employed A 600 B 360 C 660 D 840 E 540 Total 3,000 Average number of employees per workshop: 600 The average is a secondary statistic. 182 Introduction to Statistics and Data Analysis © ABE and RRC For another example, we may take Table 6.3 and calculate the proportion of employees in each wage group, as follows: Table 6.6: Wage distribution Wages group Number of employees Proportion of employees £40.00 – £59.99 105 0.035 £60.00 – £79.99 510 0.170 £80.00 – £99.99 920 0.307 £100.00 – £119.99 1,015 0.338 £120.00 – £139.99 300 0.100 £140.00 – £159.99 150 0.050 Total 3,000 1.000 The proportions given here are also secondary statistics. In commercial and business statistics it is more usual to use percentages than proportions, and so in Table 6.6 these would be 3.5%, 17%, 30.7%, 33.8%, 10% and 5%. Of course secondary statistics are not confined to simple tables. They are used in complex tables too, as in the example presented in Table 6.7. Table 6.7: Inspection results for a factory product in two successive years Year 1 Year 2 Machine no. Output No. of rejects % of rejects Output No. of rejects % of rejects 1 800 40 5.0 1,000 100 10.0 2 600 30 5.0 500 100 20.0 3 300 12 4.0 900 45 5.0 4 500 10 2.0 400 20 5.0 Total 2,200 92 4.2 2,800 265 9.5 Average per machine 550 23 4.2 700 66.2 9.5 The percentage columns and the average row show secondary statistics. All the other figures are primary statistics. Note carefully that percentages cannot be added or averaged to get the percentage of a total or of an average. You must work out such percentages on the totals or averages themselves. Another danger in the use of percentages has to be watched: you must not forget the size of the original numbers. Take the case of two doctors treating a particular disease. One doctor has only one patient and the doctor cures the patient – 100% success! The other doctor has 100 patients of whom 80 are cured – only 80% success! You can see how very unfair it would be to make a comparison based on the percentages alone. Introduction to Statistics and Data Analysis 183 © ABE and RRC Rules for Tabulation There are no absolute rules for drawing up statistical tables, but there are a few general principles which, if borne in mind, will help you to present your data in the best possible way. Try not to include too many features in any one table (say, not more than four or five) as otherwise it becomes rather clumsy. It is better to use two or more separate tables. Each table should have a clear and concise title to indicate its purpose. It should be very clear what units are being used in the table (tonnes, £, people, £000, etc.). Appropriate data groups (or class intervals) for continuous and discrete data should be chosen to ensure that all data can be included within a single group (or class). Blank spaces and long numbers should be avoided, the latter by a sensible degree of approximation. Columns should be numbered to facilitate reference. Try to have some order to the table, for example using size, time, geographical location or alphabetical order. Figures to be compared or contrasted should be placed as close together as possible. Percentages should be placed near to the numbers on which they are based. Use a ruler to draw the tables neatly – scribbled tables with freehand lines nearly always result in mistakes and are difficult to follow. However, it is useful to draw a rough sketch first so that you can choose the best layout and decide on the widths of the columns. Insert totals where these are meaningful, but avoid "nonsense totals". Ask yourself what the total will tell you before you decide to include it. An example of such a "nonsense total" is given in Table 6.8. Table 6.8: Election results Party Year 1 Seats Year 2 Seats Total Seats A 350 200 550 B 120 270 390 Total 470 470 940 The totals (470) at the foot of the two columns make sense because they tell us the total number of seats being contested, but the totals in the final column (550, 390, 940) are nonsense totals for they tell us nothing of value. If numbers need to be totalled, try to place them in a column rather than along a row for easier computation. If you need to emphasise particular numbers, then underlining, significant spacing or bold type can be used. If data is lacking in a particular instance, then insert an asterisk in the empty space and give the reasons for the lack of data in a footnote. Footnotes can also be used to indicate, for example, the source of secondary data, a change in the way the data has been recorded, or any special circumstances which make the data seem odd. 184 Introduction to Statistics and Data Analysis © ABE and RRC It is not always possible to obey all of these rules on any one occasion, and there may be times when you have a good reason for disregarding some of them. But only do so if the reason is really good – not just to save you the bother of thinking! Now study the layout of the following table (based on our previous example of 3,000 employees) and check through the list of rules to see how they have been applied. Table 6.9: ABC & Co. – Wage structure of labour force Numbers of persons in specified categories Wage group (£ per week) Workshop 40 – 59.99 60 – 79.99 80 – 99.99 100 – 119.99 120 – 139.99 140 - 159.99 Total no. employed % See note (a) (1) (2) (3) (4) (5) (6) (7) (8) (9) A 20 101 202 219 29 29 600 20.0 B 11 52 90 120 29 58 360 12.0 C 19 103 210 200 88 40 660 22.0 D 34 167 303 317 18 1 840 28.0 E 21 87 115 159 136 22 540 18.0 Total no. in wage group 105 510 920 1,015 300 150 3,000 100.0 % See note (b) 3.5 17.0 30.7 33.8 10.0 5.0 100.0 – Note (a): Total number employed in each workshop as a percentage of the total workforce. Note (b): Total number in each wage group as a percentage of the total workforce. This table can be called a "twofold" table as the workforce is broken down by wage and workshop. Questions for Practice 1 These questions are typical of those likely to be set in an examination. 1. The human resources/personnel manager of a firm submits the following report on labour and staff to the managing director: "The total number of employees is 136, of whom 62 are male. Of these men, 43 are industrial labourers and the rest are staff. The age distribution of the 43 is 'under 18' (seven), '18–45' (22) and the others are over 45. Our total industrial labour force is 113, of whom 25 are 'under 18' and 19 are 'over 45'. The total number of 'under-18s' on the site is 29 (10 male, 19 female). There are 23 'over-45s', of whom only 4 are male staff." The managing director, quite rightly, thinks that this is gibberish and instructs you to present it to him in a tabular form which he can readily understand. Show the table you would produce. Introduction to Statistics and Data Analysis 185 © ABE and RRC 2. Consider the following report. "In the period Year 1 – Year 6 there were 1,775 stoppages of work arising from industrial disputes on construction sites. Of these, 677 arose from pay disputes, 111 from conflict over demarcation, 144 on working conditions and 843 were unofficial walk-outs. 10% of all stoppages lasted only one day. Of these, the corresponding proportions were: 11% for pay disputes, 6% in the case of demarcation and 12½% in the case of working conditions. In the subsequent period Year 7 – Year 12, the number of stoppages was 2,807 higher. But, whereas the number of stoppages arising from pay disputes fell by 351 and those concerning working conditions by 35, demarcation disputes rose by 748. The number of pay disputes resolved within one day decreased by 45 compared with the earlier period, but only by 2 in the case of working conditions. During the later period 52 stoppages over demarcation were settled within the day, but the number of stoppages from all causes which lasted only one day was 64 greater in the later period than in the earlier." You are required to: (a) Tabulate the information in the above passage, making whatever calculations are needed to complete the entries in the table. (b) Write a brief comment on the main features disclosed by the completed table. Now check your answers with those given at the end of the unit. D. FREQUENCY DISTRIBUTIONS A frequency distribution is a tabulation which shows the number of times (i.e. the frequency) each different value occurs. For example, the following figures are the times (in minutes) taken by a shop-floor employee to perform a given repetitive task on 20 specified occasions during the working day: 3.5 3.8 3.8 3.4 3.6 3.6 3.8 3.9 3.7 3.5 3.4 3.7 3.6 3.8 3.6 3.7 3.7 3.7 3.5 3.9 If we now assemble and tabulate these figures, we can obtain a frequency distribution as follows: 186 Introduction to Statistics and Data Analysis © ABE and RRC Table 6.10: Time taken for one employee to perform task Length of time (minutes) Frequency 3.4 2 3.5 3 3.6 4 3.7 5 3.8 4 3.9 2 Total 20 Simple Frequency Distributions A useful way of preparing a frequency distribution from raw data is to go through the records as they stand and mark off the items by the "tally mark" or "five-bar gate" method. First look at the figures to see the highest and lowest values so as to decide the range to be covered and then prepare a blank table. Then mark the items on your table by means of a tally mark. To illustrate the procedure, Table 6.11 shows the procedure for the data in Table 6.10 after all 20 items have been entered. Table 6.11: Time taken for one employee to perform task Length of time (minutes) Tally marks Frequency 3.4 | | 2 3.5 | | | 3 3.6 | | | | 4 3.7 | | | | 5 3.8 | | | | 4 3.9 | | 2 Total 20 Grouped Frequency Distributions Sometimes the data is so extensive that a simple frequency distribution is too cumbersome and, perhaps, uninformative. Then we make use of a grouped frequency distribution. In this case, the "length of time" column consists not of separate values, but of groups of values: Introduction to Statistics and Data Analysis 187 © ABE and RRC Table 6.12: Time taken for one employee to perform task Length of time (minutes) Tally marks Frequency 3.4 to 3.5 | | | | 5 3.6 to 3.7 | | | | | | | | 9 3.8 to 3.9 | | | | | 6 Total 20 Grouped frequency distributions are only needed when there is a large number of values and, in practice, would not have been required for the small amount of data in our example. Table 6.13 shows a grouped frequency distribution used in a more realistic situation, when an ungrouped table would not have been of much use. Table 6.13: Age distribution of employees in an office Age group (years) Number of employees 16 to < 21 10 21 to < 26 17 26 to < 31 28 31 to < 36 42 36 to < 41 38 41 to < 46 30 46 to < 51 25 51 to < 56 20 56 to < 60 10 Total 220 The various groups (for example, "26 to < 31") are called "classes" and the range of values covered by a class (five years in this example) is called the "class interval". The number of items in each class (for example, 28 in the 26 to < 31 class) is called the "class frequency" and the total number of items (in this example, 220) is called the "total frequency". The term "class boundary" is used to denote the dividing line between adjacent classes, so in the age group example (see Table 6.13) the class boundaries are 16, 21, 26, ... years. In the length of time example (see table 6.12), as grouped earlier in this section, the class boundaries are 3.35, 3.55, 3.75, 3.95 minutes. This needs some explanation. As the original readings were given correct to one decimal place, we assume that is the precision to which they were measured. If we had had a more precise stopwatch, the times could have been measured more precisely. In the first group of 3.4 to 3.5 we put times which could in fact be anywhere between 3.35 and 3.55 if we had been able to measure them more precisely. A time such as 3.57 minutes would not have been in this group as it equals 3.6 minutes when corrected to one decimal place and it goes in the 3.6 to 3.7 group. The term "class limits" is used to stand for the lowest and highest values that can actually occur in a class. In the age group example, these would be 16 years and 20 years 364 days 188 Introduction to Statistics and Data Analysis © ABE and RRC for the first class, 21 years and 25 years 364 days for the second class and so on, assuming that the ages were measured correct to the nearest day. In the length of time example, the class limits are 3.4 and 3.5 minutes for the first class and 3.6 and 3.7 minutes for the second class. You should make yourself quite familiar with these terms. Choice of Class Interval When compiling a frequency distribution you should, if possible, make the length of the class interval equal for all classes. This is so that fair comparisons can be made between one class and another. Sometimes, however, this rule has to be broken – for example, some grouped frequency distributions bring together the first few classes and the last few classes because there are few instances in each of the individual classes. In such cases, before using the information, it is as well to make the classes comparable by calculating a column showing "frequency per interval of so much", as in the example for wage statistics in Table 6.14. Table 6.14: Annual income statistics Annual income to the nearest £ No. of persons Frequency per £200 interval 4,801 – 5,000 50,000 50,000 5,001 – 5,200 40,000 40,000 5,201 – 5,600 55,000 27,500 5,601 – 6,000 23,000 11,500 6,001 – 6,400 12,000 6,000 6,401 – 7,200 12,000 3,000 Total 192,000 – Notice that the intervals in the first column are: 200, 200, 400, 400, 400, 800. These intervals let you see how the last column was compiled. A superficial look at the original table (first two columns only) might have suggested that the most frequent incomes were at the middle of the scale, because of the appearance of the figure 55,000. But this apparent preponderance of the middle class is due solely to the change in the length of the class interval, and column three shows that, in fact, the most frequent incomes are at the bottom end of the scale, i.e. the top of the table. You should remember that the purpose of compiling a grouped frequency distribution is to make sense of an otherwise troublesome mass of figures. It therefore follows that we do not want to have too many groups or we will be little better off; nor do we want too few groups or we will fail to see the significant features of the distribution. As a practical guide, you will find that somewhere between about five and 10 groups will usually be suitable. Introduction to Statistics and Data Analysis 189 © ABE and RRC Cumulative Frequency Distributions Very often we are not especially interested in the separate class frequencies, but in the number of items above or below a certain value. When this is the case, we form a cumulative frequency distribution as illustrated in column three of Table 6.15. Table 6.15: Time taken for one employee to perform task Length of time (minutes) Frequency Cumulative frequency 3.4 2 2 3.5 3 5 3.6 4 9 3.7 5 14 3.8 4 18 3.9 2 20 The cumulative frequency tells us the number of items equal to or less than the specified value, and it is formed by the successive addition of the separate frequencies. A cumulative frequency column may also be formed for a grouped distribution. The example in Table 6.15 gives us the number of items "less than" a certain amount. However we may wish to know, for example, the number of persons having more than some quantity. This can easily be done by doing the cumulative additions from the bottom of the table instead of the top, and as an exercise you should now compile the "more than" cumulative frequency column for Table 6.15. Relative Frequency Distributions All the frequency distributions which we have looked at so far in this study unit have had their class frequencies expressed simply as numbers of items. However, remember that proportions or percentages are useful secondary statistics. When the frequency in each class of a frequency distribution is given as a proportion or percentage of the total frequency, the result is known as a "relative frequency distribution" and the separate proportions or percentages are the "relative frequencies". The total relative frequency is, of course, always 1.0 (or 100%). Cumulative relative frequency distributions may be compiled in the same way as ordinary cumulative frequency distributions. As an example, the distribution used in Table 6.14 is set out as a relative frequency distribution in Table 6.16 for you to study. 190 Introduction to Statistics and Data Analysis © ABE and RRC Table 6.16: Annual income statistics Annual income to the nearest £ Frequency Relative frequency (%) Cumulative frequency Relative cumulative frequency (%) 4,801 – 5,000 50,000 26 50,000 26 5,001 – 5,200 40,000 21 90,000 47 5,201 – 5,600 55,000 29 145,000 76 5,601 – 6,000 23,000 12 168,000 88 6,001 – 6,400 12,000 6 180,000 94 6,401 – 7,200 12,000 26 192,000 100 Total 192,000 100 – – This example is in the "less than" form, and you should now compile the "more than" form in the same way as you did for the non-relative distribution. Questions for Practice 2 The following figures show the number of faulty items made on successive days by a machine in an engineering workshop: 2 0 2 4 3 2 11 4 0 1 1 4 1 2 1 4 1 7 3 2 4 5 5 6 5 2 4 2 1 2 2 4 0 6 1 0 1 5 1 2 3 1 5 3 2 2 1 3 5 3 1 1 10 1 2 3 4 3 10 6 2 6 5 3 1 2 5 7 2 3 4 4 1 3 7 4 1 7 3 7 1 1 2 6 1 0 8 5 8 1 5 6 1 3 2 2 1 2 6 4 Tabulate these figures as a frequency distribution, giving columns containing the frequency, the relative frequency, the cumulative frequency and the relative cumulative frequency. Now check your answers with those given at the end of the unit. Introduction to Statistics and Data Analysis 191 © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. XYZ Company: Age of Staff and Labour Industrial Staff Total Age group Male Female Total Male Female Total Male Female Total (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) Under 18 7 18 25 3 1 4 10 19 29 18 – 45 22 47 69 12 3 15 34 50 84 Over 45 14 5 19 4 0 4 18 5 23 TOTAL 43 70 113 19 4 23 62 74 136 2. (a) The complete table is as follows: Number of Stoppages of Work Arising from Industrial Disputeson Construction Sites Year 1 – Year 6 Year 7 – Year 12 Cause 1 day More than 1 day Total 1 day More than 1 day Total Pay 74.47 602.53 677 29.47 296.53 326 Demarcation 6.66 104.34 111 52 807 859 Working conditions 18 126 144 16 93 109 Unofficial* 78.37 764.63 843 144.03 3,143.97 3,288 All 177.5 1,597.5 1,775 241.5 4,340.5 4,582 * Note that we have given the figures as worked out from the information supplied. Obviously, though, fractional numbers of stoppages cannot occur, and you should round off each figure to the nearest whole number. The subtotals of the columns will then be 177; 1,598; 241; and 4,341. (b) The overall number of stoppages in the later period was more than 2½ times greater than in the earlier period. In both periods, the number of unofficial stoppages was the highest category. This was particularly remarkable in the later period, when they amounted to more than twice the number of all other stoppages put together. Demarcation disputes rose in number dramatically, although stoppages over pay and working conditions both declined. About 95% of stoppages in the later period lasted more than one day, compared with 90% in the previous period, with those concerning working conditions being less likely to last more than one day. 192 Introduction to Statistics and Data Analysis © ABE and RRC Questions for Practice 2 Faulty items made on successive days by a machine in an engineering workshop Number of faulty items Number of days with these faulty items Relative frequency Cumulative frequency Relative cumulative frequency 0 5 0.05 5 0.05 1 23 0.23 28 0.28 2 20 0.20 48 0.48 3 13 0.13 61 0.61 4 12 0.12 73 0.73 5 10 0.10 83 0.83 6 7 0.07 90 0.90 7 5 0.05 95 0.95 8 2 0.02 97 0.97 9 0 0.00 97 0.97 10 2 0.02 99 0.99 11 1 0.01 100 1.00 Total 100 1.00 – – 193 © ABE and RRC Study Unit 7 Graphical Representation of Information Contents Page Introduction 194 A. Graphical Representation of Frequency Distributions 194 Frequency Dot Diagram 195 Frequency Bar Chart 195 Frequency Polygon 196 Histogram 197 The Ogive 200 Stem and Leaf Diagram 203 B. Pictograms 204 Limited Form 204 Accurate Form 205 C. Pie Charts 206 D. Bar Charts 207 Component Bar Chart 207 Horizontal Bar Chart 208 E. Scatter Diagrams 209 F. General Rules for Graphical Presentation 210 Answers to Questions for Practice 215 194 Graphical Representation of Information © ABE and RRC INTRODUCTION In this study unit we shall move on and look at further methods of summarising and presenting data as a means of providing useful information. Diagrams are particularly effective as a way of conveying quite complex information. Presenting information visually is easy to understand, and enables broad distributions and trends to be taken in quickly. However, diagrams are rarely as accurate as the figures themselves, and details are likely to be lost. In the first part of this unit we shall examine the specific graphical forms used for presenting frequency distributions of the type considered in the previous study unit. We then move on to discuss a number of popular graphical presentations that are used in everyday life, presentations that are not necessarily used by statisticians (e.g. those used in business journals and the mass media of newspapers and television). Objectives When you have completed this study unit you will be able to: use frequency dot diagrams, frequency bar charts, frequency polygons, histograms, ogives and stem and leaf diagrams to represent frequency distributions prepare pictograms, pie charts, bar charts and scatter diagrams from given data, and discuss the circumstances in which each form of presentation is most effective describe the general rules and principles for using graphical representations of data. A. GRAPHICAL REPRESENTATION OF FREQUENCY DISTRIBUTIONS Tabulated frequency distributions are sometimes more readily understood if represented by a diagram. Graphs and charts are normally much superior to tables (especially lengthy complex tables) for showing general states and trends. The methods of presenting frequency distributions graphically are as follows: frequency dot diagram frequency bar chart frequency polygon histogram ogive. However, such graphs and charts cannot usually be used for accurate analysis of data. The following is a diagram that to some extent overcomes this limitation: stem and leaf diagram. We will now examine each of these graphical methods in turn, initially using the data on the times (in minutes) taken by an employee to perform a given task on 20 specified occasions during the working day which we previously considered in Study Unit 6 (Table 6.10). The tabulated frequency distribution for this data is as follows: Graphical Representation of Information 195 © ABE and RRC Table 7.1: Time taken for one employee to perform task Length of time (minutes) Frequency 3.4 2 3.5 3 3.6 4 3.7 5 3.8 4 3.9 2 Total 20 Frequency Dot Diagram This is a simple form of graphical representation for the frequency distribution of a discrete variate. (In statistics, the term "variate" is used when you might have expected the term "variable". Whilst the two terms do not mean exactly the same thing, roughly speaking, at this level, they can be used interchangeably. Loosely, a variate is a quantity that can have any of a set of numerical values. Do not worry too much about the difference between these two terms.) A horizontal scale is used for the variate and a vertical scale for the frequency. Above each value on the variate scale we mark a dot for each occasion on which that value occurs. Thus, a frequency dot diagram of the information in Table 7.1 would be as shown in Figure 7.1. Figure 7.1: Frequency dot diagram to show length of time taken by operator to complete a given task Frequency Length of time (minutes) 3.4 3.5 3.6 3.7 3.8 3.9 196 Graphical Representation of Information © ABE and RRC Frequency Bar Chart We can avoid the business of marking every dot in such a diagram by drawing instead a vertical line, the length of which represents the number of dots which should be there. The frequency dot diagram in Figure 7.1 now becomes a frequency bar chart, as in Figure 7.2. Figure 7.2: Frequency bar chart to show length of time taken by operator to complete a given task Frequency Polygon Instead of drawing vertical bars as we do for a frequency bar chart, we could merely mark the position of the top end of each bar and then join up these points with straight lines. When we do this, the result is a frequency polygon, as in Figure 7.3. Figure 7.3: Frequency polygon to show length of time taken by operator to complete a given task Note that we have added two fictitious classes at each end of the distribution, i.e. we have marked in groups with zero frequency at 3.3 and 4.0. This is done to ensure that the area enclosed by the polygon and the horizontal axis is the same as the area under the corresponding histogram which we shall consider in the next section. Frequency Length of time (minutes) 0 1 2 3 4 5 6 3.4 3.5 3.6 3.7 3.8 3.9 Frequency Length of time (minutes) 0 1 2 3 4 5 6 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4 Graphical Representation of Information 197 © ABE and RRC These three kinds of diagram are all commonly used as a means of making frequency distributions more readily comprehensible. They are mostly used in those cases where the variate is discrete and where the values are not grouped. Sometimes, frequency bar charts and polygons are used with grouped data, by drawing the vertical line (or marking its top end) at the centre point of the group. Histogram A histogram is the best way of showing a grouped frequency distribution. To illustrate this, we shall use an example of a grouped frequency distribution similar to one from Study Unit 6 (Table 6.13), as set out in Table 7.2. The information in the table may be shown graphically by the histogram in Figure 7.4. Table 7.2: Age distribution of employees in an office Age group (years) Number of employees 15 to < 20 10 20 to < 25 17 25 to < 30 28 30 to < 35 42 35 to < 40 38 40 to < 45 30 45 to < 50 25 50 to < 55 20 55 to < 60 10 Total 220 Figure 7.4: Histogram showing age distribution of employees in an office Frequency Age (years) 15 20 25 30 35 40 45 50 60 0 10 20 30 40 55 198 Graphical Representation of Information © ABE and RRC In a histogram, the frequency in each group is represented by a rectangle and – this is a very important point – it is the area of the rectangle, not its height, which represents the frequency. When the lengths of the class intervals are all equal, then the heights of the rectangles represent the frequencies in the same way as do the areas (this is why the vertical scale has been marked in this diagram). If however the lengths of the class intervals are not all equal, then you must remember that the heights of the rectangles have to be adjusted to give the correct areas. (Do not stop at this point if you have not quite grasped the idea, because it will become clearer as you read on.) Look once again at the histogram of ages given in Figure 7.4. Note particularly how it illustrates the fact that the frequency falls off towards the higher age groups – any form of graph which did not reveal this fact would be misleading. Now let us imagine that the original table had not used equal class intervals but, for some reason or other, had given the last few groups as those in Table 7.3. Table 7.3: Age distribution of employees in an office (amended extract) Age group (years) Number of employees 40 to < 45 30 45 to < 50 25 50 to < 60 30 The last two groups have been lumped together as one. A wrong form of histogram, using heights instead of areas, for the amended data would look like Figure 7.5. Figure 7.5: Histogram showing age distribution of employees in an office (amended data) Frequency Age (years) 15 20 25 30 35 40 45 50 60 0 10 20 30 40 This column is the wrong size Graphical Representation of Information 199 © ABE and RRC Now, this clearly gives an entirely wrong impression of the distribution with respect to the higher age groups. In the correct form of the histogram, the height of the last group (50 to < 60) would be halved because the class interval is double all the other class intervals. The histogram in Figure 7.6 gives the right impression of the falling off of frequency in the higher age groups. The vertical axis has been labelled "Frequency density per 5-year interval" as five years is the "standard" interval on which we have based the heights of our rectangles. Figure 7.6: Amended histogram showing age distribution of employees in an office (amended data) Often it happens, in published statistics, that the last group in a frequency table is not completely specified. For example, the last few groups in our age distribution table may have been specified as follows: Table 7.4: Age distribution of employees in an office (further amended extract) Age group (years) Number of employees 40 to < 45 30 45 to < 50 25 50 and over 30 A distribution of this kind is called an "open-ended" distribution. How would we draw this on a histogram? If the last (open-ended) group has a very small frequency compared with the total frequency (say, less than about 1% or 2%), then nothing much is lost by leaving it off the histogram altogether. If the last group has a larger frequency than about 1% or 2%, then you should try to judge from the general shape of the histogram how many class intervals to spread the last frequency over, in order not to create a false impression of the extent of the distribution. In Frequency density per 5-year interval Age (years) 15 20 25 30 35 40 45 50 60 0 10 20 30 40 200 Graphical Representation of Information © ABE and RRC the example given, you would probably spread the last 30 people over two or three class intervals, but it is often simpler to assume that an open-ended class has the same length as its neighbour. Whatever procedure is adopted, it is essential to state clearly what you have done and why. The Ogive This is the name given to the graph of the cumulative frequency. It can be drawn in either the "less than" or the "or more" form, but the "less than" form is the usual one. Ogives for two of the distributions already considered are now given as examples: Figure 7.7 is for ungrouped data (as per Table 7.5) and Figure 7.8 is for grouped data (as per Table 7.6). Study these two diagrams so that you are quite sure that you know how to draw them. There is only one point which you might be tempted to overlook in the case of the grouped distribution – the points are plotted at the ends of the class intervals and not at the centre point. Look at the example Table 7.6 and Figure 7.8 and see how the 168,000 point is plotted against the upper end of the 5,601 – 6,000 group (or class interval) and not against the midpoint, 5,750. If we had been plotting a "or more" ogive, the plotting would have been against the lower end of the group (or class interval). Table 7.5: Time taken for one employee to perform task Length of time (minutes) Frequency Cumulative frequency 3.4 2 2 3.5 3 5 3.6 4 9 3.7 5 14 3.8 4 18 3.9 2 20 Table 7.6: Annual income statistics Annual income to nearest £ Frequency Cumulative frequency 4,801 – 5,000 50,000 50,000 5,001 – 5,200 40,000 90,000 5,201 – 5,600 55,000 145,000 5,601 – 6,000 23,000 168,000 6,001 – 6,400 12,000 180,000 6,401 – 7,200 12,000 192,000 Total 192,000 – Graphical Representation of Information 201 © ABE and RRC Figure 7.7: Ogive to show length of time taken by operator to complete a given task Figure 7.8: Ogive to show annual income distribution 0 5 10 15 20 3.2 3.4 3.6 3.8 4 Cumulative frequency (less than or equal to) Length of time (mins) 0 50 100 150 200 4.8 5.2 5.6 6.0 6.4 6.8 7.2 Less than cumulative frequency (000s) Annual income (£000s) 202 Graphical Representation of Information © ABE and RRC As an example of an "or more" ogive, we can compile the data from our example of annual income statistics as a "more than" cumulative frequency, shown in Table 7.7. Table 7.7: Annual income statistics Annual income to nearest £ Frequency Cumulative frequency (more than) 4,801 – 5,000 50,000 192,000 5,001 – 5,200 40,000 142,000 5,201 – 5,600 55,000 102,000 5,601 – 6,000 23,000 47,000 6,001 – 6,400 12,000 24,000 6,401 – 7,200 12,000 12,000 Total 192,000 – The ogive for this now appears as shown in Figure 7.9. Figure 7.9: Ogive to show annual income distribution Check that you see how the plotting has been made against the lower end of the group and notice how the ogive has a reversed shape. In each of Figures 7.8 and 7.9 we have added a fictitious group of zero frequency at one end of the distribution. 0 50 100 150 200 4.8 5.2 5.6 6.0 6.4 6.8 7.2 More than cumulative frequency (000s) Annual income (£000s) Graphical Representation of Information 203 © ABE and RRC It is common practice to call the cumulative frequency graph a "cumulative frequency polygon" if the points are joined by straight lines, and a "cumulative frequency curve" if the points are joined by a smooth curve. (NB. Unless you are told otherwise, always compile a "less than" cumulative frequency.) Of course all of these diagrams may be drawn from the original figures or on the basis of relative frequencies. In more advanced statistical work the latter are used almost exclusively, and you should practise using relative frequencies whenever possible. Stem and Leaf Diagram A stem and leaf diagram is the final method of presenting frequency distributions graphically that we will study in this section. It differs from the frequency dot diagram, frequency bar chart, frequency polygon, histogram and ogive that we have already studied in that it has the advantage of preserving the original data. Thus, in addition to providing a visual representation of the data, stem and leaf diagrams can also allow further analysis of the data as it is possible to obtain the original data points from the diagram (i.e. there is no loss of information). Similar to the histogram, a stem and leaf diagram is a way of grouping your original data into classes. Like the histogram, it provides a partial sorting of the original data to allow a visual identification of the distribution pattern of the data. Unlike the histogram, it provides extra detail regarding individual values. In a stem and leaf diagram, each data value is split into a stem and a leaf. The leaf is usually the last digit of the number and the other digits to the left of the leaf form the stem. The stem and leaf diagram is plotted by listing the first digits (i.e. the "stems") in a column (and sorted in numerical order), then listing the second digit (i.e. the "leaves") in a second column to the right of the appropriate "stem" (and again sorted in numerical order). It is probably easiest to understand the stem and leaf diagram through an example, so let us construct a stem and leaf diagram using the following data of a student's exam marks: 79 80 82 68 65 59 61 57 50 62 61 70 69 64 67 70 62 65 65 73 76 87 80 82 83 79 77 71 80 77 Before you begin, you might want to arrange the data in numerical order, but is not a required step as ordering can be done later. But what is necessary is that you separate each number into a stem and a leaf. Since these are two digit numbers, the tens digit is the stem and the units digit is the leaf (e.g. for the number 79, the stem is "7" and the leaf is "9"). Next, group the numbers with the same stems (e.g. 79 and 77 are in the same group as both have a stem of 7). Then, list the stems in numerical order (in our example, the stems range from 5 to 8 as the lowest value is 50 and the highest value is 87) in a column. In a second column list the leaves to the right of the appropriate stem values. If your leaf values are not in increasing order, order them now.) Finally, add a title to your diagram. If you have successfully followed these steps, then you should have finished up with a stem and leaf diagram identical to the one shown in Figure 7.10. As you will see, for the stem value of 5 we have leaf values of 0, 7 and 9, which correspond to the original values of 50, 57 and 59. 204 Graphical Representation of Information © ABE and RRC Figure 7.10: Stem and leaf diagram to show the distribution of exam marks Stem Leaf 5 0 7 9 6 1 1 2 2 4 5 5 5 7 8 9 7 0 0 1 3 6 7 7 9 9 8 0 0 0 2 2 3 7 Key: 50 means 50 marks B. PICTOGRAMS Pictograms are the simplest method of presenting information visually. These diagrams are variously called "pictograms", "ideograms", or "picturegrams" – the words all refer to the same thing. Their use is confined to the simplified presentation of statistical data for the general public. Pictograms consist of simple pictures which represent quantities. There are two types and these are illustrated in the following examples. The data we will use is shown in Table 7.8. Table 7.8: Cruises organised by a shipping line between Year 1 and Year 3 Year Number of cruises No. of passengers carried 1 200 100,000 2 250 140,000 3 300 180,000 Limited Form We could represent the number of cruises by showing pictures of ships of varying size, as in Figure 7.11. Figure 7.11: Number of cruises Years 1-3 (Source: Table 7.8) Year 1 Year 2 Year 3 200 cruises 250 cruises 300 cruises Graphical Representation of Information 205 © ABE and RRC Although these diagrams show that the number of cruises has increased each year, they can give false impressions of the actual increases. The reader can become confused as to whether the quantity is represented by the length or height of the pictograms, their area on the paper, or the volume of the object they represent. It is difficult to judge what increase has taken place. (Sometimes you will even find pictograms in which the sizes shown are actually wrong in relation to the real increases.) Accurate Form If the purpose of the pictogram is to covey anything but the most minimal of information, to avoid confusion it is better to try and give it a degree of accuracy in representing the numerical data. This is usually done by representing specific quantities of the variable with pictures or graphics of some sort – usually related to the subject of the variable. So, we could represent the numbers of passengers carried by using pictures of people as in Figure 7.12. Figure 7.12: Number of passengers carried Years 1-3 (Source: Table 7.8) Each matchstick man is the same height and represents 20,000 passengers, so there can be no confusion over size. These diagrams have no purpose other than generally presenting statistics in a simple way. They are not capable of providing real accurate detail. Consider Figure 7.13 – it is difficult to represent a quantity of less than 10m barrels – so we are left wondering what "[" means. Figure 7.13: Imports of crude oil Year 1: Year 2: Year 3 Key: 20,000 passengers Year 2: Year 1: Key: represents 10m barrels of oil 206 Graphical Representation of Information © ABE and RRC C. PIE CHARTS Pie charts, known also as circular diagrams, are used to show the manner in which various components add up to a total. Like pictograms, they are only used to display very simple information to non-expert readers. They are popular in computer graphics. Suppose that we wish to illustrate the sales of gas in Great Britain in a certain year. The data is taken from the Annual Abstract of Statistics as shown in Table 7.9. Table 7.9: Gas sales in Great Britain in one year (Source: Annual Abstract of Statistics) Uses Million therms % Domestic 1,383 51 Industrial 843 31 Commercial 437 16 Public* 60 2 Total 2,723 100 * Central and local government uses, including public lighting The data is illustrated as a pie chart or circular diagram in Figure 7.14. Figure 7.14: Example of a pie chart – gas sales in Great Britain (Source: Table 7.9) There are a number of simple steps to creating a pie chart. (a) Tabulate the data and calculate the percentages. (b) Convert the percentages into degrees, e.g. 51% of 360 100 51 360 183.6 (c) Construct the diagram by means of a pair of compasses and a protractor. Don't overlook this point, because inaccurate and roughly drawn diagrams do not convey information effectively. Commercial 437m therms Domestic 1,383m therms Industrial 843m therms Public 60m therms Graphical Representation of Information 207 © ABE and RRC 0 2 4 6 8 10 12 14 16 Year 1 Year 2 (d) Label the diagram clearly, using a separate legend or key if necessary. (A key is illustrated in Figure 7.15 below.) You can insert the exact values of the data for each segment if more detail is required, since it is not possible to read this exactly from the diagram itself. (e) If you have the choice, don't use a diagram of this kind with more than four or five component parts. The main use of a pie chart is to show the relationship each component part bears to the whole. Pie charts are sometimes used side by side to provide comparisons, but this is not really to be recommended. Unless the whole diagram in each case represents exactly the same total amount, other diagrams such as bar charts (see next section) are much clearer. D. BAR CHARTS We have already met one kind of bar chart in the course of our studies of frequency distributions, namely the frequency bar chart. The bars there were simply thick lines representing, by their length, the frequencies of different values of the variate. However the idea of a bar chart can be extended beyond the field of frequency distributions, and we will now illustrate a number of the types of bar chart in common use. We say "illustrate" because there are no rigid and fixed types, but only general ideas which are best studied by means of examples. You can supplement the examples in this study unit by looking at the commercial pages of newspapers and magazines. Component Bar Chart This first type of bar chart serves the same purpose as a circular diagram and, for that reason, is sometimes called a "component bar diagram" (see Figure 7.15). Figure 7.15: Component bar chart showing cost of production of ZYX Co. Ltd Note that the lengths of the components represent the amounts, and that the components are drawn in the same order so as to facilitate comparison. These bar charts are preferable to circular diagrams because: they are easily read, even when there are many components they are more easily drawn it is easier to compare several bars side by side than several circles. Wages Materials Royalties Key: 208 Graphical Representation of Information © ABE and RRC Bar charts with vertical bars are sometimes called "column charts", to distinguish them from those in which the bars are horizontal (see Figure 7.16). Figure 7.16: Horizontal bar chart of visitors arriving in the UK in one year This figure is also an example of a percentage component bar chart, i.e. the information is expressed in percentages rather than in actual numbers of visitors. If you compare several percentage component bar charts, you must be careful. Each bar chart will be the same length, as they each represent 100%, but they will not necessarily represent the same actual quantities – for example, 50% might have been 1 million, whereas in another year it may have been nearer to 4 million and in another to 8 million. Horizontal Bar Chart A typical case of presentation by a horizontal bar chart is shown in Figure 7.17. Note how a loss is shown by drawing the bar on the other side of the zero line. Figure 7.17: Horizontal bar chart for the So & So Company Ltd to show profits made by branches in Year 1 and Year 2 Pie charts and bar charts are especially useful for "categorical" variables as well as for numerical variables. The example in Figure 7.17 shows a categorical variable – i.e. the different branches form the different categories, whereas in Figure 7.15 we have a numerical variable, namely, time. Figure 7.17 is also an example of a multiple or compound bar chart as there is more than one bar for each category. European 51% Commonwealth 23% USA 22% O t h e r s 4 % 0 15 10 20 25 5 5 Profit (£000s) Loss (£000s) London B London A Newcastle Bristol Year 1 Year 1 Key: Graphical Representation of Information 209 © ABE and RRC E. SCATTER DIAGRAMS So far we have learnt about various methods used to present data graphically. The final graphical method that we shall learn about here is called a scatter diagram. A scatter diagram (also known as a scattergraph, scattergram or scatter plot) differs fundamentally from the other graphical forms of presenting data that we have considered so far. Scatter diagrams are used as a visual aid to examine potential relationships between two sets of (quantitative) variables, rather than simply providing a visual summary of one (or more) set of variables (as in the case of frequency distribution graphs, pictograms, pie charts and bar charts). Specifically, scatter diagrams are the appropriate method of presenting data when you want to examine the relationship between two continuous variables, such as the relationship that may exist between sales revenue and the level of advertising expenditure. Scatter diagrams are plotted with one variable (e.g. sales revenue) on the x-axis (the horizontal axis) and the other variable (e.g. level of advertising expenditure) on the y-axis (vertical axis). It is convention to plot the independent variable on the x-axis and the dependent variable on the y-axis. So using our example of sales revenue and the level of advertising expenditure, we are examining whether sales revenue (the dependent variable) is dependent on the level of advertising (the independent variable): does an increase in the level of advertising expenditure lead to an increase in sales revenue? Let us look at an example. Table 7.10 presents sales data for a manufacturing company against its expenditure on advertising over a given period. Looking at the data, it is not immediately obviously whether a relationship exists or not and if so, whether this relationship is particularly strong. Table 7.10: Advertising expenditure and sales for a manufacturing company Advertising expenditure (£000) Sales revenue (£000) 23 123 25 134 28 150 31 179 31 187 33 191 36 200 37 202 43 203 49 210 20 85 12 62 8 49 19 106 17 91 15 89 210 Graphical Representation of Information © ABE and RRC However, if we plot this data on a scatter diagram, the relationship between the level of advertising expenditure and sales revenue for this manufacturing company becomes immediately clear. The scatter diagram presented in Figure 7.18 confirms that there appears to be a positive relationship. As the company spends more on advertising, its sales revenue increases (although sales revenue does seem to flatten out after advertising reaches £350,000). Figure 7.18 Scatter diagram of advertising expenditure against sales revenue for a manufacturing company While scatter diagrams are used to study possible relationships between two variables, it is important to stress that these diagrams cannot prove that one variable causes the other. However, they may indicate the existence and strength of an association between two variables. This association is known as correlation. Thus, the relationship shown in Figure 7.18, where both variables are increasing together, is described as positive correlation (both variables are positively correlated with each other). In contrast, if one variable increases while the other decreases, the relationship is described as negative correlation. For example, we would expect the efficiency of the machines in this manufacturing company to decline as they get older, suggesting that the efficiency and age of machinery are negatively correlated. If the scatter diagram does not show any relationship between the variables then no correlation exists. For example, we would not expect any correlation between the cost of raw materials and employee output. These three types of correlation are shown in Figure 7.19. 0 50 100 150 200 250 0 10 20 30 40 50 60 Sales revenue (£000) Advertising expenditure (£000) Graphical Representation of Information 211 © ABE and RRC Figure 7.19: Scatter diagram examples showing different types of correlation Positive correlation Negative correlation No correlation Years of post school education 1 2 Annual Income (£000s) 10 20 3 4 50 40 30 0 0 250 Age of machinery (months) 10 20 Machinery efficiency (output per day) 50 100 30 40 200 150 0 0 Employee output (items per day) 20 40 Cost of raw materials (£ per tonne) 20 40 60 80 100 80 60 0 0 212 Graphical Representation of Information © ABE and RRC F. GENERAL RULES FOR GRAPHICAL PRESENTATION There are a number of general rules which must be borne in mind when planning and using the graphical methods covered in this unit. Graphs and charts must be given clear but brief titles. The axes of graphs must be clearly labelled, and the scales of values clearly marked. Diagrams should be accompanied by the original data, or at least by a reference to the source of the data. Avoid excessive detail, as this defeats the object of diagrams. Wherever necessary, guidelines should be inserted to facilitate reading. Try to include the origins of scales. Obeying this rule sometimes leads to rather a waste of paper space. In such a case the graph could be "broken" as shown in Figure 7.20, but take care not to distort the graph by overemphasising small variations. Figure 7.20: Sales of LMN Factory A Months Sales (£000s) M F J M J J A S O N D 50 60 70 Graphical Representation of Information 213 © ABE and RRC Questions for Practice These questions are typical of those likely to be set in an examination. 1. The following figures show the number of faulty items made on successive days by a machine in an engineering workshop: 2 0 2 4 3 2 11 4 0 1 1 4 1 2 1 4 1 7 3 2 4 5 5 6 5 2 4 2 1 2 2 4 0 6 1 0 1 5 1 2 3 1 5 3 2 2 1 3 5 3 1 1 10 1 2 3 4 3 10 6 2 6 5 3 1 2 5 7 2 3 4 4 1 3 7 4 1 7 3 7 1 1 2 6 1 0 8 5 8 1 5 6 1 3 2 2 1 2 6 4 Tabulate these figures as a frequency distribution, giving columns containing the frequency, the relative frequency, the cumulative frequency and the relative cumulative frequency. Illustrate the distribution by means of a frequency bar chart of the relative figures. 2. Illustrate the following data by means of a histogram, a frequency polygon and an ogive: Weekly wage (£) No. of employees 31.01 to 36.00 6 36.01 to 41.00 8 41.01 to 46.00 12 46.01 to 51.00 18 51.01 to 56.00 25 56.01 to 61.00 30 61.01 to 66.00 24 66.01 to 71.00 14 3. Construct a stem and leaf diagram to illustrate the age distribution of customers purchasing clothing from a retail store over a one-hour period given the following data on customer ages: 8 13 16 25 26 29 30 32 37 38 40 41 44 47 49 51 54 55 58 61 63 67 75 78 82 86 95 214 Graphical Representation of Information © ABE and RRC 4. Construct a circular diagram to illustrate the following figures: Source Amount (£000) Income Tax 1,863,400 Estate Duties 164,500 Customs & Excise 1,762,500 Miscellaneous 99,450 Sources of revenue in Ruritania for year ending 31 March 5. Illustrate by means of a bar chart the following data relating to the number of units sold of a particular kind of machine: Year Home sales Commonwealth sales Foreign sales 1 6,500 5,000 5,800 2 6,250 5,400 5,700 3 5,850 6,600 5,500 6. The unit price of a commodity and its demand over a period of time are shown in the following table: Price (£ per item) Number of items sold (000) 1.20 41 2.39 38 3.30 34 4.31 29 5.27 26 6.29 30 7.26 25 8.24 22 9.18 19 10.16 15 Draw a scatter diagram of the data and comment on the relationship between price and the amount sold. Now check your answers with those given at the end of the unit. Graphical Representation of Information 215 © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE 1. Number of faulty items Number of days with these items Relative frequency Cumulative frequency Relative cumulative frequency 0 5 0.05 5 0.05 1 23 0.23 28 0.28 2 20 0.20 48 0.48 3 13 0.13 61 0.61 4 12 0.12 73 0.73 5 10 0.10 83 0.83 6 7 0.07 90 0.90 7 5 0.05 95 0.95 8 2 0.02 97 0.97 9 0 0.00 97 0.97 10 2 0.02 99 0.99 11 1 0.01 100 1.00 Total 100 1.00 – – Frequency bar chart – faulty items produced by machine in workshop Relative frequency No. of faulty items per day 0 0.05 0.1 0.15 0.2 0.25 0 1 2 3 4 5 6 7 8 9 10 11 216 Graphical Representation of Information © ABE and RRC 2. Combined histogram and frequency polygon Workings for ogive: Wage Group (£) Frequency Cumulative frequency 31.01 – 36.00 6 6 36.01 – 41.00 8 14 41.01 – 46.00 12 26 46.01 – 51.00 18 44 51.01 – 56.00 25 69 56.01 – 61.00 30 99 61.01 – 66.00 24 123 66.01 – 71.00 14 137 The ogive follows: Weekly wage (£) No. of employees 31 36 41 46 51 56 61 66 0 10 20 30 71 Graphical Representation of Information 217 © ABE and RRC Ogive for wages distribution 3. Stem and leaf diagram to show the age distribution of retail customers for a clothing store over a one-hour period Stem Leaf 0 8 1 3 6 2 5 6 9 3 0 2 7 8 4 0 1 4 7 9 5 1 4 5 8 6 1 3 7 7 5 8 8 2 6 9 5 Key: 08 means 8 years old Cumulative frequency Weekly wage (£) 0 20 40 60 80 100 120 140 31 36 41 46 51 56 61 66 71 218 Graphical Representation of Information © ABE and RRC 0 2,000 4,000 6,000 8,000 10,000 12,000 14,000 16,000 18,000 Year 1 Year 2 Year3 4. The amounts must first be turned into percentages and degrees. Income tax 47.9% 172 Estate duties 4.2% 15 Customs & excise 45.3% 163 Miscellaneous 2.6% 9 Total 100.0% 359 Note: The angles do not add up to 360 exactly because of rounding errors. Sources of revenue in Ruritania 5. The bar chart is shown in below. Shading has been added for clarity. You can, of course, draw bar charts on a percentage basis. Bar chart of machine sales Income tax Customs & excise Estate duties Miscellaneous Foreign sales Home sales Commonwealth sales Key: Graphical Representation of Information 219 © ABE and RRC 6. The scatter diagram showing the relationship between product price and the number of items sold for a company follows. In general, the scatter diagram shows that as price increases demand for the product (measured in terms of number of items sold) falls. This suggests that there is a negative correlation between the price of this particular item and product demand. 0 5 10 15 20 25 30 35 40 45 0 2 4 6 8 10 12 Sales (000s) Price (£ per item) 220 Graphical Representation of Information © ABE and RRC 221 © ABE and RRC Study Unit 8 Measures of Location Contents Page Introduction 222 A. The Arithmetic Mean 223 The Mean of a Simple Frequency Distribution 224 The Mean of a Grouped Frequency Distribution 225 Simplified Calculation 226 Calculating the Mean from a Stem and Leaf Diagram 230 Characteristics of the Arithmetic Mean 231 B. The Mode 232 Mode of a Simple Frequency Distribution 232 Mode of a Grouped Frequency Distribution 232 Calculating the Mode from a Stem and Leaf Diagram 234 Characteristics of the Mode 235 C. The Median 236 Median of a Simple Frequency Distribution 236 Median of a Grouped Frequency Distribution 237 Calculating the Median from a Stem and Leaf Diagram 238 Answers to Questions for Practice 243 222 Measures of Location © ABE and RRC INTRODUCTION We have looked at frequency distributions in detail in Study Unit 6 and you should, by means of a quick revision, make sure that you have understood them well before proceeding. A frequency distribution may be used to give us concise information about its variate, but often we wish to compare two or more distributions. Consider for example the distribution of the weights of eggs from two different breeds of poultry, which is a topic in which you would be interested if you worked for an egg marketing company. Having weighed a large number of eggs from each breed, we might have compiled frequency distributions and graphed the results. The two frequency polygons might well look something like Figure 8.1. Figure 8.1: Frequency polygon showing weight of eggs from two breeds of hen Examining these distributions you will see that they look much alike except for one thing – they are located on different parts of the scale. In this case the distributions overlap and, although some eggs from Breed A are of less weight than some eggs from Breed B, eggs from Breed A are, in general, heavier than those from Breed B. Remember that one of the objects of statistical analysis is to condense unwieldy data so as to make it more readily understood. Drawing the frequency curves has enabled us to make an important general statement concerning the relative egg weights of the two breeds of poultry, but we would now like to take the matter further. We would like to calculate some figure which will serve to indicate the general level of the variate under discussion. In everyday life, we commonly use such a figure when we talk about the "average" value of something or other. We might have said, in reference to the two kinds of egg, that those from Breed A had a higher average weight than those from Breed B. Distributions with different averages indicate that there is a different general level of the variate in the two groups. The single value which we use to describe the general level of the variate is called a "measure of location" or a "measure of central tendency". There are three such measures with which you need to be familiar: the arithmetic mean the mode the median. Weight of eggs Frequency Breed A Breed B Measures of Location 223 © ABE and RRC Objectives When you have completed this study unit you will be able to: calculate the arithmetic mean and mode of grouped and ungrouped distributions, including stem and leaf diagrams find the median of grouped and ungrouped data (including stem and leaf diagrams) by calculation and, where appropriate, graphical methods utilise the assumed mean method and the "class interval" method in calculating the arithmetic mean list the relative advantages and disadvantages of the arithmetic mean, median and mode, and so decide on the appropriate measure of location to use in common situations. A. THE ARITHMETIC MEAN The arithmetic mean is what we normally think of as the "average" of a set of values. It is obtained by adding together all the values and then dividing the total by the number of values involved. Take, for example, the following set of values which are the heights, in centimetres, of seven men: Table 8.1: Height of seven men Man Height (cms) A 187 B 160 C 163 D 180 E 180 F 168 G 187 Total 1225 The arithmetic mean of these heights is 1225 7 175 cms. Notice that some values occur more than once, but we still add them all. At this point we must introduce a little algebra. We don't always want to specify what particular items we are discussing, whether it is heights, egg weights, wages, etc. For this reason in general discussion we will use (as you will recall from algebra) some general letter, usually x. Also, we indicate the sum of a number of values of x by the Greek letter sigma, written Σ. Thus, in our example, we may write: Σx 1225 We shall be using sigma notation extensively in this unit, so make sure you understand the meaning of the expression. 224 Measures of Location © ABE and RRC We indicate the arithmetic mean by the symbol x (called "x bar") and the number of items by the letter n. The calculation of the arithmetic mean can then be described by formula: n Σx x or Σx n 1 x The latter expression is customary in statistical work. Applying it to the example above, we have: 175 1225 7 1 x ) ( cms. You will often find the arithmetic mean simply referred to as "the mean" when there is no chance of confusion with other means (which we are not concerned with here). The Mean of a Simple Frequency Distribution When there are many items (that is when n is large), the arithmetic can be eased somewhat by forming a frequency distribution. Consider the following example. Example 1: Table 8.2: Height distribution data Height in cms (x) No. of men at this height (f) Product (fx) 160 1 160 163 1 163 168 1 168 180 2 360 187 2 374 Total Σf 7 Σ(fx) 1225 Indicating the frequency of each value by the letter f, you can see that Σf n and that, when the x's are not all the separate values but only the different ones, the formula becomes: Σf fx Σ x ) ( which is 1225/7 175 cms as before. Of course, with only seven items it would not be necessary in practice to use this method. However if we had a much larger number of items the method would save a lot of additions. Example 2: Consider the following table and, as practice, complete the (fx) column and calculate the value of the arithmetic mean, x . Do this before moving on. Measures of Location 225 © ABE and RRC Table 8.3: Data for calculation of arithmetic mean Value of x Number of items (f) Product (fx) 3 1 3 4 4 16 5 7 35 6 15 7 35 8 24 9 8 10 6 Total You should have obtained the following answers: the total numbers of items, Σf 100 the total product, Σ(fx) 713 the arithmetic mean, 100 713 x 7.13 Make sure that you understand everything we have covered so far. Revise it if necessary, before going on to the next paragraph. It is most important that you do not get muddled about calculating arithmetic means. The Mean of a Grouped Frequency Distribution Suppose now that you have a grouped frequency distribution. In this case, you will remember, we do not know the actual individual values, only the groups in which they lie (unless you have a stem and leaf diagram, which we will discuss later). How then can we calculate the arithmetic mean? The answer is that we cannot calculate the exact value of x , but we can make an approximation sufficiently accurate for most statistical purposes. We do this by assuming that all the values in any group are equal to the midpoint (or mid-value) of that group. The procedure is very similar to that for a simple frequency distribution (which is why we stressed the need for revision) and is shown in this example: 226 Measures of Location © ABE and RRC Table 8.4: Data for calculation of arithmetic mean of a grouped frequency distribution Group Mid-value (x) Frequency (f) Product (fx) 0 to < 10 5 3 15 10 to < 20 15 6 90 20 to < 30 25 10 250 30 to < 40 35 16 560 40 to < 50 45 9 405 50 to < 60 55 5 275 60 to < 70 65 1 65 Total Σf 50 Σ(fx) 1,660 1,660 50 1 Σf Σfx x 33.2 Provided that Σf is not less than about 50 and the number of groups is not less than about 12, the arithmetic mean thus calculated is sufficiently accurate for all practical purposes. There is one pitfall to be avoided when using this method – if all the groups should not have the same class interval, be sure that you get the correct mid-values! Table 8.5 is part of a larger table with varying class intervals, to illustrate the point: Table 8.5: Data for grouped frequency distribution Group Mid-value (x) 0 to < 10 5 10 to < 20 15 20 to< 40 30 40 to < 60 50 60 to < 100 80 You will remember that in discussing the drawing of histograms we had to deal with the case where the last group was not exactly specified. The same rules for drawing the histogram apply to the calculation of the arithmetic mean. Simplified Calculation It is possible to simplify the arithmetic involved in calculating an arithmetic mean still further by the following two devices: working from an assumed mean in the middle of one convenient class working in class intervals instead of in the original units. Let us consider the first device. Measures of Location 227 © ABE and RRC (a) Working from an assumed mean in the middle of one convenient class If you go back to our earlier examples, you will discover after some arithmetic that if you add up the differences in value between each reading and the true mean, then these differences add up to zero. Take first the height distribution (Table 8.1) considered at the start of this section: Table 8.6: Height distribution data, x 175 cms Man Height, x (cms) (x – x ) (cms) A 187 12 B 160 15 C 163 12 D 180 5 E 180 5 F 168 7 G 187 12 ) x Σ(x 34 34 0 That is, 0 ) x Σ(x . Second, consider the grouped frequency distribution (Tables 8.3, 8.4) given earlier in this section: Table 8.7: Data for grouped frequency distribution, x 33.2 Group Mid-value (x) Frequency (f) (x – x ) f(x – x ) 0 to < 10 5 3 28.2 84.6 10 to < 20 15 6 18.2 109.2 20 to < 30 25 10 8.2 82 30 to < 40 35 16 1.8 28.8 40 to < 50 45 9 11.8 106.2 50 to < 60 55 5 21.8 109 60 to < 70 65 1 31.8 31.8 Total Σf 50 Σ(f(x – x )) 275.8 275.8 0 That is, 0 ) x Σ(x . If we take any value other than x and follow the same procedure, the sum of the differences (sometimes called deviations) will not be zero. In our first example, let us assume the mean to be 170 cms and label the assumed mean x o . The differences between each reading and this assumed value are: 228 Measures of Location © ABE and RRC Table 8.8: Differences between x and x o for height distribution data Man Height, x (cms) d = (x – x o ) (cms) A 187 17 B 160 10 C 163 7 D 180 10 E 180 10 F 168 2 G 187 17 Σ(x x o ) 54 19 35 That is Σ(x – x o ) 35 cms or Σd 35 cms. We can make use of this property and use it as a short cut for finding x . We start by choosing some value of x as an assumed mean. We try to choose it near to where we think the true mean ( x ) will lie, and we always choose it as the midpoint of one of the groups when we are dealing with a grouped frequency distribution. In the example just given, the total deviation (d) does not equal zero, so 170 cannot be the true mean. As the total deviation is positive, we must have underestimated in our choice of x o , so the true mean is higher than 170. As there are seven readings, we need to adjust x o upwards by one seventh of the total deviation, i.e. by (35) 7 5. Therefore the true value of x is: 175 5 170 7 ) 35 ( 170 cms. We know this to be the correct answer from our earlier work. Let us now illustrate the short cut method for the grouped frequency distribution. We shall take x o as 35 as this is the mid-value in the centre of the distribution. Table 8.9: Differences between x and x o for grouped frequency distribution data Group Mid-value (x) Frequency (f) d = (x – x o ) f(x – x o ) = fd 0 to < 10 5 3 30 90 10 to < 20 15 6 20 120 20 to < 30 25 10 10 100 30 to < 40 35 16 0 0 40 to < 50 45 9 10 90 50 to < 60 55 5 20 100 60 to < 70 65 1 30 30 Total Σf 50 Σfd 310 220 90 This time Σfd 90. Measures of Location 229 © ABE and RRC This means we must have overestimated x o , as the total deviation (Σfd) is negative. As there are 50 readings altogether, the true mean must be th 50 1 of (90) lower than 35, i.e. 2 . 33 8 . 1 35 50 ) 90 ( 35 x which is as we found previously. (b) Working in class intervals instead of in the original units This method of calculation can be used with a grouped frequency distribution to work in units of the class interval instead of in the original units. In the fourth column of Table 8.9, you can see that all the deviations are multiples of 10, so we could have worked in units of 10 throughout and then compensated for this at the end of the calculation. Let us repeat the calculation using this method. The result (with x o 35) is set out in Table 8.10. Note that here the symbol used for the length of the class interval is c, although you may also come across the symbol i used for this purpose. Table 8.10: Differences between x and x o for grouped data using class intervals Group Mid-value (x) Frequency (f) (x – x o ) d = c x x o fd 0 to < 10 5 3 –30 3 9 10 to < 20 15 6 –20 2 12 20 to < 30 25 10 10 1 10 30 to < 40 35 16 0 0 0 40 to < 50 45 9 10 1 9 50 to < 60 55 5 20 2 10 60 to < 70 65 1 30 3 3 Total Σf 50 9 31 22 c x x Σ o Remembering that we need to compensate, by multiplying by 10, for working in class interval units in the table: 10 50 ) 9 ( 35 x 33.2 8 . 1 35 50 90 35 x as before. The general formula for x that applies for all grouped frequency distributions having equal class intervals can be written as: c n Σfd x x o where d c x x o 230 Measures of Location © ABE and RRC As we mentioned at an earlier stage, you have to be very careful if the class intervals are unequal, because you can only use one such interval as your working unit. Table 8.11 shows how to deal with this situation. Table 8.11: Differences between x and x o using unequal class intervals Group Mid-value (x) d = c x x o Frequency (f) Product (fd) 0 to < 10 5 3 3 9 10 to < 20 15 2 6 12 20 to < 30 25 1 10 10 30 to < 40 35 0 16 0 40 to < 50 45 1 9 9 50 to < 70 60 2½ 6 15 Total 50 24 (31) 7 The assumed mean is 35 as before, and the working unit is a class interval of 10. Notice how d for the last group is worked out: the midpoint is 60, which is 2½ times 10 above the assumed mean. The required arithmetic mean is, therefore: 6 . 33 4 . 1 35 50 70 35 50 10 7 35 x We have reached a slightly different figure from before because of the error introduced by the coarser grouping in the "50 to < 70" region. The method just described is of great importance. By using it correctly, you can often do the calculations for very complicated-looking distributions by using mental arithmetic and pencil and paper. With the advent of electronic calculators, the time saving on calculations of the arithmetic mean is not great, but the method is still preferable because: The numbers involved are smaller and thus you are less likely to make slips in arithmetic. The method can be extended to enable us to find easily the standard deviation of a frequency distribution (which we shall examine in Study Unit 9). Calculating the Mean from a Stem and Leaf Diagram You will recall that a stem and leaf diagram presents frequency distributions of grouped data graphically. However because these diagrams preserve the original data, they can be used for further analysis to calculate the mean relatively straightforwardly, as if the data was ungrouped. Consider the following example of a stem and leaf diagram in Figure 8.2, which shows the distribution of marks awarded to an ice skater in a competition. Note that the stem values represent the units of a mark awarded and the leaf values represent the tenths of a mark awarded (i.e. a mark of 5.8 is expressed in the stem and leaf diagram as 5|8). Measures of Location 231 © ABE and RRC Figure 8.2: Stem and leaf diagram to show the marks awarded to a competition skater Stem Leaf 0 0 1 2 3 5 9 4 0 1 1 3 5 0 5 8 Note: 0|0 0.0 marks Although the stem and leaf diagram presents grouped data, the individual data is preserved so that the mean can be calculated using the simple formula: n Σx x Accordingly: 10 ) 8 . 5 5 . 5 0 . 5 3 . 4 1 . 4 1 . 4 0 . 4 9 . 3 5 . 3 0 . 0 ( Σ x 0 . 4 10 40.2 x Characteristics of the Arithmetic Mean There are a number of characteristics of the arithmetic mean which you must be aware of: It is not necessary to know the value of every item in a distribution in order to calculate the arithmetic mean. Only the total and the number of items are needed. For example, if you know the total wages bill and the number of employees, you can calculate the arithmetic mean wage without knowing the wages of each person. The arithmetic mean is fully representative because it is based on all, and not just some, of the items in the distribution. One or two extreme values can make the arithmetic mean somewhat unrepresentative by their influence on it. For example, if a millionaire came to live in a country village, the inclusion of his or her income in the arithmetic mean for the village might make the place seem very much better off than it really was! The arithmetic mean is reasonably easy to calculate and to understand. In more advanced statistical work, it has the advantage of being amenable to algebraic manipulation. 232 Measures of Location © ABE and RRC B. THE MODE The first alternative to the mean which we will discuss is the mode. This is the name given to the most frequently occurring value. Mode of a Simple Frequency Distribution Consider the following frequency distribution: Table 8.12 Accident distribution data Number of accidents per day (x) Number of days with the stated number of accidents (f) 0 27 1 39 2 30 3 20 4 7 Total 123 In this case the most frequently occurring value is 1 (it occurred 39 times) and so the mode of this distribution is 1. Note that the mode, like the mean, is a value of the variate x, not the frequency of that value. A common error is to say that the mode of the distribution in Table 8.12 is 39. This is wrong. The mode is 1. Watch out, and do not fall into this trap! For comparison, calculate the arithmetic mean of the distribution – it works out at 1.52. The mode is used in those cases where it is essential for the measure of location to be an actually occurring value. An example is the case of a survey carried out by a clothing store to determine what size of garment to stock in the greatest quantity. The average size of garment in demand might turn out to be, let us say, 9.3724. This is not an actually occurring value and doesn't help us to answer our problem. However, the mode of the distribution obtained from the survey would be an actual value (perhaps size 8) and it would provide the answer to the problem. Mode of a Grouped Frequency Distribution When the data is given in the form of a grouped frequency distribution, it is not quite so easy to determine the mode (unless you have a stem and leaf diagram, which we will discuss later). What, you might ask, is the mode of the following distribution shown in Table 8.13? Measures of Location 233 © ABE and RRC Table 8.13: Data for determining the mode of a grouped frequency distribution Group Frequency 0 to < 10 4 10 to < 20 6 20 to < 30 10 30 to < 40 16 40 to < 50 24 50 to < 60 32 60 to < 70 38 70 to < 80 40 80 to < 90 20 90 to < 100 10 100 to < 110 5 110 to < 120 1 All we can really say is that "70 to < 80" is the modal group (the group with the largest frequency). You may be tempted to say that the mode is 75, but this is not true, nor even a useful approximation in most cases. The reason is that the modal group depends on the method of grouping, which can be chosen quite arbitrarily to suit our convenience. The distribution could have been set out with class intervals of five instead of 10, and would then have appeared as follows (only the middle part is shown, to illustrate the point): Table 8.14: Revised data for determining the mode of a grouped frequency distribution Group Frequency 60 to < 65 16 65 to < 70 22 70 to < 75 21 75 to < 80 19 80 to < 85 12 85 to < 90 8 The modal group is now "65 to < 70". Likewise, we will get different modal groups if the grouping is by 15 or by 20 or by any other class interval, and so the midpoint of the modal group is not a good way of estimating the mode. In practical work, this determination of the modal group is usually sufficient, but you may occasionally be asked for the mode to be determined from a grouped distribution. A number of procedures based on the frequencies in the groups adjacent to the modal group can be used, and we shall now describe one procedure. You should note, however, that these procedures are only mathematical devices for finding the most likely position of the mode – it is not possible to calculate an exact and true value in a grouped distribution. 234 Measures of Location © ABE and RRC We saw that the modal group of our original distribution was "70 to < 80". Now examine the groups on each side of the modal group. The group below ("60 to < 70") has a frequency of 38, and the one above ("80 to < 90") has a frequency of 20. This suggests to us that the mode may be some way towards the lower end of the modal group rather than at the centre. A graphical method for estimating the mode is shown in Figure 8.3. This method can be used when the distribution has equal class intervals. Draw that part of the histogram which covers the modal class and the adjacent classes on either side. Draw in the diagonals AB and CD as shown in Figure 8.3. From the point of intersection draw a vertical line downwards. Where this line crosses the horizontal axis is the mode. In our example the mode is just less than 71. Figure 8.3: Using a histogram to determine the most likely value of the mode Calculating the Mode from a Stem and Leaf Diagram Although a stem and leaf diagram presents frequency distributions of grouped data graphically, as we have remarked the original data is preserved. This means that these diagrams can be used to calculate the mode relatively straightforwardly as if the data was ungrouped. Let us consider our previous example of a stem and leaf diagram which shows the distribution of marks awarded to an ice skater in a competition (Figure 8.4, previously as Figure 8.2)). Remember that the stem values represent the units of a mark awarded and the leaf values represent the tenths of a mark awarded. Variate Class frequency A 20 30 40 60 80 90 70 B C D Mode Measures of Location 235 © ABE and RRC Figure 8.4: Stem and leaf diagram to show the marks awarded to a competition skater Stem Leaf 0 0 1 2 3 5 9 4 0 1 1 3 5 0 5 8 Note: 0|0 0.0 marks As the mode is defined as the data value that occurs most often, all we need to do is look for the leaf (number) that occurs the most often on one stem of the diagram. In our example, there are two '1' leaves on the '4' stem (i.e. two data points of value 4.1), so the mode is 4.1. Characteristics of the Mode Some of the characteristics of the mode are worth noting, particularly as they compare with those of the arithmetic mean. The mode is very easy to find with ungrouped distributions, since no calculation is required. It can only be determined approximately with grouped distributions. It is not affected by the occurrence of extreme values. Unlike the arithmetic mean, it is not based on all the items in the distribution, but only on those near its own value. In ungrouped distributions the mode is an actually occurring value. It is not amenable to the algebraic manipulation needed in advanced statistical work. It is not unique, i.e. there can be more than one mode. For example, in the set of numbers, 6, 7, 7, 7, 8, 8, 9, 10, 10, 10, 12, 13, there are two modes, namely 7 and 10. This set of numbers would be referred to as having a bimodal distribution. The mode may not exist. For example, in the set of numbers 7, 8, 10, 11, 12 each number occurs only once so this distribution has no mode. 236 Measures of Location © ABE and RRC C. THE MEDIAN The desirable feature of any measure of location is that it should be near the middle of the distribution to which it refers. Now, if a value is near the middle of the distribution, then we expect about half of the distribution to have larger values, and the other half to have smaller values. This suggests to us that a possible measure of location might be that value which is such that exactly half (50%) of the distribution has larger values and exactly half has lower values. The value which divides the distribution into equal parts is called the median. Look at the following set of values: 6, 7, 7, 8, 8, 9, 10, 10, 10, 12, 13 The total of these eleven numbers is 100 and so the arithmetic mean is therefore 100/11 9.091 (three decimal places), while the mode is 10 because that is the number which occurs most often (three times). The median however is 9, because there are five values above and five values below 9. Our first rule for determining the median is, therefore, as follows: Arrange all the values in order of magnitude and the median is then the middle value. Note that all the values are to be used – even though some of them may be repeated, they must all be put separately into the list. In the example just dealt with, it was easy to pick out the middle value because there was an odd number of values. But what if there is an even number? Then, by convention, the median is taken to be the arithmetic mean of the two values in the middle. For example, take the following set of values: 6, 7, 7, 8, 8, 9, 10, 10, 11, 12 The two values in the middle are 8 and 9, so the median is 8½. Median of a Simple Frequency Distribution Of course statistical data is rarely in such small groups and, as you have already learned, we usually deal with frequency distributions. How then do we find the median if our data is in the form of a distribution? Let us take the example of the frequency distribution of accidents already used in discussing the mode (Table 8.13). The total number of values is 123, and so when those values are arranged in order of magnitude, the median will be the 62 nd item because that will be the middle item. To see what the value of the 62 nd item will be, let us draw up the distribution again. Table 8.15 repeats the data from Table 8.12, but adds a column setting out the cumulative frequency. Table 8.15: Accident distribution data Number of accidents per day (x) Number of days with the stated number of accidents (f) Cumulative frequency (F) 0 27 27 1 39 66 2 30 96 3 20 116 4 7 123 Total 123 Measures of Location 237 © ABE and RRC You can see from the last column that, if we were to list all the separate values in order, the first 27 would all be 0's and from then up to the 66 th would be 1's. It therefore follows that the 62 nd item would be a 1 and that the median of this distribution is 1. Median of a Grouped Frequency Distribution The final problem connected with the median is how to find it when our data is in the form of a grouped distribution. The solution to the problem, as you might expect, is very similar to the solution for an ungrouped distribution – we halve the total frequency and then find, from the cumulative frequency column, the corresponding value of the variate. Because a grouped frequency distribution nearly always has a large total frequency, and because we do not know the exact values of the items in each group, it is not necessary to find the two middle items when the total frequency is even – just halve the total frequency and use the answer (whether it is a whole number or not) for the subsequent calculation. Table 8.16: Data for determining the median of a grouped frequency distribution Group Frequency (f) Cumulative frequency (F) 0 to < 10 4 4 10 to < 20 6 10 20 to < 30 10 20 30 to < 40 16 36 40 to < 50 24 60 50 to < 60 32 92 60 to < 70 38 130 70 to < 80 40 170 80 to < 90 20 190 90 to < 100 10 200 100 to < 110 5 205 110 to < 120 1 206 Total 206 The total frequency is 206 and therefore the median is the 103 rd item which, from the cumulative frequency column, must lie in the "60 to < 70" group. But exactly where in the "60 to < 70" group? Well, there are 92 items before we get to that group and we need the 103 rd item, so we obviously need to move into that group by 11 items. Altogether in our "60 to < 70" group there are 38 items, so we need to move 11/38 of the way into that group, that is 11/38 of 10 above 60. Our median is therefore: 60 38 110 60 2.89 62.89 The use of the cumulative frequency distribution will no doubt remind you of its graphical representation, the ogive. In practice, a convenient way to find the median of a grouped distribution is to draw the ogive. Then, against a cumulative frequency of half the total frequency, read off the median. In our example the median would be read against 103 on the cumulative frequency scale (see Figure 8.5). If the ogive is drawn with relative frequencies, then the median is always read off against 50%. 238 Measures of Location © ABE and RRC Figure 8.5: Reading the median from an ogive Calculating the Median from a Stem and Leaf Diagram Similarly, we can use a stem and leaf diagram to calculate the median value of a distribution of grouped data as the original data is preserved. Again, let us consider our previous example of a stem and leaf diagram which shows the distribution of marks awarded to an ice skater in a competition (Figure 8.6). Again, remember that the stem values represent the units of a mark awarded and the leaf values represent the tenths of a mark awarded. Figure 8.6: Stem and leaf diagram to show the marks awarded to a competition skater Stem Leaf 0 0 1 2 3 5 9 4 0 1 1 3 5 0 5 8 Note: 0|0 0.0 marks As noted, the stem and leaf diagram presents grouped data, but the individual data is preserved. This means that the median can be calculated relatively straightforwardly. As the median is the middle value in the set, we can simply go through the data presented in our stem and leaf diagram and cross off pairs of high and low leaves. If we start with the 0 20 40 60 80 100 120 140 160 180 200 220 0 20 40 60 80 100 120 Value of the variate Cumulative frequency Median 63 103 Measures of Location 239 © ABE and RRC rightmost leaf on the bottom stem and the leftmost leaf on the top stem, we can work our way towards the middle of the values crossing off the pairs until we have only one (or two) leave(s) left that have not been crossed out. That value (or the average of the two values) is the thus median. In our example, we are left with two middle values, namely the two '1' leaves on the '4' stem. The average of these values, and hence the median, is thus 4.1. Characteristic features of the median, which you should compare with those of the mean and the mode, are as follows: It is fairly easily obtained in most cases, and is readily understood as being the "half- way point". It is less affected by extreme values than the mean. The millionaire in the country village might alter considerably the mean income of the village but he or she would have almost no effect at all on the median. It can be obtained without actually having all the values. For example, if we want to know the median height of a group of 21 men, we do not have to measure the height of every single one. It is only necessary to stand the men in order of their heights and then only the middle one (number 11) need be measured, for his height will be the median height. Thus median is of value when we have open-ended classes at the edges of the distribution, as its calculation does not depend on the precise values of the variate in these classes, whereas the value of the arithmetic mean does. The median is not very amenable to further algebraic manipulation. 240 Measures of Location © ABE and RRC Questions for Practice 1. The following table shows the consumption of electricity of 100 householders during a particular week. Calculate the arithmetic mean consumption of the 100 householders. Consumption of electricity of 100 householders (one week) Consumption (kilowatt hours) Number of householders 0 – under 10 5 10 – under 20 8 20 – under 30 20 30 – under 40 29 40 – under 50 20 50 – under 60 11 60 – under 70 6 70 – under 80 1 80 or over 0 2. As part of a health study the heights of several dozen men in a particular location are measured as a sample. After this has been done, it is found that the sample contains two brothers who are dwarfs. What sort of "average" would you calculate from the sample, and why? 3. The rate of effluent discharge from two plants based on different systems was sampled at four-hourly intervals and the results are given below: System A 10 12 10 11 9 12 14 8 9 12 7 13 10 12 11 13 8 10 9 12 13 10 16 12 7 6 5 10 11 14 System B 10 17 18 6 5 10 11 13 6 15 14 10 12 9 8 7 4 18 19 10 12 13 16 9 12 3 4 12 4 9 Calculate the mean rate of discharge for each system. 4. Using the short-cut methods given in this study unit, calculate the arithmetic mean (i.e. the average) size of shareholding, from the grouped distribution in the following table. Measures of Location 241 © ABE and RRC Distribution of shareholding Number of shares Number of persons with this size of holding Less than 50 23 50 – 99 55 100 – 199 122 200 – 299 136 300 – 399 100 400 – 499 95 500 – 599 81 600 – 699 63 700 – 799 17 800 – 1,000 8 5. A garage keeps two cars for daily hire. The number of cars asked for in a day has varied as follows: Cars wanted in a day Proportion of days 0 0.36 1 0.38 2 0.18 3 0.06 4 0.02 5 or more 0 (a) What is the mean daily demand? (b) What is the mean number supplied (i.e. hired out) per day? (c) On what fraction of days is some demand turned away? 6. The following distribution gives the weight in kilograms of 100 schoolboys. Find the median weight. Weight in kg No. of schoolboys 45 – 49 2 50 – 54 15 55 – 59 37 60 – 64 31 65 – 69 11 70 – 74 4 100 242 Measures of Location © ABE and RRC 7. The holdings of shares (nominal value £1) in DEF plc are given in the following table. Calculate (a) the arithmetic mean holding, (b) the median holding and (c) the modal holding. No. of shares held No. of persons with these shares No. of shares held No. of persons with these shares Under 50 80 300 – 349 28 50 – 99 142 350 – 399 15 100 – 149 110 400 – 449 9 150 – 199 75 450 – 499 7 200 – 249 52 500 – 549 3 250 – 299 38 550 – 599 2 8. The age distribution of employees working for a manufacturing company is given in the following stem and leaf diagram. Stem Leaf 1 8 8 9 2 0 3 5 5 7 9 3 1 3 8 4 2 3 5 Note: 18 18 years old Based on the data presented in this diagram, calculate the: (a) mean (b) median (c) mode. Now check your answers with those given at the end of the unit. Measures of Location 243 © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE 1. Consumption of electricity of 100 householders (one week) Consumption (kwH) Mid-value (x) Number of householders (f) fx 0 – under 10 5 5 25 10 – under 20 15 8 120 20 – under 30 25 20 500 30 – under 40 35 29 1,015 40 – under 50 45 20 900 50 – under 60 55 11 605 60 – under 70 65 6 390 70 – under 80 75 1 75 80 or over 0 0 100 3,630 x 3 36 100 630 3 f Σ fx Σ . , i.e. the arithmetic mean of the householders' consumption is 36.3 kilowatt hours. 2. In order to give the correct impression of the mean in the sample location, it would be best to use the median. Normally one thinks of the arithmetic mean as the best measure of average, but in this case it would be too much affected by the two extremely low values to be representative. The mode would not be so affected, but with such a small sample (a dozen or so) it would be erratic, or might not exist if all the heights were different. 3. You can group the data first if you wish, but it is probably quicker just to add the numbers as they stand. System A: total number of readings 30 total of rates of discharge 316 mean rate of discharge 53 . 10 30 316 System B: total number of readings 30 total of rates of discharge 316 mean rate of discharge 53 . 10 30 316 244 Measures of Location © ABE and RRC 4. Take c 100, the most common class interval, and x o 250, the middle of the modal group. Distribution of shareholding No. of shares Mid-value (x) No. of persons with a holding of x (f) d = c x x o fd Less than 50 25.0 23 2.3 -51.8 50 – 99 74.5 55 1.8 -96.5 100 – 199 149.5 122 1.0 -122.6 200 – 299 249.5 136 0.0 -0.7 300 – 399 349.5 100 1.0 99.5 400 – 499 449.5 95 2.0 189.5 500 – 599 549.5 81 3.0 242.6 600 – 699 649.5 63 4.0 251.7 700 – 799 749.5 17 5.0 85.0 800 – 1,000 900.0 8 6.5 52.0 Total 700 648.7 100 700 7 . 648 250 x 250 92.7 343 to the nearest whole number. Therefore, the mean shareholding is 343 shares. 5. (a) Demand (x 1 ) Frequency (as a proportion) (f) fx 1 0 0.36 0 1 0.38 0.38 2 0.18 0.36 3 0.06 0.18 4 0.02 0.08 5 or more 0 0 1.00 1.00 Mean demand 1 00 . 1 00 . 1 Σf Σfx 1 car per day Measures of Location 245 © ABE and RRC (b) Supply (x 2 ) Frequency (as a proportion) (f) fx 2 0 0.36 0 1 0.38 0.38 2 (0.18 0.06 0.02) 0.26 0.52 1.00 0.90 Mean supply 90 . 0 00 . 1 90 . 0 Σf Σfx 2 cars per day (c) Demand is turned away whenever three or more cars are requested in one day. This proportion of days is 0.06 0.02 0.08. Therefore, on eight days in every 100 demand is turned away. 6. Weight in kg No. of schoolboys Cumulative frequency 45 – 49 2 2 50 – 54 15 17 55 – 59 37 54 60 – 64 31 85 65 – 69 11 96 70 – 74 4 100 100 Half the total frequency is 50 so the median lies in the "55 – 59 kg" group, and in the (50 17) 33 rd item within it. Median weight 5 37 33 55 55 4.46 59.46 60 kg to the nearest kg. 246 Measures of Location © ABE and RRC 7. Take c 50, x o 224.5 Distribution of the sizes of shareholdings in DEF plc Number of shares held Mid-value No. of persons with these shares d = c x x o fd Cumulative frequency Under 50 24.5 80 4 320 80 50 – 99 74.5 142 3 426 222 100 – 149 124.5 110 2 220 332 150 – 199 174.5 75 1 75 407 200 – 249 224.5 52 0 0 459 250 – 299 274.5 38 1 38 497 300 – 349 324.5 28 2 56 525 350 – 399 374.5 15 3 45 540 400 – 449 424.5 9 4 36 549 450 – 499 474.5 7 5 35 556 500 – 549 524.5 3 6 18 559 550 – 599 574.5 2 7 14 561 Total 561 1,041 242 799 (a) x 224.5 50 561 799 224.5 (1.424 50) 224.5 71.2 153.3 153 shares to the nearest whole number. (N.B. You may have chosen a different x o , but the result will be the same.) (b) Half of the total frequency is 280.5 and therefore the median lies in the "100 – 149" group and it is the 58.5 th (280.5 222) item within it. Median 99.5 1 . 126 110 50 5 . 58 126 shares to the nearest whole number. (c) The modal group is the "50 – 99" group and the estimated mode is given by drawing this group, together with the surrounding class groups, on a histogram. This is shown on the next page. Reading from this, the estimated mode 83 shares. Measures of Location 247 © ABE and RRC 8. (a) x 15 45) 43 42 38 33 31 29 27 25 25 23 20 19 18 (18 Σf Σfx 29.1 i.e. the arithmetic mean age of manufacturing employees is 29.1 years. (b) The median age of manufacturing employees is 27 years. (c) In our example, there are two 8 leaves on the 1 stem (i.e. two data points of value 18) and two 5 leaves on the 2 stem (i.e. two data points of value 25). Thus, the data set is bimodal with the mode age of manufacturing employees being 18 years and 25 years. Shareholding Frequency (no of persons) 80 140 100 150 50 Under 50 group 120 100 100 – 50 group Mode 83 248 Measures of Location © ABE and RRC 249 © ABE and RRC Study Unit 9 Measures of Dispersion Contents Page Introduction 250 A. Range 251 B. The Quartile Deviation, Deciles and Percentiles 251 The Quartile Deviation 251 Calculation of the Quartile Deviation 253 Deciles and Percentiles 254 C. Standard Deviation 255 The Variance and Standard Deviation 255 Alternative Formulae for Standard Deviation and Variance 257 Standard Deviation of a Simple Frequency Distribution 258 Standard Deviation of a Grouped Frequency Distribution 259 Characteristics of the Standard Deviation 262 D. The Coefficient of Variation 262 E. Skewness 263 Answers to Questions for Practice 268 250 Measures of Dispersion © ABE and RRC INTRODUCTION In order to get an idea of the general level of values in a frequency distribution, we have studied the various measures of location that are available. However, the figures which go to make up a distribution may all be very close to the central value, or they may be widely dispersed about it – for example, the mean of 49 and 51 is 50, but the mean of 0 and 100 is also 50. You can see that these two distributions may have the same mean, but the individual values may be spread about the mean in vastly different ways. When applying statistical methods to practical problems, knowledge of this spread, which we call "dispersion" or "variation", is of great importance. Examine the figures in the following table: Table 9.1: Output from two factories over 10 weeks Weekly output Week Factory A Factory B 1 94 136 2 100 92 3 106 110 4 100 36 5 90 102 6 101 57 7 107 108 8 98 81 9 101 156 10 98 117 Total 995 995 Mean output 99.5 99.5 Although the two factories have the same mean output, they are very different in their week- to-week consistency. Factory A achieves its mean production with only very little variation from week to week, whereas Factory B achieves the same mean by erratic ups and downs from week to week. This example shows that a mean (or other measure of location) does not by itself tell the whole story. We therefore need to supplement it with a measure of dispersion. As was the case with measures of location, there are several different measures of dispersion in use by statisticians. Each has its own particular merits and demerits, which will be discussed later. The measures in common use are: range quartile deviation standard deviation. We will discuss each of these in this unit. Measures of Dispersion 251 © ABE and RRC Objectives When you have completed this study unit you will be able to: calculate the range of a distribution from given data find, by both calculation and graphical methods, the quartile deviation, deciles and percentiles of grouped data calculate the standard deviation and variance of grouped and ungrouped distributions, using the appropriate formulae calculate the standard deviation using the "corrected/approximate variance" method calculate the coefficient of variation and explain its use calculate Pearson's first and second coefficient of skewness from the appropriate data explain the nature of skewness and describe the relationship between the mean, median and mode in a skewed distribution identify the advantages and disadvantages of the range, quartile deviation and standard deviation as measures of dispersion. A. RANGE The range is the simplest measure of dispersion: it is simply the difference between the largest and the smallest. In the example just given, we can see that the lowest weekly output for Factory A was 90 and the highest was 107; the range is therefore 107 ÷90, i.e.17. For Factory B the range is 156 ÷ 36 = 120. The larger range for Factory B shows that it performs less consistently than Factory A. The advantage of the range as a measure of the dispersion of a distribution is that it is very easy to calculate and its meaning is easy to understand. For these reasons it is used a great deal in industrial quality control work. Its disadvantage is that it is based on only two of the individual values, and takes no account of all those in between. As a result, one or two extreme results can make it quite unrepresentative. Consequently, the range is not much used in statistical analysis except in the case just mentioned. B. THE QUARTILE DEVIATION, DECILES AND PERCENTILES The Quartile Deviation This measure of dispersion is sometimes called the "semi-interquartile range". To understand it, you must cast your mind back to the method of obtaining the median from the ogive. The median, you remember, is the value which divides the total frequency into two halves. The values which divide the total frequency into quarters are called quartiles and they can also be found from the ogive, as shown in Figure 9.1. 252 Measures of Dispersion © ABE and RRC Figure 9.1: Identifying quartiles from an ogive This is the same ogive that we drew in Study Unit 8 when finding the median of a grouped frequency distribution. You will notice that we have added the relative cumulative frequency scale to the right of the graph. 100% corresponds to 206, i.e. the total frequency. It is then easy to read off the values of the variate corresponding to 25%, 50% and 75% of the cumulative frequency, giving the lower quartile (Q 1 ), the median and the upper quartile (Q 3 ) respectively. Q 1 = 46.5 median = 63 (as found previously) Q 3 = 76 The difference between the two quartiles is the interquartile range and half of the difference is the semi-interquartile range or quartile deviation: Quartile deviation = 2 Q Q 1 3 ÷ = 2 5 . 46 76 ÷ = 2 5 . 29 = 14.75 0 20 40 60 80 100 120 140 160 180 200 220 0 20 40 60 80 100 120 Value of the variate Cumulative frequency Relative cumulative frequency (%) 100 75 50 25 Measures of Dispersion 253 © ABE and RRC Alternatively, you can work out: 25% of the total frequency, i.e. , 5 . 51 4 206 = and 75% of the total frequency, i.e. 154.5 and then read from the ogive the values of the variate corresponding to 51.5 and 154.5 on the cumulative frequency scale (i.e. the left-hand scale). The end result is the same. Calculation of the Quartile Deviation The quartile deviation is not difficult to calculate in situations where the graphical method is not acceptable. Remember that graphical methods are never quite as accurate as calculations. We shall again use the same example, and the table of values is reproduced for convenience as Table 9.2: Table 9.2: Data for finding the quartile deviation of a grouped frequency distribution Group Frequency (f) Cumulative frequency (F) 0 to < 10 4 4 10 to < 20 6 10 20 to < 30 10 20 30 to < 40 16 36 40 to < 50 24 60 50 to < 60 32 92 60 to < 70 38 130 70 to < 80 40 170 80 to < 90 20 190 90 to < 100 10 200 100 to < 110 5 205 110 to < 120 1 206 Total 206 The lower quartile is the 25% point in the total distribution and the upper quartile the 75% point. As the total frequency is 206: the 25% point = 4 206 = 51½ and the 75% point = 206 4 3 × = 154½. We can make the calculations in exactly the same manner as we used for calculating the median. 254 Measures of Dispersion © ABE and RRC Looking at Table 9.2, the 51½ th item comes in the "40 to < 50" group and will be the (51½ ÷ 36) = 15½ th item within it. So, the lower quartile = 40 + 10 24 ½ 15 × = 40 + 6.458 = 46.458 Similarly, the upper quartile will be the 154½ th item which is in the "70 to < 80" group and is the (154½ ÷ 130) = 24½ th item within it. So, the upper quartile = 70 + 10 40 ½ 24 × = 70 + 6.125 = 76.125 Remember that the units of the quartiles and of the median are the same as those of the variate. The quartile deviation = 2 Q Q 1 3 ÷ = 2 458 . 46 125 . 76 ÷ = 2 667 . 29 = 14.8335 = 14.8 to 1 decimal place. The quartile deviation is unaffected by an occasional extreme value. However it is not based on the values of all the items in the distribution and, to this extent, it is less representative than the standard deviation (which we discuss later). In general, when a median is the appropriate measure of location, then the quartile deviation should be used as the measure of dispersion. Deciles and Percentiles It is sometimes convenient, particularly when dealing with wages and employment statistics, to consider values similar to the quartiles, but which divide the distribution more finely. Such partition values are deciles and percentiles. From their names you will probably have guessed that the deciles are the values which divide the total frequency into tenths and the percentiles are the values which divide the total frequency into hundredths. Obviously it is only meaningful to consider such values when we have a large total frequency. The deciles are labelled D 1 , D 2 ... D 9 . For instance, the second decile (D 2 ) is the value below which 20% of the data lies, and the sixth decile (D 6 ) is the value below which 60% of the data lies. The percentiles are labelled P 1 , P 2 ... P 99 . Again we can say that, for example, P 5 is the value below which 5% of the data lies and P 64 is the value below which 64% of the data lies. Using the same example as previously, let us calculate as an illustration the third decile D 3 . The method follows exactly the same principles as the calculation of the median and quartiles. Measures of Dispersion 255 © ABE and RRC The total frequency is 206 and D 3 is the value below which 30% of the data lies: 30% of 206 is 8 . 61 206 100 30 = × We are therefore looking for the value of the 61.8 th item. A glance at the cumulative frequency column shows that the 61.8 th item lies in the "50 to < 60" group, and is the (61.8 ÷ 60) = 1.8 th item within it. So: D 3 = 50 + 10 32 8 . 1 × = 50 + 32 18 = 50.6 to one decimal place. Therefore 30% of our data lies below 50.6. We could also have found this result graphically – check that you agree with the calculation by reading D 3 from the graph in Figure 9.1. You will see that the calculation method enables us to give a more precise answer than is obtainable graphically. C. STANDARD DEVIATION The most important of the measures of dispersion is the standard deviation. Except for the use of the range in statistical quality control and the use of the quartile deviation in wages statistics, the standard deviation is used almost exclusively in statistical practice. The standard deviation is defined as the square root of the variance, and so we need to know how to calculate the variance first. The Variance and Standard Deviation We start by finding the deviations from the mean, and then squaring them, which removes the negative signs in a mathematically acceptable fashion. The variance is then defined as the mean of the deviations from the mean, squared: variance = n ) x x ( Σ 2 ÷ , where n is the total frequency. Consider the following example in Table 9.3. 256 Measures of Dispersion © ABE and RRC Table 9.3: Output from Factory A over 10 weeks Week Weekly output (x) Deviation ) ( x x ÷ Deviation squared 2 x x ) ( ÷ 1 94 ÷5.5 30.25 2 100 +0.5 0.25 3 106 +6.5 42.25 4 100 +0.5 0.25 5 90 ÷9.5 90.25 6 101 +1.5 2.25 7 107 +7.5 56.25 8 98 ÷1.5 2.25 9 101 +1.5 2.25 10 98 ÷1.5 2.25 Total 995 +18.0 ÷18.0 228.50 Mean output 99.5 Variance = 22.85 The mean of the squared deviations is the variance: variance = n x x Σ 2 ) ( ÷ = 10 28.50 2 = 22.85 The formula for the standard deviation, which is defined as the square root of the variance, is: SD = n ) x x Σ 2 ÷ ( , again where n is the total frequency. In this example, then: SD = 85 . 22 = 4.78. The symbol usually used to denote standard deviation is "σ" (theta), although you may sometimes find "s" or "sd" used. Measures of Dispersion 257 © ABE and RRC Alternative Formulae for Standard Deviation and Variance The formula we have used to calculate the standard deviation is: n ) x Σ(x σ 2 ÷ = Formula (a) where: x = value of the observations x = mean of the observations n = number of observations. As the standard deviation is defined as the square root of the variance, therefore: variance = σ 2 = n ) x x ( Σ 2 ÷ By expanding the expression Σ(x ÷ x ) 2 we can rewrite the formula for standard deviation in two alternative and sometimes more useful forms, as follows: 2 2 x n Σx σ ÷ = Formula (b) or 2 2 n x Σ n x Σ σ | . | \ | ÷ = Formula (c) The choice of the formula to use depends on the information that you already have and the information that you are asked to give. In any calculation, you should aim to keep the arithmetic as simple as possible and the rounding errors as small as possible. If you already know x and it is a small integer, there is no reason why you should not use formula (a). If you already know x but it is not an integer (as in the previous example), then formula (b) is the best to use. If you do not know x , then you should use formula (c) – particularly if you are not asked to calculate x . When you are dealing with data in the form of a simple or grouped frequency distribution, then the formula for calculating the standard deviation needs to be expressed slightly differently: Σf ) x x f Σ σ 2 ÷ = ( Formula (d) or | | . | \ | ÷ = n ) fx Σ ( fx Σ n 1 σ 2 2 , where n = Σf Formula (e) Formula (d), like formula (a), is derived directly from the definition of the standard deviation. Formula (e) is obtained by using the sigma notation as in formula (c) and this is the one to use in calculations, as shown in the following sections. 258 Measures of Dispersion © ABE and RRC Standard Deviation of a Simple Frequency Distribution If the data in Table 9.3 had been given as a frequency distribution (as is often the case), then only the different values would appear in the "x" column and we would have to remember to multiply each result by its frequency. This is shown in Table 9.4 where we set out the information for calculating the standard deviation by applying formula (d): Table 9.4: Determining the standard deviation of a simple frequency distribution Weekly output (x) Frequency (f) fx Deviation ) ( x x ÷ Deviation squared 2 x x ) ( ÷ f(x – x ) 2 90 1 90 ÷9.5 90.25 90.25 94 1 94 ÷5.5 30.25 30.25 98 2 196 ÷1.5 2.25 4.50 100 2 200 +0.5 0.25 0.50 101 2 202 +1.5 2.25 4.50 106 1 106 +6.5 42.25 42.25 107 1 107 +7.5 56.25 56.25 Total 10 995 228.50 Mean 99.5 Variance = 22.85 The standard deviation is then worked out in the same way as previously, as follows: Σf ) x x f Σ σ 2 ÷ = ( σ = 10 5 . 228 = 85 . 22 (22.85 is the variance) σ = 4.78 Measures of Dispersion 259 © ABE and RRC However, we could have applied formula (e) and reduced the calculation as follows: Table 9.5: Determining the standard deviation of a simple frequency distribution Weekly output (x) Frequency (f) fx x 2 fx 2 90 1 90 8,100 8,100 94 1 94 8,836 8,836 98 2 196 9,604 19,208 100 2 200 10,000 20,000 101 2 202 10,201 20,402 106 1 106 11,236 11,236 107 1 107 11,449 11,449 10 995 99,231 Therefore: n = Σf = 10 Σfx = 995 Σfx 2 = 99,231 Calculating the standard deviation proceeds as follows: | | . | \ | ÷ = n ) fx Σ ( fx Σ n 1 σ 2 2 = | | . | \ | ÷ 10 995 231 , 99 10 1 2 = | . | \ | ÷ 10 025 , 990 231 , 99 10 1 = ( ) 5 . 002 , 99 231 , 99 10 1 ÷ = 5 . 228 10 1 × = 85 . 22 (the variance) = 4.78. Standard Deviation of a Grouped Frequency Distribution When we come to the problem of finding the standard deviation of a grouped frequency distribution, we again assume that all the readings in a given group fall at the midpoint of the group. Using formula (e) and applying it to the distribution shown in Table 9.6, the standard deviation is calculated as follows. 260 Measures of Dispersion © ABE and RRC Table 9.6: Determining the standard deviation of a grouped frequency distribution Class Mid-value (x) Frequency (f) fx x 2 fx 2 10 to < 20 15 2 30 225 450 20 to < 30 25 5 125 625 3,125 30 to < 40 35 8 280 1,225 9,800 40 to < 50 45 6 270 2,025 12,150 50 to < 60 55 5 275 3,025 15,125 60 to < 70 65 3 195 4,225 12,675 70 to < 80 75 1 75 5,625 5,625 Total 30 1,250 58,950 Therefore: n = Σf = 30 Σfx = 1,250 Σfx 2 = 58,950 Calculating the standard deviation proceeds as follows: | | . | \ | ÷ = n ) fx Σ ( fx Σ n 1 σ 2 2 = | | . | \ | ÷ 30 250 , 1 950 , 58 30 1 2 = | . | \ | ÷ 30 500 , 562 , 1 950 , 58 30 1 = ( ) 33 . 083 , 52 950 , 58 30 1 ÷ = 67 . 866 , 6 30 1 × = 89 . 228 (the variance) = 15.13 The arithmetic for this table is rather tedious even with a calculator, but we can extend the short-cut method which we used for finding the arithmetic mean of a distribution to find the standard deviation as well. In that method we: worked from an assumed mean worked in class intervals applied a correction to the assumed mean. Table 9.7 applies this approach to working out the standard deviation. Measures of Dispersion 261 © ABE and RRC Table 9.7: Determining the standard deviation of a grouped frequency distribution (short cut) Class Mid-value (x) Frequency (f) d = c x x o ÷ fd d 2 fd 2 10 to < 20 15 2 ÷2 ÷4 4 8 20 to < 30 25 5 ÷1 ÷5 1 5 30 to < 40 35 8 0 0 0 0 40 to < 50 45 6 +1 +6 1 6 50 to < 60 55 5 +2 +10 4 20 60 to < 70 65 3 +3 +9 9 27 70 to < 80 75 1 +4 +4 16 16 Total 30 +29 ÷9 82 +20 x o = 35, c = 10 The standard deviation is calculated in four steps from this table, as follows: (a) The approximate variance is obtained from 2 Σfd n 1 × which in our case is equal to: 30 1 × 82 = 30 82 = 2.7333 (b) The correction is ÷ 2 Σfd n 1 | . | \ | × It is always subtracted from the approximate variance to get the corrected variance. In our case the correction is: ÷(20/30) 2 = ÷0.4444. (c) The corrected variance is thus: 2.7333 ÷ 0.4444 = 2.2889 (d) The standard deviation is then the class interval times the square root of the corrected variance: SD = 2889 . 2 = 1.513 class intervals × 10 = 15.13 This may seem a little complicated, but if you work through the example a few times, it will all fall into place. Remember the following points: Work from an assumed mean at the midpoint of any convenient class. The correction is always subtracted from the approximate variance. As you are working in class intervals, it is necessary to multiply by the class interval as the last step. 262 Measures of Dispersion © ABE and RRC The correction factor is the same as that used for the short-cut calculation of the mean, but for the SD it has to be squared. The column for d 2 may be omitted since fd 2 = fd multiplied by d. But do not omit it until you have really grasped the principles involved. The formula for calculating the standard deviation in this way can be written as: 2 n Σfd n Σfd c SD | . | \ | ÷ = 2 The assumed mean should be chosen from a group with the most common interval and c will be that interval. If the intervals vary too much, we revert to the basic formula. Characteristics of the Standard Deviation In spite of the apparently complicated method of calculation, the standard deviation is the measure of dispersion used in all but the very simplest of statistical studies. It is based on all of the individual items. It gives slightly more emphasis to the larger deviations but does not ignore the smaller ones and, most important, it can be treated mathematically in more advanced statistics. D. THE COEFFICIENT OF VARIATION Suppose that we are comparing the profits earned by two businesses. One of them may be a fairly large business with average monthly profits of £50,000, while the other may be a small firm with average monthly profits of only £2,000. Clearly, the general level of profits is very different in the two cases, but what about the month-by-month variability? We can compare the variability of the two firms by calculating the two standard deviations. Let us suppose that both standard deviations come to £500. Now, £500 is a much more significant amount in relation to the small firm than it is in relation to the large firm. So although they have the same standard deviations, it would be unrealistic to say that the two businesses are equally consistent in their monthly profits. To overcome this difficulty, we can express the standard deviation as a percentage of the mean in each case. We call the result the "coefficient of variation". Applying the idea to the figures which we have just quoted, we get coefficients of variation (usually indicated in formulae by V or CV) as follows: for the large firm: V = 100 000 , 50 500 × = 1.0% for the small firm: V = 100 000 , 2 500 × = 25.0% This shows that, relatively speaking, the small firm is more erratic in its earnings than the large firm. Note that although a standard deviation has the same units as the variate, the coefficient of variation is a ratio and thus has no units. Another application of the coefficient of variation comes when we try to compare distributions the data of which are in different units – for example, when we try to compare a French business with an American business. To avoid the trouble of converting the dollars to euros (or vice versa) we can calculate the coefficients of variation in each case and thus obtain comparable measures of dispersion. Measures of Dispersion 263 © ABE and RRC E. SKEWNESS When the items in a distribution are dispersed equally on each side of the mean, we say that the distribution is symmetrical. Figure 9.2. shows two symmetrical distributions. Figure 9.2: Symmetrical distributions When the items are not symmetrically dispersed on each side of the mean, we say that the distribution is skewed or asymmetric. A distribution which has a tail drawn out to the right is said to be positively skewed, while one with a tail to the left is negatively skewed. Two distributions may have the same mean and the same standard deviation but they may be differently skewed. This will be obvious if you look at one of the skew distributions in Figure 9.3 and then look at the same one through from the other side of the paper! Figure 9.3: Skewed distributions What does skewness tell us? It tells us that we are to expect a few unusually high values in a positively skewed distribution or a few unusually low values in a negatively skewed distribution. If a distribution is symmetrical, the mean, mode and median all occur at the same point, i.e. right in the middle. But in a skew distribution, the mean and the median lie somewhere along the side of the "tail", although the mode is still at the point where the curve is highest. The more skew the distribution, the greater the distance from the mode to the mean and the median, but these two are always in the same order: working outwards from the mode, the median comes first and then the mean – see Figure 9.4. Variate Frequency Variate Frequency (a) Positively skewed (b) Negatively skewed 264 Measures of Dispersion © ABE and RRC Figure 9.4: Positions of the mean, mode and median in skew distributions For most distributions, except for those with very long tails, the following approximate relationship holds: mean ÷ mode = 3(mean ÷ median) The more skewed the distribution, the more spread out are these three measures of location, and so we can use the amount of this spread to measure the amount of skewness. The most usual way of doing this is to calculate Pearson's first coefficient of skewness: Pearson's first coefficient of skewness = SD mode mean÷ However as we have seen the mode is not always easy to find, and so we can use an equivalent formula for Pearson's second coefficient of skewness: Pearson's second coefficient of skewness = SD median) ean m ( 3 ÷ You need to be familiar with these formulae as the basis of calculating the skewness of a distribution (or its "coefficient of skewness", as it is also called). When you do the calculation, remember to get the correct sign (+ or ÷) when subtracting the mode or median from the mean. You will get negative answers from negatively skewed distributions, and positive answers for positively skewed distributions. The value of the coefficient of skewness is between ÷3 and +3, although values below ÷1 and above +1 are rare and indicate very skewed distributions. Examples of variates with positive skew distributions include size of incomes of a large group of workers, size of households, length of service in an organisation and age of a workforce. Negative skew distributions occur less frequently. One such example is the age at death for the adult population in the UK. Variate Frequency M e a n M e d i a n M o d e M e a n M e d i a n M o d e Positively skewed Negatively skewed Measures of Dispersion 265 © ABE and RRC Questions for Practice 1. (a) Explain the advantages of using the coefficient of variation as a measure of dispersion. (b) A small finance company has loaned money to 10,000 customers. Each loan is a multiple of £1,000 and the details are given below: Loan Amount (£000) Number of customers 1 535 2 211 3 427 4 1,360 5 1,650 6 1,504 7 1,280 8 1,615 9 1,012 10 406 11 or more 0 10,000 Calculate: (i) The arithmetic mean of the loan amounts (£). (ii) The standard deviation of the loan amounts (£). (iii) The coefficient of variation of the loan amounts. 2. Calculate the arithmetic mean and standard deviation for the following data using the short-cut method. Explain carefully what is measured by the standard deviation. Income of borrower £ No. of borrowers Under 1,800 1 1,800 – 2,400 6 2,400 – 3,000 9 3,000 – 3,600 14 3,600 – 4,200 17 4,200 – 4,800 23 4,800 – 5,400 28 5,400 – 6,000 35 6,000 – 6,600 39 6,600 – 7,200 48 266 Measures of Dispersion © ABE and RRC 3. Consider the following table: Percentages of owners with assets covered by estate duty statistics Percentages Assets over (£) Assets not over (£) Year 1 Year 11 – 1,000 48.3 18.8 1,000 3,000 29.5 25.1 3,000 5,000 10.9 11.9 5,000 10,000 6.2 21.3 10,000 25,000 3.4 17.9 25,000 100,000 1.5 4.4 100,000 – 0.2 0.6 Total number of owners (000s) 18,448 19,140 You are required to: (a) Derive by graphical methods the median, lower and upper quartile values for Year 1 and Year 11. (b) Calculate for the two distributions the median values and explain any difference between these results and those obtained in (a). (c) Estimate for each year the number of people whose assets are less than (i) £2,500, and (ii) £25,000. (d) Write a brief report on the data, using your results where relevant. 4. Items are produced to a target dimension of 3.25 cm on a single machine. Production is carried out on each of three shifts, each shift having a different operator – A, B and C. It is decided to investigate the accuracy of the operators. The results of a sample of 100 items produced by each operator are as follows: Accuracy of operators Dimension of item (cm) Operator 3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 A 1 6 10 21 36 17 6 2 1 B 0 0 7 25 34 27 6 1 0 C 3 22 49 22 4 0 0 0 0 Measures of Dispersion 267 © ABE and RRC Required: (a) State, with brief explanation, which operator is: (i) The most accurate in keeping to the target. (ii) The most consistent in his or her results. (b) Illustrate your answers to part (a) by comparing the results obtained by each operator as frequency polygons on the same graph. (c) Why is it that a measure of the mean level of the dimension obtained by each operator is an inadequate guide to his or her performance? (d) Calculate the semi-interquartile range of the results obtained by operator A, to 3 decimal places. 5. Calculate the coefficient of skewness for a distribution with a mean of 10, median of 11, and standard deviation of 5. What does your answer tell you about this frequency distribution? Now check your answers with those given at the end of the unit. 268 Measures of Dispersion © ABE and RRC ANSWERS TO QUESTIONS FOR PRACTICE 1. (a) The coefficient of variation is the standard deviation expressed as a percentage of the arithmetic mean. It enables the relative variation of two sets of data to be compared. This is of use when comparing sets of data which have widely differing means, a situation which makes absolute comparisons difficult. The coefficient of variation has no units and thus the relative variation of sets of data expressed in different units can be compared. This is particularly useful when dealing, say, with profits expressed in different currencies. (b) We need to draw up the following table: Loan amount (£000) (x) Number of customers (f) fx (x – x ) (x – x ) 2 f(x – x ) 2 1 535 535 –5 25 13,375 2 211 422 –4 16 3,376 3 427 1,281 –3 9 3,843 4 1,360 5,440 –2 4 5,440 5 1,650 8,250 –1 1 1,650 6 1,504 9,024 0 0 0 7 1,280 8,960 1 1 1,280 8 1,615 12,920 2 4 6,460 9 1,012 9,108 3 9 9,108 10 406 4,060 4 16 6,496 10,000 60,000 51,028 (i) 10,000 60,000 x = = £6,000 (ii) SD = 1028 . 5 Σf ) x x Σf 2 = ÷ ( = 2.259 thousand pounds = £2,259 to nearest £. (iii) coefficient of variation = % 100 000 , 6 259 , 2 × = 37.65% Measures of Dispersion 269 © ABE and RRC 2. We need to draw up the following table, with x o = £5,100 and c = £600 Income of borrower £ Mid-value (x) No. of borrowers (f) d = c x x o ÷ fd fd 2 Under 1,800 1,500 * 1 ÷6 ÷6 36 1,800 to < 2,400 2,100 6 ÷5 ÷30 150 2,400 to < 3,000 2,700 9 ÷4 ÷36 144 3,000 to < 3,600 3,300 14 ÷3 ÷42 126 3,600 to < 4,200 3,900 17 ÷2 ÷34 68 4,200 to < 4,800 4,500 23 ÷1 ÷23 23 4,800 to < 5,400 5,100 28 0 0 0 5,400 to < 6,000 5,700 35 1 35 35 6,000 to < 6,600 6,300 39 2 78 156 6,600 to 7,200 6,900 48 3 144 432 220 86 1,170 * Assume the first class covers the range £1,200 – £1,800 so that its mid-value is £1,500. x = £5,100 + 600 220 86 × = £5,100 + 234.55 = £5,334.55 = £5,335 to the nearest £ variance = 2 220 86 220 170 , 1 | . | \ | ÷ in class interval units squared = 5.3182 – 0.1528 = 5.1654 SD = 1654 . 5 = 2.2727 class interval units = £1,363.6 = £1,364 to the nearest £. The standard deviation measures the dispersion or spread of a set of data. It does this by taking every reading into account. We take the deviation of each reading from the mean and square it to remove any negative signs in a mathematically acceptable fashion. We then add all the squared deviations, find their mean and finally take the square root. 270 Measures of Dispersion © ABE and RRC 3 (a) We need first to draw up the following table Cumulative % frequency (less than) Assets over (£) Year 1 Year 11 1,000 48.3 18.8 3,000 77.8 43.9 5,000 88.7 55.8 10,000 94.9 77.1 25,000 98.3 95.0 100,000 99.8 99.4 From this, we can draw ogives for both years: Ogive to show cumulative percentage of owners with assets covered by estate duty statistics 0 10 20 30 40 50 60 70 80 90 100 0 5 10 15 20 25 Cumulative percentage frequency Assets (£000s) Year 1 Year 11 Measures of Dispersion 271 © ABE and RRC Reading off the graph at a cumulative percentage frequency of 50 we obtain the median: Year 1: median = £1,100 Year 11: median = £4,000 Reading off the graph at cumulative percentage frequencies of 25 and 75 we obtain the lower and upper quartile figures: Year 1: lower quartile = £500, upper quartile = £2,800 Year 11: lower quartile = £1,500, upper quartile = £9,500. (b) Year 1: The 50% mark is in the £1,000 – £3,000 group: median = £1,000 + 2,000 £ 5 . 29 3 . 48 50 × ÷ = £1,000 + 2,000 £ 5 . 29 7 . 1 × = £1,000 + 115.25 = £1,115.25 Year 11: The 50% mark is in the £3,000 – £5,000 group: median = £3,000 + 2,000 £ 9 . 11 9 . 43 50 × ÷ = £3,000 + 2,000 £ 9 . 11 1 . 6 × = £3,000 + 1,025.21 = £4,025.21 These figures are more precise than the results in (a). This is to be expected as graphical methods are approximate and are limited by the size of the graph paper. (c) Reading from the ogive at a value on the horizontal scale corresponding to £2,500, we find that in Year 1, 70% of owners had assets less than £2,500. In Year 11 the corresponding figure was 37½%. Thus, the number with assets less than £2,500 was: Year 1: 18,448,000 100 70 × = 12,913,600 Year 11: 19,140,000 100 5 . 37 × = 7,177,500. From the table of values in (a), we note that in Year 1, 98.3% of owners had assets less than £25,000. In Year 11 the corresponding percentage was 95%. Therefore the number of those with assets less than £25,000 was 18,134,384 in Year 1 and 18,183,000 in Year 11. (d) The total number of owners with assets covered by estate duty statistics rose by 692,000 from Year 1 to Year 11. 272 Measures of Dispersion © ABE and RRC As well as an increase in total wealth, there appears to have been some redistribution of wealth in this period. Whereas the modal group in Year 1 was under £1,000, in Year 11 it was £1,000 – £3,000 closely followed by the £5,000 – £10,000 group. In Year 1, 1.7% had assets over £25,000 whereas 5% had assets over that amount in Year 11. Although inflation might be responsible for lifting some owners to higher groups, it is unlikely to account for a change in percentage as large as this. The actual number of people with assets less than £25,000 was greater in Year 11 than in Year 1, presumably because more owners are covered by the statistics in Year 11. As a typical value for assets, we note that in Year 1, 50% of the owners had assets less than £1,115 but in Year 11, 50% had assets less than £4,025. The interquartile range for Year 1 was £500 – £2,800 but in Year 11 it was much larger, £1,500 – £9,500, indicating a greater spread of assets in the middle range in the later year. 4. (a) As this question says "with brief explanation", it implies that no calculations are to be made. So we could make the following observations: (i) B is most accurate in keeping to the target of 3.25 cm as his/her items are spread about 3.25 cm, but with a narrower spread than A. (ii) C is the most consistent in his/her results as almost 50% of his/her items are 3.23 cm and his/her maximum spread either way is 0.02 cm. You could back up your assertions with calculations of the mean and standard deviation in each case if time permits. It is good practice to work them out, so details of the calculations are now given. You need to draw up the following table. Note that f A and d A refer to operator A. Similarly for operators B and C. Dimension (cm) (x) f A f B f C d A = d B d C f A d A f B d B f C d C f A d A 2 f B d B 2 f C d C 2 3.21 1 - 3 ÷4 ÷2 ÷4 - ÷6 16 - 12 3.22 6 - 22 ÷3 ÷1 ÷18 - ÷22 54 - 22 3.23 10 7 49 ÷2 0 ÷20 ÷14 0 40 28 0 3.24 21 25 22 ÷1 1 ÷21 ÷25 22 21 25 22 3.25 36 34 4 0 2 0 0 8 0 0 16 3.26 17 27 - 1 - 17 27 - 17 27 - 3.27 6 6 - 2 - 12 12 - 24 24 - 3.28 2 1 - 3 - 6 3 - 18 9 - 3.29 1 - - 4 - 4 - - 16 - - Total 100 100 100 ÷24 3 2 206 113 72 d A = d B = 01 . 0 25 . 3 x ÷ d C = 01 . 0 23 . 3 x ÷ Measures of Dispersion 273 © ABE and RRC x A = 3.25 + 100 ) 24 (÷ × 0.01 = 3.25 ÷ 0.0024 = 3.2476 cm x B = 3.25 + 100 ) 3 (+ × 0.01 = 3.25 + 0.0003 = 3.2503 cm x C = 3.23 + 100 2 × 0.01 = 3.23 + 0.0002 = 3.2302 cm SD A = 0.01 2 100 24 100 206 | . | \ | ÷ ÷ = 0.01 0576 . 0 06 . 2 ÷ = 0.01 0024 . 2 = 0.0142 cm SD B = 0.01 2 100 3 100 113 | . | \ | ÷ = 0.01 009 . 0 13 . 1 ÷ = 0.01 1291 . 1 = 0.0106 cm SD C = 0.01 2 100 2 100 72 | . | \ | ÷ = 0.01 0004 . 0 72 . 0 ÷ = 0.01 7196 . 0 = 0.00848 cm This confirms our earlier assertion as x B is the mean closest to 3.25 cm and C has the smallest standard deviation. (b) Frequency polygons to show output of three operators (c) The mean on its own gives no indication of the spread of the readings about the mean. The mean dimension produced by A is close to 3.25 cm, but his/her output varies from 3.21 cm to 3.29 cm. 0 10 20 30 40 50 60 3.2 3.22 3.24 3.26 3.28 3.3 Operator A Operator C Operator B Dimension of item (cm) 274 Measures of Dispersion © ABE and RRC (d) Assume that each dimension has been measured to the nearest 0.01 cm – i.e. we assume that the one item produced by A in the 3.21 cm category is, in fact, less than 3.215 cm. Cumulative frequency Less than (cm) 1 3.215 7 3.225 17 3.235 38 3.245 74 3.255 91 3.265 97 3.275 99 3.285 100 3.295 The lower quartile Q 1 is the value below which 25% of the data lies, i.e. it is the value of the 25th item. Q 1 = 3.235 + 01 . 0 21 ) 17 25 ( × ÷ = 3.235 + 21 01 . 0 8 × = 3.2388 cm. Similarly Q 3 = 3.255 + 01 . 0 17 ) 74 75 ( × ÷ = 3.255 + 01 . 0 17 1 × = 3.2556 cm. The semi-interquartile range = 2 Q Q 1 3 ÷ = 2 0168 . 0 = 0.008 cm to three decimal places. This tells us something about the spread of the middle 50% of the data. It could be of use to the manufacturer for predicting the variation in A's production, so that possible wastage when items are made too far from the target dimension could be estimated. 5. Coefficient of skewness = deviation standard median) 3(mean÷ = 5 ) 11 10 ( 3 ÷ = ÷0.6 Thus we can say that the distribution is slightly negatively skewed – i.e. asymmetric with the peak of the curve being to the right. 275 © ABE and RRC Study Unit 10 Probability Contents Page Introduction 276 A. Estimating Probabilities 277 Theoretical Probabilities 277 Empirical Probabilities 278 Subjective Probabilities 279 B. Types of Event 279 C. The Two Laws of Probability 280 Addition Law for Mutually Exclusive Events 280 Addition Law for Non-Mutually Exclusive Events 282 Multiplication Law for Independent Events 283 Multiplication Law for Dependent Events 285 Listing Equally Likely Outcomes 286 D. Conditional Probability 286 E. Tree Diagrams 288 F. Sample Space 294 G. Expected Value 295 H. The Normal Distribution 298 Properties of the Normal Distribution 300 Using Tables of the Normal Distribution 301 Further Probability Calculations 302 Answers to Questions for Practice 305 Appendix 311 276 Probability © ABE and RRC INTRODUCTION Probability is one of those ideas about which we all have some notion, but it may not be a very definite one. Initially, we will not spend our time trying to get an exact definition, but will confine ourselves to the task of grasping the idea generally and seeing how it can be used. Other words which convey much the same idea as probability are "chance" and "likelihood". Just as there is a scale of temperature, because some things are hotter than others, so there is a scale of probability. Some things are more probable than others – for example, snow is more likely to fall in winter than in summer, or a healthy person has more chance of surviving an attack of influenza than an unhealthy person. There is, you will note, some uncertainty in these matters. Most things in real life are uncertain to some degree or other, and it is for this reason that the theory of probability is of great practical value. It is the branch of mathematics which deals specifically with matters of uncertainty. For the purpose of learning the theory, it is necessary to start with simple things like coin- tossing and die-throwing. This may seem a bit remote from business and industrial life, but it will help you understand the more practical applications. We shall then move on to consider the application of probability to real situations. This study unit also introduces the concept of the normal distribution. This is a particular aspect of frequency distributions which we considered previously. The normal distribution has a number of properties which can be used to work out the probability of particular event happening. Objectives When you have completed this study unit you will be able to: explain the meaning of a priori, empirical and subjective probability explain the probability terms mutually exclusive, non-mutually exclusive, independent, dependent, non-independent, and complementary explain what is meant by conditional probability solve probability problems using the addition law for mutually exclusive and non- mutually exclusive events, and the multiplication law for dependent and independent events construct and use tree diagrams, contingency tables and sample spaces to solve probability problems calculate, from appropriate data, the expected value of an outcome describe, in simple terms, the significance of the normal frequency distribution explain the properties of the normal distribution in terms of the location of observations, the mean and the standard deviation use normal distribution tables to find probabilities that observations within specified limits will occur. Probability 277 © ABE and RRC A. ESTIMATING PROBABILITIES Suppose someone tells you: "there is a 50-50 chance that we will be able to deliver your order on Friday". This statement means something intuitively, even though when Friday arrives there are only two outcomes. Either the order will be delivered or it will not. Statements like this are trying to put probabilities or chances on uncertain events. Probability is measured on a scale between 0 and 1: the probability limits are thus: 0 ≤ P(A) ≤ 1 where P(A) is the probability of an event A occurring. Any event which is impossible has a probability of 0, and any event which is certain to occur has a probability of 1. Accordingly, the probability of all outcomes for a particular event does not exceed 1: the total probability rule is thus: ΣP = 1 (for all outcomes). For example, the probability that the sun will not rise tomorrow is 0; the probability that a light bulb will fail sooner or later is 1. For uncertain events, the probability of occurrence is somewhere between 0 and 1. The 50-50 chance mentioned previously is equivalent to a probability of 0.5. Try to estimate probabilities for the following events. Remember that events which are more likely to occur than not have probabilities which are greater than 0.5, and the more certain they are the closer the probabilities are to 1. Similarly, events which are more likely not to occur have probabilities which are less than 0.5. The probabilities get closer to 0 as the events get more unlikely. The events are: (a) The probability that a coin will fall heads when tossed. (b) The probability that it will snow next January. (c) The probability that sales for your company will reach record levels next year. (d) The probability that your car will not break down on your next journey. (e) The probability that the throw of a die will show a six. The probabilities are as follows: (a) The probability of heads is one in two or 0.5. (b) This probability is quite low. It is somewhere between 0 and 0.1. (c) You can answer this one yourself. (d) This depends on how frequently your car is serviced. For a reliable car it should be considerably greater than 0.5. (e) The probability of a six is one in six or 6 1 or 0.167. Theoretical Probabilities Sometimes probabilities can be specified by considering the physical aspects of the situation. For example, consider the tossing of a coin. What is the probability that it will fall heads? There are two sides to a coin. There is no reason to favour either side as a coin is symmetrical. Therefore, the probability of heads, which we call P(H), is: P(H) = 0.5 278 Probability © ABE and RRC Another example is throwing a die. A die has six sides. Again, assuming it is not weighted in favour of any of the sides, there is no reason to favour one side rather than another. Therefore, the probability of a six showing uppermost, P(6), is: P(6) = 6 1 = 0.167 As a third and final example, imagine a box containing 100 beads of which 23 are black and 77 white. If we pick one bead out of the box at random (blindfold and with the box well shaken up), what is the probability that we will draw a black bead? We have 23 chances out of 100, so the probability is: P(B) = 100 23 = 0.23 Probabilities of this kind, which can be assessed from our prior knowledge of a situation, are also called a priori probabilities. In general terms, we can say that if an event E can happen in h ways out of a total of n possible ways, then the probability of that event occurring (called a success) is given by: P(E) = outcomes possible of number total occurring E of ways possible of number = n h In the preceding examples, we introduced the concept that some outcomes are equally likely, while others are not equally likely. In probability, when there are the same chances for more than one outcome, we say that the outcomes are equally likely to occur. For example: If someone tosses a coin, the chances of getting a heads or tails are the same, i.e. all outcomes are equally likely. The probability of getting a heads is P(H) = 0.5 and the probability of getting a tails is P(T) = 0.5. If someone throws a die, the chances of getting a six are no different to getting any other number, i.e. all outcomes are equally likely. The probability of getting a one is P(1) = 0.167, the probability of getting a two is P(2) = 0.167, etc. But not all outcomes are equally likely to occur. In the example about the box containing 100 beads, of which 23 are black and 77 white, the probability of getting a black bead is P(B) = 0.23, whereas the probability of getting a white bead is P(R) = 0.77. Thus, the chances of getting a red bead or a black bead are not the same, i.e. all outcomes for this particular event are not equally likely. Empirical Probabilities Often it is not possible to give a theoretical probability of an event. For example, what is the probability that an item on a production line will fail a quality control test? This question can be answered either by measuring the probability in a test situation or by relying on previous results (i.e. empirically). If 100 items are taken from the production line and tested, then: probability of failure, P(F) = tested items of no. total fail which items of no. So, if 5 items actually fail the test P(F) = 100 5 = 0.05 Sometimes it is not possible to set up an experiment to calculate an empirical probability. For example, what are your chances of passing a particular examination? You cannot sit a series Probability 279 © ABE and RRC of examinations to answer this. Previous results must be used. If you have taken 12 examinations in the past, and failed only one, you might estimate: probability of passing, P(pass) = 12 11 = 0.92 Subjective Probabilities A subjective probability describes an individual's personal judgement or instinct about how likely a particular event is to occur. It is not based on any precise theoretical or empirical foundation, but is often a reasonable assessment by a knowledgeable person. Thus, a person's subjective probability of an event describes his or her degree of belief in the outcome of that event. Like all probabilities, a subjective probability is conventionally expressed on a scale from 0 to 1; a rare event has a subjective probability close to 0, a very common event has a subjective probability close to 1. In business, managers will often make strategic decisions based on the subjective probability of the outcome of an event – i.e. on their instinct, given their knowledge of the business environment in which they operate. For example, investors on the stock market may base many of their decisions to buy and sell shares on their instinct. Successful investors are those who know which stocks to buy at a relatively low price, and similarly they know at what price to sell them prior to a downturn in the market. If these investment decisions were based on a precise theoretical or empirical foundation, then we would all be doing it! But they are not, they are based on instinct: investors attach (subjective) probabilities to the likelihood that share prices will move in a particular direction and they make their decisions accordingly. B. TYPES OF EVENT In probability, we can identify five types of event: mutually exclusive non-mutually exclusive independent dependent or non-independent complementary. (a) Mutually exclusive events If two events A and B are mutually exclusive then the occurrence of one event precludes the possibility of the other occurring. Thus: P(A and B) = 0. For example, the two sides of a coin are mutually exclusive since, on the throw of the coin, heads automatically rules out the possibility of tails. On the throw of a die, a six excludes all other possibilities. In fact, all the sides of a die are mutually exclusive; the occurrence of any one of them as the top face automatically excludes any of the others. (b) Non-mutually-exclusive events These are events which can occur together. For example, in a pack of playing cards hearts and queens are non-mutually exclusive, since there is one card, the queen of hearts, which is both a heart and a queen and so satisfies both criteria for success. 280 Probability © ABE and RRC (c) Independent events These are events which are not mutually exclusive and where the occurrence of one event does not affect the occurrence of the other. We can express this as: P(A) = P(A|B) and/or P(B) = P(B|A) where the vertical line is a shorthand way of writing "given that". For example, the tossing of a coin in no way affects the result of the next toss of the coin; each toss has an independent outcome. (d) Dependent or non-independent events These are situations where the outcome of one event is dependent on another event. For example, the probability of a car owner being able to drive to work in her car is dependent on her being able to start the car. The probability of her being able to drive to work given that the car starts is a conditional probability. We can express this as: P(drive to work|car starts) again where the vertical line is a shorthand way of writing "given that". (e) Complementary events An event either occurs or it does not occur, i.e. we are certain that one or other of these situations holds. For example, if we throw a die and denote the event where a six is uppermost by A, and the event where either a one, two, three, four or five is uppermost by A (or not A), then A and A are complementary, i.e. they are mutually exclusive with a total probability of 1. Thus: P(A) + P( A ) = 1 This relationship between complementary events is useful, as it is often easier to find the probability of an event not occurring than to find the probability that it does occur. Using this formula, we can always find P(A) by subtracting P( A ) from 1. C. THE TWO LAWS OF PROBABILITY In this section we will define the two laws of probability, namely the addition law and the multiplication law, and discuss the situations when each can be used. Addition Law for Mutually Exclusive Events Consider again the example of throwing a die. You will remember that: P(6) = 6 1 Similarly: P(1) = 6 1 P(2) = 6 1 P(3) = 6 1 P(4) = 6 1 P(5) = 6 1 Probability 281 © ABE and RRC What is the chance of getting 1, 2 or 3? From the symmetry of the die you can see that P(1 or 2 or 3) = 0.5. But also, from the equations shown here you can see that: P(1) + P(2) + P(3) = 6 1 + 6 1 + 6 1 = 0.5 This illustrates that P(1 or 2 or 3) = P(1) + P(2) + P(3) This result is a general one and it is called the addition law of probabilities for mutually exclusive events. It is used to calculate the probability of one of any group of mutually exclusive events. It is stated more generally as: P(A or B or ... or N) = P(A) + P(B) + ... + P(N) where A, B ... N are mutually exclusive events. If all possible mutually exclusive events are listed, then it is certain that one of these outcomes will occur. For example, when the die is tossed there must be one number showing afterwards. P(1 or 2 or 3 or 4 or 5 or 6) = 1 Using the addition law for mutually exclusive events, this can also be stated as P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 Again this is a general rule. The sum of the probabilities of a complete list of mutually exclusive events will always be 1. Example: An urn contains 100 coloured balls. Five of these are red, seven are blue and the rest are white. One ball is to be drawn at random from the urn. What is the probability that it will be red? Probability that ball is red P(R) = 100 5 = 0.05 What is the probability that it will be blue? Probability that ball is blue P(B) = 100 7 = 0.07 What is the probability that it will be red or blue? P(R or B) = P(R) + P(B) = 0.05 + 0.07 = 0.12 This result uses the addition law for mutually exclusive events since a ball cannot be both blue and red. What is the probability that it will be white? The ball must be either red or blue or white. This is a complete list of mutually exclusive possibilities. Therefore: P(R) + P(B) + P(W) = 1 Then the probability of one ball being white is given as: P(W) = 1 ÷ P(R) ÷ P(B) = 1 ÷ 0.05 ÷ 0.07 = 0.88 282 Probability © ABE and RRC Addition Law for Non-Mutually Exclusive Events Events which are non-mutually exclusive are, by definition, capable of occurring together. To determine the probability of one of the events occurring, we can still use the addition law, but the probability of the events occurring together must be deducted: P(A or B or both) = P(A) + P(B) ÷ P(A and B) Example 1: If one card is drawn from a pack of 52 playing cards, what is the probability (a) that it is either a spade or an ace, or (b) that it is either a spade or the ace of diamonds? (a) Let event B be "the card is a spade" and event A be "the card is an ace". We need to determine the probability of the event being either a spade or an ace. We also need to take account of the probability of the card being the ace of spades (i.e. the event being both A and B). However, this probability will be included in the probability of it being a spade and in the probability of it being an ace. Thus, we need to deduct that probability from one of the other probabilities. This is given by: P(spade or ace [or both]) = P(A or B) = P(A) + P(B) ÷ P(A and B) P(A) = 52 4 pack in no. aces of no. = P(B) = 52 13 pack in no. spades of no. = P(A and B) = 52 1 pack in no. spades of aces of no. = P(spade or ace) = 13 4 52 16 52 1 52 13 52 4 = = ÷ + (b) Let event B be "the card is a spade" and event A be "the card is the ace of diamonds". Here, the conditions are not met by both events occurring, so the probability is given by: P(spade or ace of diamonds) = P(A) + P(B) P(A) = 52 1 pack in no. diamonds of aces of no. = P(B) = 52 13 pack in no. spades of no. = P(spade or ace of diamonds) = 26 7 52 14 52 13 52 1 = = + Probability 283 © ABE and RRC Example 2: At a local shop 50% of customers buy unwrapped bread and 60% buy wrapped bread. What proportion of customers buy at least one kind of bread if 20% buy both wrapped and unwrapped. Let: T represent those customers buying unwrapped bread and W represent those customers buying wrapped bread. Then: P(buy at least one kind of bread) = P(buy wrapped or unwrapped or both) = P(T or W) = P(T) + P(W) ÷ P(T and W) = 0.5 + 0.6 ÷ 0.2 = 0.9 So, 10 9 of the customers buy at least one kind of bread. Multiplication Law for Independent Events Consider an item on a production line. This item could be defective or acceptable. These two possibilities are mutually exclusive and represent a complete list of alternatives. Assume that: probability that it is defective, P(D) = 0.2 probability that it is acceptable, P(A) = 0.8 Now consider another facet of these items. There is a system for checking them, but only every tenth item is checked. This is shown as: probability that it is checked, P(C) = 0.1 probability that it is not checked, P(N) = 0.9 Again these two possibilities are mutually exclusive and they represent a complete list of alternatives. An item is either checked or it is not. Consider the possibility that an individual item is both defective and not checked. These two events can obviously both occur together so they are not mutually exclusive. They are, however, independent. That is to say, whether an item is defective or acceptable does not affect the probability of it being tested. There are also other kinds of independent events. If you toss a coin once and then again a second time, the outcome of the second test is independent of the results of the first one. The results of any third or subsequent test are also independent of any previous results. The probability of heads on any test is 0.5 even if all the previous tests have resulted in heads. To work out the probability of two independent events both happening, you use the multiplication law for independent events. This can be stated as: P(A and B) = P(A) × P(B) if A and B are independent events. Again this result is true for any number of independent events. So: P(A and B ... and N) = P(A) × P(B) ... × P(N) 284 Probability © ABE and RRC Consider the last example: for any item probability that it is defective: P(D) = 0.2 probability that it is acceptable: P(A) = 0.8 probability that it is checked: P(C) = 0.1 probability that it is not checked: P(N) = 0.9 Using the multiplication law to calculate the probability that an item is both defective and not checked, we get: P(D and N) = 0.2 × 0.9 = 0.18 The probabilities of the other combinations of independent events can also be calculated. P(D and C) = 0.2 × 0.1 = 0.02 P(A and N) = 0.8 × 0.9 = 0.72 P(A and C) = 0.8 × 0.1 = 0.08 Example 1: A machine produces two batches of items. The first batch contains 1,000 items of which 20 are damaged. The second batch contains 10,000 items of which 50 are damaged. If one item is taken from each batch, what is the probability that both items are defective? For the item from the first batch: probability that it is defective: P(D 1 ) = 02 . 0 1,000 20 = For the item taken from the second batch: probability that it is defective: P(D 2 ) = 005 . 0 10,000 50 = Since these two probabilities are independent: P(D 1 and D 2 ) = P(D 1 ) × P(D 2 ) = 0.02 × 0.005 = 0.0001 Example 2: A card is drawn at random from a well shuffled pack of playing cards. What is the probability that the card is a heart? What is the probability that the card is a three? What is the probability that the card is the three of hearts? Probability of a heart: P(H) = 52 13 = 4 1 Probability of a three: P(3) = 52 4 = 13 1 Probability of the three of hearts, since the suit and the number of a card are independent: P(H and 3) = P(H) × P(3) = 52 1 13 1 4 1 = × Probability 285 © ABE and RRC Example 3: A die is thrown three times. What is the probability of one or more sixes in these three throws? Probability of no six in first throw = 6 5 Similarly, probability of no six in second or third throw = 6 5 The result of each throw is independent, so: probability of no six in all three throws = 216 125 6 5 6 5 6 5 = × × Since no sixes and one or more sixes are mutually exclusive and cover all possibilities: probability of one or more sixes = 216 91 216 125 1 = ÷ Multiplication Law for Dependent Events To work out the probability of two dependent events both happening, you use the multiplication law for dependent events. This can be stated as: P(A and B) = P(A) × P(B|A) where: A and B are dependent events and the vertical line is a shorthand way of writing "given that". Example 1: For example, suppose there are 10 marbles in a bag, and 3 are defective. Two marbles are to be selected one after the other without replacement. What is the probability of selecting a defective marble followed by another defective marble? Probability that the first marble selected is defective: P(A) 10 3 = Probability that the second marble selected is defective: P(B) 9 2 = Thus, P(A and B) 7% 9 2 10 3 = × = This means that if this experiment were repeated 100 times, in the long run 7 experiments would result in defective marbles on both the first and second selections. Example 2: Select one card at random from a deck of cards and find the probability that the card is an 8 and a diamond. We know that there are four '8' cards (i.e. P(8) ) 4 1 = in the pack and that a quarter of the pack are diamonds (i.e. P(D) ) 52 4 = , thus: P(D and 8) 52 1 4 1 52 4 = × = 286 Probability © ABE and RRC Listing Equally Likely Outcomes Another matter about which you must be careful is the listing of equally likely outcomes. Be sure that you list all of them. For example, we can list the possible results of tossing two coins: First Coin Second Coin Heads Heads Tails Heads Heads Tails Tails Tails There are four equally likely outcomes. Do not make the mistake of saying, for example, that there are only two outcomes (both heads or not both heads) – you must list all the possible outcomes. (In this case "not both heads" can result in three different ways, so the probability of this result will be higher than "both heads".) In this example, the probability that there will be one heads and one tails (heads-tails, or tails- heads) is 0.5. This is a case of the addition law at work: the probability of heads-tails (¼) plus the probability of tails-heads (¼). Putting it another way, the probability of different faces is equal to the probability of the same faces – in both cases 0.5. D. CONDITIONAL PROBABILITY The probability of an event A occurring where that event is conditional upon event B having also occurred is expressed as: P(A|B) The events A and B must be causally linked in some way for A to be conditional upon B. For example, we noted earlier the case where a car owner's ability to drive to work in her car (event A) is dependent on her being able to start the car (event B). There is self evidently a causal link between the two events. However, if we consider the case of her being able to drive to work given that she is married, there is no necessary link between the two events – and there is no conditional probability. In a case involving equally likely outcomes for both events (i.e. both events A and B may or may not occur), the conditional probability of event A occurring, given that B also occurs, is calculated by the following formula: P(A|B) = ) B ( n B) and n(A where n = the number of possible outcomes. Similarly, the conditional probability of event B occurring, given that A also occurs, is calculated by the following formula: P(B|A) = n(A) B) and n(A where n = the number of possible outcomes. One approach to determining probability in such circumstances is to construct a table of the possible outcomes – sometimes called a contingency table. Consider the case of 100 candidates attempting their driving test for the first time. Suppose 80 of the candidates pass and 50 of them have had 10 or more lessons. Of those who have had 10 or more lessons, 45 of them passed. What is the probability of passing at the first attempt if you have had only 6 lessons? Probability 287 © ABE and RRC We can construct a table which identifies the possible outcomes as follows in Table 10.1. Table 10.1: A partially completed contingency table Passed Failed Total 10 plus lessons 45 50 Less than 10 lessons Total 80 100 We will call event A "passed" and event B "having had less than 10 lessons". Make sure you understand the construction of this table: We have listed the possible outcomes of event A along the top – these outcomes are either to have passed or failed the test. Thus, the columns will show the number of outcomes for either possibility. We have listed the possible outcomes of event B along the side – these are either to have had 10 or more lessons, or less than 10 lessons. Thus, the rows will show the number of outcomes for either possibility. We have entered in the information that we know: that there are a total of 100 possible outcomes; that the total number of outcomes in which event A was "passed" is 80; that the total number of outcomes in which event B is "had 10 or more lessons" is 50; and that the total number of outcomes in which event A was "passed" and event B was "had 10 or more lessons" is 45. We can complete the rest of the table by subtraction as in Table 10.2. Table 10.2: Contingency table Passed Failed Total 10 plus lessons 45 5 50 Less than 10 lessons 35 15 50 Total 80 20 100 We can now work out the total outcomes applicable to the question "what is the probability of passing at the first attempt if you have had only 6 lessons?" Remember that event A is "passed" and event B is "having had less than 10 lessons". Therefore: the total number of outcomes of B = 50 the total number of outcomes of both A and B = 35. Thus the probability is: P(A|B) = n(B) B) and n(A = 50 35 = 0.7 288 Probability © ABE and RRC We could consider the question: "what is the probability of failing at the first attempt if you have had more than 10 lessons?". Here event A is "failed" and event B is "having had more than 10 lessons". The calculation is: P(A|B) = n(B) B) and n(A = 50 5 = 0.1 Questions for Practice 1 See if you can work out the answer to the following questions using the preceding table: "What is the probability of failing at the first attempt if you have only had 8 lessons?" Remember to be clear about which is event A and event B, then read off the total number of outcomes for event B and for event A and B together, and then apply the formula. Now check your answers with those given at the end of the unit. E. TREE DIAGRAMS A compound experiment, i.e. one with more than one component part, may be regarded as a sequence of similar simple experiments. For example, the rolling of two dice can be considered as the rolling of one followed by the rolling of the other; and the tossing of four coins can be thought of as tossing one after the other. A tree diagram enables us to construct an exhaustive list of mutually exclusive outcomes of a compound experiment. Furthermore, a tree diagram gives us a pictorial representation of probability. By exhaustive, we mean that every possible outcome is considered. By mutually exclusive we mean, as before, that if one of the outcomes of the compound experiment occurs then the others cannot. In the following examples, we shall consider the implications of conditional probability – the probability of one event occurring if another event has also occurred. Example 1: The concept can be illustrated using the example of a bag containing five red and three white billiard balls. If two are selected at random without replacement, what is the probability that one of each colour is drawn? Let: R indicate a red ball being selected W indicate a white ball being selected When the first ball is selected, the probabilities are: P(R) = 8 5 P(W) = 8 3 Probability 289 © ABE and RRC When the second ball is selected, there are only seven balls left and the mix of red and white balls will be determined by the selection of the first ball – i.e. the probability of a red or a white ball being selected is conditional upon the previous event. Thus: If the first ball selected was red, there will be four red and three white balls left. The probabilities for the selection of the second ball are then: P(R) = 7 4 , and P(W) = 7 3 If the first ball selected was white, there will be five red and two white balls left. The probabilities for the selection of the second ball are then: P(R) = 7 5 , and P(W) = 7 2 This process of working out the different conditional probabilities can be very complicated and it is easier to represent the situation by means of a tree diagram as in Figure 10.1. The probabilities at each stage are shown alongside the branches of the tree. Figure 10.1: Example 1 – tree diagram The diagram is constructed as follows, working from left to right. At the start we take a ball from the bag. This ball is either red or white so we draw two branches labelled R and W, corresponding to the two possibilities. We then also write on the branch the probability of the outcome of this simple experiment being along that branch. We then consider drawing a second ball from the bag. Whether we draw a red or a white ball the first time, we can still draw a red or a white ball the second time. So we mark in the two possibilities at the end of each of the two branches of our existing tree diagram. We enter on these second branches the conditional probabilities associated with them. Thus, on the uppermost branch in the diagram we must insert the probability of obtaining a second red ball given that the first was red. Each complete branch from start to tip represents one possible outcome of the compound experiment and each of the branches is mutually exclusive. We can see that there are four different mutually exclusive outcomes possible, namely RR, RW, WR and WW. 2nd Ball 1st Ball Start R R W W W R 8 5 8 3 7 3 7 4 7 5 7 2 290 Probability © ABE and RRC To obtain the probability of a particular outcome of the compound experiment occurring, we multiply the probabilities along the different sections of the branch, using the general multiplication law for probabilities. We thus obtain the probabilities shown in Table 10.3. The sum of the probabilities should add up to 1, as we know one or other of these mutually exclusive outcomes is certain to happen. Table 10.3: Example 1 – outcomes from compound (conditional) events Outcome Probability RR 56 20 7 4 8 5 = × RW 56 15 7 3 8 5 = × WR 56 15 7 5 8 3 = × WW 56 6 7 2 8 3 = × Total 1 Example 2: A bag contains three red balls, two white balls and one blue ball. Two balls are drawn at random (without replacement). Find the probability that: (a) Both white balls are drawn. (b) The blue ball is not drawn. (c) A red then a white are drawn. (d) A red and a white are drawn. To solve this problem, let us build up a tree diagram. Stage 1 will be to consider the situation when the first ball is drawn: Figure 10.2: Example 2 – tree diagram of first event Start R1 W1 B1 6 3 6 2 6 1 Probability 291 © ABE and RRC We have given the first ball drawn a subscript of 1 – for example, red first = R 1 . We shall give the second ball drawn a subscript of 2. The probabilities in the first stage will be: P(R 1 ) = 6 3 balls of no. total balls red of no. = P(W 1 ) = 6 2 (P(B 1 ) = 6 1 The possibilities in stage 2 may now be mapped on the tree diagram. When the second ball is drawn, the probability of it being red, white or blue is conditional on the outcome of the first event. Thus, note there was only one blue ball in the bag, so if we picked a blue ball first then we can have only a red or a white second ball. Also, whatever colour is chosen first, there are only five balls left as there is no replacement. Figure 10.3: Example 2 – tree diagram of first and second events We can now list all the possible outcomes, with their associated probabilities. 2nd Ball 1st Ball B2 W2 R2 5 2 5 2 5 1 B2 W2 R2 5 1 5 3 5 1 W2 R2 5 3 5 2 Start R1 W1 B1 6 3 6 2 6 1 292 Probability © ABE and RRC Table 10.4: Example 2 – outcomes from compound (conditional) events Outcome Probability R 1 R 2 5 1 5 2 6 3 = × R 1 W 2 5 1 5 2 6 3 = × R 1 B 2 10 1 5 1 6 3 = × W 1 R 2 5 1 5 3 6 2 = × W 1 W 2 15 1 5 1 6 2 = × W 1 B 2 15 1 5 1 6 2 = × B 1 R 2 10 1 5 3 6 1 = × B 1 W 2 15 1 5 2 6 1 = × Total 1 It is possible to read off the probabilities we require from Table 10.4. (a) Probability that both white balls are drawn: P(W 1 W 2 ) = 15 1 (b) Probability the blue ball is not drawn: P(R 1 R 2 ) + P(R 1 W 2 ) + P(W 1 R 2 ) + P(W 1 W 2 ) = 3 2 15 1 5 1 5 1 5 1 = + + + (c) Probability that a red then a white are drawn: P(R 1 W 2 ) = 5 1 (d) Probability that a red and a white are drawn: P(R 1 W 2 ) + P(W 1 R 2 ) = 5 2 5 1 5 1 = + Example 3: A couple decide to continue having children, to a maximum of four, until they have a son. Draw a tree diagram to find the possible family size and calculate the probability that they have a son. We can assume that any one child is equally likely to be a boy or a girl, i.e. P(B) = P(G) = 0.5. Note that once they have produced a son, they do not have any more children. The tree diagram will be as in Figure 10.4. Probability 293 © ABE and RRC Figure 10.4: Example 3 – Tree diagram The table of probabilities is as follows. Table 10.5: Example 3 – probabilities of outcomes Possible Families Probability 1 Boy 2 1 1 Girl, 1 Boy 4 1 2 1 2 = | . | \ | 2 Girls, 1 Boy 8 1 2 1 3 = | . | \ | 3 Girls, 1 Boy 16 1 2 1 4 = | . | \ | 4 Girls 16 1 2 1 4 = | . | \ | Total 1 The probability that the family will have a son before their fourth child is therefore: 16 15 16 1 8 1 4 1 2 1 = + + + Start ½ B ½ B B B G G G ½ ½ ½ ½ ½ ½ G 2nd Child 1st Child 4th Child 3rd Child 294 Probability © ABE and RRC F. SAMPLE SPACE You need a clear head to perform probability calculations successfully. It helps, then, to have some diagrammatic form of representation. We have seen the value of tree diagrams in the last section, and now we consider another method of drawing the possible outcomes of an event. When we toss a coin three times and note the outcome, we are performing a statistical experiment. If we make a list of all the possible outcomes of our experiment, we call this a sample space. The sample space in the coin-tossing experiment is: H H H H T T T H H T H T H T H T T H H H T T T T where, for example, T H H means that on the first toss we obtained a tail, on the second a head, and on the third a head. Let us consider another example. Suppose we have five people – A, B, C, D, E – and we wish to select a random sample of two for interview, i.e. each of A, B, C, D, E must have the same chance of being chosen. What is the sample space, i.e. the list of all possible different samples? The sample space is: A B B C C D D E A C B D C E A D B E A E In this example the order of the sample, i.e. whether we choose A followed by B or B followed by A, is not relevant as we would still interview both A and B. Having written down our sample space, we might be interested in a particular section of it. For instance, in our first example we might want to see in how many of the outcomes we obtained only one head: i.e. we are interested in the event of obtaining exactly one head. Looking back at the sample space, we see that there are three outcomes making up this event. If we have a fair coin, the probability of obtaining a head or a tail at any toss is ½ and all the outcomes in the sample space are equally likely. There are eight outcomes in all, so we can now deduce that: Probability of obtaining exactly one head in 3 tosses of a fair coin = space sample in outcomes of no. total head one with space sample in outcomes of no. = 8 3 Alternatively, we could write: P(event A) = outcomes of no. total A in outcomes of no. = 8 3 Probability 295 © ABE and RRC Now consider the example of an experiment involving the rolling of one fair die. What is the probability of obtaining: (a) a score greater than 3 (b) an odd number (c) both a score greater than 3 and an odd number The sample space is a listing of all the possible outcomes as follows: 1, 2, 3, 4, 5, 6 The total number of possible outcomes is 6, and it is now easy to read off the probabilities from the sample space: (a) a score greater than 3 – three possibilities: P = 2 1 6 3 = (b) an odd number – three possibilities: P = 2 1 6 3 = (c) both a score greater than 3 and an odd number – one possibility: P = 6 1 G. EXPECTED VALUE Probabilities may also be used to assist the decision-making process in business situations. Before a new undertaking is started, for example, it may be possible to list or categorise all the different possible outcomes. On the basis of past experience it may also be possible to attribute probabilities to each outcome. These figures can then be used to calculate an expected value for the income from the new undertaking. Example: Suppose a company is considering opening a new retail outlet in a town where it currently has no branches. On the basis of information from its other branches in similar towns, it can categorise the possible revenue from the new outlet into categories of high, medium or low. Table 10.6 shows the probabilities that have been estimated for each category. You will note that the probabilities sum to 1. Table 10.6: Probabilities of likely revenue returns Likely Revenue Probability High 0.4 Medium 0.5 Low 0.1 Using these figures it is possible to calculate the expected value for the revenue from the new branch. The expected value is an average for the revenue, weighted by the probabilities. It is the sum of each value for the outcome multiplied by its probability. Thus, the expected value E of variable x with associated probabilities P(x) is calculated by the following formula: E(x) = ΣxP(x) where x is the value of the outcome. 296 Probability © ABE and RRC In our example, if we give values to the outcomes of each category (i.e. figures for the high, medium and low revenues) and then multiply each by its probability and sum them, we get an average revenue return for new branches. This expected revenue can then be compared with the outlay on the new branch in order to assess the level of profitability. Table 10.7: Expected value of revenue returns Category Revenue (£000) Probability Prob. × Rev. (£000) High 5,000 0.4 2,000 Medium 3,000 0.5 1,500 Low 1,000 0.1 100 Total: £3,600 The concept of expected value is used to incorporate the consequences of uncertainty into the decision-making process. Questions for Practice 2 1. If three coins are thrown, what is the probability that all three will show tails? 2. (a) (i) What is the probability of getting a total of six points in a simultaneous throw of two dice? (ii) What is the probability of getting at least one six from a simultaneous throw of two dice? (b) Out of a production batch of 10,000 items, 600 are found to be of the wrong dimensions. Of the 600 which do not meet the specifications for measurement, 200 are too small. What is the probability that, if one item is taken at random from the 10,000: (i) It will be too small? (ii) It will be too large? If two items are chosen at random from the 10,000, what is the probability that: (iii) Both are too large? (iv) One is too small and one too large? 3. (a) Calculate, when three dice are thrown together, the probability of obtaining: (i) three sixes; (ii) one six; (iii) at least one six. (b) Calculate the probability of drawing in four successive draws the ace and king of hearts and diamonds suits from a pack of 52 playing cards. Probability 297 © ABE and RRC (c) Calculate the probability that three marbles drawn at random from a bag containing 24 marbles, six of each of four colours (red, blue, green and white), will be: (i) all red; (ii) all of any one of the four colours. 4. An analysis of 72 firms which had introduced flexitime working showed that 54 had used some form of clocking in system whilst the rest relied on employees completing their own time sheets. It was found that for 60 of the firms output per employee rose and in all other firms it fell. Output fell in only 3 of the firms using employee-completed forms. Construct a contingency table and then: (a) Calculate the probability of output per employee falling if an automatic clocking in system is introduced. (b) Determine under which system it is more probable that output per employee will rise. 5. The probability that machine A will be performing a useful function in five years' time is 0.25, while the probability that machine B will still be operating usefully at the end of the same period is 0.33. Find the probability that, in five years' time: (a) Both machines will be performing a useful function. (b) Neither will be operating. (c) Only machine B will be operating. (d) At least one of the machines will be operating. 6. Three machines, A, B and C, produce respectively 60%, 30% and 10% of the total production of a factory. The percentages of defective production for each machine are respectively 2%, 4% and 6%. (a) If an item is selected at random, find the probability that the item is defective. (b) If an item is selected at random and found to be defective, what is the probability that the item was produced by machine A? 7. Using a sample space diagram, calculate the probability of getting, with two throws of a die: (a) A total score of less than 9. (b) A total score of 9. Now check your answers with those given at the end of the unit. 298 Probability © ABE and RRC H. THE NORMAL DISTRIBUTION Earlier in the course, we considered various graphical ways of representing a frequency distribution. We considered a frequency dot diagram, a bar chart, a polygon and a frequency histogram. For example, a typical histogram will take the form shown in Figure 10.5. You can immediately see from this diagram that the values in the centre are much more likely to occur than those at either extreme. Figure 10.5: Typical frequency histogram Consider now a continuous variable in which you have been able to make a very large number of observations. You could compile a frequency distribution and then draw a frequency bar chart with a very large number of bars, or a histogram with a very large number of narrow groups. Your diagrams might look something like those in Figure 10.6. Figure 10.6: Frequency bar chart and histogram Frequency Value 80 90 100 110 120 130 140 0 10 20 30 40 150 Probability 299 © ABE and RRC If you now imagine that these diagrams relate to a relative frequency distribution and that a smooth curve is drawn through the tops of the bars or rectangles, you will arrive at the idea of a frequency curve. We can then abstract the concept of a frequency distribution by illustrating it without any actual figures as a frequency curve which is just a graph of frequency against the variate (variable): Figure 10.7: Frequency curve The "normal" or "Gaussian" distribution was discovered in the early eighteenth century and is probably the most important distribution in the whole of statistical theory. This distribution seems to represent accurately the random variation shown by natural phenomena – for example: heights of adult men from one race weights of a species of animal distribution of IQ levels in children of a certain age weights of items packaged by a particular packing machine life expectancy of light bulbs. The typical shape of the normal distribution is shown in Figure 10.8. You will see that it has a central peak (i.e. it is unimodal) and that it is symmetrical about this centre. The mean of this distribution is shown as "m" on the diagram, and is located at the centre. The standard deviation, which is usually denoted by "σ", is also shown. Figure 10.8: The normal distribution Variate Frequency m Variate Frequency o 300 Probability © ABE and RRC Properties of the Normal Distribution The normal distribution has a number of interesting properties. These allow us to carry out certain calculations on it. For normal distributions, we find that: 68% of the observations are within ± 1 standard deviation of the mean and 95% are within ± 2 standard deviations of the mean. We can illustrate these properties on the curve itself: Figure 10.9: Properties of the normal distribution These figures can be expressed as probabilities. For example, if an observation x comes from a normal distribution with mean m and standard deviation σ, the probability that x is between (m ÷ σ) and (m + σ) is: P(m ÷ σ < x < m + σ) = 0.68 Also P(m ÷ 2σ < x < m + 2σ) = 0.95 m s.d. s.d. Shaded area = 68% m s.d. s.d. s.d. s.d. Shaded area = 95% Probability 301 © ABE and RRC Using Tables of the Normal Distribution Because the normal distribution is perfectly symmetrical, it is possible to calculate the probability of an observation being within any range, not just (m ÷ σ) to (m + σ) and (m ÷ 2σ) to (m + 2σ). These calculations are complex, but fortunately, there is a short cut available. There are tables which set out these probabilities. We provide one such table in the appendix to this study unit. It relates to the area in the tail of the normal distribution. Note that the mean, here, is denoted by the symbol "μ" (the Greek letter mu). Thus, the first column in the table represents the position of an observation, x, in relation to the mean (x ÷ μ), expressed in terms of the standard deviation by dividing it by σ. Figure 10.10: Proportion of distribution in an area under the normal distribution curve The area under a section of the curve represents the proportion of observations of that size in relation to the total number of observations. For example, the shaded area shown in Figure 10.10 represents the chance of an observation being greater than m + 2σ. The vertical line which defines this area is at m + 2σ. Looking up the value 2 in the table gives the proportion of the normal distribution which lies outside of that value: P(x > m + 2σ) = 0.02275 which is just over 2%. Similarly, P(x > m + 1σ) is found by looking up the value 1 in the table. This gives: P(x > m + 1σ) = 0.1587 which is nearly 16%. You can extract any value from P(x = m) to P(x > m + 3σ) from this table. This means that you can find the area in the tail of the normal distribution wherever the vertical line is drawn on the diagram. Negative distances from the mean are not shown in the tables. However, since the distribution is symmetrical, it is easy to calculate these. P(x < m ÷ 5σ) = P(x > m + 5σ) So P(x > m ÷ 5σ) = 1 ÷ P(x < m ÷ 5σ) This is illustrated in Figure 10.11. m m + 2σ m ÷ 2σ m ÷ σ m + σ 302 Probability © ABE and RRC Figure 10.11: Using the symmetry to determine the area under the normal curve Further Probability Calculations It is possible to calculate the probability of an observation being in the shaded area shown in Figure 10.12, using values from the tables. This represents the probability that x is between m ÷ 0.7σ and m + 1.5σ: i.e. P(m ÷ 0.7σ < x < m + 1.5σ) Figure 10.12: Probability of an observation being in a particular area m ÷ 0.7σ m + 1.5σ m m + 2σ m ÷ 2σ m ÷ σ m + σ B A Area A is read from tables. Area B = Area A m m + 2σ m ÷ 2σ m ÷ σ m + σ C Area C = 1 ÷ Area B. This is because Area C and Area B are mutually exclusive and cover the whole range. Probability 303 © ABE and RRC First find P(x > m + 1.5σ) = 0.0668 from the table. Then find P(x < m ÷ 0.7σ) or P(x > m + 0.7σ) = 0.2420 since the distribution is symmetrical. The proportion of the area under the curve which is shaded in Figure 10.12 is then: 1 ÷ 0.0668 ÷ 0.2420 = 0.6912 Hence P(m ÷ 0.7σ < x < m + 1.5σ) = 0.6912 Example: A production line produces items with a mean weight of 70 grams and a standard deviation of 2 grams. Assuming that the items come from a normal distribution, find the probability that an item will weigh 65 grams or less. 65 grams is 5 grams below the mean. Since the standard deviation is 2 grams, this is 2.5 standard deviations below the mean. Let x be the weight of an individual item. Therefore: P(x < 65) = P(x < m ÷ 2.5σ) = P(x > m = 2.5σ) = 0.00621 from the tables. Now find the probability that an item will weigh between 69 and 72 grams. 69 grams is 0.5 standard deviations below the mean, and 72 grams is 1 standard deviation above the mean. Therefore we need to find P(m ÷ 0.5σ < x < m + σ): P(x > m + σ) = 0.1587 P(x < m ÷ 0.5σ) = P(x > m + 0.5σ) = 0.3085 So P(m ÷ 0.5σ< x < m + σ) = 1 ÷ 0.1587 ÷ 0.3085 or P(69 < x < 72) = 0.5328 304 Probability © ABE and RRC Questions for Practice 3 1. A factory which makes batteries tests some of them by running them continuously. They are found to have a mean life of 320 hours and a standard deviation of 20 hours. By assuming that this distribution is normal, estimate: (a) The probability that a battery will last for more than 380 hours. (b) The probability that it will last for less than 290 hours. (c) The probability that it will last for between 310 and 330 hours. (d) The value below which only 1 in 500 batteries will fall. 2. A machine produces 1,800 items a day on average. The shape of the distribution of items produced in one day is very similar to a normal distribution with a standard deviation of 80 items. On any one day: (a) If more than 2,000 items are produced, the employees get a bonus. What is the probability that a bonus will be earned? (b) When less than 1,700 items are produced, overtime is available. What is the probability that overtime will be worked? (c) What is the probability that the workforce neither gets a bonus nor works overtime? Now check your answers with those given at the end of the unit. Probability 305 © ABE and RRC ANSWERS TO QUESTIONS Questions for Practice 1 Event A is "failed" and event B is "having less than 10 lessons". The calculation then is: P(A|B) = ) B ( n ) B and A ( n = 50 15 = 0.3 Questions for Practice 2 1. The probability of tails is 0.5 for each coin. The events are independent. Therefore: P(3T) = 8 1 2 1 2 1 2 1 = × × 2. (a) (i) Six points can be obtained from the following combinations: 1st die 2nd die 1 5 2 4 3 3 4 2 5 1 Since each die can fall with any of its faces uppermost, there are a total of 36 possible combinations, i.e. 6 × 6 = 36. Therefore: P(six points) = 36 5 (ii) P(no six) = 36 25 6 5 6 5 = × P(at least one six) = 1 ÷ P(no six) = 36 11 (b) (i) P(too small) 02 . 0 000 , 10 200 = = (ii) P(too large) 04 . 0 000 , 10 400 = = (iii) Note that, if the first item is too large, it will be removed from the calculation of the second item. Therefore: P(both too large) = P(1 st too large) × P(2 nd too large) = 0.04 × 9,999 399 = 0.001596 (iv) The first item can be either too large or too small. Therefore: P(one too large, one too small) = 0.04 × 9,999 399 02 . 0 9,999 199 × + = 0.001594 306 Probability © ABE and RRC 3. (a) (i) Each die must show a six. Therefore: P(three sixes) = 216 1 6 1 3 = | . | \ | (ii) One die must show a six and the others must not show a six. Any one of the three dice could be the one that shows a six. Number of ways the die showing six can be chosen = 3 probability of six on one die = | . | \ | 6 1 probability of not obtaining six on the other two dice = 2 6 5 | . | \ | Therefore: P(one six) = 3 × | . | \ | 6 1 × 2 6 5 | . | \ | = 72 25 (iii) Probability of obtaining at least one six = 1 ÷ probability of not obtaining any sixes = 1 ÷ probability that all three dice show 1, 2, 3, 4 or 5 = 1 ÷ 216 91 216 125 1 6 5 3 = ÷ = | . | \ | (b) Required probability= 49 1 50 2 51 3 52 4 × × × = 0.000003694 to 4 significant figures. (c) (i) Probability that first is red = 24 6 probability that second is red = 23 5 probability that third is red = 22 4 probability that all three are red = = × × 22 4 23 5 24 6 0.009881 to 4 sig. figs. (ii) Probability that the three will be all of any one of the four colours = probability that all three are red + probability that all three are blue + probability that all three are green + probability that all three are white = 22 4 23 5 24 6 4 × × × = 0.03953 to 4 sig. figs. Probability 307 © ABE and RRC 4. The contingency table will be as follows: Output up Output down Total Clocking in system 45 9 54 Employee completion system 15 3 18 Total 60 12 72 (a) The probability of output per employee falling (event A) given that an automatic clocking in system is used (event B) is: P(A|B) = ) B ( n ) B and A ( n = 54 9 = 6 1 (b) The probability of output per employee rising (event A) given that an automatic clocking in system is used (event B) is: P(A|B) = ) B ( n ) B and A ( n = 54 45 = 6 5 The probability of output per employee rising (event A) given that an employee completion system is used (event B) is: P(A|B) = ) B ( n ) B and A ( n = 18 15 = 6 1 The probabilities are the same. 5. Let: A = Machine A performing usefully, and A = Machine A not performing usefully B = Machine B performing usefully, and B = Machine B not performing usefully. We can draw a tree diagram to map the situation as follows: Joint probabilities 3 1 x 4 3 = 12 3 3 2 x 4 3 = 12 6 3 2 x 4 1 = 12 3 3 1 x 4 1 = 12 1 4 1 A A B B 3 1 3 2 B B 3 1 3 2 4 3 Total = 1 308 Probability © ABE and RRC (a) Probability of both machines operating is 12 1 (b) Probability of neither operating is 12 6 or 2 1 (c) Probability of only B operating is 12 3 or 4 1 (d) Probability of at least one machine operating is 2 1 12 3 12 2 12 1 = + + 6. Let: A = produced on machine A B = produced on machine B C = produced on machine C D = defective D = not defective. The tree diagram for the situation is as follows. (a) P(D) = 0.012 + 0.012 + 0.006 = 0.03 (b) P(defective that came from machine A) = defective of y probabilit A from defective a of y probabilit = 03 . 0 012 . 0 = 0.4 Joint probabilities 0.30 x 0.04 = 0.012 0.60 x 0.98 = 0.588 0.60 x 0.02 = 0.012 A Total = 1.000 0.30 x 0.96 = 0.288 0.10 x 0.06 = 0.006 0.10 x 0.94 = 0.094 0.94 0.06 0.04 0.96 0.98 0.02 B 0.10 D D D D D D C 0.30 0.60 Probability 309 © ABE and RRC 7. The sample space diagram sets out all the possible total scores obtainable when two dice are thrown. There are 36 equally likely outcomes: 1st die: 1 2 3 4 5 6 2nd die 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 (a) 26 outcomes give scores less than 9. Therefore: P(total score less than 9) = 722 . 0 18 13 36 26 = = (b) 4 outcomes give a score of 9. Therefore: P(total score of 9) = 111 . 0 9 1 36 4 = = Questions for Practice 3 1. (a) 380 hours is 60 hours or 3 standard deviations above the mean. Therefore: P(observation > m + 3σ) = 0.00135 from the tables. (b) 290 hours is 30 hours or 1.5 standard deviations below the mean. Therefore: P(observation < m ÷ 1.5σ) = P(observation > m + 1.5σ) = 0.0668 (c) 310 hours is 0.5 standard deviations below the mean and 330 hours is 0.5 standard deviations above the mean. P(observation > m + 0.5σ) = 0.3085 from the tables Similarly, P(observation < m ÷ 0.5σ) = 0.3085 Therefore: P(m ÷ 0.5σ < observation < m + 0.5σ) = P(310 < observation < 330) = 1 ÷ 0.3085 ÷ 0.3085 = 0.383 Alternatively, using the symmetry of the normal distribution: P(m < observation < m + 0.5σ) = 0.5σ ÷ 0.3085 = 0.1915. Hence, P(m ÷ 0.5σ < observation < m + 0.5σ) = 2 × 0.1915 = 0.383 (d) Expressing 1 in 500 as a proportion: 1 in 500 = 002 . 0 500 1 = 310 Probability © ABE and RRC Using the tables in reverse, and looking up the value 0.002 in the body of the tables, the nearest we can get is 0.00199. This is when: 88 . 2 σ μ x = | . | \ | ÷ Therefore, 1 in 500 batteries will be more than 2.88 standard deviations below the mean. Answer: = 320 ÷ 2.88 × 20 = 320 ÷ 57.6 = 262.4 Therefore 1 in 500 batteries will have a life of less than 262.4 hours. 2. Mean, m = 1,800 Standard deviation, σ = 80 (a) In a normal distribution, 2,000 items is: 80 1,800) (2,000÷ = 2.5 standard deviations from the mean Therefore, the probability of 2,000 items being produced is: P(x > m + 2.5σ) = 0.00621 (b) In a normal distribution, 1,700 items is: 80 1,800) (1,700÷ = ÷1.5 standard deviations from the mean Therefore, the probability of 1,700 items being produced is: P(x < m ÷ 1.25σ) = P(x > m + 1.25σ) = 0.1056 (c) Probability of no bonus and no overtime: = 1 ÷ 0.1056 ÷ 0.00621 = 0.88819 Probability 311 © ABE and RRC Appendix: Areas in the Tail of the Normal Distribution The table of values of the standard normal distribution set out below provides a means of determining the probability of an observation (x) lying within specified standard deviations (o) of the mean of the distribution (μ). σ μ) - (x .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .4960 .4920 .4880 .4840 .4801 .4761 .4721 .4681 .4641 0.1 .4602 .4562 .4522 .4483 .4443 .4404 .4364 .4325 .4286 .4247 0.2 .4207 .4168 .4129 .4090 .4052 .4013 .3874 .3936 .3897 .3859 0.3 .3821 .3783 .3745 .3707 .3669 .3632 .3594 .3557 .3520 .3483 0.4 .3446 .3409 .3372 .3336 .3300 .3264 .3228 .3192 .3156 .3121 0.5 .3085 .3050 .3015 .2981 .2946 .2912 .2877 .2843 .2810 .2776 0.6 .2743 .2709 .2676 .2643 .2611 .2578 .2546 .2514 .2483 .2451 0.7 .2420 .2389 .2358 .2327 .2296 .2266 .2236 .2206 .2177 .2148 0.8 .2119 .2090 .2061 .2033 .2005 .1977 .1949 .1922 .1894 .1867 0.9 .1841 .1814 .1788 .1762 .1736 .1711 .1685 .1660 .1635 .1611 1.0 .1587 .1562 .1539 .1515 .1492 .1469 .1446 .1423 .1401 .1379 1.1 .1357 .1335 .1314 .1292 .1271 .1251 .1230 .1210 .1190 .1170 1.2 .1151 .1131 .1112 .1093 .1075 .1056 .1038 .1020 .1003 .0985 1.3 .0968 .0951 .0934 .0918 .0901 .0885 .0869 .0853 .0838 .0823 1.4 .0808 .0793 .0778 .0764 .0749 .0735 .0721 .0708 .0694 .0681 1.5 .0668 .0655 .0643 .0630 .0618 .0606 .0594 .0582 .0571 .0559 1.6 .0548 .0537 .0526 .0516 .0505 .0495 .0485 .0475 .0465 .0455 1.7 .0446 .0436 .0427 .0418 .0409 .0401 .0392 .0384 .0375 .0367 1.8 .0359 .0351 .0344 .0336 .0329 .0322 .0314 .0307 .0301 .0294 1.9 .0287 .0281 .0274 .0268 .0262 .0256 .0250 .0244 .0239 .0233 2.0 .02275 .02222 .02169 .02118 .02068 .02018 .01970 .01923 .01876 .01831 2.1 .01786 .01743 .01700 .01659 .01618 .01578 .01539 .01500 .01463 .01426 2.2 .01390 .01355 .01321 .01287 .01255 .01222 .01191 .01160 .01130 .01101 2.3 .01072 .01044 .01017 .00990 .00964 .00939 .00914 .00889 .00866 .00842 2.4 .00820 .00798 .00776 .00755 .00734 .00714 .00695 .00676 .00657 .00639 2.5 .00621 .00604 .00587 .00570 .00554 .00539 .00523 .00508 .00494 .00480 2.6 .00466 .00453 .00440 .00427 .00415 .00402 .00391 .00379 .00368 .00357 2.7 .00347 .00336 .00326 .00317 .00307 .00298 .00289 .00280 .00272 .00264 2.8 .00256 .00248 .00240 .00233 .00226 .00219 .00212 .00205 .00199 .00193 2.9 .00187 .00181 .00175 .00169 .00164 .00159 .00154 .00149 .00144 .00139 3.0 .00135 312 Probability © ABE and RRC i Certificate in Business Management INTRODUCTION TO QUANTITATIVE METHODS Contents Unit Title Page iii v vii 1 3 4 5 7 11 13 22 26 29 31 39 40 40 43 50 55 63 64 64 68 70 76 79 89 90 91 96 98 106 Introduction to the Manual Syllabus Formulae, rules and tables provided with examination paper 1 Basic Numerical Concepts Introduction Number Systems Numbers – Approximation and Integers Arithmetic Dealing with Negative Numbers Fractions Decimals Percentages Ratios Further Key Concepts Algebra, Equations and Formulae Introduction Some Introductory Definitions Algebraic Notation Solving Equations Formulae Basic Business Applications Introduction Currency and Rates of Exchange Discounts and Commission Simple and Compound Interest Depreciation Pay and Taxation Simultaneous, Quadratic, Exponential and Logarithmic Equations Introduction Simultaneous Equations Formulating Problems as Simultaneous Equations Quadratic Equations Logarithmic and Exponential Functions 2 3 4 ii Unit 5 Title Introduction to Graphs Introduction Basic Principles of Graphs Graphing the Functions of a Variable Graphs of Equations Gradients Using Graphs in Business Problems Introduction to Statistics and Data Analysis Introduction Data for Statistics Methods of Collecting Data Classification and Tabulation of Data Frequency Distributions Graphical Representation of Information Introduction Graphical Representation of Frequency Distributions Pictograms Pie Charts Bar Charts Scatter Diagrams General Rules for Graphical Presentation Measures of Location Introduction The Arithmetic Mean The Mode The Median Measures of Dispersion Introduction Range The Quartile Deviation, Deciles and Percentiles Standard Deviation The Coefficient of Variation Skewness Probability Introduction Estimating Probabilities Types of Event The Two Laws of Probability Conditional Probability Tree Diagrams Sample Space Expected Value The Normal Distribution Page 119 120 121 124 126 141 147 171 172 173 175 177 185 193 194 194 204 206 207 209 210 221 222 223 232 236 249 250 251 251 255 262 263 275 276 277 279 280 286 288 294 295 298 6 7 8 9 10 iii Introduction to the Study Manual Welcome to this study manual for Introduction to Quantitative Methods. The manual has been specially written to assist you in your studies for the ABE Certificate in Business Management and is designed to meet the learning outcomes specified for this module in the syllabus. As such, it provides a thorough introduction to each subject area and guides you through the various topics which you will need to understand. However, it is not intended to "stand alone" as the only source of information in studying the module, and we set out below some guidance on additional resources which you should use to help in preparing for the examination. The syllabus for the module is set out on the following pages and you should read this carefully so that you understand the scope of the module and what you will be required to know for the examination. Also included in the syllabus are details of the method of assessment – the examination – and the books recommended as additional reading. Following the syllabus, we also set out the additional material which will be supplied with the examination paper for this module. This is provided here for reference only, to help you understand the scope of the syllabus, and you will find the various formulae and rules given there fully explained later in the manual. The main study material then follows in the form of a number of study units as shown in the contents. Each of these units is concerned with one topic area and takes you through all the key elements of that area, step by step. You should work carefully through each study unit in turn, tackling any questions or activities as they occur, and ensuring that you fully understand everything that has been covered before moving on to the next unit. You will also find it very helpful to use the additional reading to develop your understanding of each topic area when you have completed the study unit. Additional resources ABE website – www.abeuk.com. You should ensure that you refer to the Members Area of the website from time to time for advice and guidance on studying and preparing for the examination. We shall be publishing articles which provide general guidance to all students and, where appropriate, also give specific information about particular modules, including updates to the recommended reading and to the study units themselves. Additional reading – It is important you do not rely solely on this manual to gain the information needed for the examination on this module. You should, therefore, study some other books to help develop your understanding of the topics under consideration. The main books recommended to support this manual are included in the syllabus which follows, but you should also refer to the ABE website for further details of additional reading which may be published from time to time. Newspapers – You should get into the habit of reading a good quality newspaper on a regular basis to ensure that you keep up to date with any developments which may be relevant to the subjects in this module. Your college tutor – If you are studying through a college, you should use your tutors to help with any areas of the syllabus with which you are having difficulty. That is what they are there for! Do not be afraid to approach your tutor for this module to seek clarification on any issue, as they will want you to succeed as much as you want to. Your own personal experience – The ABE examinations are not just about learning lots of facts, concepts and ideas from the study manual and other books. They are also about how these are applied in the real world and you should always think how the topics under consideration relate to your own work and to the situation at your own workplace and others with which you are familiar. Using your own experience in this iv way should help to develop your understanding by appreciating the practical application and significance of what you read, and make your studies relevant to your personal development at work. It should also provide you with examples which can be used in your examination answers. And finally … We hope you enjoy your studies and find them useful not just for preparing for the examination, but also in understanding the modern world of business and in developing in your own job. We wish you every success in your studies and in the examination for this module. The Association of Business Executives September 2008 v Unit Title: Introduction to Quantitative Methods Level: 3 Learning Outcomes and Indicative Content: Candidates will be able to: 1. Demonstrate the rules of numeracy 1.1 1.2 1.3 2. Unit Code: IQM Learning Hours: 100 Apply the four rules to whole numbers, fractions and decimals Express numbers in standard form Multiply and divide negative numbers Apply calculations 2.1 2.2 Compare numbers using ratios, proportions and percentages; Obtain values for simple financial transactions involving purchases, wages, taxation, discounts p g Determine values for simple discounts and compound interest, and for purchases, wages, taxation, depreciation of an asset using the straight line method Determine values for simple and compound interest and the reducing balance method depreciation Convert foreign currency Make calculations using a scientific calculator including roots and powers; logarithms and exponential values Evaluate terms involving a sequence of operations and use of brackets Interpret, transpose and evaluate formulae Approximate data using rounding, significant figures 2.3 2 2.3 2.4 2.5 2.6 2.7 2.8 3. Use algebraic methods 3.1 3.2 3.3 3.4 3.5 Solve linear and simultaneous equations Solve quadratic equations using factorisation and formulae Solve equations using roots or logarithms Determine the equation of a straight line through two points and also when given one point and its gradient Determine the gradient and intercepts on the x or y axes for a straight line 4. Construct and use: graphs, charts and diagrams 4.1 4.2 4.3 4.4 Draw charts and diagrams derived from tabular data: eg bar charts, pie charts, scatter diagrams Plot graphs, applying general rules and principles of graphical construction including axes, choice of scale and zero Plot and interpret mathematical graphs for simple linear, quadratic, exponential and logarithmic equations Identify points of importance on graphs eg maximum, minimum and where they cut co-ordinate axes vi . The expression of a linear function. usually annual) n = Number of time periods.vii INTRODUCTION TO QUANTITATIVE METHODS FORMULAE FOR BUSINESS MATHEMATICS AND STATISTICS INTEREST The formula for calculating compound interest: ⎛ r ⎞ A = P ⎜1 + ⎝ 100 ⎟ ⎠ n where: A = Accrued amount P = Original principal r = Rate of interest (for a particular time period. takes the following form: y = mx + c Note that: c = the y intercept (the point where the line crosses the y axis) m = the gradient (or slope) of the line . DEPRECIATION ● Straight-line method: Cost of asset Annual depreciation = –––––––––––– Useful life (Cost of asset) – (Value at end of useful life) or Annual depreciation = –––––––––––––––––––––––––––––––––––– Useful life ● Reducing balance method: D = B (1 – i) n where: D = Depreciated value at the end of the n th time period B = Original value at beginning of time period i = Depreciation rate (as a proportion) n = Number of time periods (normally years) STRAIGHT LINE A linear function is one for which. a straight line is obtained. and hence the formula of a straight line. when the relationship is plotted on a graph. ⎛ p⎞ log ⎜ ⎟ = log p – log q ⎝q⎠ 3. If y = axn then n = (log y – log a) ÷ log x PROBABILITY ● Probability rules: Probability limits: Total probability rule: For complementary events: For two mutually exclusive events: For independent events: 0 ≤ P(A) ≤ 1 – P(A) + P(A ) = 1 P(A and B) = 0 P(A) = P(A|B) and/or P(B) = P(B|A) ΣP = 1 (for all outcomes) ● Multiplication rules: For independent events: For dependent events: P(A and B) = P(A)P(B) P(A and B) = P(A)P(B|A) ● Additional rules: For mutually exclusive events: For non-mutually exclusive events: P(A or B) = P(A) + P(B) P(A or B) = P(A) + P(B) – P(A and B) ● Conditional rules: P(A and B) P(A|B) = –––––––––– P(B) P(A and B) and P(B|A) = ––––––––– P(A) Expected value of variables x with associated probabilities P(x) is E(x) = ΣxP(x) . log(p × q) = log p + log q 2.viii QUADRATIC EQUATION A quadratic equation of the form ax 2 + bx + c = 0 can be solved using the following formula: x= – b ± b 2 – 4ac 2a RULES FOR LOGARITHMS 1. log pn = n log p 4. ● ● Range = Largest value – Smallest value. where the number of ⎝ 2 ⎟ ⎠ = observations is odd and where n is the number of observations. ● Median for ungrouped data: ⎛ n + 1⎞ = Value of the ⎜ th observation in a ranked data set. ● – Mean for grouped data: μ = Σmf/N and x = Σmf/n where m is the midpoint and f is the frequency of a class. ● Standard deviation for grouped data: σ= ● ( ∑ mf ) ∑m f – 2 2 N N and s = ( ∑ mf ) ∑m f – 2 2 n n –1 Pearson’s measure of skewness: Mean – Mode 3(Mean – Median) Psk = ––––––––––––––– or –––––––––––––––– Standard deviation Standard deviation . Standard deviation for ungrouped data: σ= ∑x 2 (∑ x ) – N N 2 and s = ∑x 2 (∑ x ) – n n –1 2 where σ and s are the population and sample standard deviations respectively.ix STATISTICAL MEASURES ● Mean for ungrouped data: μ = ∑ x x and x = ∑ n N – where N and x are the population and sample means respectively. 00 σ 0.02222 .01831 .00889 .2206 .0694 .2611 .1446 .00181 .1271 .4 2.00379 .0344 .0721 .3409 .3015 .2090 .00139 .4 0.1056 .00144 .0 .0606 .1210 .1190 .3594 .0418 .02275 .00842 .07 .1075 .4247 .00248 .9 3.0934 .02068 .0735 .02018 .2358 .7 1.0281 .00391 .2005 .0571 .0409 .00326 .3874 .01786 .6 0.00714 .0427 .00776 .02 .2 0.00798 .05 .00621 .0516 .08 .00415 .00604 .8 2.1711 .4562 .0314 .2810 .0526 .0392 .01539 .00939 .0465 .00199 .0475 .2483 .4129 .00317 .0367 .0853 .00508 .2420 .0495 .01287 .1335 .3483 .3372 .4207 .4602 .2981 .3745 .01017 .1562 .1170 .0901 .00539 .2266 .01044 .1131 .5 1.1867 .1230 .00272 .00466 .3085 .3192 .0 1.00587 .0436 .00990 .0594 .1020 .00368 .1539 .0262 .3050 .00193 .1841 .3783 .2236 .0244 .01426 .0655 .00427 .3707 .00336 .1 1.1423 .0401 .00289 .1038 .5 0.3936 .00233 .01970 .0322 .3859 .00402 ..0708 .8 0.00347 .4960 .1112 .2946 .3228 .01876 .0351 .1003 .4840 .7 2.0869 .01 .00159 .00570 .0256 .0793 .01072 .2177 .1 2.1314 .03 .04 .1611 .4801 .01500 .3897 .4364 .2296 .01578 .4 1.00523 .2578 .00734 .9 2.01700 .00154 .01923 .0233 .0268 .0749 .0985 .0307 .3446 .01101 .1814 .0618 .2843 .00755 .2061 .1357 .0 2.00820 .3300 .1151 .00357 .01160 .6 1.4443 .2327 .3821 .0918 .00240 .0838 .2 2.2389 .4090 .1788 .2033 .3 0.00226 .0885 .00175 .7 0.2776 .4286 .4168 .00639 .00298 .0681 .0294 .2514 .0287 .00187 .0485 .0951 .00307 .1685 .4404 .3557 .00964 .01130 .1949 . μ x AREAS IN TAIL OF THE STANDARD NORMAL DISTRIBUTION (x – μ) –––––.0455 .0250 .1469 .4920 .0764 .0582 .1401 .0778 .2 1.0329 .00453 .0643 .00219 .01321 .01618 .1635 .00480 .1292 .01659 .0808 .4013 .00494 .00212 .00256 .00440 .0239 .1 0.01191 .1922 .2643 .1587 .0630 .0559 .0548 .2119 .0274 .01222 .01390 .1660 .0823 .01743 .0384 .0505 .0 0.4522 .3632 .1762 .4721 .2546 .8 1.4325 .00657 .0446 .3520 .00264 .1894 .00164 .0359 .06 .01355 .4483 .0537 .0336 .4761 .3 1.1515 .2451 .4052 .1093 .0301 .0375 .0668 .1251 .5000 .2743 .4880 .01255 .2709 .00866 .2912 .3121 .00695 .00554 .02169 .0968 .2877 .00135 .1977 .2676 .01463 .1736 .5 2.6 2.02118 .9 1.4641 .00205 .3336 .3264 .2148 .00280 .3156 .00914 .09 .x STANDARD NORMAL DISTRIBUTION The table of values of the standard normal distribution set out below provides a means of determining the probability of an observation (x ) lying within specified standard deviations (σ) of the mean of the distribution (μ ).00676 .3669 .1492 .3 2.00149 .00169 .4681 .1379 . Number Systems Denary Number System Roman Number System Binary System Numbers – Approximation and Integers Approximation Integers Arithmetic Addition (symbol ) Subtraction (symbol ) Multiplication (symbol ) Division (symbol ) The Rule of Priority Brackets D. (Continued over) © ABE and RRC . F. Dealing with Negative Numbers Addition and Subtraction Multiplication and Division Fractions Basic Rules for Fractions Adding and Subtracting with Fractions Multiplying and Dividing Fractions Decimals Arithmetic with Decimals Limiting the Number of Decimal Places Page 3 4 4 4 4 5 5 6 7 7 8 8 9 9 10 11 11 13 13 14 17 20 22 23 25 B. E.1 Study Unit 1 Basic Numerical Concepts Contents Introduction A. C. I. Percentages Arithmetic with Percentages Percentages and Fractions Percentages and Decimals Calculating Percentages Ratios Further Key Concepts Indices Expressing Numbers in Standard Form Factorials 26 27 27 27 28 29 31 31 32 33 35 H.2 Basic Numerical Concepts G. Answers to Questions for Practice © ABE and RRC . as well as being essential to understanding and solving many business problems. However. subtract. multiplication and division manipulate negative numbers cancel fractions down to their simplest terms change improper fractions to mixed numbers or integers. it is accepted practice in examinations. and vice versa add. To get the most out of the material. in charts and diagrams. You should therefore get to know how to use one quickly and accurately. subtraction. Throughout the course. carry out calculations using percentages and divide a given quantity according to a ratio explain the terms index. Sometimes we can express numerical information more clearly in pictures – for example. reciprocal and factorial. there will be plenty of practice questions for you to work through. we recommend strongly that you do attempt all these questions. Firstly though. One initial word about calculators.Basic Numerical Concepts 3 INTRODUCTION The skill of being able to handle numbers well is very important in many jobs. © ABE and RRC . multiply and divide fractions and mixed numbers carry out calculations using decimals calculate ratios and percentages. but it is essential to ensure that you fully understand the basics before moving on to some of the more advanced applications in later units. We shall consider this later in the course. root. power. we need to examine the basic processes of manipulating numbers themselves. Understanding how the principles work is essential if you are to use the calculator correctly. It is perfectly acceptable to use calculators to perform virtually all arithmetic operations – indeed. In this study unit. you should also ensure that you know exactly how to perform all the same operations manually. Objectives When you have completed this study unit you will be able to: identify some different number systems round-up numbers and correct them to significant figures carry out calculations involving the processes of addition. we shall start by reviewing a number of basic numerical concepts and operations. as this is a practical subject. You may often be called upon to handle numbers and express yourself clearly in numerical form – and this is part of effective communication. It is likely that you will be familiar with at least some of these. Roman numerals are still in occasional use in. This means that. we can note that it is possible to have negative numbers – numbers which are less than zero. etc.e. the dates on film productions. This is illustrated in the following table. In this system. we call this number system the base ten system or denary system. counting from 0 to 1 uses up all the symbols. Here are some Roman numbers and their denary equivalent: XIV IX 14 9 XXXVI LVlll 36 58 Binary System As we have said. 20°C below zero). © ABE and RRC . for example. Because ten figures (i.4 Basic Numerical Concepts A. Numbers which are more than zero are positive. in this number. say. so a new column has to be started when counting to 2. or the binary system. Although we invariably use the normal denary system for calculations. Before we look at other number systems. there are: two hundreds (289) eight tens (289) nine units (289). 73 means 73. 20°C (i. but in practice this is not necessary and we adopt the convention that. We could indicate a positive number by a plus sign (). A common example of the use of negative numbers is in the measurement of temperature – for example. 0 and 1. tabulation. This is equally true of the Roman numerical system. but this system uses letters instead of figures. By contrast. house numbers. and similarly the third column at 4. This uses just two symbols (figures). Figure 1.e.1: The denary system Seventh column Millions Sixth column Hundreds of thousands Fifth column Tens of thousands Fourth column Thousands Third column Hundreds Second column Tens First column Units The number of columns indicates the actual amounts involved. our numbering system is a base ten system. symbols 0 to 9) are used. Consider the number 289. We indicate a negative number by a minus sign (). Roman Number System We have seen that position is very important in the number system that we normally use. computers use base two. NUMBER SYSTEMS Denary Number System We are going to begin by looking at the number system we use every day. 000.836 and asked to quote it to the nearest thousand.Basic Numerical Concepts 5 Figure 1. to the nearest thousand. NUMBERS – APPROXIMATION AND INTEGERS Approximation Every figure used in a particular number has a meaning. B. round down for figures of five or more. There are strict mathematical rules for deciding whether to round up or down to the nearest figure: Look at the digit after the one in the last required column. The halfway point between the two is 27. When a number is transferred to the computer. However. on some occasions. When rounding off 3. Then: for figures of four or less. Clearly. you need only express the distance to the nearest mile. when rounding off 236 to the nearest hundred. corresponding to "components" of the computer being either switched off or on. © ABE and RRC .500. if you are calculating the cost of the journey.000. the figure lies between 27.000 and 28.2: The binary system Denary 0 1 2 3 4 5 6 7 8 9 Binary 0 1 10 11 100 101 110 111 1000 1001 This system is necessary because computers can only easily recognise two figures – zero or one. Thus. the digit after the one in the last required column is 3 (the last required column being the hundreds column) and the answer would be to round down to 200. the last required column is the tens column and the digit after that is 5. For example.455 to the nearest ten.836 is more than this and so.460. so we would round up to 3. it is translated into the binary system. precision is not needed and may even hinder communication. (a) Rounding off Suppose you are given the figure 27. It is important to understand the methods of approximation used in number language. round up. the answer would be 28. The number 27. to make the number language easier to understand – and is particularly important when we consider decimals later in the unit. we obtain an integer. 1½.25 By adjusting any number to the nearest whole number. Round off the following the figures to the nearest thousand: (a) (b) (c) 2. the Customs and Excise rule is always to round down to the nearest one pence.290 5. in VAT calculations. 3. 0. giving 2.570.8.820 8. we consider the second significant figure. We do this by using the rounding off rule – round down from 4 or round up from 5 – so 2. Do not forget to include the insignificant zeros. To write 2. and apply the rule for rounding off.017.236. Integers Integers are simply whole numbers. are integers.002 has six significant figures – the zeros here are significant as they occur in the middle of the number.45. the 7.6 Basic Numerical Concepts (However. so that the digits keep their correct place values. 0. 7¼.) (b) Significant figures By significant figures we mean all figures other than zeros at the beginning or end of a number. a number which is not an integer can be approximated to become one by rounding off. The following are not integers: ¾.000 © ABE and RRC . To write it correct to three significant figures. 3. This has four significant figures.017 correct to three significant figures becomes 2. we need to drop the fourth figure – i.231 6. 1. etc. Thus. We can round off a number to a specified number of significant figures.e. as are all the negative whole numbers.746 8. Thus: 632.501 Round off the following figures to the nearest million: (a) (b) (c) 6.632 also has only three significant figures 630. Consider the figure 2. Questions for Practice 1 1. This is likely to be done to aid the process of communication – i.000 has three significant figures as the zeros are not significant 000. the zero. 2.480.e. Thus. 1.000.017 correct to one significant figure. 20. you should note that different rules may apply in certain situations – for example.020. 4. 322 16 3. Where the sum of the numbers in a column amount to a figure greater than nine (for example. 5. 3.901 26. in the above calculation. it is 1 or. Where you have several columns of numbers to add (as may be the case with certain statistical or financial information). only the last number is inserted into the total and the first number is carried forward to be added into the next column. and although in practice it is likely you will perform many arithmetic operations using a calculator.836 3. ARITHMETIC We will now revise some basic arithmetic processes. C.886 9.Basic Numerical Concepts 7 3. more precisely. Correct the following to three significant figures: (a) (b) (c) (d) 2. then those in the next column to the left and so on.727 Total The figures in the right hand column (the units column) are added first.100 and 43. to add 246. the first task is to set the figures out in columns: 246 5. the sum of the units column comes to 17). So. If the numbers are written underneath each other in the correct position. so layout and neatness matter. Addition (symbol ) When adding numbers. one ten which is carried forward into the "tens" column. Here. there is less chance of error.100 43 8. it is important that you know how to do the calculations manually and understand all the rules that apply and the terminology used. How many significant figures are in the following? (a) (b) (c) (d) 1. remember that the position of each digit in a number is important.736 22.894 7.000 Now check your answers with those given at the end of the unit. the columns can be cross-checked to produce check totals which verify the accuracy of the addition: © ABE and RRC . 16.322. it is important to add each column correctly. When carrying out arithmetic operations. These may well be familiar to you.726 4.300 1. to subtract 21 from 36. in fact. Moving on to the next column. Subtraction (symbol ) Again. the second column now becomes 3 (1 1). it is again very helpful to set out the calculation in columns: 246 23 In practice. 2 4 8. it is essential to put back the one borrowed and this is added to the number to be subtracted. When larger numbers are involved. we shall note the names given to the various elements involved in a multiplication operation: © ABE and RRC . the first task is to set the sum out: 36 21 15 Total As with addition. column by column. starting with the units column. and subtracting 8 from that allows you to put 6 into the total. it is likely that you would use a calculator for this type of "long multiplication". or 3 2. but we shall briefly review the principles of doing it manually.8 Basic Numerical Concepts Columns A 246 211 29 300 47 833 B 210 203 107 99 102 721 C 270 109 13 7 199 598 D 23 400 16 3 91 533 Totals 749 923 165 409 439 2. starting from the right – the units column. you must work column by column.685 Rows Totals Note that each row is added up as well as each column. thus providing a full cross-check. Thus. The one that you borrow is. it is important to set out the figures underneath each other correctly in columns and to work through the calculation. Multiplication (symbol ) The multiplication of single numbers is relatively easy – for example. So. which leaves 1. Now consider this calculation: 34 18 16 Total Where it is not possible to subtract one number from another (as in the units column where you cannot take 8 from 4). the procedure is to "borrow" one from the next column to the left and add it to the first number (the 4). one ten – so adding that to the first number makes 14. Before doing so though. Returning to our example of long multiplication.920 5. Then we multiply the multiplicand by the next number of the multiplier (the number of tens. the rule of priority in arithmetical operations requires that: division and multiplication are carried out before addition and subtraction. it is 360 the divisor is the number by which the dividend is to be divided – here it is 24 and the answer to a division operation is called the quotient – here the quotient for our example is 15. it is 246 the multiplier is the number by which the multiplicand is to be multiplied – here it is 23 and the answer to a multiplication operation is called the product – so to find the product of two or more numbers. the subtotals are added to give the product for the whole operation: 246 23 738 4.Basic Numerical Concepts 9 the multiplicand is the number to be multiplied – in this example. you multiply them together.920. © ABE and RRC . The Rule of Priority Calculations are often more complicated than a simple list of numbers to be added and/or subtracted. which is 3) and enter the product as a subtotal – this shown here as 738. Thus. the calculation may be set out in a number of ways: 360 24. here we have to remember that we are multiplying by tens and so we need to add a zero into the units column of this subtotal before entering the product for this operation (492). which is 2). the actual product of 246 20 is 4. or most commonly 360 24 Just as there were three names for the various elements involved in a multiplication operation. 16 2 8. the task becomes more complicated and you should use a calculator for all long division. or 360/24. Division involving a single number divisor is relatively easy – for example. However. Finally. They may involve a number of different operations – for example: 863 8 2 When faced with such a calculation. so there are three names or terms used in division: the dividend is the number to be divided – in this example.658 Subtotal (units) Subtotal (tens) Total product Division (symbol ) In division. the first operation is to multiply the multiplicand by the last number of the multiplier (the number of units. For larger numbers. Brackets Brackets are used extensively in arithmetic. If there are brackets within brackets. or even multiply the eight by "six minus three plus four" (i. we could have written the above calculation as follows: (8 6) 3 8 2 This makes it perfectly clear what numbers are to be multiplied. the contents of the brackets are evaluated first. by seven).10 Basic Numerical Concepts Thus. The rule of priority explained above now needs to be slightly modified. Consider another example: 12 3 3 6 1 Again there are a number of ways in which this calculation may be interpreted. Example: 2 3 (6 8 2) 2 3 (6 4) 232 26 8 Thus we can see that brackets are important in signifying priority as well as showing how to treat particular operations. for example: (12 3) (3 6) 1 12 (3 3) (6 1) (12 3) 3 (6 1) (12 [3 3]) 6 1 Note that we can put brackets within brackets to further separate elements within the calculation and clarify how it is to be performed. It must be strictly followed except when brackets are included. or multiply the eight by "six minus three" (i. In that case. in the above example. It is not really clear whether we should: (a) (b) (c) multiply the eight by six. Thus. by three).e. we can do the addition and subtraction: 48 3 4 49. then the innermost brackets are evaluated first. (multiplication before addition) (within the brackets. We use brackets to separate particular elements in the calculation so that we know exactly which operations to perform on which numbers. 8 4 2 then.e. putting the results of the first step into the calculation. the procedure would be: (a) (b) first the multiplication and division: 8 6 48. the division first) © ABE and RRC . and we shall meet them again when we explore algebra. Look at the previous calculation again. say. as follows: 0 1 2 3 4 5 6 7 8 This brings us to the expected answer of seven (3 4). Work out the answers to the following questions. Work out the answers to the following. DEALING WITH NEGATIVE NUMBERS Addition and Subtraction We can use a number scale. you can visualise the addition of. (a) (b) (c) (d) (e) (f) (6 3) 6 4 (6 3) (6 4) 10 (5 2) (10 5) 24 24 (2 2) 6 (24 [4 2]) (15 [3 2] 6 [4 3] 2) 11 Now check your answers with those given at the end of the unit. as a simple picture of addition and subtraction. using a calculator if necessary. A 322 219 169 210 Totals B 418 16 368 191 C 42 600 191 100 D 23 46 328 169 Totals 3. 6 minus 4 is represented as follows: © ABE and RRC .944 36 Add each of the four columns in the following table and cross-check your answer. D.Basic Numerical Concepts 11 Questions for Practice 2 1. Similarly we can view subtraction as moving to the left on the scale. three and four by finding the position of three on the scale and counting off a further four divisions to the right. (a) (b) (c) (d) 2. Thus. Thus.460 159 59. 937 61 1.976 63 19. like the markings on a ruler. (Note that. the process is slightly different. so we can also write: (3) 4 4 (3) 1 We can illustrate the second sum as follows: 3 2 1 0 1 2 3 4 5 This is the same as 4 3.12 Basic Numerical Concepts 0 1 2 3 4 5 6 7 8 This illustrates the sum 6 4 2. For example: 4 (3) 4 3 7 To sum up: a plus and minus make a minus two minuses make a plus. Start at 3 on the scale and count four divisions to the right: 4 3 2 1 0 1 2 3 4 Therefore. (Draw your own number scales to help visualise the sums if you find it helpful. a plus and a minus together make a minus. here. In effect. therefore.) The order of the figures of these calculations does not matter. All we do is extend the scale to the left of zero: 6 5 4 3 2 1 0 1 2 3 4 5 6 We can now visualise the addition and subtraction of negative numbers in the same way as above. Consider the sum (3) 4. (3) 4 1.) 7 (3) 7 3 10 2 (6) 2 6 4 © ABE and RRC . we change the subtraction sign to a plus and treat the number as positive. Consider the following examples. With subtraction. To subtract a minus number. Negative numbers can also be represented on the number scale. we have shown the figure of "minus 3" in brackets to avoid confusion with the addition and subtraction signs. using a number scale if necessary: 1. 8 (6) (8) (2) (7) (7) Now check your answers with those given at the end of the unit. or 5 © ABE and RRC . the product is always another whole number. with a remainder of 2 parts of 5. with a remainder of 2 10 3 3. but a part of a whole number – so. (4) 7 (9) 3 (3) (5) 3. 8. The fraction is expressed as the remainder divided by the original divisor and the way in which it is shown. 5. Consider the following two examples: 12 5 2. In contrast. 4. or two divided by five. the result of the division is a whole number and a fraction of a whole number. in each case. then. 7. E. is in the same form as a division term: 2 12 5 2. the division of one whole number by another does not always result in a whole number. When two or more whole numbers are multiplied together. 10. 7 (4) 6 (8) (5) (4) 10 (10) 2.Basic Numerical Concepts 13 Multiplication and Division The rules for multiplication and division are as follows: Type of calculation Minus times plus Minus times minus Minus divide by plus Minus divide by minus Plus divide by minus Product/Quotient minus plus minus plus minus Example 3 7 21 7 (3) 21 21 3 7 21 (3) 7 21 (3) 7 Questions for Practice 3 Solve the following sums. 6. 9. with a remainder of 1 The remainder is not a whole number. FRACTIONS A fraction is a part of a whole number. is always to reduce the fraction to the smallest possible whole numbers for both its numerator and denominator. etc. When two numbers are divided. 5 50 93 9 The rule in expressing fractions. For example: 5 551 5 Fractions are often thought of as being quite difficult. Examples of proper fractions are: 1 2 4 4 . The result of dividing out an improper fraction is a whole number plus a part of one whole. the fraction is known as a proper fraction. The following are all valid fractions: 2 20 47 6 . However. the expression is not a true fraction. the fraction is known as an improper fraction. The result of dividing out a proper fraction is only a part of one whole and therefore a proper fraction has a value of less than one. The result of dividing out is one whole part and no remainder. . Basic Rules for Fractions Cancelling down Both the numerator and the denominator in a fraction can be any number. or Each part of the fraction has a specific terminology: the top number is the numerator and the bottom number is the denominator. .14 Basic Numerical Concepts 10 3 3. (a) The numerator is larger than the denominator In this case. The resulting fraction is then in its lowest possible terms. the numerators are 2 and 1. 1 3 Thus. or 1 24 24 (b) The denominator is larger than the numerator In this case. there are three possible outcomes. or 4 9 9 30 6 1 whole part and a remainder of 6 parts of 24. . . for the above fractions. We shall start by looking at these rules before going on to examine their application in performing arithmetic operations on fractions. and the denominators are 5 and 3. or one divided by three. as in the two examples above. . 5 3 7 5 (c) The numerator is the same as the denominator In this case. though. Taking two more examples: 42 6 4 whole parts and a remainder of 6 parts of 9. they are not really hard as long as you learn the basic rules about how they can be manipulated. © ABE and RRC . with a remainder of 1 part of 3. The answer must be 1. 2 8 could be expressed as by multiplying both terms by 4.Basic Numerical Concepts 15 The process of reducing a fraction to its lowest possible terms is called cancelling down. Thus: 20 2 50 5 A fraction which has not been reduced to its lowest possible terms is called a vulgar fraction. For example. 20 5 If we wanted to change the denominator to a specific number – for example. so we should use 10. you would probably quickly see that both are divisible by 3. we can do that by multiplying both the numerator and denominator by the same number. and 10. so you could start the process of cancelling down by dividing by three: 165 3 55 77 231 3 This gives us a new form of the same fraction. just the way in which the parts of the whole are expressed. large vulgar fractions can often be reduced in stages to their lowest possible terms. and we can now see that both terms can be further divided – this time by 11: 55 11 5 77 11 7 So. both terms can be divided by 33. Cancelling down can be a tricky process since it is not always clear what the largest whole number to use might be. Changing the denominator to required number The previous process works in reverse. but you are very unlikely to see that straight away. Divide both the numerator and denominator by the largest whole number which will divide into them exactly. However. However the rule states that we use the largest whole number. To do this. If we wanted to express a given fraction in a different way. to express ½ in eighths – the rule is to divide the required denominator by the existing denominator (8 2 4) and then multiply both terms of the original fraction by this result: 1 4 4 24 8 © ABE and RRC . Dividing both parts of the fraction by 10 reduces it to: 2 5 Note that we have not changed the value of the fraction. for example. consider the following fraction: 165 231 In fact. we follow another simple rule. 5. So. Consider the following fraction from those listed previously: 20 50 Which whole number will divide into both 20 and 50? There are several: 2. Changing a mixed number into an improper fraction This is the exact reverse of the above operation. to write 3 (a) (b) (c) 2 as an improper fraction: 3 first multiply the whole number (3) by the denominator (3) 3 3 9. Changing an improper fraction into a whole or mixed number Suppose we are given the following improper fraction: 35 8 The rule for converting this is to divide the numerator by the denominator and place any remainder over the denominator. eighths) are left over: 35 3 35 8 4 and a remainder of three 4 8 8 This result is known as a mixed number – one comprising a whole number and a fraction. For example. then add the numerator of the fraction to this product and place the sum over the denominator.16 Basic Numerical Concepts Note again that we have not changed the value of the fraction. just the way in which the parts of the whole are expressed. changing the three whole ones into nine thirds then add the numerator (2) to this 2 9 11. Thus we find out how many whole parts there are and how many parts of the whole (in this case. Reduce the following fractions to their lowest possible terms: (a) (c) (e) (g) (i) 9 36 21 70 14 49 36 81 24 192 8 64 18 54 25 150 45 135 33 88 (b) (d) (f) (h) (j) © ABE and RRC . The rule is to multiply the whole number by the denominator in the fraction. thus adding the two thirds to the nine thirds from the first operation finally place the sum over the denominator 11 3 Questions for Practice 4 1. Adding and Subtracting with Fractions (a) Proper fractions Adding or subtracting fractions depends upon them having the same denominator. we can change the expression of the first fraction by multiplying both the numerator and denominator by 2. In this case. For example: 3 2 5 7 7 7 5 1 4 1 8 8 8 2 Where the denominators are not the same.Basic Numerical Concepts 17 2. Change the following improper fractions to mixed numbers: (a) (c) (b) (d) 4. we need to change the form of expression so that they become the same. This process is called finding the common denominator. Where the denominators are the same. thus making its denominator 10 – the same as the second fraction: 4 3 7 10 10 10 © ABE and RRC . It is not possible to do these operations if the denominators are different. Change the following mixed numbers to improper fractions: (a) (c) 2 4 (b) (d) 3 9 Now check your answers with those given at the end of the unit. Change the following fractions as indicated: (a) (c) 1 to 10ths 5 4 to 81sts 9 12 5 25 6 2 3 1 8 (b) (d) 3 to 20ths 5 9 to 88ths 11 17 8 31 3 3 8 3 4 3. adding or subtracting is just a simple case of applying the operation to numerators. Consider the following calculation: 2 3 10 5 We cannot add these two fractions together as they stand – we have to find the common denominator. the same principles are applied to subtracting mixed numbers. 1 1 3 5 2 4 2 4 Consider the following calculation: 34 1 1 3 (3 4 5 2) 4 2 4 1 2 3 14 4 4 4 14 6 4 1 2 14 1 15 1 2 In general. Consider the following calculation: 3 1 1 10 5 4 The lowest common denominator for these three fractions is 20. For example: 4 5 1 2 1 5 1 5 3 1 (4 1) 3 3 3 6 2 6 3 6 2 6 6 © ABE and RRC . a common denominator may be found by multiplying the denominators together change all the fractions in the calculation to this common denominator then add (or subtract) the numerators and place the result over the common denominator. The procedure is as follows: first add all the integers together then add the fractions together as explained above. We must then convert all three to the new denominator by applying the rule explained above for changing a fraction to a required denominator: 3 1 1 12 5 2 15 3 10 20 20 20 20 5 4 4 The rule is simple: (b) find the lowest common denominator – and where this is not readily apparent. and then add together the sum of the integers and the fractions. we deal with the integers and the fractions separately.18 Basic Numerical Concepts Not all sums are as simple as this. Mixed numbers When adding mixed numbers. It is often the case that we have to change the form of expression of all the fractions in a sum in order to give them all the same denominator. 17. convert it to a fraction with the same common denominator and add it to the fraction in the mixed number. 18.Basic Numerical Concepts 19 However. 15. Consider this calculation: 4 1 5 1 2 6 Here. 16. 14. 5. 9. 8. 1 1 3 4 1 2 8 3 2 3 5 4 5 1 6 3 5 1 7 3 1 1 10 6 1 1 5 6 2 1 3 4 7 1 8 5 3 3 4 8 2. 11. Thus: 4 1 1 3 1 3 2 2 9 6 3 3 6 6 6 We can now carry out the subtraction as above: 4 1 5 9 5 4 2 9 5 1 3 1 (3 1) 2 2 2 6 6 6 6 3 6 6 Questions for Practice 5 Carry out the following calculations: 1. 13. 10. 7. the procedure is slightly different where the second fraction is larger than the first. 2 1 3 1 2 4 4 1 6 4 3 3 1 2 1 8 4 2 3 2 4 3 1 1 2 8 3 3 2 1 1 2 4 3 2 2 1 1 31 2 3 2 5 4 2 5 1 2 9 3 1 1 1 2 2 8 3 6 Now check your answers with those given at the end of the unit. it is not possible to subtract the fractions even after changing them to the common denominator: 3 5 6 6 To complete the calculation we need to take one whole number from the integer in the first mixed number. 12. © ABE and RRC . 4. 3. 6. For example: 3 2 32 6 1 12 4 3 43 2 3 3 33 9 40 8 5 85 (d) Dividing by a fraction In this case. ¼ of £12. This is just another way of expressing multiplication. For example: 5 4 1 4 11 4 44 2 11 2 11 22 2 3 1 19 2 38 1 9 4 4 2 1 4 2 © ABE and RRC . we must complete one further step – convert the mixed number to an improper fraction. (b) Dividing a fraction by a whole number In this case. Before we can apply the same methods to mixed numbers. For example: 2 25 10 1 5 3 3 3 3 3 3 37 21 1 7 5 4 4 4 4 We often see the word "of" in a calculation involving fractions – for example.000. we simply multiply the denominator by the whole number and place the old numerator over the product (which is now the new denominator). For example: 3 3 3 5 20 4 45 2 2 2 5 3 35 15 (c) Multiplying one fraction by another In this case. we multiply the numerators by each other and the denominators by each other. we simply multiply the numerator by the whole number and place the product (which is now the new numerator) over the old denominator.20 Basic Numerical Concepts Multiplying and Dividing Fractions (a) Multiplying a fraction by a whole number In this case. and then reduce the result to the lowest possible terms. we turn the divisor upside down and multiply. For example: 4 3 4 16 1 4 5 3 4 3 3 1 1 1 3 3 1 6 3 6 1 6 2 (e) Dealing with mixed numbers All that we have said so far in this section applies to the multiplication and division of proper and improper fractions. Some calculations can be very long. If you make a mistake somewhere. rather than in the steps you have gone through. you can use cancellation during a division calculation at the step when. that you cannot cancel across addition and subtraction signs. However. Thus. it is much easier to identify the error if all the workings are shown. in an examination. they will not be able to follow the process and see that the error is purely mathematical. even if you make a mistake in the arithmetic. repeating the calculation involving mixed numbers from above.Basic Numerical Concepts 21 (f) Cancelling during multiplication We have seen how it is possible to reduce fractions to their lowest possible terms by cancelling down. for example: 6 3 cancels down to by dividing both terms by 2.) Cancellation can also be used in the multiplication of fractions. too. 8 4 We often show the process of cancelling down during a calculation by crossing out the cancelled terms and inserting the reduced terms as follows: 2 8 2 12 3 3 (It is always helpful to set out all the workings throughout a calculation. This involves dividing both the numerator and the denominator by the same number. we could show it as follows: 4 3 1 2 1 19 1 19 9 4 2 4 2 1 2 2 © ABE and RRC . However. having turned the divisor upside down. you are multiplying the two fractions. For example: 3 1 1 1 12 4 16 4 Remember that every cancelling step must involve a numerator and a denominator. It is also the case that. involving many steps. In this case it can be applied across the multiplication sign – dividing the numerator of one fraction and the denominator of the other by the same number. if you do not show all the workings. examiners will give "marks for workings". So. the following operation would be wrong: 3 3 1 4 1 12 3 3 Note. Unlike fractions though. so 0. 4. etc. However. 7 5 3 4 3 6 2 3 3 7 2.02 is two-hundredths of a whole and 0. and decimals are effectively fractions expressed in tenths.1 is one-tenth part of a whole and 0. so 0.6 1 10 10 10 1 67 81 .001 We can see that any fraction with a denominator of ten. 7. 67 hundredths) of a whole the next number to the right. Decimals do not have a numerator or a denominator. Rather. DECIMALS Decimals are an alternative way of expressing a particular part of a whole. ten thousand. 8. The term "decimal" means "in relation to ten". 8 6 2 6 3 7 3 8 3 6 4 5 4 7 Now check your answers with those given at the end of the unit. so 0.67 is six-tenths and seven-hundredths (i. if there is one.1 1 3 6 . the part of the whole is shown by a number following a decimal point: the first number following the decimal point represents the number of tenths. one hundred.054 . hundredths.e. 3. © ABE and RRC . again if there is one. decimals are written completely differently and this makes carrying out arithmetic operations on them much easier. 6. We can see the relationship between decimals and fractions as follows: 0.67 . represents the number of thousandths.3 . thousandths.01 0. represents the number of hundredths.382 5 1000 1000 1000 0.134 is 134 thousandths of a whole and so on up to as many figures as are necessary. 1.22 Basic Numerical Concepts Questions for Practice 6 Carry out the following calculations: 1. 0. fractions with a different denominator are also relatively easy to convert by simple division.81 3 100 100 100 1 54 382 . 0. 5. one thousand. etc. F.005 is five-thousandths of a whole and 0. 3. 0. can be easily converted into a decimal.3 is three-tenths of a whole the next number to the right. 5. 172 20. we count the number of figures after a decimal point in the original sum – there is just one. For example. To get it in the right place. what would the answer be? © ABE and RRC .857 44. we can multiply the numbers as follows: 40 5 200 Now we need to put the decimal point into the answer. arrange the numbers in columns with the decimal points all being in the same vertical line.6 7. subtraction. The one key point to bear in mind is to get the position of the decimal point in the right place.84 9. ignoring the decimal points.7 – 18.5.857 from 63. subtraction is easy provided that you follow the same rule of ensuring that the decimal points are aligned vertically. Layout is all important in achieving this. subtract 18. add the following: 3. The trick is to get the position of the decimal point right! So.612 Total Similarly.5. multiply the numbers out as if the figures were whole numbers.7: 63.0 or just 20 If the sum had been 0. Then you can carry out the arithmetic in the same way as for whole numbers. For example. There is a simple rule for this – the answer should have the same number of decimal places as there were in the question. although you should ensure that you take care to get the decimal point in the correct place when both inputting the figures and when reading the answer. The principles of multiplying decimals manually are the same as those considered earlier in the study unit for whole numbers. and then put back the decimal point in the answer.843 Total (b) Multiplying decimals It is easier to use a calculator to multiply decimals.Basic Numerical Concepts 23 Arithmetic with Decimals Performing addition.5) – and place the decimal point that number of places from the right in the answer: 20. Consider the calculation of 40 0.4 0.84 9. Again. in the multiplier (0.6 7. multiplication and division with decimals is exactly the same as carrying out those operations on whole numbers. (a) Adding and subtracting To add or subtract decimals.172 The layout for this is as follows: 3. it is important to get the decimal point in the right place. The simplest way of carrying this out is to ignore the decimal points in the first instance. 24 Basic Numerical Concepts The multiplication would be the same. For example: 3.62 10 36. For example: 0.6 However. we insert the decimal point two places to the right in the answer: 0. we can use the technique noted above in respect of multiplication to find the answer quickly without using a calculator: (d) to divide by ten. but in most instances it is easier to use a calculator. etc. You can do this manually. but this time there are 2 figures after decimal points in the original sum. to multiply by 100. This type of decimal. 68. hundredths.3333333 or however many 3s for which there is room on the paper or the calculator screen. thousandths. is known as a recurring decimal. depending upon the size of the decimal. and vice versa Fractions other than tenths. For example. can also be converted into decimals. simply move the decimal point one place to the left – for example. it is far easier to use a calculator when dividing with decimals. you simply move the decimal point one place to the right. Consider the fraction 1 .32 100 0.736 3 © ABE and RRC . not all fractions will convert so easily into decimals in this way. when dividing by ten or by one hundred (or even by a thousand or a million). So. where one or more numbers repeat infinitely.62 100 362 This can be a very useful technique. Thus.e.6832. enabling quick manual calculations to be done with ease. to change 3 into decimal form. 3 5). 5 Using a calculator the answer is easily found: 0. 3 Dividing this out (or using a calculator) gives the answer 0. we simply place the decimal over ten or one hundred etc. to multiply a decimal by ten. It is simply a process of dividing the numerator by the denominator and expressing the answer in decimal form.32 10 6. we do not need such a degree of precision. you move the decimal point two places to the right: 3. However.832 to divide by one hundred. simply move the decimal point two places to the left – for example. In most business applications. Converting fractions to decimals.36 36 9 25 100 736 92 3 1000 125 3.2 Likewise. the decimal becomes the numerator and the denominator is 10 or 100 etc.2 One final point to note about multiplying decimals is that. (c) Division with decimals Again. To change a decimal into a fraction. 68. so we abbreviate the decimal to a certain size – as we shall discuss in the next section. you divide 3 by 5 (i. respectively. The most common examples are one-third and two-thirds.209 (b) (d) (f) (h) 8. In such cases. We then speak of a number or an answer as being "correct to two (or three) decimal places". For example. (When referring to these types of recurring decimals. but we have shown two-thirds as being 0.16 3. So. we would normally just quote the decimal as. subtracting.2 7. multiplying or dividing several decimals which have been rounded – something which we shall see later in the course in respect of a set of figures which should add up to 100. in business applications. that level of precision may be critical. there are also plenty of other fractions which do convert into exact decimals.33 and 0.Basic Numerical Concepts 25 Limiting the Number of Decimal Places We noted that certain fractions do not convert into exact decimals. we look at the digit in the fourth decimal place: 7.6 3.6666666.43 © ABE and RRC . it is not easy to add the above two decimals together and multiplying them does not bear thinking about.2857142) correct to three decimal places. these would be 0.66. say.3 recurring".. In most business circumstances then. we need to draw attention to the problem by stating that "errors are due to rounding".. but you should note that the level of accuracy suffers.263 8. For decimals with a larger number of decimal places than the required number. we look at the digit in the decimal after the one in the last required place and then: for figures of five or more.792 7.) However.36 2.67.81 3.. or 7 11 0.6470568 17 This gives us a problem: how much detail do we want to know? It is very rare that..286. If we were to express these correct to two decimal places.182 12. but only at the level of thousandths or greater. although in engineering. so we are only interested in the level of detail down to hundredths or thousandths.36 4. These types of small error can increase when adding. Carry out the following calculations: (a) (c) (e) (g) 3. round down by leaving the figure in the last required place unchanged. Thus. Following the rule for rounding. we would then round up the number in the third decimal place to give 0. we limit the number of decimal places to a manageable number. we would want to know an answer precise to the millionth part of a whole.07 8. but do not.14 8. which convert into 0.84 3.8 4. if we want to express the decimal form of two-sevenths (0.3333333. round up the figure in the last required place for figures of four or less.67 respectively. Working to only two or three decimal places greatly simplifies decimals. This is usually set at two or three decimal places.33 by two gives us 0.2857142. Now consider the decimal forms of one-third and two-thirds.32 11. "0. we use the principles of rounding discussed earlier in the study unit to reduce them to the correct size. Such large decimals also present problems in carrying out arithmetic procedures – for example.08 11. For example: 2 0.79 6. Questions for Practice 7 1.67 4.42 10. multiplying 0. and 0. The per cent sign is %. or four parts of a hundred 4 100 28% means twenty-eight-hundredths of a whole.8 4.8 3. PERCENTAGES A percentage is a means of expressing a fraction in parts of a hundred – the words "per cent" simply mean "per hundred".26 Basic Numerical Concepts (i) (k) (m) (o) 2.125 1.2 2.2 11. © ABE and RRC . 3. so when we say "percentage".052 Change the following fractions into decimal form.3 0.592 2.750 Now check your answers with those given at the end of the unit. Some examples: 4% means four-hundredths of a whole.6 0. correct to the number of decimal places (dp) specified: (a) (c) (e) (g) 2 to 2 dp 5 3 to 2 dp 4 5 to 3 dp 8 4 to 3 dp 7 (b) (d) (f) (h) 3 to 3 dp 8 7 to 3 dp 8 5 to 3 dp 6 5 to 3 dp 16 3.75 (b) (d) 0. so 20 per cent is written as 20%. G.07 (j) (l) (n) (p) 4.8 5. Change the following decimals into fractions and reduce them to their lowest possible terms: (a) (c) 0.7 2.72 1. It is therefore important that you are fully conversant with working with them. or twenty-eight parts of a hundred 28 100 100% means one hundred-hundredths of a whole.60 0.62 2.63 5. or a hundred parts of a hundred 100 1 100 Percentages are used extensively in many aspects of business since they are generally more convenient and straightforward to use than fractions and decimals.1 154 2. The expression "20 per cent" means 20 out of 100. we mean "out of a hundred". Take an example: if Peter spends 40% of his income on rent and 25% on household expenses. For example: 60% 24% 60 2 100 5 24 6 25 100 Changing a fraction into a percentage is the reverse of this process – simply multiply by 100. so we need only be concerned with their addition and subtraction. this is easy – you simply move the decimal point two places to the left.67% 3 3 As you become familiar with percentages you will get to know certain equivalent fractions well. To change a decimal to a percentage.3% 1 8 1 3 20% 50% 1 5 1 2 25% 75% 1 4 3 4 Percentages and Decimals To change a percentage to a decimal. to change a percentage to a fraction you simply divide it by 100. consider this example: if Alan eats 20% of his cake and 10% of his pie. what percentage of his income remains? Here both percentages are parts of the same whole – Peter's income – so we can calculate the percentage remaining as follows: 100% (40% 25%) 35% However. again you divide it by 100. Thus: 11. but you need to remember that we can only add or subtract percentages if they are parts of the same whole. it is not possible to add the two together. what percentage remains? Here the two percentages are not parts of the same whole – one refers to cake and the other to pie. These processes are quite straightforward. For example: 3 300 100 % 37½% or 37. Here are some of them: 12½% 33.5% 0.36% 36% © ABE and RRC .115 Working the conversion the other way is equally easy. Therefore. move the decimal point two places to the right: 0.Basic Numerical Concepts 27 Arithmetic with Percentages Multiplication and division have no meaning when applied to percentages. As we saw in the last section. In effect all this means is that you place the percentage over 100 and then reduce it to its lowest possible terms. Percentages and Fractions Since a percentage is a part of one hundred.5% 8 8 2 200 100 % 66. 2 1. as given above. to show 23 pence as a percentage of £2. we work the calculation through the other way.5% 2 There are occasions when we deal with percentages greater than 100%.000) £1.200. For example. This could be both as pence or both as pounds: 23 100 11.000 and she sold it at 20% profit.23 100 11. change the percentage rate into a fraction or a decimal and then multiply the given number by that fraction/decimal.28 Basic Numerical Concepts Calculating Percentages To find a particular percentage of a given number.200 and says that she made 20% profit. we need to multiply the selling price by Cost 1. For example.200. to find 15% of £260: £260 or 15 £3900 £39 100 100 £260 0.5% 200 or 0.000 £(1. so the selling price is: £1. For example.000 120 100 : 120 To prove that this is correct. how much did she buy it for? The cost plus the profit 120% of cost £1. is 100% To work out the cost. we convert the numbers into a fraction with the first number as the numerator and the second the denominator. The cost was £1. The cost. we need to express both quantities in the same units before working out the percentage. © ABE and RRC .200 100 £1. therefore. and then change the fraction into a percentage by multiplying by 100.000 20% of £1. if a woman sells a car for £1.000 0. For example.15 £39 To express one number as a percentage of another. what percentage of cars have been sold if 27 have gone from a total stock of 300? 27 2700 100 9% 300 300 Do not forget that percentages refer to parts of the same whole and therefore can only be calculated where the units of measurement of the quantities concerned are the same. RATIOS A ratio is a way of expressing the relationship between two quantities. It is essential that the two quantities are expressed in the same units of measurement – pence.790 4. For example. Calculate the following: (a) (c) 60 as a percentage of 300 £25 as a percentage of £1.000 (b) (d) 25 as a percentage of 75 30 as a percentage of 20 Now check your answers with those given at the end of the unit. etc. To do this. © ABE and RRC .5% of £138 (b) (d) 25% of 360° 27% of £1. if we wanted to give the ratio of the width of a table to its depth where the width is 2 m and the depth is 87 cm. Change the following fractions into percentages: (a) 2. we would have to convert the two quantities to the same units before giving the ratio. H. express the ratio as a fraction by putting the first figure over the second and cancel the resulting fraction.Basic Numerical Concepts 29 Questions for Practice 8 1. number of people. Then re-express the ratio in the form of numerator : denominator. 3 100 3 8 (b) Change the following percentages into fractions and reduce them to their lowest possible terms (a) (c) 2% 7½% (b) (d) 8% 18% 3. so the correct form of showing the above ratio would be: 200: 87 (pronounced "200 to 87") There are two general rules in respect of ratios: They should always be reduced to their lowest possible terms (as with vulgar fractions). Note that the ratio itself is not in any particular unit – it just shows the relationship between quantities of the same unit. Calculate the following: (a) (c) (e) 15% of £200 12½% of 280 bottles 17. Expressing both in centimetres would give the ratio of 200 to 87. We use a special symbol (the colon symbol : ) in expressing ratios. – or the comparison will not be valid. If the profit for a year is £18. how much does each partner receive? To divide a quantity according to a given ratio: first add the terms of the ratio to find the total number of parts then find what fraction each term of the ratio is to the whole. The profit is divided into 3 1 5 9 parts. For example. Boddington and Devenish partnership. and finally divide the total quantity into parts according to the fractions. Therefore: Ansell receives 3 of the profits 9 1 of the profits and 9 For the Ansell. where the respective figures are 33½ mpg and 47 mpg.000. staying the same or deteriorating. If we look at the ratio of profits to sales across two years.000 9 1 £18. The ratio would be 85 : 17.000 £2.30 Basic Numerical Concepts For example. (This will be examined elsewhere in your studies. There should not be any fractions or decimals in them. For example. a common way of analysing business performance is to compare profits with sales. we may be able to see if this aspect of business performance is improving. this would be calculated as follows: Boddington receives Devenish receives Therefore: Ansell gets 5 of the profits. consider the ratio of girls to boys in a group of 85 girls and 17 boys. Boddington and Devenish – who share the profits of their partnership in the ratio of 3 : 1 : 5. here. multiplying by 2 to give: 67 : 94 Ratios are used extensively in business to provide information about the way in which one element relates to another.000 £10.) Another application is to work out quantities according to a particular ratio.000 9 Boddington gets Devenish gets 5 £18. ratios are often used to express the relationship between partners in a partnership – sharing profits on a basis of. This can be reduced as follows: 85 1 17 5 The ratio of 85 to 17 is. but convert the figures to whole numbers by.000 9 © ABE and RRC . Ratios should always be expressed in whole numbers. 9 3 £18. 5 : 1. We would not express this as 33½ : 47. say. 50 : 50 or 80 : 20.000 £6. therefore. Consider the case of three partners – Ansell. Consider the ratio of average miles per gallon for cars with petrol engines to that for cars with diesel engines. Thus in the case of 26 we refer to the term as "two raised to the sixth power" or "two to the power of six". I. Thus: 2 2 is written as 22 (the little "2" raised up is the superscript number) 2 2 2 is written as 23 2 2 2 2 2 2 2 is written as 27 etc.000 7.) The power of a number is the result of multiplying that number a specified number of times.000 A B C D 2. Total profit is £7. Now check your answers with those given at the end of the unit. Indices Indices are found when we multiply a number by itself one or more times. The number of times that the multiplication is repeated is indicated by a superscript number to the right of the number being multiplied.000 75. and you will get the opportunity to practice their use in later study units. we do include practice in relation to the expression of numbers in standard form. © ABE and RRC .200.000 63. The number being multiplied (here. Express the following sales and profits figures as profit to sales ratios for each firm: Firm Sales £ 100. 2) is termed the base and the superscript number is called the exponent or index of its power. 22 should be referred to as "two to the power of two". G and H share profits in the ratio of 2 : 3 : 3.000 30.000. but we normally call it "2 squared".Basic Numerical Concepts 31 Questions for Practice 9 1. similarly. F. Thus. Here we concentrate mostly on what these concepts mean. (Strictly.000 15.000 10. 8 is the third power of 2 (23) and 81 is the fourth power of 3 (34). Total profit is £60. Y and Z share profits in the ratio of 3 : 4: 5. 23 is referred to as "2 cubed".000 Profits £ 25. FURTHER KEY CONCEPTS In this last section we shall briefly introduce a number of concepts which will be used extensively later in the course. However. Calculate the distribution of profits for the following partnerships: (a) (b) X. 000. The reciprocal of any number is one divided by that number.3825 1013 and 8. Thus 36 refers to the square root of 36.43605 1014. though.3825 (10 10 10 10 10 10 10 10 10 10 10 10 10). is equal to the first number. A root is the number that produces a given number when raised to a specified power.000.605.000. can you even tell which is the bigger? Probably not. the fourth root of 16 is 2 (2 2 2 2 16). A square root of any number is the number which.000. Standard form provides such a way.000. Expressing Numbers in Standard Form The exploration of indices leads us on to this very useful way of expressing numbers. customary to omit the 2 from the root sign when referring to the square root. as discussed above. and it is helpful if we can find a different way of expressing them which will make them easier to deal with. it would be very difficult indeed. Thus 2 9 3 (3 3 9) The root index (the little 4 or little 2 in the above examples) can be any number.825. The above two numbers.000. as in the case of 16¼.125 3 8 2 Indices also do not have to be whole numbers – they can be fractional. The method of expressing numbers in standard form reduces the number to a value between 1 and 10. Working with very large (or very small) numbers can be difficult. © ABE and RRC . Thus. For example: the reciprocal 5 is 23 represents 1 and 5 1 1 0. It is made a little easier by the two numbers being very close to each other on the page. However. In fact. It is. It is more usual though to refer to this as the fourth root of the base and to write it as follows: 4 16 The symbol indicates a root. as in the case of 23. the first number would be: 9. A negative exponent indicates a reciprocal.32 Basic Numerical Concepts It is quite possible to have a negative exponent. It is a very big number and you would probably have great difficulty stating it. The 13 and 14 after the 10 are index numbers. which is 6. This indicates that the base is raised to the power of a quarter. help is at hand. under this system are: 9. when multiplied by itself. Written out in full. Consider the number 93. followed by the number of 10s you need to multiply it by in order to make it up to the correct size. let alone multiplying it by 843. but if one was on a different page. etc.43 102. You may encounter factorials in certain types of statistical operations.000 104 10.001. it indicates a reciprocal – i.01 and 103 0. the second number from above. Before we consider this in more detail. For example. 4 factorial is written as 4! (and is sometimes read as "four bang") and is equal to 4 3 2 1.43605 1014 is larger than 9. 2. We know that 10 10 100. © ABE and RRC . 10 843. We can go on multiplying 10 by itself to get: 103 1. We also know from the above that. Thus.605. so an easier way of dealing with a number expressed in standard form is to remember that the index number after the 10 tells you how many places the decimal point has to be moved to the right in order to create the full number (remembering also to put in the 0s). etc. Again.000 We could go on to say that 102 0. Thus: 101 1 0.000. Just as with indices we can also have numbers expressed in standard form using a negative index – for example. 4 3 6 0 5 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Converting numbers in this way makes arithmetic operations much easier – something which we shall return to later in your studies – and also allows us to compare numbers at a glance. it would be helpful to recap exactly what the index number means.000.43 102 0.Basic Numerical Concepts 33 This is not particularly useful. Factorials A factorial is the product of all the whole numbers from a given number down to one.0243.e. when there is a minus number in the index. you will note that a minus index number after the 10 tells you how many places the decimal point has to be moved to the left in order to create the full number (remembering also to put in the 0s as well). So.1. we can see immediately that 8.000. one divided by the number as many times as the index specifies. written out in full would be: 8 .3825 1013. Now you should be able to work out that 2. so we can say that 102 100. 68 102 9.853 106 6. (a) (b) (c) (d) (e) (f) 4. Express the following numbers in standard form: (a) (b) (c) (d) (e) (f) 2.35 102 1.02 105 7.16 1010 Now check your answers with those given at the end of the unit.000 0. 234 0.420 0.00000000000000003792 Write out in full the following numbers which are expressed in standard form.34 Basic Numerical Concepts Questions for Practice 10 1.300.0005032 845.045 6.653 1012 8. © ABE and RRC . 730 Three Two (b) 8. (a) (b) (c) (d) (e) (f) (6 3) 6 4 9 6 4 54 4 58 (6 3) (6 4) 9 10 90 10 (5 2) (10 5) 10 7 2 70 2 68 24 24 (2 2) 24 24 4 24 6 30 6 (24 [4 2]) 6 (24 8) 6 3 18 (15 [3 2] 6 [4 3] 2) 11 (15 5 6 12 2) 11 (3 6 24) 11 33 11 3 Questions for Practice 3 1. 3 2 6 15 100 © ABE and RRC .412 Totals 3. 3.890 Four Four (b) (d) (b) (d) 3.000 7.000. A 322 219 169 210 920 B 418 16 368 191 993 C 42 600 191 100 933 D 23 46 328 169 566 Check 805 881 1. 2.000 Questions for Practice 2 1. 4.000 7.000 5.000. 5. 4.Basic Numerical Concepts 35 ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1.840 7. (a) (c) 57.140 554 The check on your answers is achieved by adding a column into which you put the sum of each row. 7. 9.000.000 2. 3 14 6 9 1 2. 8. The total of the check column should equal the sum of the four column totals. (a) (c) (a) (c) (a) (c) (a) (c) 4.056 670 3. 10. 2. 3.000 (b) 8. 6.157 952 (b) (d) 232.880 9. 8. (a) (c) 4. 14. 5. (a) (c) 3. 6. 13. 9. 4 15 11 30 5 12 27 40 3 8 1 3 20 1 2 8 21 4 7 1 4 11 30 8 1 5 8 1 12 © ABE and RRC . 11. 12. (a) (c) (e) (g) (i) 2. 4.36 Basic Numerical Concepts Questions for Practice 4 1. 3. 7. 7 12 19 24 2. 10. (a) (c) 1 4 3 10 2 7 4 9 1 8 2 10 36 81 (b) (d) (f) (h) (j) (b) (d) (b) (d) (b) (d) 1 8 1 3 1 6 1 3 3 8 12 20 72 88 2 4 2 5 1 6 2 1 8 1 3 10 8 3 33 8 27 8 39 4 Questions for Practice 5 1. 40 0. 3 3 19 24 11 12 Questions for Practice 6 1.47 23.68 26. 3. 18. 6.276 1.75 0. 4.84 3.833 0. 21 1 5 4 4 15 1 2 6 2 2 1 18 9 7 24 2.413 5. 5.6 0. (a) (c) (e) (g) (i) (k) (m) (o) 2.092 5.571 125 1 1000 8 175 3 1 100 4 (b) (d) (f) (h) (j) (l) (n) (p) (b) (d) (f) (h) (b) (d) 19.591 10.6 50 0. 16 1 5 3 3 18 4 2 7 7 3 1 24 8 5 28 Questions for Practice 7 1.178 9.875 0. 8.313 60 3 100 5 750 3 1000 4 © ABE and RRC . (a) (c) 11.136 23. (a) (c) (e) (g) 3. 7. 2 2 2 9 7 24 16.17 15.375 0.42 6.78 70 5. 17.625 0.Basic Numerical Concepts 37 15. (a) (c) 3% 1 50 3 40 (b) (b) (d) (b) (d) (b) (d) 37.000.800 G: £2.0935 0. (a) (c) (e) 2.15 20% 2.000 (f) © ABE and RRC . (a) (c) (e) 2.000 Y: £20.5% 2 25 9 50 £30 35 bottles £24.000.700 Questions for Practice 10 1.000000000816 7. 2.34 102 6.42 103 8.33% 150% Questions for Practice 9 1.000 (b) (d) (f) (b) (d) 4.000 (b) F: £1.700 H: £2.5% 90° £483.653.453 108 468 1. (a) (a) (c) 3.30 33. (a) (c) (e) 4.032 104 3. Firm A: 1 : 4 Firm B: 1 : 9 Firm C: 1 : 3 Firm D: 1 : 5 2.38 Basic Numerical Concepts Questions for Practice 8 1.853. (a) X: £15.5 102 5.0000602 0.000 Z: £25.792 1015 0. Powers and Roots Brackets in Algebra Solving Equations Equality of Treatment to Both Sides of the Equation Transposition Equations with the Unknown Quantity on Both Sides Formulae Formulating a Problem Constants and Variables Substitution Rearranging Formulae Page 40 40 40 42 42 43 43 44 45 45 48 50 50 52 54 55 55 56 56 57 59 B. C. Answers to Questions © ABE and RRC . D.39 Study Unit 2 Algebra. Some Introductory Definitions Algebra Equations Formulae Algebraic Notation Addition and Subtraction Multiplication Division Indices. Equations and Formulae Contents Introduction A. Finally. A. There are formulae which describe a great many physical situations or mathematical problems and you will encounter these throughout your studies. algebra enables general expressions to be developed and general results to be obtained. Algebra essentially involves the substitution of letters for numbers in calculations. in the previous expression. but less than 6 – whereas a letter can be used to represent any number. and these will be examined in full. There are various rules and procedures which can be applied to equations to help this process. that we substitute letters for the first two numbers. In this sense then. We all know that 2 3 5. It enables us to work out an unknown quantity or value provided we know certain specific quantities or values. we use letters to represent numbers. a number is a specific quantity – for example. both in this course and many others. that sets out a rule which may be applied in particular situations. but later in the course we shall consider more complex forms. A formula is one particular form of generalised mathematical statement. usually expressed in the form of an equation. Suppose. we look briefly at formulae. so it is important to understand these at the outset. instead of using numbers. and is generally used to enable one or more unknown quantities to be worked out. so that: 2a 3b We can then write: ab5 All that has happened is that we have replaced the numbers with letters. 5 is more than 4. An equation is simply a statement of equality between two mathematical expressions. We shall then go on to introduce the subject of equations. Objectives When you have completed this study unit you will be able to: outline the basic principles of algebra apply the basic arithmetic operations of addition. However. and simplify algebraic expressions by the process of collecting like terms define equations and formulae and outline their uses find unknown quantities and values from simple equations by using transposition outline the principles of formulae and how they are constructed rearrange the terms in a formula to isolate different unknowns. subtraction. the same basic rules are applied. SOME INTRODUCTORY DEFINITIONS Algebra Algebra is a branch of mathematics in which. though. Here we shall be concerned with "simple" equations. Thus.40 Algebra. However. so that we can establish rules and procedures for carrying out mathematical operations which can be applied whatever the actual numbers involved. multiplication and division to algebraic notation. Equations and Formulae INTRODUCTION In this study unit we continue our review of basic concepts by looking at algebra. "a" could be 4 and "b" © ABE and RRC . Example 1: Suppose you have a piece of wood which is 7 metres long and from it you wish to cut a piece 4 metres long. this is a general expression which can be made specific by putting in particular values for "l" and "w". Substituting the letters "l" and "w" to represent the actual length and width. Example 2: To find the area of a floor measuring 10 m long and 9 m wide. Arithmetic is specific. letters may be substituted to give us a general expression: std where: speed s mph time t hours distance travelled d miles. The length of the remaining piece is 3 m. Equations and Formulae 41 could be 1. calculated as follows: 743 This is a specific arithmetic statement relating to cutting a specific amount from this particular piece of wood. we multiply one dimension by the other: area 10 9 square metres 90 sq metres. we can reformulate the expression as: area l w Again. whereas arithmetic (using numbers) uses definite numbers and gives definite results. The calculation can now be shown as: xy3 This is now a general statement for cutting one length of wood from another to leave a piece 3 metres in length. We only know that they are 2 and 3 respectively because we defined them as such before. The main consequence of this is that algebra uses general expressions and gives general results. Let us consider some examples to illustrate this further. © ABE and RRC . Example 3: The distance travelled by a train in 3 hours at a speed of 60 miles per hour is easily calculated as: 3 30 180 miles. We could translate this into an algebraic expression by substituting letters for the specific lengths: let the original length of the piece of wood x metres and the length of the piece cut off y metres. whereas algebra is general. Here again.Algebra. Don't worry. The two sides must always be in equality for the statement to be an equation. we have the makings of an extremely useful mathematical tool. This is not particularly helpful in itself. whatever the actual values. the statement that "2 2 4" is an equation. Note that an equation has two sides. speed etc. Thus. width. equations can get very complicated. work out the unknown) very easily.e. To save writing "the required number" (or the " unknown" number). A certain number is multiplied by 4 and the result is 20. It provides a set of rules which can be used in a particular situation in order to solve the problem. If we now introduce the concept of algebra into this. x 4 20 x 4 20 x45 x31 The correct equations are: The point of constructing equations in this way is that it provides a means by which you can work out the unknown value (here. If a certain number is divided by 3 the result is 1. © ABE and RRC . However. Equations and Formulae Thus. we can write this as: 3 the required number 6. Equations An equation is simply a mathematical statement that one expression is equal to another. using algebra – working with letters instead of numbers – allows us to construct general mathematical expressions. For example. are. So. but is very important when we come to consider equations. Here. though – there are a number of simple rules which can be used to help solve them. "x"). they are "2 2" and "4". Formulae A mathematical formula (plural "formulae") is a special type of equation which can be used for solving a particular problem. it is more convenient to call it "x". we have already introduced two formulae in the preceding section on algebra: area length width distance travelled speed time. The essence of these two statements is that they are always true. whatever the actual values of length. so that we can then write: 3x6 Here are a few simple statements which you can easily write as equations for practice: (a) (b) (c) (d) (a) (b) (c) (d) A certain number is added to 4 and the result is 20. If 4 is taken from a certain number the result is 5. a formula is an equation which always applies to a particular mathematical problem. given that a certain number multiplied by 3 is 6.42 Algebra. These examples are very simple and you can probably solve them (i. for example. In fact. the operations of addition. by using the general rules governing equations. Thus. just as: 33333336 so y y y y y y y 6. all the terms to be added above were "y". we could put them into the equation. You will meet a good many in your studies – principally in this subject and in financial subjects. can easily be used to solve the problem faced. subtraction. we can easily work out the distance travelled by substituting 60 for "s" and 2 for "t" in the above formula: d st 60 2 120 miles. Again. dst where: speed s mph time t hours distance travelled d miles. or 6y You should note that this holds good only for like terms. do not worry too much about them. if we know we have been driving for 2 hours at an average speed of 60 mph. ALGEBRAIC NOTATION As algebraic letters simply represent numbers. we would write simply "ab" (or sometimes "a. They are usually quite straightforward and. multiplication and division are still applicable in the same way. There are many formulae used in mathematics and other physical sciences such as chemistry or physics. Having a formula for a particular problem set out as a general equation using algebra enables us to work out the answer by substituting actual values back into it for the unknowns. instead of "a b". However. but if not we would have to treat them as unknown. © ABE and RRC . For example. Addition and Subtraction Addition and subtraction follow the same rules as in arithmetic.Algebra. B. we can formulate a general statement using algebra as follows: Alw where: A area l length w width. So. using the full stop to represent multiplication). Thus. in algebra it is not always necessary to write the multiplication sign. Equations and Formulae 43 If we know the actual values.b". Like terms are those of the same symbol or value – for example. 44 Algebra, Equations and Formulae When unlike terms are to be added, we can simplify an arithmetical statement, but not an algebraic statement. For example: 3 6 7 16 yxzyxz The three numbers, when added, can be reduced to a simpler form, i.e. 16. However, there is not a simpler means of expressing "x y 3". We can, though, simplify expressions when they involve like terms. Just as the result of adding two numbers does not depend upon their order, so the result of addition in algebra is not affected by order. Thus: 27729 and x y 3 3 y x y 3 x, etc. In simplifying an algebraic expression, we can collect like terms together whenever possible. For example: 5x 2y x 3y 2 (5x x) (2y 3y) 2 4x 5y 2 We have collected together all the "x"s (5x x) and all the "y"s (2y 3y), but cannot do anything with the "2" which is unlike the other terms. This distinction between like and unlike terms, and the way in which they can be treated, is an important point. Multiplication As we have seen, the product of a number and a letter, for example 3 a, may be written as 3a. Always place the number before the letter, thus 3a, not a3. Since multiplication by 1 does not alter the multiplicand, the expression 1a is not used; you just write a. To multiply unlike terms, we simply write them together with no sign. Thus: a b c d abcd If numerical factors are involved then they are multiplied together and placed in front of the simplified term for the letters: 3a 2b 4c d 24abcd That this is correct can be shown by writing it in full and changing the order of multiplication: 3a2b4cd 324abcd 24abcd Multiplying like terms (for example, x x) gives a result of the term being raised to the power of the multiplier (here, x2). We shall consider this in more detail later. © ABE and RRC Algebra, Equations and Formulae 45 Division Division should present little difficulty if we follow the rules used in arithmetic. Consequently, just as we write: 4 for 4 2 2 so we can write: x for x y y Remember that any number goes into itself once. This will prevent you from saying that 6 6 0 instead of 6 6 1. In the same way x x 1. It is then easy to cancel algebraic fractions. For example: 2xy 2y can be cancelled by dividing both the numerator and denominator by x to give 4x 4 2y can be further cancelled by dividing both the numerator and denominator by 2 to 4 y give 2 2xy or we could simply say we can cancel by dividing both parts by 2x. 4x We can always cancel like terms in the numerator and denominator. Indices, Powers and Roots You will remember that indices are found when we multiply the same number by itself several times. The same rules apply in algebra, but we can investigate certain aspects further by using algebraic notation. To recap: xx xxx xxxx and so on. In, for example, x5, x is termed the base and 5 is called the exponent or index of the power. Note two particular points before we go on: x x1 x0 1 (a) Multiplication of indices You will note that, for example: x2 x4 (x x) (x x x x) xxxxxx x6 x2 4 is written as x2 is written as x3 is written as x 4 (referred to as "x squared") (referred to as "x cubed") (referred to as "x to the power four" or "x raised to the fourth power") © ABE and RRC 46 Algebra, Equations and Formulae and x3 x5 (x x x) (x x x x x) xxxxxxxx x8 x3 5 There is in fact a general rule that, when we have two expressions with the same base, multiplication is achieved by adding the indices. Thus: xm xn xm + n (b) Division of indices When we divide expressions with the same base, we find we can achieve this by subtracting the indices. For example: xxxxxx x6 3 xxx x (Rule 1) xxx x3 x6 3 and x2 xx x x x x1 x2 1 The general rule is: xm xm n n x (Rule 2) (c) Showing reciprocals In Rule 2 above, let us put m 0, so that the expression becomes : x0 x0 n n x We know that x0 1, so 1 x0 n x n n x Thus, negatives denote reciprocals. (d) Raising one power by another To work out an expression such as (x2)3 – i.e. x squared all cubed – we have x2 x2 x2 as it is the cube of x2. © ABE and RRC Algebra, Equations and Formulae 47 Thus (x2)3 x2 x2 x2 xxxxxx x6 x2 3 In general, we have (xm)n xmn (e) Roots Indices do not have to be whole numbers. They can be fractional. Let us consider x½ (i.e. x raised to the power half). We know from Rule (3) that: (x½)2 x½ 2 x Thus, x½ when multiplied by itself gives x. However, this is exactly the property that defines the square root of x. Therefore, x½ is the square root of x which you will sometimes see denoted as: 2 (Rule 3) x or, more commonly, as x. Therefore, x½ 1 x . (The symbol "" means "identical with".) 1 Similarly, ( x 3 )3 x. Thus, x 3 is the cube root of x which is denoted by the symbol Similarly, x¼ is the fourth root of x denoted by Thus, in general, we can say that: x 1 n 3 x. 4 x. is the nth root of x or n x. Note that in all these expressions the number to the left of the root sign – for example the 3 in 3 x – is written so that it is level with the top of the root sign. A number written level with the foot of the root sign – for example, 3 x – would imply a multiple of the square root of x – in this example, 3 times the square root of x. We shall often need to use square roots in later work, particularly in statistics and in stock control. (f) Roots of powers There is one more type of index you may meet – an expression such as: or x 3 , or in general x n . To interpret these, remember that the numerator of a fractional index denotes a power and that the denominator denotes a root. Thus, x 3 2 5 m is (x3)½, i.e. the square root of (x cubed) or 2 x 3 . Alternatively, it is (x½)3, i.e. the cube of (the square root of x) or ( x )3. Let us consider its value when x 4: 4 3 2 2 43 444 64 8 (since 8 8 64) or 4 3 2 ( 4 ) 3 23 2 2 2 8 Therefore it does not matter in which order you perform the two operations, and we have: x m n n x m ( n x )m © ABE and RRC 48 Algebra, Equations and Formulae (g) Collecting like terms We saw earlier how we can collect like items. Let us look at a couple of examples where indices are involved. It is no more complicated, even though it may look that way. a2 2a2 a2 a2 a2 3a2 These are like terms and can be added. b2 2b2 2b3 b2 b2 b2 b3 b3 3b2 2b3 Note that b2 and b3 are not like terms, so they cannot be combined. Brackets in Algebra The rules for brackets are exactly the same in algebra as in arithmetic. However, if we have unlike terms inside the brackets, it is not possible to collect them together before removing the brackets. In such cases, we need to be very careful when removing them and always obey the following three rules If the bracket has a term (a number or a letter) in front of it, everything within the bracket must be multiplied by that term. If a bracket has a sign before it, or before the term by which the terms inside the brackets are to be multiplied, the signs or within the brackets remain unchanged. If a bracket has a sign before it, or before the term by which the terms inside the brackets are to be multiplied, the signs or within the brackets are changed so that becomes and vice versa. These rules need a little explanation, so let us consider some examples. Example 1: 4x 2(x y) The terms inside the brackets are unlike and cannot be further collected. All the terms within the brackets must be multiplied by the term outside before removing the brackets, so this gives: 4x 2x 2y 6x 2y Example 2: x (y 3z) Again, the terms inside the brackets are unlike and cannot be further collected, so we can proceed to remove the brackets as the next step. Since there is a sign before the brackets, we must reverse the signs of the terms inside when removing them. Thus: x y 3z We can clarify this by imagining that there is an unwritten "1" before the bracket, making the expression effectively: x 1(y 3z) So "1" is multiplying the whole bracket: x 1 (y 3z) © ABE and RRC Algebra, Equations and Formulae 49 This gives us three separate terms: x, 1 y, and 1 3z Thus, we have: x y 3z Example 3: 3a(a 3) 4b(b 4) The first bracket is multiplied by 3a and the second by 4b. This gives us: 3a2 9a 4b2 16b Note that the sign before the second bracket changes the sign inside to . Often we find it necessary to insert brackets into an algebraic expression. This is known as factorisation because, when we include the brackets, we place any common factor outside the brackets. Thus: x 5y 5z x 5(y z) x 5y 5z x 5(y z) Notice how we must keep an eye on the signs as well. Questions for Practice 1 Simplify the following expressions, writing "no simpler form" if you think any cannot be further simplified: 1. 3. 5. 7. 9. 11. 13 15. 17. aaaa 3x 2x x 5x 3x x 3a 2a x y xy a 2b 3a 4b 5 x + 2(y + z) x (y 3z) a(b c) b(c a) 2. 4. 6. 8. 10. 12 14. 16. 18. ab ab 3x 2y y 7xy 2x y aaba3 xxxyyy 3x (x 3) 4 3(2x y) (x y) 4x 2(x y) 3a(a 3) 4b(b 4) Now check your answers with those given at the end of the unit. © ABE and RRC we mean finding the value of the unknown or unknowns using the other numbers in the equation. then we get: x4464 x 10 (b) The same number may be subtracted from both sides of the equation For example: x 6 18 If we deduct 6 from each side. we can see a number of possible operations which may be carried out without changing the equality of the equation. Looking at the rule in more detail. anything we do to the equation must maintain this equality. you must also do to the other. So. What you do to one side of the equation. or 10 3 3 © ABE and RRC . For example: 1 3 of a number 10. in: 4x 20 the value of the unknown (x) can be discovered because of the stated relationship between 4 and 20. these are based on the simple rules which we set out here. In a later study unit we shall examine more complex equations – those with two or more unknowns – and consider additional rules. Equality of Treatment to Both Sides of the Equation This is the golden rule that you must always follow when handling equations and there are no exceptions. SOLVING EQUATIONS When we talk about solving an equation. There are a number of rules which can be used to help solve equations and we shall consider these next in relation to equations which have only one unknown. You can easily see that if 4 times the unknown 20 then the unknown must be 5. For example. However. (a) The same number may be added to both sides of the equation For example: x46 If we add 4 to each side. or 1 x x 10. Equations and Formulae C. Another way of thinking about this is that an equation is like a balance. or take an equal weight from each side. Remember that an equation has two sides and those two sides must always be in equality for the whole statement to remain an equation. You can add an equal weight to each side.50 Algebra. we get: x 6 6 18 6 x 12 (c) Both sides of the equation may be multiplied by the same number. and it will remain "in balance". If the weights of a balance are equal on both sides it is "in balance". © ABE and RRC . 2. we get: x 3 10 3 3 3x 30 3 x 30 (d) Both sides of the equation may be divided by the same number. 3. x. writing "no simpler form" if you think any cannot be further simplified: 1. on one side of the equals sign. Equations and Formulae 51 If we multiply both sides by 3.Algebra. For example: 2 times a number is 30. we get: 2x 30 2 2 x 15 You should have noticed that the object of such operations is to isolate a single unknown. 5. x 7 21 x 3 26 x + 9 21 x 11 5 x 7 8 3x 30 Now check your answers with those given at the end of the unit. 4. 6. Questions for Practice 2 Simplify the following expressions. or 2x 30 If we divide both sides by 2. a divisor may be transposed from one side of an equation by changing it to the multiplier on the other. Similarly. This is done so as to isolate an unknown quantity on one side. Equations and Formulae Transposition When you understand the rules in the previous subsection you may shorten your working by using transposition instead. A multiplier may be transposed from one side of an equation by changing it to the divisor on the other. Consider the following examples: © ABE and RRC .52 Algebra. or from to . Consider the following examples: (a) x46 Transfer the 4 to the right-hand side of the equation and reverse its sign: x64 x 10 (b) x 6 18 Transfer the 6 to the right-hand side and reverse its sign: x 18 6 x 12 (c) 12 x 25 Transpose the 12 and reverse its sign: x 25 12 x 13 (d) 18 x 14 First transpose the x and reverse its sign: 18 14 x Then transpose the 14 and reverse its sign: 18 14 x x4 Note that we can keep transposing terms from one side to the other until we have isolated the unknown quantity. You must observe the following rules: An added or subtracted term may be transposed from one side of an equation to the other if its sign is changed from to . Transposition is a process of transferring a quantity from one side of an equation to another by changing its sign of operation. move 4 across the sign and change it from divisor to multiplier: x 12 4 x 48 (b) 2x 20 To get x by itself.Algebra. © ABE and RRC . we transpose the multiplier (6) and change it to a divisor: 24 x 6 x4 This rule is called cross multiplication. Equations and Formulae 53 (a) x 12 4 To get x by itself. to get x by itself. we need to transpose x and change it from divisor to multiplier: 24 6x Then. 4. 2. transpose the 2 and change it from multiplier to divisor: x 20 2 x 10 (c) 24 6 x In this case. 3. x 36 6 72 24 x 3x 3 21 5x 25 2x 3 15 21 x 15 Now check your answers with those given at the end of the unit. 5. Questions for Practice 3 Find the value of x by transposition: 1. 6. We simply keep transposing terms as necessary until we have collected all the unknown terms on one side of the equation. and thus isolating x on the left: x 14 2 x7 Questions for Practice 4 Find the value of x: 1. Equations and Formulae Equations with the Unknown Quantity on Both Sides These equations are treated in the same way as the other equations we have met so far. 2. 3. 4. changing its sign: 3x x 8 6 2x 14 Finally. consider 3x 6 x 8 First move the 6 to the right-hand side.54 Algebra. 5. For example. © ABE and RRC . making it 6: 3x x 8 6 Then transpose x from right to left. x 3 8 2x 2x 24 x 1 2 x 15 x 5x 3 27 2x 18 x 5x 2 2x 24 6x 3 Now check your answers with those given at the end of the unit. move the 2 to the right-hand side changing it from multiplier to divisor. 6. The easiest way to explain this is to give an example. Equations and Formulae 55 D. and then cycles at a certain number of kph for a certain number of hours. taking into account the time over which the loan is repaid. Formulating a Problem When a person borrows money.Algebra. the calculation is as follows: Interest the principal multiplied by the percentage rate of interest. we can say he covers (xy pq) kilometres in (y q) hours. it is possible to develop formulae to model any situation involving mathematical relationships. If we want to work out the amount of interest repayable on a particular loan. FORMULAE A formula is a mathematical model of a real situation. For example. Therefore. multiplied by the number of time periods the interest is to apply. what we have done is express the real situation as a mathematical model using an algebraic equation. whilst others describe key mathematical relationships which you will meet time and again throughout the course. We could model this situation by using algebraic notation and then show the whole calculation as an equation: let:: I simple interest P principal r rate of interest (%) n number of periods then: IP r n 100 This equation is the formula for calculating simple interest. he or she has to pay interest on the loan (called the "principal"). What is his average speed? let: x the speed at which he runs (in kph) y the number of hours he runs p the speed at which he cycles (in kph) q the number of hours he cycles. However. Thus. consider the following situation: A man runs at a certain number of kilometres per hour for a certain number of hours. The amount of interest payable is determined by the annual rate of interest charged on the money borrowed. provided of course that we know all the others. This particular formula is very important in financial studies and can be applied in many situations. It can be used for any simple interest calculation and enables us to find the value of any one of the elements. Some are just useful for helping to understand how to work out a particular problem or situation you are faced with. then: average speed xy + pq kph y+q © ABE and RRC . e. in the formula for simple interest. we said "let x the speed at which he cycles (in kph)". The letters on the other hand are variables.000 invested at 10% per annum (i. If she sells the mixture at v pence per kg. Doing this is the process of substitution. does she make? Bring your answer to £. The last step is to test the formula by going back to the problem and check that it works by using real number values in the formula through the process of substitution. no matter what the values given to the P. we have to give the variables the values which apply in that particular situation. ensuring that all terms are expressed in the same units. see if you can work the following problem out by developing a formula. The numbers in a formula always stay the same and are known as constants. for example. numbers. Now check your answer with that given at the end of the unit. it only represents that quantity and the units must be exactly specified. what profit. Activity As practice. the constant will always be 100. For example. The value that they have in working out the formula in any particular situation may vary – it depends on the particular circumstances of the problem. Constants and Variables As we have seen. consider the following example: What is the simple interest on £1. in pence. Substitution We defined a formula as a mathematical model of a real situation.56 Algebra. then. A grocer mixes m kg of tea at x pence per kg with n kg of tea at y pence per kg. to have said "let x the speed at which he cycles" would have been inaccurate. Turning back to the formula for simple interest. These are combined together in particular ways to express the mathematical relationships between the terms such that a particular result is obtained. r and n in a particular situation. So. are as follows: select a letter to represent each of the quantities required convert each block of information from the problem into an algebraic term or statement. In order to use that model in a real situation. sometimes. Equations and Formulae Note that when we use a letter such as x to represent an unknown quantity. We shall look at substitution shortly. using the letters selected form an equation by combining the symbolic terms. The steps to take in formulating the problem. formulae consist of letters (algebraic notation) and. per year) for two years? First of all we should restate the formula: IP r n 100 © ABE and RRC . Note that the unknown quantities have already been given letters. but by using the rules we have discussed in relation to simple equations. Equations and Formulae 57 Now we substitute the values from the problem into the formula: P £1. In the next study unit you will get much more practice in this. : 100 100 To find n. So for example.000 10 2 100 £1. we should note that a formula can be rearranged to find a different unknown quantity. we may wish to know how long it would take to earn a specified amount of interest from a loan made to someone else.000 10 20 100 £200 This explains the principle of substitution. Rearranging Formulae Finally. we can rearrange the formula so that it does. We can do this by dividing both sides by P Pr n I 100 Pr Pr 100 100 n I Pr 100 Pr 100 I 100 Pr nI n 100 I Pr © ABE and RRC .Algebra. as it may also be expressed. Again. we need to rearrange the formula so that n is isolated on one side of the equation. The way in which the formula is expressed at present does not allow us to do that. we start by restating the original formula: IP r n 100 Note that we can also express this as: I P rn 100 r Pr or.000 r 10 n2 So: I £1. when considering simple interest. at a particular rate of interest. 2. To find P – the principal which would provide a certain amount of interest at a particular rate of interest over a specified period. Questions for Practice 5 Rearrange the simple interest formula to allow the following calculations to be made: 1. © ABE and RRC . Now check your answers with those given at the end of the unit. However. Equations and Formulae Don't worry too much about the mechanics of this for now – again you will have plenty of practice at these types of operation as you work through the course. see if you can rearrange the simple interest formula to isolate the other two unknowns. To find r – the rate of interest which would have to be charged to earn a particular amount of interest on a specified principal over a certain period of time.58 Algebra. 3x 3y 4a 4b 5 3x x 3 2x 3 x 2y 2z 4 6x 3y x y 4 5x 4y x y 3z 4x 2x 2y 6x 2y ab ac bc ab 2ab ac bc 3a2 9a 4b2 16b Questions for Practice 2 1. 15. x 3 24 18 . 4a 2ab 6x 3x 3y 3x No simpler form. a 3a b 3 No simpler form. 5. 2. x 28 x 29 x 12 x 55 x 56 x 10 Questions for Practice 3 1. 18. 72 x. 5. 10.x6 3 3x 21 3.Algebra. 9. 17. 3. 12 13 14. 2. 3. 11. 2. 7. 6. 8. 4. 16. 3. 4. 6. x 36 6 218 72 24x. x © ABE and RRC . Equations and Formulae 59 ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. we divide both sides by rn P I 100 rn rn 100 100 I P rn 100 PI So 2. x 8 1 2 5 2 1 3 3 x x 15. x 21 15 x. start with the original formula IP To find r P rn n. 4. 3. we divide both sides by P n: 100 © ABE and RRC . x 10 18 2 5x x. 21 15 x. x 2x 8 3. or 100 100 To isolate P. x 5 24 6x 2x 1 1 3 72 9 5 x. We always start by restating the original formula: IP r P rn n or I 100 100 r rn n. x 24 5 24 4½ 3 3 3 16 16 2 Questions for Practice 5 1. Equations and Formulae 4. 6. or I 100 100 r .60 Algebra. x 25 5 5 18 9 2 2x 15 3. x 2x x 24. rn 100 I 100 rn 100 I rn P Once again. x 10 5x 2x 27 3. 6. x 6 Questions for Practice 4 1. 5. 5. 2. which is what we want to find 100 so r% I Pn Activity The solution would be as follows: profit selling price cost price selling price quantity sold price v(m n) pence cost price mx ny pence profit v(m n) (mx ny) pence To convert this to £. Equations and Formulae 61 r Pn I 100 Pn Pn I r Pn 100 r r%. we simply divide the result by 100: profit £ v (m + n) (mx + ny) 100 © ABE and RRC .Algebra. Equations and Formulae © ABE and RRC .62 Algebra. Answers to Questions for Practice © ABE and RRC .63 Study Unit 3 Basic Business Applications Contents Introduction A. Currency and Rates of Exchange Rates of Exchange Currency Conversion Discounts and Commission Discounts Commission Simple and Compound Interest Simple Interest Compound Interest Depreciation Straight-Line Depreciation Reducing Balance Depreciation Pay and Taxation Make Up of Pay Taxation Page 64 64 64 66 68 68 68 70 71 72 76 76 77 79 79 81 84 B. E. D. C. changes in government policies. one unit of British currency (£1) might cost 2. These values vary from day-to-day and are quoted in most daily papers. the principal. the relationship between the value of exports and imports. To give you some idea of the scale of changes over time. etc. The intention here is not to go into detail about the nature of these problems. For example. Such value is measured by how much one unit of currency costs in another currency. and present British currency comprises the pound sterling (£) and the penny.64 Basic Business Applications INTRODUCTION In this study unit we shall apply some of the numerical concepts introduced in the previous two units to some of the basic problems encountered regularly in the business world. which is one-hundredth of a pound. Thus. taking into account non-taxable items and personal tax allowances.e. CURRENCY AND RATES OF EXCHANGE Each country has its own system of coinage and this is known as its currency.11). They are invariably based on the decimal number system – for example. The process of finding the value of the currency of one country in terms of the currency of another country is called exchange. in most cases. At first sight. The relative values of the different currencies used to depend upon the amount of gold which a country possessed. you should not find them difficult in practice. applying the correct formulae and setting out the calculations clearly. but to concentrate on the principles of calculation involved in solving them. before the Second World War the value of the dollar was about 21p (i. Rates of Exchange You are no doubt familiar with various currency systems. America uses the dollar ($) and the cent. The changes in value are brought about by such things as changes in government. and use them to calculate any of the amount of interest payable (or receivable). Europe has the euro (€) and the cent. given information about the other variables explain depreciation and calculate it according to the straight-line method and the reducing-balance method outline the way in which gross pay is made up and the main deductions made from gross pay in order to leave net or take-home pay calculate the income tax payable on pay. $1 £0.11 units of US currency ($2. the rate of interest or the time period covered. some of these business applications may seem quite hard to work out. Objectives When you have completed this study unit you will be able to: explain the rate of exchange between different currencies and convert values in one currency into another explain discounts and commissions and calculate them from given information state the formulae for simple interest and compound interest. which is onehundredth of a euro. whereas at the time of writing (2008) it is © ABE and RRC . their values depend on the political and economic circumstances of the countries concerned. This relationship of one currency to another is known as the rate of exchange. the national budget.21). but if you follow the logical steps to breaking down the problem shown here. we shall use very simplified situations and limited amounts of information. A. but now that we are no longer on the gold standard. which is equivalent to one-hundredth of a dollar. one side of the ratio is always "1". or how much a particular amount of one currency is worth in another. undertaking the exchange to make a profit.) As an example of the two different prices.0076 Swiss francs £1 $1. It shows the value of one unit of currency in terms of another currency. Thus. and a small change in the value of those currencies can produce substantial losses or gains.00 Swiss francs or 1 Swiss franc £0.286 (euros) Selling rate 2. their effects can be very large when you consider that large amounts of money could be involved.50). rather than "x : y".e. $1 £0. (The difference between the two enables the bank or bureau de change etc. although it is a ratio. the changes are very small. and vice versa. although they can take place quite rapidly. Here are some examples of typical rates of exchange of the British pound: £1 2. Companies may have many millions of pounds sterling invested in overseas currencies. For any relationship between two different currencies. although the changes may be very slight. there are two further points to note about the rate of exchange. the rate of exchange can be expressed in two ways – how much one unit of one currency is worth in terms of the other.7 Japanese yen Note that. The selling price determines the amount of foreign currency you will be sold for your pounds.Basic Business Applications 65 about 50p (i.9832 (US dollars) £1 €1.98 (US dollars) £1 €1. the rates quoted might be: Buying rate £1 2.9830 (US dollars) €1.43 (euros) £1 237. and the buying price determines the amount of pounds you will receive in return for your foreign currency. Even then.00 Swiss francs £1 $1. However. In general.50 The rate of exchange is usually quoted in two forms – a selling price and a buying price. such large changes usually only occur over a long period or because of some particular economic problem. depending on whether the value of the currency rises or falls against the pound. Thus. Before we look at the relevant calculations involved in converting one currency to another.0069 Swiss francs $1. the rate of exchange between the British pound and the Swiss franc could be shown as: £1 2. The rate of exchange is expressed as a ratio.2845 (euros) © ABE and RRC . the rate of exchange is usually shown as "x y". It is this ratio that determines how much of one currency can be bought with another currency. we simply divide each side of the equation by e – the rate of exchange: y xe e e x y e We can now apply these formulae to any given problem. but we round down to express it correct to two decimal places as required for currency.275 euros if the rate of exchange is £1 1. £x) into the second currency (say $y).37 1. Example 1: What is the cost in £ of an invoice for €1. The rate of exchange is expressed as 2.98. Substituting in the first formula (to find y) we get: y SF 63. we would substitute the following values into the formula: x 100 e 1. again x is the amount in pounds and y the amount in Swiss francs.02 units of the second currency (SF) is equivalent to 1 unit of the first (£).40 in Swiss francs if the rate of exchange is £1 SF2. dollars) equivalent to one unit of a first currency (e.66 Basic Business Applications Currency Conversion To find the value of a sum of money in one currency in terms of another currency. to convert £100 into $ at the rate of £1 $1. To convert a certain amount of the first currency (say.068 128. Thus the formula would be: yxe So. © ABE and RRC . euros are the second currency and pounds are the first. we could develop a simple formula using the rate of exchange. Let: e rate of exchange expressed as the number of units of a second currency (e.02 128. So. Substituting in the second formula (to find x) we get: £x £ 1.g. To do this. we multiply the amount of the first currency by the rate of exchange.98 £1) x amount in first currency y amount in second currency.98 $198 If we wanted to know how much a particular sum of the second (say $y) currency would be worth in the first (say £x) currency. we would have to rearrange the above formula to isolate x. So.40 2.07 Swiss francs (rounded up). as in $1.29 (Note that the actual result of the calculation is £988.) Example 2: Find the value of £63.275 £988.g.02. pounds.29 units of one currency is equivalent to 1 unit of the other.37209.98 The formula gives us the following calculation: $y 100 1. and x is the amount in pounds and y the amount in euros. as follows.29? The rate of exchange is expressed as 1. 05 Swiss francs to the pound and 0.03 (rounded down). again x is the amount in pounds and y the amount in dollars. While in Switzerland they spent 720 Swiss francs and then changed the remaining francs into euros as they crossed the border into Italy. The exchange rate was 1.000 £512. 4.05 kroner to the pound and sells at 9. John and Paul ate a meal in a restaurant whilst on holiday in Cyprus. © ABE and RRC . (a) (b) Mr Watson changed £100 into Swedish Kroner for a trip to Sweden. (a) (b) 3. the Jones family changed £800 into Swiss francs. While travelling they used 200 litres of petrol which cost €1. Substituting in the second formula (to find x) we get: £x £ Example 4: Convert 100.507.70. So.3 yen.000 to pounds given that £1 $1. Before going on holiday to Switzerland and Italy. how much did the petrol cost them in pounds? what was the price per litre of the petrol in pence? 2.29 euros to the pound.01 $y 250 2. £x £ Example 5: Find the value of £250 in dollars if the exchange rate is £1 $2. A bank buys Swedish kroner (SEK) at 10. The meal for two cost £ (Cyprus) 35. The rate of exchange is expressed as 1. Calculate the cost of the meal in pounds.35 per litre.3 3.99. The exchange rate was 0.54 (rounded up). The procedure should be becoming clear to you now.000 Japanese yen to pounds given that £1 195.Basic Business Applications 67 Example 3: Convert $3.01 £502. Calculate: (a) (b) the number of Swiss francs they received the number of euros they bought.99 Questions for Practice 1 1. The Smiths spent a holiday touring in France.50 100.000 £1.7888 £ Cyprus to the British pound.642 euros to the Swiss franc. How much money did he lose because of the cancellation? Now check your answers with those given at the end of the unit. The exchange rate was 2. 1.45 kroner to the pound. 195.99 units of the second currency ($) is equivalent to 1 unit of the first (£). How many Kroner did he receive? Sadly the excursion was cancelled and so he changed all his Kroner back into pounds. 400 12. but was sold at a discount of 15%. What is the net price? discount £720 15 £108 100 net price £720 £108 £612 Example 2: A car was listed at £3.400 £6. Let us look in more detail at their exact meanings and the way in which they are calculated. DISCOUNTS AND COMMISSION You've probably heard the terms "discount" and "commission".600 20 £720 100 net price £3.68 Basic Business Applications B. Consider the following example. To find the commission.5 25 £6. She is paid commission of 12½%. broker or salesman for buying or selling goods or services.600 £720 £2. commission 12½% of £6. To calculate a discount (D). Example 1: A three-piece suite was listed as £720. Discounts A discount is a reduction in the selling price of an item. B) by the rate of the discount (R) and divide by 100. Calculate the amount due to her. What do they mean to you? Discounts and commissions are a practical use of percentage calculations. Thus: DB R 100 Consider the following examples.600 but was sold for cash less a discount of 20%. What was the final selling price? discount £3. This is expressed as a percentage of the value of the goods/services traded and is called the rate of commission. The discount is expressed as a percentage of the list price (the list price is the original price) and what the buyer actually pays is the net price – the list price minus the discount. we simply multiply the value of the goods/services traded (known as the principal) by the rate of commission.800 Commission A commission is the amount of money paid to an agent.400 £800 100 200 © ABE and RRC . Pauline obtained £6. usually offered as an inducement for quantity buying or for prompt payment. you use the following rule or formula: multiply the list price (or base.400 of new business for the PR agency for which she works. the commission payable on converting £200 into $ traveller's cheques is either £3 or £2 (1% of £200).47.000? Calculate the commission earned by a shop assistant who sold goods to the value of £824 if the rate of commission is 3%. In these cases. the person exchanging the currency or buying traveller's cheques is charged either: a fixed cost – £3 per transaction or a variable cost – the value of 1% or 2% of the amount of the transaction. 4. how many dollars would you get when converting: (a) (b) £100 to American dollars in cash? £200 to American dollars in cash and in traveller's cheques? In each case. and given that £1 $1. The commission charged on traveller's cheques is commonly £3 or 1% whichever is larger. The fixed rate is the larger so that would be applied: £100 £3 £97 £97 1. We shall consider fixed and variable elements in more detail later. The fixed rate is the larger so that would be applied: £200 £3 £197 £197 1.47 $289.12 However. and then apply whichever is the larger. travel agents and bureaux de change to cover their costs when carrying out a currency exchange.5% of the value of each house she sells.000 car with a discount of 15%? An estate agent charges a commission of 1.47 $288. The commission on changing cash is commonly £3 or 2% whichever is larger. When going on holiday. 3.50 (b) The commission payable on converting £200 into dollars cash is either £3 or £4 (2% of £200). but consider the following example in relation to currency exchanges.47 $142. most people take a limited amount of cash in the currency of each country to be visited and take the remainder that they will need in the more secure form of traveller's cheques. we would need to work out the commission at both the fixed and the variable rate. © ABE and RRC . The variable rate is the larger so that would be applied: £200 £4 £196 £196 1.59 Questions for Practice 2 1. (a) The commission payable on converting £100 into dollars cash is either £3 or £2 (2% of £100). What is the price paid for a £310 washing machine with a discount of 10%? What is the price paid for a £7. Using the rates of commission quoted above. 2. How much commission is earned by selling a house for £104.Basic Business Applications 69 Commissions are also usually charged by banks. this is the bank or building society paying interest to you on money which you have loaned to them. building society. Note that when you receive interest on savings from a bank or a building society. a family of four decided to take the equivalent of £900 abroad.520 and £10. credit card company. £100 into Australian dollars and £120 into Hong Kong dollars.798 New Zealand dollars (NZD) £1 2. but which they feel sure they will have in the future. he sold two cars for £7. So you can receive interest on a loan you make to someone and you can also be asked to pay interest on a loan which you take out. Australia and Hong Kong. How much of each currency did they receive after commission was deducted? After buying the foreign currencies. (i) (ii) (c) How much did they exchange for travellers cheques? How much commission did they pay for the cheques? 6. In one particular week. The smallest value of traveller's cheque which can be bought is £10. C. Let us consider the position from the point of view of the borrower. etc. The cost charged by the lender can vary a great deal depending on the method of obtaining credit. again the rate varying according to the amount of the © ABE and RRC . A car salesman is paid 2.640. SIMPLE AND COMPOUND INTEREST We introduced the subject of interest in the last study unit and will now consider it in more detail.65 Hong Kong dollars (HKD) Use the following exchange rates: Now check your answers with those given at the end of the unit.70 Basic Business Applications 5. What was his commission for that week? On a holiday touring New Zealand. This cost is then quoted as an annual percentage rate of interest to be charged on the loan (known as the principal). The travel agent charges commission on each currency exchange at the rate of £3 or 2% (whichever is the larger) for cash and £3 or 1% (whichever is the larger) for travellers cheques. the amount borrowed and the time over which the loan is repaid. It is often the case that individuals or companies wish to spend money which they do not currently have. Interest calculations involve the application of simple formulae to specific problems.458 Australian dollars (AUD) £1 19. (a) They changed £110 into New Zealand dollars.5% commission on his weekly sales over £6. Interest is what we call the amount of money someone will give you for letting them borrow your money or what you pay for borrowing money from someone else.000. they changed as much as possible of the remaining money into traveller's cheques. (b) What was the total cost per person of the foreign currencies and travellers cheques? £1 . Therefore they borrow that money from a lender – a bank. – and pay the amount borrowed back to the lender over a particular period. Usually finance companies quote a particular rate of interest applicable to either loans taken out or savings placed with them. SI £270 and R 9. Thus: P 100SI RT SI PT R% T 100SI PR Consider the following examples.000 9 18. Here we need to work out T where P £2. There are two ways of looking at interest – simple interest and compound interest. Note.90 100 Example 2: Calculate the length of time taken for £2. we can also rearrange this formula to allow us to work out any of the variables. T. there is a simple formula to work out the amount of interest your money will earn: SI PRT 100 This means that the interest is found by multiplying the principal by the rate. we get: T 100 270 27. (Note that the way in which this formula is expressed is slightly different from that used in the previous unit. but is stated in a rather easier fashion. with a rate of interest.) As we have seen. and then dividing by 100.40 9 5 £5.40 in an account that paid simple interest at a rate of 9%. R.Basic Business Applications 71 loan/savings and also the period of the loan (or the amount of time you guarantee to leave the savings with them).000 to earn £270 if invested at 9%. say P. Simple Interest Simple interest (SI) is calculated on the basis of having a principal amount. that the notation is also slightly different. the rate (R) is 9% and the time (T) is 5 years.000 © ABE and RRC . So. It is often the case that formulae can be expressed in slightly different ways. As we have seen. The principal (P) is £16. too.000 1. Substituting in the formula. in the bank for a number of years. It works in exactly the same way. then by the time.40.000. substituting in the formula. Calculate how much interest would be paid to John if he kept the money in the account for 5 years. Example 1: John had £16. we get: SI 16.5 years 2. 5% per annum. in the second year. how much interest will you receive? £10. 2. At what annual rate of interest was the money invested? Now check your answers with those given at the end of the unit.080 which is £86. the interest is 8% of £1. 4. After 4 years.75% interest rate? If you invest £6.000 and the interest is 8% per year.240 in interest had been withdrawn.08 each time. Adding this to the original principal. However. Now. 3. how much interest will you earn and how much will you have in your account after 4 years? If you invest £5. usually annual) n number of time periods. the amount invested at the end of the first year is £1.000 at 9. you are finding 100% 8% 108% of the amount at the beginning of the year. a total of £3. then the principal increases and so the interest earned gets bigger each year.080. The amount accrued at the end of the second year is £1.72 Basic Business Applications Questions for Practice 3 1. What is the simple interest on £250 over 3 years at 8. The interest in the first year is 8% of £1.000 which is £80. So you can multiply by 1. Consider the following examples.750 at 8. Compound Interest Let us now look at how compound interest differs from simple interest.25% per annum for 6 months.40. © ABE and RRC . If an amount of money is invested and the interest is added to the investment.166.800 was invested and at the end of each year the interest was withdrawn.50. as follows: r A P 1 100 n where: A accrued amount P original principal r rate of interest (for a particular time period. to find the amount at the end of any year. Suppose you begin with £1. There is a standard formula which can be used to calculate compound interest. You can continue working out the amounts in this way. 20 interest at 12% (of £3.000 interest at 12% (of £3.51 interest at 5.51 value of investment after one year £2. Example 2: £2.215.105) £110.000 is invested in an account paying 12% compound interest per year.331.78 Alternatively.105 £110.000 1.Basic Business Applications 73 Example 1: £3.763. we get: investment at the beginning of year 1 £2.000 r 12% n3 Substituting these values in the formula gives: r A P 1 100 n 12 3 £3.5% per annum.000 is invested in a high interest account for 1½ years.000 £105 £2.25% (of £2. The rate of interest for 1 year 10.25% (of £2. we get: investment at the beginning of year 1 £3.214. Working this out year by year.404928 100 3 £4.000 £360 £3.51 £2.25%. Find the value of the investment after 3 years.214.360 interest at 12% (of £3.000) £360 value of the investment after 1 year £3.215.000 1.763.12 £3.25% (of £2.215. Calculate the value of the investment after this time and the amount of interest earned. Therefore the rate of interest for 6 months 5.360) £403. and it is paid into the account every six months.5%.20 value of investment after 2 years £3.763.000 1 £3.20) £451.58 value of the investment after 3 years £3.31 value of the investment after 1½ years £2. we could use the formula: r A P 1 100 n where: P £3.20 £451.51 £116.82 © ABE and RRC .360 £403.000 interest at 5.105 interest at 5.51) £116.78 (to the nearest p).58 £4.20 £3. Working this out by each six month period.31 £2.000) £105 value of the investment after 6 months £2. The interest rate is 10. In other words.000 £331. the replacement of an asset – such as machinery and equipment – where the future cost of the asset is known). we could use the formula: r A P 1 100 n where: P £2.82 Note that at each stage the amount of interest has been rounded to the nearest penny.000 £331. the original principal P) would be worth at some point in the future (i.e. Consider the following example: A manufacturing company is considering replacing its production line in three years' time.82 £2. looking at our formula.e.331. (Interest earned each year will be reinvested and no money will be withdrawn).25% n3 Substituting these values in the formula gives: r A P 1 100 n 5. Alternatively. the original principal (P). the accrued amount (A)). a financial manager might want to know how much to invest now in a bank account in order to accrue a certain sum of money for a known future expenditure (e. the rate of interest (r) and the number of time periods (n) are known and the uncertain variable is the accrued amount (A).25 3 £2.331.0525 £2. the accrued amount (A)).000. the rate of interest (r) and the number of time periods (n) are known with certainty and the uncertain variable this time is the original principal (P).83 £2.331.g. For example.83 (to the nearest p).1659134 100 3 £2. the original principal (P)) to achieve a certain level of known return at some point in the future (i. The cost of this is expected to total £40. the accrued amount (A). In other words.000 1. looking at our formula. if the interest rate is 12% per annum.000 1.e.74 Basic Business Applications Total compound interest earned £2. Total compound interest earned £2. Calculate how much must be invested now in a bank account to purchase the new production line in three years' time. © ABE and RRC . there are some situations when we might want to know how much we would need to invest today (i.83 So far we have used the compound interest formula r A P 1 100 n to work out how much a sum of money today (i.000 1 £2.000 r 5. However.e. 21 (to the nearest p) Thus. we need to rearrange our compound interest formula.0% n3 Substituting these values in the formula gives: 12 40.000 P 1 100 3 To solve for P. © ABE and RRC .000 12 1 100 3 40. Thus: r A P 1 becomes: 100 n P A r 1 100 n So 12 40.000 1.000 required to replace its production line in three year's time.000 1. the manufacturing company needs to invest £28.Basic Business Applications 75 Using our compound interest formula: r A P 1 100 n We know that: P £40.471.000 P 1 becomes: 100 3 P 40.404928 £28.000 r 12.21 today at an interest rate of 12% per annum if it is to accrue the £40.123 40.471. You will recall from Study Unit 2 Section D how we go about rearranging formulae. Find out how much must be invested now. as long as the method chosen is applied consistently from year to year. A second offers 8. The formula would then be: annual depreciation cost of asset . We can express this as a formula as follows: annual depreciation cost of asset useful life Thus. over the period of years selected. to buy this machinery in three years' time. If you had £2. 3.000 is invested at 10% per annum compound interest which is paid annually.5% per annum payable half-yearly. £6. if the interest rate which can be earned on investments is 10%. How much is in the account at the end of 1½ years? One building society offers an interest rate of 8. such as cars.000. but the most commonly used are the straight-line method (using fixed instalments) and the reducing-balance method (applying a fixed percentage).76 Basic Business Applications Questions for Practice 4 1. 2. The amount by which an asset depreciates each year may also have important consequences.75% payable annually. In the case of certain types of asset. Straight-Line Depreciation In the straight-line method. it may be anticipated that the asset will be used for say three years and then sold. 4. (Interest earned each year will be reinvested and no money will be withdrawn). the amount of depreciation is the difference between the cost of the asset and its anticipated value at the end of its useful life.000 to invest. Now check your answers with those given at the end of the unit. This is an important concept in business where certain types of asset need to be accurately valued in order to help determine the overall value of the business. There are several methods for calculating depreciation. consumption or other loss of value of an asset arising from use. divided by the number of years over which it will be used. How much is in the account after 3 years? £2. passage of time. or obsolescence through technology or market changes. Give your answer to the nearest pound. depreciation is calculated by dividing the cost of the asset by the number of years the asset is expected to be used in the business. DEPRECIATION Depreciation is a measure of the wearing out. a car loses value over time and is worth less each year. D.value at end of useful life useful life Consider the following examples. © ABE and RRC . For example. which savings account should you choose? An office has to replace some of its machinery in three years' time which it anticipates will cost £9. In this case.000 is invested at 10% per annum payable every 6 months. It is acceptable to use different methods for different categories of asset. the value of the asset will be reduced to zero. 000 £500 £500 end of year 3: value £500 £500 £0 Example 2: A machine cost £4.Basic Business Applications 77 Example 1: A computer costing £1.646.40 Amount of depreciation in year 3 12% of £4. Reducing Balance Depreciation In the reducing-balance method.500 £500 3 end of year 1: value £1.646.000 0. and how many years will it be before the machine is valueless? Depreciation after 3 years will be 3 £600 £1. such as motor vehicles. Consider the following example.12 £720 100 Therefore. value of machine at the end of year 2 £5.000 £6. will have a small percentage (say 2½%). the period over which the machine is depreciated to zero will be: £4.000 end of year 2: value £1. assets having long lives. How much will it be worth after 3 years if it depreciates at £600 per year.280 12 £5. will bear a large percentage (say 15%).800 £3. the total provision for depreciation will never equal the cost price at the end of the asset's useful life (i. value of machine at the end of year 1 £6. depreciation is calculated each year by applying a fixed percentage to the diminishing balance.280 0.12 £557.280 £5. annual depreciation £1.40 12 £4.40 £4.e.12 £633.500 £500 £1. Unlike the straight-line method then. The percentage to be applied will depend on the type of asset. For example.60 100 Therefore.800. the final balance never reaches zero).000 12 £6. A company bought a machine for £6000.280 Amount of depreciation in year 2 12% of £5.800 £600 8 years.646. but assets which quickly depreciate.500 will have a useful life of 3 years.57 100 © ABE and RRC .800 £1. Show the process of depreciation.000 £720 £5. such as buildings.40 0.646.800 The value of the machine will be £4.60 £4. If it depreciates at 12% per annum what will it be worth after 3 years? Amount of depreciation in year 1 12% of £6.000 If the annual depreciation is £600.280 £633. 3. Repeatedly multiplying by 0. A quicker and simpler way is to realise that 12% is 12/100.88 £4. which is 0. (a) (b) 2.000 0.40 0. The rate of depreciation is 20% per annum. If you are reducing the amount by a factor of 0. Questions for Practice 5 1.12 and then subtracting this from the original amount. you are left with 0. Thus: At the end of the year 2.88 gives the same result as multiplying the amount by 0.280 This is the same as: value of machine at the end of year 1 £6. If the machine will be valueless after 4 years. How much will it be worth after 2 years if it depreciates at £400 per year? How many years before the machine is valueless? A factory buys a machine for £7.40 £557. What is its value after 3 years? © ABE and RRC .280 x 0.000.78 Basic Business Applications Therefore.88 (88%) of the amount. A launderette buys a washing machine which cost £2.83 We can express this method by the following formula: D B(1 i)n where: D depreciated value at the end of the nth time period B original value at beginning of time period i depreciation rate (as a proportion) n number of time periods (normally years). We can demonstrate this in respect of the above example as follows: Amount of depreciation in year 1 12% of £6.000 £6.000 88% £6. value of machine at the end of year 3 £4.000.646.280 This is both quicker and easier and you are less likely to make a mistake.57 £4.12 (12%).000 £720 £5.646. the value of the machine £5. value of machine at the end of year 1 £6.12 £720 100 Therefore.646.83 However this is obviously a time-consuming method.88 will give the amount left each year. So.088. what should be the amount of depreciation per year? A small business buys a computer costing £4.88 £5.500.000 12 £6. the value of the machine £4.12.40 At the end of the year 3.000 0.88 £4. multiplying the amount by 0.088. using only a basic construction of how pay is made up and the way in which the taxation system works. Thus. the most common rates being time and a half (i.30 40 £212 An employee can often increase a basic wage by working longer than the basic week – i.84 per hour. Company B bought an exactly similar fleet.30 per hour for a 40 hour week.500 every month (£18. For example. A second-hand motorbike is bought for £795. This is also an area where it is very important to adopt clear layout when working through calculations. we shall be applying a number of the basic numerical concepts we have considered previously – including the distinction between fixed and variable values. an office clerk is paid £5. Now check your answers with those given at the end of the unit. this is paid in the form of either a salary or wages. A higher hourly rate is usually paid for these additional hours. PAY AND TAXATION Pay and taxation are complex issues and it is not our intention here to explore them in any detail. by doing overtime.000 12). but it decided to write-off 20% of the value each year. Its value depreciates by 11% per year.e. What is her weekly wage? weekly wage rate of pay hours worked £5. Thus: monthly pay annual salary 12 For example. Salaries are generally paid monthly and are based on a fixed annual amount. In addition.000 would receive £1. E.Basic Business Applications 79 4. Again. Make Up of Pay All employees receive payment for their labour.e. Consider a factory worker who works a basic 36 hour week for which she is paid a basic rate of £5. At its most basic.250. For how much could it be sold two years later? (Give your answer to the nearest pound. we shall concentrate on the principles of their calculation. she works 5 hours overtime at time and a half and 3 hours overtime at double time. © ABE and RRC . many wage earners are required to work a fixed number of hours in a week. Wages are generally paid weekly and are usually based on a fixed rate for working a certain number of hours. Calculate the value of the cars to Company A and to Company B at the end of the 3rd year. percentages and simple formulae. an employee with an annual salary of £18. It decided to write-off equal instalments of the capital cost over 5 years. and they are paid for these hours at a basic hourly rate.) Company A bought a fleet of 20 cars for its sales force. 1½ times the basic hourly rate) and double time (twice the basic hourly rate). each costing £15. Calculate her weekly wage. 5. Rather. Calculate the weekly wage of a factory worker if she works 42 hours per week at a rate of pay of £5.05 £4. 6. 40-hour week. and some shop assistants.5 hour week at an hourly rate of £4. Overtime worked at the weekend is paid at double time. he is paid a commission of 18% on the excess. 2. What would be his income for a month in which he sold goods to the value of £9. and the commission on their sales forms the largest part of their pay.84 1.5) £43.750 how much is she paid per month? A basic working week is 36 hours and the weekly wage is £306. Commission is paid at 8% on all insurance sold up to a value of £4.280 per year plus a monthly commission of 5% on all sales over £5.500. Calculate her wage for a week when she worked five hours overtime during the week and four hours overtime on Saturday.100 and £2. If she works overtime during the week. In some cases.000) worth of sales: commission 5% of (£9. or even non-existent.400 £5. If the value of insurance sold exceeds £4. 4. What is the basic hourly rate? A trainee mechanic works a basic 37.700.900.84 £210. £5.000. and she works a basic 5-day.400? basic monthly pay annual salary £8. she is paid at time and a half. £4.24 5 hours (£5. are often paid a basic salary or wage plus a percentage of the value of the goods they have sold.25 per hour If an employee's annual salary is £18. 3.80 Basic Business Applications Basic pay: Overtime at time and a half: Overtime at double time: Total pay: 36 hours £5. Now check your answers with those given at the end of the unit. their basic salary or wage can be quite small.000 per week. 5.20 per hour.280 £690 12 12 He earns commission on (£9.90.400 £220 Total monthly pay basic pay commission £690 £220 £910 Questions for Practice 6 1. An insurance representative is paid purely on a commission basis. © ABE and RRC . Calculate his total pay for the 4 weeks in which his sales were £3. He is paid overtime at time and half.000) 0.04 £289.400 £5.84 2) £35.000. a salesman earns a basic salary of £8.08 People who are employed as salespersons or representatives. For example.80 3 hours (£5. How much does he earn in a week in which he does 9 hours overtime? An employee's basic wage is £6. Income tax is a major source of finance for the government and is used to fund all types of public expenditure. the UK system divides taxable pay into two bands of income. The amount of gross income on which tax is payable (known as taxable pay) may be affected by the provision of certain allowances within the tax system and also by certain elements of pay or deductions being exempted from tax (non-taxable items). Most people have income tax deducted from their pay by their employer before they receive it. are decided each year by the government as part of the Budget.000 20% £4. the total tax liability would be: £20.000 20% (£45.000) 40% £7.Basic Business Applications 81 Taxation The total earnings made up in the way just discussed is known as gross pay. Certain amounts of money are deducted from gross pay and the pay the employee actually ends up with is the net pay or take-home pay. From April 2008. The percentage rate at which the tax is applied varies according to the level of income.200 £3. the total tax liability would be: £36. The two main deductions are: income tax and pension contributions. based on the amount a person earns in a tax year that begins on 6th April and ends on 5th April the following year.800 These annual tax bands can be converted into monthly or weekly amounts for calculating tax on a PAYE basis and deducting the correct amount from an employees gross pay: © ABE and RRC . This method of paying income tax is called PAYE (Pay As You Earn).000 £36.1: Taxation bands Annual taxable income Tax rate 0 – £36.600 £10.000 in a year. and the levels of allowances. very few employees actually receive all the money they have earned. There are often other deductions – some being further national or local taxes on pay (such as the UK system of National Insurance) and others being deductions made by the employer for work-specific reasons (such as repayments of a loan for a car or for a public transport season ticket). The rates of tax applied to particular bands of income. Unfortunately.000 in a year.000 For a person with a total taxable pay of £45. The tax is a percentage of gross pay.001 and above 20% 40% This means that. the employer then pays the tax to the government. with the following rates of tax applied to each: Table 3. for a person with a total taxable pay of £20.000 £36. ) Now we can develop a framework for calculating the amount of tax payable annually.560 The monthly income tax deducted will be £2. then you would pay no income tax at all. so: gross taxable pay gross pay non-taxable items £18.800 £12.001 and above Monthly taxable income 0 – £3. using the tax bands set out here. Where pension contributions are not paid through the employer. This amount (sometimes called free-pay) depends on the individual's circumstances – for example. where they are deducted from pay by the employer.800 tax liability £12.2: Annual. £213 rounded down. Personal allowances – the amount of money an individual is allowed to earn each year before he or she starts to pay tax.000 £36. such as an employer's occupational pension scheme. monthly and weekly taxable income Tax rate 20% 40% Annual taxable income 0 – £36.800 20% £2.800? What is the monthly tax deduction? There are no non-taxable items specified.82 Basic Business Applications Table 3. © ABE and RRC . using the bands and tax rates set out here.e. The level of personal allowances are generally set as an annual amount. an adjustment may be made to a person's personal allowances to allow for them. etc. i.600 per annum and has personal allowances of £5. Here we shall consider just two.600 £5. Pension contributions – the amounts paid by an individual into a pension scheme. may also be deducted from gross pay before calculating tax.560 divided by 12. (Note that this does not necessarily apply to any other types of deductions from pay. pension contributions. Therefore.001 and above Weekly taxable income 0 – £692 £693 and above There are many possible allowances and non-taxable items which affect the level of taxable pay. but this can be converted into monthly or weekly amounts of "free-pay".000 of net taxable pay 20% plus remainder of net taxable pay 40% Consider the following examples. If your personal allowances are greater than your actual pay. are not taxable. number of children.000 £3.600 net taxable pay gross taxable pay personal allowances £18. whether they are single or married. gross taxable pay gross pay non-taxable items net taxable pay gross taxable pay personal allowances tax liability first £36. Example 1: How much annual tax will be paid by someone who earns £18. Basic Business Applications 83 Example 2: How much annual tax will be paid by someone who earns £43. based on the information supplied Tax liability £2. but wages remain the same. He has personal allowances amounting to £4.000 12 Questions for Practice 7 1.200 £180 £7. and at double time for any additional hours worked.850. in the next Budget. Overtime is paid at a rate of time and a half for the first five hours. She pays £3. the Chancellor changes the rate of income tax from 24% to 23%. For example: How much tax will be paid each month by someone who earns £24. 5. making contributions at a rate of 6% of basic earnings only. If tax is payable at the rate of 24%.000.000 20% (£36.450 £36. As a single mother. the pension contributions are non-taxable.000 and personal allowances of £5. so: gross taxable pay £43. 2.800 is paid a weekly wage of £80.000 £2.e. 4.000 per annum if the tax rate is 25%? Net taxable pay per month. calculate his monthly pay. © ABE and RRC . What is the rate of income tax? A weekly paid worker receives a basic wage of £5. Always answer the question using the information supplied. 3. or without reference to deductions from gross pay or personal allowances. An employee is paid a salary of £24. If. has personal allowances of £4.000) 40% £7.450 tax liability £36. Calculate her net pay in a week when she works 48 hours if the tax bands and rates are as set out above (i.5 hour week.000 25% £500 per month.000 5%) £40. as per 2008 in the UK). she has a personal taxation allowance of £5.400 £36.492 per annum which he receives in 12 monthly payments.380 Note that taxation questions are often set without reference to income bands with different taxation rates applying.850 £4.50 per hour for a 37.000 per annum. Now check your answers with those given at the end of the unit.800. how much less tax will be paid by the employee in Question 1 above. Would this person have to pay tax? An employee has a gross income of £23. £24.850 net taxable pay £40. Here.400 and pays pension contributions through his employer at a rate of 5% of his annual salary. She also belongs to an occupational pension scheme.600 tax. Someone whose personal tax allowance is £4.000 (£43. 2.70 Total cost is the £900 worth of foreign currency plus the commission of £9 for the cash transactions and £5.26 (a) (b) (a) (b) (a) (b) £209. 6.986 £2.30 105p (rounded).05 HKD. (a) They would be charged £3 for each cash transaction.63 £2. 2. £45. 945 Swedish kroner £5. cost per person is £228.39 NZD £97 into AUD 238. 2. Therefore.295 interest and £9.64 euros.315.560 £24.950 £1.25 7. 4. Spread between the four members of the family.68. 1640 Swiss francs 590. 2.97 Questions for Practice 2 1.72 £304 There are three parts to this question.70. 3. £31 £5. £231.43 AUD £117 into HKD 2299.045 in the account. 4. 4. they would be changing: £107 into NZD 299.5% Questions for Practice 4 1. Questions for Practice 3 1. £7.25 © ABE and RRC .70 for the travellers cheques £914. 3. 3. (b) (c) (i) (ii) £570 £5. £65. 5.84 Basic Business Applications ANSWERS TO QUESTIONS Questions for Practice 1 1. 75 A £2.000 1.25 2 £2. work out the accrued amount after one year in each case: For the first building society. (a) (b) £1.5% payable six monthly equates to 4.00 Therefore. using the formula for calculating the accrued amount over two six month periods is: r A P 1 100 n 4.304. To compare the effects of the two rates.000 1 £2. Questions for Practice 5 1.175.880 value at the end of year 3 £2.331 £6.600 80% £2.000 1 £2.1659134 100 2 £2.10 9.200 5 years £1. 3. 2. if the office is to accrue the £9.000 1.25% as a six-monthly rate. is the better investment.880 80% £2.600 value at the end of year 2 £3. the rate of 8. © ABE and RRC .75% paid annually.75% paid annually is: 8.Basic Business Applications 85 3. 4.000 1.762 Therefore.173. £6. with the rate of 8. Thus: value at the end of year 1 £4. Thus.750 The value at the end of each year is 80% of the value at the beginning of the year.61 (to the nearest p) The calculation for the second building society rate of 8.000 required to replace some of its machinery in three year's time.000 3 9.500 80% £3.762 would need to be invested today at an interest rate of 10% per annum.000 1. Using the formula: P A r 1 100 9.000 10 1 100 3 n 1.0425 £2.0875 100 1 £2. the second building society. 25 6% £12. 5.562.5 hours at the basic rate of £5.200 value at the end of year 3 £195. 4.50 per hour 5 hours at (£5.00 Non-taxable items are pension contributions amounting to: © ABE and RRC . Thus: value at the end of year 1 £795 89% £707.90 £344. 2. 3.000 3 £183. Questions for Practice 6 1.86 Basic Business Applications 4. £220.5) per hour 5.50 2) per hour gross pay £206.000 value at the end of year 2 £244.247. 6.50 £308.000 £122.25 £41.50 £8.55 89% £629. The value at the end of each year is 89% of the value at the beginning of the year. 5.456 Questions for Practice 7 1. 4. 2.000 5 after 3 years. total depreciation £61.10 £1.000 80% £195.72 £630 to the nearest pound.50 1.50 £249.000 £61.000 Company B: The value at the end of each year is 80% of the value at the beginning of the year.38 £206. Thus: value at the end of year 1 £305.200 80% £156.41 per month No 20% Gross pay is made up as follows: 37.000 value of the fleet after 3 years £305.250 20 £305.55 value at the end of year 2 £707.5 hours at (£5.16 per month £16. Company A: annual amount of depreciation £305. Total cost of the fleets to both companies £15. £1.25 £60.000.160.50 £1.000 80% £244.000 £183. 5. 3. 38 £295.62 £5.81 20% £48.850 52 £295.16 £247.62 £54.62 Net taxable pay gross taxable pay weekly personal allowances £295.81 Tax liability is based on the weekly income tax bands.38 £48.Basic Business Applications 87 Gross taxable pay gross pay non-taxable items £308. tax liability £240.00 £12.81 £240. and this level of weekly net taxable income falls within the 20% liability band (0 – £692).00 £12.16 Net pay gross pay pension contribution tax £308.46 © ABE and RRC . 88 Basic Business Applications © ABE and RRC . C. Quadratic. Simultaneous Equations Solution by Elimination Formulating Problems as Simultaneous Equations Quadratic Equations Solving a Quadratic Equation by Factorisation Solving Quadratic Equations with no Terms in x Solving Quadratic Equations by Formula Logarithmic and Exponential Functions The Exponential Number e Logarithms Relationship between Logarithms and the Exponential Number Laws of Logs and Exponents Solving Equations using Logarithms and the Exponential Number Page 90 91 91 96 98 99 102 103 106 106 106 108 108 110 111 B. Exponential and Logarithmic Equations Contents Introduction A. D.89 Study Unit 4 Simultaneous. Answers to Questions for Practice © ABE and RRC . Before we start though. and both sides may be multiplied or divided by the same number without changing its equality. we will examine the principles and rules of logarithms and exponential values.90 Simultaneous. We shall also consider the distinction between linear and quadratic equations. Quantities may be transposed – transferred from one side of the equation to the other – by changing their signs. Similarly. so if you are unsure of any points. you must also do to the other. To do this. Finally. It has two sides and they must always be equal. What you do to one side of the equation. Solving an equation means finding the value of the unknown quantity (or quantities) in the equation – the variable – by using the other numbers. the constants. finding out the value of the unknown quantity. the expressions which form the equation may be manipulated with the objective of isolating the unknown on one side of the equation. It is essential that you fully understand these before studying this unit. and learn how to solve equations using them. using the same rules. we considered simple equations where there was only one unknown quantity. go back to Study Unit 2 and revise the section on equations there. a difference which will be developed in the next study unit when we look at drawing graphs to represent equations. or from to . an added or subtracted term may be transposed if its sign is changed from to . in order to collect all the unknown terms on the same side of the equation. we should recap the key principles already examined in relation to equations. Thus. Thus. Objectives When you have completed this study unit you will be able to: use the elimination method to solve simultaneous equations including two unknowns formulate problems as simultaneous equations define a quadratic equation and distinguish it from a linear equation solve quadratic equations by the factorisation method simplify and solve equations using both base 10 and natural logarithms. The golden rule in manipulating equations is equality of treatment to both sides of the equation. Unknowns may also be transposed.e. a multiplier may be transferred from one side of an equation by changing it to the divisor on the other. An equation is a statement of equality between two expressions. © ABE and RRC . Exponential and Logarithmic Equations INTRODUCTION In Study Unit 2 we introduced the concept of equations and looked at the basic principles of solving them – i. the same number may be added or subtracted from both sides of the equation without changing its equality. and a divisor may be transferred from one side by changing it to the multiplier on the other. Now we shall take our study of equations further by examining equations where there is more than one unknown quantity – simultaneous equations. Quadratic. In that study unit. many situations will arise when there is more than one unknown. in practice. Example 1: Consider the following pair of equations: 3x y 11 x y 3 (1) (2) We can add the two equations together to derive a third equation which will eliminate y: © ABE and RRC . any other equation obtained from them will also be true. and all pairs of values which satisfy the one will satisfy the other. By adding or subtracting suitable multiples of the given equations. However. The following examples explain the method. To determine a particular pair of unknown values then. and then the other unknown may be discovered by substitution in the original equation. One way of getting more information is to have another equation – for example: x y 7 and 2x y 12 Now there is only one solution pair: x 5 and y 2. SIMULTANEOUS EQUATIONS As we said in the introduction. there is only one solution: x 4½ and y 2½. Once we have done that. we had only one unknown quantity to find. we may find any number of pairs of values to satisfy it. Exponential and Logarithmic Equations 91 A. Similarly.3. up until now. For example: the equations x y 7 and 2x 2y 5 are inconsistent – they cannot be true simultaneously. Consider the following example: xy7 This may be satisfied by: x 3 and y 4 x 2½ and y 4½ x 1. These groups of equations are known as simultaneous equations. Quadratic. when solving equations. we need more information. and no pair of values which satisfies one will satisfy the other the equations 3x 3y 21 and 2x 2y 14 are not independent – they are really the same equations. Solution by Elimination If two equations are true simultaneously. etc. Therefore to find two unknowns we need two equations. the resulting equation can be solved using the methods we already know. that for equations to be simultaneous they must satisfy two conditions – they must be consistent and independent. Study them carefully. The method of solution by elimination depends on this fact. etc. Note though. to find three unknowns we need three equations. if x y 7 and x y 2. If an equation involves two unknown quantities.3 and y 8.Simultaneous. we can eliminate one of the unknowns and obtain an equation containing only one unknown. 5 © ABE and RRC . so the coefficient of the term 3x is 3. we derive a changed version of the same equation by multiplying (1) by 3 and then subtract (2) from this equation: Multiplying (6x 5y 6) by 3 gives: 18x 15y 18 Note that. Therefore. we can see that the coefficient of x in (2) is a multiple of the coefficient of x in (1). independent equation. y) was possible because both the "y"s in the original two equations had a coefficient of 1 – i. but simply a different way of expressing the same equation. we get: 6x 6 15 9 x 1. It is not a new. If we now substitute that value for x in (1).92 Simultaneous. To make the coefficients equal. x 2.) If we want to eliminate one of the unknowns by adding or subtracting the two equations. However. this is not the case here. Exponential and Logarithmic Equations 3x y 11 x y 3 4x 8 Thus. the coefficients were equal. the coefficients of the unknowns must be equal.e. providing we apply the same proportionate change to each term in the equation (by multiplying or dividing each by the same number). for example. Example 2: Consider the following pair of equations: 6x 5y 6 18x 7y 6 (1) (2) We want one of the unknowns to have equal coefficients. Quadratic. We can now subtract equation (2) from this: 18x 15y 18 18x 7y 6 8y 24 y 3 To find x. We can check that the values we have determined for x and y are correct by substituting both in equation (2): x y 2 5 3 Note that eliminating one of the unknowns (in this case. we are left with 4x 8. we substitute the value of y 3 in (1) so that 5y becomes 15: 6x (15) 6 By transposition. the equation has not been changed. so that 3x becomes 6: 6 y 11 y 5 (the symbol means "therefore"). by adding the two equations. but unlike in Example 1. (The coefficient is simply the factor or number applied to an algebraic term. we substitute the value of x 2 in (2) so that 9x becomes 18: 18 2y 16 By transposition. 1. so that the coefficient of y in both equations is 14. etc. there are no equal or multiple coefficients. we get: 2y 16 18 2 y 1 Important note: When working through the processes of transposition and substitution. it is not necessary to write (7) and (15). Quadratic. Where there are no equal or multiple coefficients. 3x 4y 1 9x 5y 10 2. we need to change the way in which both equations are expressed so that we can make the coefficients of one of the unknowns equal.. We can then choose which unknown to eliminate and. Exponential and Logarithmic Equations 93 It is always good practice to check the values by substitution in the other equation. to avoid large numerical terms. We can then eliminate y by adding the two equations: 26x 14y 66 63x 14y 112 89x 178 x 2 To find y. Questions for Practice 1 Solve the following simultaneous equations by finding values for x and y. we multiply (1) by 2 and (2) by 7.Simultaneous. To do this here. we shall decide to eliminate y. 5x 3y 14 7x 5y 38 3. but be sure to pay special attention to all calculations involving negative numbers. 2x 11y 23 7x 8y 50 © ABE and RRC . Example 3: Consider the following pair of equations: 13x 7y 33 9x 2y 16 (1) (2) This time. so there is no obvious unknown term which can be easily eliminated. but we will skip that step from now on. 3x y 8 x 5y 4 4. 1 12 x y 12 x 7 15 y 21 Now check your answers with those given at the end of the unit. x 1 2 2 3 y 16 x 2y 14 3 5 7. To eliminate 1 . Quadratic.94 Simultaneous. Now we shall look at some harder examples! (a) Where the unknowns are denominators Consider the following equations: 4 3 4 x y 3 7 – 2 x y (1) (2) 1 1 and as the unknowns and then solve the problem in the usual x y Here we can treat way. 9x 10y 12 2x 15y 5 6. Exponential and Logarithmic Equations 5. multiply (1) by 7 and (2) by 3 and then subtract: y 28 x 9 x 19 x 21 28 y 21 y (1) (2) 6 22 22 19 19 22 1 x x Substituting for 1 in (1) gives: x 4 22 3 4 y 19 88 3 4 y 19 © ABE and RRC . we need to reformulate the equation as two simultaneous equations. multiply (2) by 3 and add to (1): © ABE and RRC . and rewrite the equations as: x y Alternatively. Taking the first pair of expressions. Quadratic.Simultaneous. but remember to write your answer as x ? and y ?. you may find it easier to put 4m 3n 4 3m 7n 2 (1) (2) Solve these in the usual way. we can rearrange the terms to form one equation in the usual form: x 5y 2x 4y 3 x 9y 3 Now take the second pair of expressions and rearrange them: 2x 4y 3 3x 2y 4 2x 3y 4 We now have two simultaneous equations as follows: x 9y 3 2x 3y 4 5x 15 x3 y 2 3 (1) (2) These can be solved in the usual way. (b) Forming simultaneous equations from single equations Equations are sometimes given in the form: x 5y 2x 4y 3 3x 2y 4 In order to solve this. To eliminate y. not as m and n. Exponential and Logarithmic Equations 95 88 76 12 3 y 19 19 4 1 y 19 y 19 43 4 4 1 1 m and n. Consider the following examples and try to form the equations for yourself before looking at the solutions given. but it is most advisable.) Let: x expenditure of 1 man in £ y expenditure of 1 boy in £ Then we can express the information given as equations: 10x 8y 160 (all terms must be expressed in the same unit. Example 1: The expenditure of 10 men and 8 boys amounts to £160. Exponential and Logarithmic Equations Questions for Practice 2 Solve the following equations for x and y: 3 20 13 x y 7 10 2 x y Now check your answers with those given at the end of the unit. Note too. it is very useful to develop mathematical models of real situations to help analyse and solve problems.96 Simultaneous.e. that unless you are told to find both or all of the unknowns. FORMULATING PROBLEMS AS SIMULTANEOUS EQUATIONS As we have seen previously. here pounds) and 4x 6y 18 Solving these two equations will answer the problem. In the same way. as otherwise you have no means of checking your answer. not in your own equation. We have seen that simple equations can be developed to model a situation and enable us to find the value of one unknown variable. we can develop models which will enable us to find two or more unknown variables by using simultaneous equations. how much does each man and boy spend? (Assume each man spends the same amount and each boy spends the same amount. it is not essential to do so. Having found the number for which the unknowns stand. Check your answer in the question given. B. where we know the values of all the other variables. i. If 4 men together spend £18 more than 6 boys. © ABE and RRC . give the answer to the question in words. Quadratic. The key points about formulating problems as simultaneous equations are the same as doing the same with simple equations: Always state clearly what your unknowns represent and give the units you are using. The use of algebraic notation to represent variables in a situation and their combination with numerical constants to form equations are the building blocks of such mathematical models. Exponential and Logarithmic Equations 97 Example 2: In a bag containing black and white balls. we get: x5 The required number is 54. half the number of white is equal to a third the number of black. (B) If we now multiply (A) by 4. Note how the equations are built up. The number itself is six times the sum of its digits. is such that it is diminished by 9 if the digits are reversed. we get: 1 2 x 1 3 y and 2(x y) 3y 4 Example 3: Pay special attention to the following problem. we get: 10x y 9 10y x and 10x y 6(x y) for (1): 9x 9y 9 or xy1 for (2): 4x 5y 4x 4y 4 We can then substitute the value of 4x from (B) to get: 5y 4y 4 y4 Substituting for y in (A). we get: (A) (2) Rearranging these equations to isolate the constant in (1) and simplify both. Expressing the information given as equations. A certain number. Find the number.Simultaneous. and the reversed number is 10y x. Let: x the tens figure y the units figure Then the number is 10x y. we get: (1) © ABE and RRC . and twice the total number of balls exceeds three times the number of black balls by four. of two digits. Expressing the information given as equations. How many balls did the bag contain? Let: x number of white balls y number of black balls. Quadratic. In the following year the income from advertisements was increased by 2 12½% and the income from sales decreased by 16 3 %. Find the distance from London to Birmingham and the average speed of the train. using simultaneous equations. 11 kg of tea and 6 kg of sugar cost £17. C.30. or 3y2 2y 5. the result is a straight line. 3. Find the numbers. The total income decreased by £12. to £670.) By contrast. (The types of equation studied so far are known as linear equations since. in a year. Half the sum of two numbers is 20 and three times their difference is 18. if we draw a graph of the outcome of the equation for all the possible values of the unknowns.50. For example.98 Simultaneous. A train from London to Birmingham takes 2 hours for the journey. xy does not feature in the equation. There are generally two possible values for the unknown. a quadratic equation is one that contains the square of the unknown number. as we shall see in the next study unit. A quadratic equation is obtained by making such an expression equal to a similar expression or to a number.30. where y stands for x2 Note that these are examples of quadratic expressions. usually. 2. the train would take 12 minutes longer. QUADRATIC EQUATIONS All the equations we have studied so far have been of a form where all the variables are to the power of 1 and there is no term where x and y are multiplied together – i. Find the original income from advertisements and sales using a calculation method. the following are quadratic expressions in x: 4x2 7x 2x2 x 1 1 2 x2 2 We can also have quadratic expressions in x2. Quadratic. Exponential and Logarithmic Equations Questions for Practice 3 1. 3x4 2x2 5 is a quadratic in x2. 6 kg of tea and 11 kg of sugar cost £13. If the average speed of the train were decreased by 5 miles per hour. One of the features of quadratic equations is that. but no higher power. The income from advertisements and sales for a college magazine amounted. © ABE and RRC . they cannot be solved definitively. Now check your answers with those given at the end of the unit.e. 4. For example. since it may be also be written as: 3(x2)2 2(x2) 5. Find the cost of tea and sugar per kilo. Again. x may have any value. the result is a curve. one of the factors must be zero: either: x0 or: (x 3) 0 further. then either x 0 or y 0. or (x 5) 0 x 0. reflecting this. They need not both be 0. we do not need to give it in full every time. or x 5 The important thing to remember is that this applies only to a product the value of which is zero. Example 3: Consider the following equation: (x 5)(x 2) 4 We must rearrange this so that the RHS is zero. as is shown in the following examples. we cannot tell either of their values – x may be 3. Exponential and Logarithmic Equations 99 A second feature. Having understood the principle of this procedure. x 0. or if x 3. First of all. or x may be ½. we shall examine this later. since the product of two numbers is zero only when one of them is itself zero. If x 0. while the LHS can be expressed in factors. and if y 0. is that when we draw a graph of the outcome of a quadratic equation for all the possible values of the unknown. y may have any value. The following example simplifies the steps. the product of the two factors x and (x 3) is zero. then x 3 so the equation is satisfied if x 0. the above method of solving an equation applies only when the RHS (righthand side) of the equation is zero. This is true whatever the value of the product. For example. etc. in which case y must be 4. if x and y are two quantities and all we are told about them is that their product is 12. Therefore. we cannot determine their values unless we are told something more. Example 2: Consider the following equation: x(x 5) 0 Either. with the one exception that if xy 0. Quadratic.Simultaneous. but one of them must be. if (x 3) 0. we have to expand the product of the factors: x2 7x 10 4 © ABE and RRC . This important fact provides the easiest method of solving a quadratic equation. Therefore. in which case y must be 24. Example 1: Consider the following equation: x(x 3) 0 Here. Solving a Quadratic Equation by Factorisation If an equation contains two unknown numbers. remember to change the sign of the number as you transfer it to the RHS of the equation. However.. no matter how simple an equation is. though. Then. you may give the answer at once. or (x 1) 0.100 Simultaneous. always copy it down first exactly as it is set.. x 6. You will. 3. If you do this. On the other hand.. or the terms are not already collected on one side of the equation. You should have found these questions very simple and you may even have been able to work them out in your head. 6. Quadratic. but do not just guess. or x 1 Check this in the original equation just to justify to ourselves that the method works: (x 5)(x 2) 4 if x 6: (6 5)(6 2) 1 4 4 if x 1: (1 5)(1 2) (4)(1) 4 Questions for Practice 4 Solve the following equations: 1. Exponential and Logarithmic Equations Then we rearrange the equation so that the RHS is zero: x2 7x 6 0 Next we have to re-express the LHS in factors: (x 6)(x 1) 0 Now we can say that either (x 6) 0. Commonly. x(x 1) 0 (x 1)(x 2) 0 3x(x 1) 0 4x2 0 (2x 1)(x 3) 0 (3x 2)(2x 3) 0 Now check your answers with those given at the end of the unit. 2. or x 2½ One of the reasons that the questions just given were so easy is that the LHS of the equations were already expressed in factor terms. 5. you may prefer to omit the "either . For example: (4x 3)(2x 5) 0 4x 3. have to work as follows: © ABE and RRC . or . or 2x 5 x ¾." step. 4. equations are not so simple – the factors may not already have been found. therefore. if you are sure of the working.. Factorise the LHS expression – the product of the factors is then zero. Example 4: Consider the following equation: 7x2 23x 60 First. Exponential and Logarithmic Equations 101 Collect all the terms on the LHS. Work through the following two examples carefully to make sure that you understand the process thoroughly. Equate each factor in turn to zero. Finally. Quadratic. collect all the terms on the LHS.Simultaneous. give your answers as alternative values for x: 5 x 1 7 or 5 Example 5: Solve x(5x 2) 3(2x 1) Expand the product of the factors: 5x2 2x 6x 3 Collect the terms: 5x2 8x 3 0 Factorise the LHS: (5x 3)(x 1) 0 5x 3. so that the RHS is zero: 7x2 23x 60 0 Next factorise the LHS expression: (7x 12)(x 5) 0 Then equate each factor in turn to zero: either (7x 12) 0. so that the RHS is zero. or x 5 Finally. or x 1 x 3 5 or 1 © ABE and RRC . where possible. give your answers as alternative values for x. or (x 5) 0 which may be expressed as: 7x 12. 9. there is a quicker and easier method than the factor method of solving a quadratic equation. Quadratic. 3. There is. 2. Consider the following equation: 25x2 36 0 Using the factor method. however. Exponential and Logarithmic Equations Questions for Practice 5 Solve the following equations: 1. 4.2. 8. or 5x 6 x 1. a simpler method of obtaining this answer and you should use it whenever there is no term in x in the given equation: 25x2 36 0 25x2 36 x2 36 25 © ABE and RRC .102 Simultaneous. 12.2 or 1. 5. 10. 6. 7 x2 5 x x2 4x 4 0 x2 10x 24 x(x 2) 2(2 x) x2 11x 30 0 2x2 5 11x 10x2 11x 3 0 (x 3)(x 3) 8x (x 1)2 (x 1)2 34 (x 1)2 (x 1)2 x2 9x2 25 12x2 12x 9 Now check your answers with those given at the end of the unit. Solving Quadratic Equations with no Terms in x In particular circumstances. 11. 7. this would be solved as follows: (5x 6)(5x 6) 0 5x 6. it is: b 2a 2 Remember that this needs to be added to both sides of the equation to keep it balanced. it can be used to establish a formula.2 5 Questions for Practice 6 Solve the following equations: 1. b c b2 b x x a a 4a 2 2a 2 2 © ABE and RRC . 2. 5. x2 16 4x2 9 3x2 27 0 x2 a2 0 9x2 1 Now check your answers with those given at the end of the unit. where a. Exponential and Logarithmic Equations 103 To solve this. even where no real roots can be found. In this case.Simultaneous. Rearranging the equation we get: ax2 bx c 0 ax2 bx c x2 b c x a a At this point. we need to "complete the square". take the square root of each side. Quadratic. b and c may have any numerical value. Consider the equation ax2 bx c 0. remembering that there are two square roots numerically equal. Solving Quadratic Equations by Formula Since the method just used for solving a quadratic equation applies to all cases. 3. which means adding or subtracting the expression that is needed in order to be able to factorise the LHS. 4. but opposite in sign: x2 36 25 x 6 or 1. always write it down first.718 12. b 15. paying great attention to the signs. although you should use the factor method whenever possible.72 or 1. Quadratic.236 10 10 1. then write the values of a. c in the particular case. c 5 Therefore: x 4 16 60 4 8. Exponential and Logarithmic Equations Factorising the LHS we get: b b 2 .12 (to two decimal places) © ABE and RRC .28 (to two decimal places) Example 2: Consider the following equation: 3x2 4x 5 0 x b b 2 4ac 2a where: a 3. When using it. c 11 Therefore: x 15 225 220 15 2.4ac x+ 2a 4a 2 2 x b 2 4ac b 2a 4a 2 b b 2 4ac 2a 2a x x b b 2 4ac 2a This is an important formula. It can be used to solve any quadratic equation.718 6 6 4. Example 1: Consider the following equation: 5x2 15x 11 x b b 2 4ac 2a 0 where: a 5.104 Simultaneous.79 or 2. b. b 4.718 or 6 6 0. For example: 2x2 4x 5 0 Applying the formula: x b b 2 4ac 2a where: a 2. Remember that the square of a number is always positive.Simultaneous. 3. 2x2 13x 4 0 5x2 3x 4 0 x2 5x 2 0 4x2 7x 1 0 x2 x 1 0 Now check your answers with those given at the end of the unit. Exponential and Logarithmic Equations 105 Example 3: Consider the following equation: 3 2x2 4x First arrange equation with x2 term positive: 2x2 4x 3 0 Then apply the formula as usual. then the square root will be a whole number and you should have solved the equation by the factor method. Questions for Practice 7 Solve the following by means of the formula: 1. 4. © ABE and RRC . and that no real square root can be found for a negative number. c 5 Therefore: x 4 16 40 4 4 24 4 There are no real roots to this expression. Therefore. Quadratic. b 4. If b2 4ac is a perfect square. if the term b2 4ac is negative. 5. no real roots of the equation can be found. 2. (Calculators differ.) By performing this calculation. Logarithms A logarithm (also known simply as a log) is used to write an expression involving powers. Quadratic.7182818 Questions for Practice 8 Using your scientific calculator. This value occurs so frequently when mathematics is used to model physical and economic phenomena that it is convenient to write it simply as e. Some scientific calculators may require you to enter 1 before pressing the ex button.e. 4. It is common to work with logarithms using the number 10 as the base. 5. LOGARITHMIC AND EXPONENTIAL FUNCTIONS So far in this study unit we have considered simultaneous and quadratic equations. In other words. Have a go on your scientific calculator now to obtain the value of e. Having mastered the application of powers in Study Unit 2. a logarithm is the power to which a fixed number (base) must be raised in order to produce a given number. you should have no problem handling logarithms! In essence. The Exponential Number e There is a particular constant used in many mathematical calculations whose value (to seven decimal places) is 2. you have just demonstrated that: e1 2. find the following values: 1. 102 100 (i.7182818. Using common logarithms we can rewrite this expression as: log10 100 2 © ABE and RRC . it is known as the exponential constant or exponential number.106 Simultaneous. Let us work through some examples to aid our understanding of the application of logarithms. 10 10 100). To do so. referred to as common logarithms. e0 e3 e-2 e1/2 e-3 Now check your answers with those given at the end of the unit. Let us first consider the expression 102. In this section we will now move on to consider logarithmic and exponential functions. so you may have to experiment with your own calculator to find out how to obtain the value of e. 3. From our understanding of powers. 2. 10 is the base and 2 is the power or index. Exponential and Logarithmic Equations D. simply locate and press the ex button on your calculator and enter 1. log10 50 log10 45 log10 2 log10 0.) © ABE and RRC . (Note – calculators differ.7182818) must be raised in order to produce the number 2. Using your scientific calculator. simply locate and press the ln button on your calculator and enter the number you want to find the natural logarithm of.693" or "the natural log of 2 is 0. To do so.693 is the power to which the base value of e (i. In other words. a logarithm is basically an alternative way of writing an expression involving powers. 3. have a go at confirming the above results. Thus.Simultaneous. so you may have to experiment with your own calculator to find out how to obtain the common log of a number. simply locate and press the log button on your calculator and enter the number you want to find the common logarithm of. have a go at confirming this result. Exponential and Logarithmic Equations 107 This statement can be read as the "log to the base 10 of 100 is 2".693 This statement can be read as the "log to the base e of 2 is 0. 3 is the power to which the base number 10 must be raised in order to produce the number 1000). 2 is the power to which the base number 10 must be raised in order to produce the number 100.e. 2. the value of 2. To do so.4 Now check your answers with those given at the end of the unit. 0. so you may have to experiment with your own to find out how to obtain the natural log of a number.e.e.7182818). (Note – calculators differ. 5.e. Some calculators may require you to enter the number you want to find the logarithm of before pressing the log button. Some calculators may require you to enter the number you want to find the natural logarithm of before pressing the ln button. 1 is the power to which the base number 10 must be raised in order to produce the number 10).) Questions for Practice 9 Using your scientific calculator. find the logs of the following values: 1. Using this logic. Another base that we frequently use when working with logarithms (in addition to base 10) is the exponential number e (i. Logarithms to the base e are commonly referred to as natural logarithms and are denoted by ln or loge. Using your scientific calculator. Quadratic.693". 4. For example: ln(2) 0. log10 10 1 (i.4 log10 4. 2. then: log10 1000 3 (i. In other words. 3. think of a number for y and then calculate the natural logarithm of this number to find x (i. Then. using the formula y ex). Quadratic. Use your calculator to confirm this natural logarithmic and exponential relationship. for any base: log(p q) log p log q p log log p log q q log pn n log p if y axn then n (log y log a) ÷ log x log 1 0 ln(ex) x and finally. Relationship between Logarithms and the Exponential Number From this discussion of natural logarithms and the exponential number. 2. thus demonstrating that an inverse relationship exists between the natural logarithm function f(x) ex and the exponential function f(y) ln(x). Here we give the various rules of logarithms (which can be equally applied to common logarithms (i. so if y ex. then ln(y) x. 5. do not forget that: © ABE and RRC . find the natural logarithms of the following values: 1. 4. which have already been described in Study Unit 2. The resulting value of y should be the same as the initial number you chose for y.3) ln(2002) ln(10) ln(38) Now check your answers with those given at the end of the unit. to the base 10) and natural logarithms (to the base e)). The natural logarithm function f(x) ex is the inverse of the exponential function f(y) ln(x).76) ln(0. This is not a mistake! When we take the natural logarithm of an exponential function the result is the exponent. we can conclude that: ln(ex) x Laws of Logs and Exponents Numbers expressed in logarithmic or exponential form obey the same rules applicable to powers and indices. there should be some logic to the following rules. Thus. Since logarithms are basically powers.e. First.108 Simultaneous. you may have identified that a relationship exists between the two.e. Notice that the variables x and y have been reversed. Thus. Exponential and Logarithmic Equations Questions for Practice 10 Using your scientific calculator. using this value for x. calculate the value of y using the exponential function y ex. ln(1. Exponential and Logarithmic Equations 109 Using these laws. Quadratic.Simultaneous. 2. Example 1: Simplify the following logarithm expression into a single log term: log(x) 2 log(y) Using the rule [log pn n log p] we can rearrange the second part of the expression: log(x) log(y2) Using the rule [log(p q) log p log q] we can rearrange the expression: log(xy2) Example 2: Simplify the following logarithm expression into a single log term: 3ln(A) 4ln(B) Using the rule [log pn n log p] we can rearrange both parts of the expression: ln(A3) – ln(B4) p Using the rule [log log p log q] we can rearrange the expression: q ln A4 B 3 Questions for Practice 11 Simplify the following logarithmic expressions to a single log term: 1. 3. we can simplify some expressions involving logarithms to a single log term. © ABE and RRC . log(x) log(x 4) 4 log(2) 2 log(3) log(12) ln(x2y) ln(y) 3 ln(x) Now check your answers with those given at the end of the unit. as well as how to solve equations in general. © ABE and RRC . Exponential and Logarithmic Equations Solving Equations using Logarithms and the Exponential Number Before completing this section on logarithms and exponents. Questions for Practice 12 Solve the following equations.110 Simultaneous. let us bring together everything we have learnt about them. 4x 100 e2x 10 log 3x 3 Now check your answers with those given at the end of the unit. 3. by answering the following questions. Quadratic. 2. giving your answer correct to 2 decimal places (you will need a scientific calculator to evaluate logs and natural logs): 1. Simultaneous. multiply (i) by 3 so that the coefficients are equal and then subtract (ii): 9x 12y 3 9x 5y 10 7y 7 y 1 Substituting y 1 in (i) we get: 3x 4 1 3x 5 2 x 13 (Check in (ii): 15 5 10) 2. multiply (i) by 5 and (ii) by 3 so that the coefficients are equal and then add: 25x 15y 70 21x 15y 114 46x 184 x 4 Substituting x 4 in (i) we get: 20 3y 14 3y 6 y2 (Check in (ii): 28 10 38) The full workings for the next three answers are not given. x 2 3 4 or to eliminate x. multiply (i) by 5 and add the equations: 16x 44. 3x y 8 x 5y 4 (i) (ii) To eliminate y. 3. y 1 4 © ABE and RRC . 5x 3y 14 7x 5y 38 (i) (ii) To eliminate y. 3x 4y 1 9x 5y 10 (i) (ii) To eliminate x. Quadratic. but the method is indicated. multiply (ii) by 3 and subtract (i): 6y 4. Exponential and Logarithmic Equations 111 ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. subtract 7 times (i) from 2 times (ii): y 1 5. 2x 11y 23 7x 8y 50 or (i) (ii) To eliminate y. y 3 10 substituting for y in (ii): 1 2 1 2 1 2 6 x 6 5 14 x 14 7 5 4 x 65 1 x 13 3 5 7. subtract 2 times (ii) from (i): 3 3 y 12. so divide by this to get a simpler equation: x5y4 (i) 13 15 7 2 To eliminate x. Exponential and Logarithmic Equations 4. Quadratic. 1 1 2 x 3 y 12 5 (i) (ii) x 15 y 21 1 2 1 7 Notice that (i) has a factor 3. y 15 Substituting in (i) for y gives: 1 2 1 2 x34 x7 x 14 © ABE and RRC .112 Simultaneous. 9x 10y 12 2x 15y 5 or to eliminate x. subtract 2 times (i) from 9 times (ii): y 3 5 6. y 13. subtract 2 times (ii) from 3 times (i): x 2 (i) (ii) x 2y 14 6 1 To eliminate x. subtract 2 times (i) from (ii): ( 15 5 )y 13. 10y 36. x 3 y 16 1 2 2 (i) (ii) To eliminate y. subtract 8 times (i) from 11 times (ii): x 6 to eliminate x. the numbers are 17 and 23. Quadratic. if half the sum of two numbers is 20: 1 2 1 2 (x y) 20 or x y 40 3(x y) 18 or xy6 2y y 34 17 (ii) Eliminating x by subtraction. change each amount of money to pence. if three times their difference is 18: Substituting for y in (i). and b Therefore: x 1. Let: x cost of tea per kg in pence y cost of sugar per kg in pence. Then 6x 11y 1.Simultaneous. 2. we get: x 17 40 x 23 Therefore. and b .330 11x 6y 1. we get: (i) and. Let: x first number y second number Then. Exponential and Logarithmic Equations 113 Questions for Practice 2 Let a 1 1 . As the units must all be the same.730 (i) (ii) © ABE and RRC . Then express the equations using these terms: x y 3a 20b 13 7a 10b 2 Solving these in the normal way gives: a 1. and y 2 Questions for Practice 3 1. 4.780 20x 20y 13.114 Simultaneous.330 x 1. Let: x the average speed (in miles per hour) y the distance (in miles) We can use a formula for the time taken: distance time taken speed (ii) To eliminate y from the pair of equations. Thus.250 y 50 Substituting for y in (i) we get: 6x 550 1.50 per kilo. Let: x the original income from advertisements (in £) y the original income from sales (in £). Exponential and Logarithmic Equations To eliminate x. and the cost of sugar £0. Quadratic. multiply (i) by 20: © ABE and RRC . the original income from: advertisements £340 sales £330.380 x 340 Substituting for x in (i): 340 y 670 y 330 Thus.30 per kilo. subtract 6 times (ii) from 11 times (i) to get: 85y 4.400 and subtract this from (ii) to get: 7x 2.50 (i) (x 9 8 1 8 x) (y 5 6 1 6 1 y) 657 2 x 1 y 657 2 Multiplying this by 24 to eliminate the fractions. 3. the cost of tea £1. in the first year: x y 670 In the second year: 1 2 (x 12 2 % x) (y 16 3 % y) 670 12. we get: 27x 20y 15.330 550 6 x 130 Therefore. Quadratic. 2. x 2 or 1 x 2 x 12 or 2 x 2 or 2 x 5 or 6 x 1 2 or 5 1 x 3 or 2 5 © ABE and RRC . 5. we get: y 110 Thus: average speed for the journey 55 mph distance from London to Birmingham 110 miles. we get: y y 2x (i) 1 2 5 x 11 (ii) Subtracting (ii) from (i) gives: 0 1 5 x 11 1 5 11 x 55 x x 55 Substituting for x in (i). 5. 6. 3. Exponential and Logarithmic Equations 115 So at average speed: y 2 hours x and at the slower speed: y 1 2 hours 12 minutes 2 5 hours (x 5) Rearranging these equations into standard linear form. 2. x 0 or 1 x 1 or 2 x 0 or 1 x0 x x 1 2 2 3 or 3 1 or 1 2 Questions for Practice 5 1. Questions for Practice 4 1. 3. 4. 7.Simultaneous. 6. 4. 6.4 -0.050 Questions for Practice 9 1. 10. x 9 or 1 x 4 or 4 (written as x 4) x 0 or 4 2 x 13 1 x 1 2 or 1 2 Questions for Practice 6 1. 2. Quadratic. 12.4 0. 2. 5.116 Simultaneous. 3.64 4. x 4 1 x 1 2 x 3 x a 1 x3 Questions for Practice 7 1. 3. 0. 5. log 10 50 1.16. Exponential and Logarithmic Equations 8. 9. 2. 4.44 0.643 © ABE and RRC .59 1.649 e-3 0. 5.699 log 10 45 1.56. 4.24. 3. 2.301 log 10 0. 4. 0.62. 5. 0. 4. 0.086 e-2 0.135 e1/2 1. 3. 1.653 log 10 2 0.62 Questions for Practice 8 1. e0 1 e3 20.18 1.398 log 10 4. 11.32. 2x In(10) x In10 2.32 e2x 10. Exponential and Logarithmic Equations 117 Questions for Practice 10 1. 4. Using the rule [log pn n log p]: 4 log(2) log(24) log(16). Using the rule [log(p q) log p log q]: log(x) log(x 4) log(x (x 4)) log(x2 4x) 2. 4x 10. 2. 2. 3x 10 3 . x 10 3 ÷3 x 333. ln(1. x log 100 log 4 . 5.3) 1.Simultaneous. x 1 15 log 3x 3. x 3. and 2 log(3) log(32) log(9) p Using the rules [log(p q) log p log q] and [log log p log q]: q 4 log(2) 2 log(3) log(12) log(16) log(9) log(12) 16 9 log 12 log(12) 3.76) 0.602 ln(10) 2.303 ln(38) 3. 3.638 Questions for Practice 11 1.5653 ln(0.33 © ABE and RRC .204 ln(2002) 7. Using the rules [log(p q) log p log q] and [log pn n log p]: ln(x2y) ln(y) 3ln(x) ln(x2) ln(y) ln(y) ln(x3) ln(x5) Questions for Practice 12 1. 3. Quadratic. 118 Simultaneous. Exponential and Logarithmic Equations © ABE and RRC . Quadratic. Answers to Questions for Practice © ABE and RRC . D. C.119 Study Unit 5 Introduction to Graphs Contents Introduction A. E. Basic Principles of Graphs Framework of a Graph Drawing a Graph Graphing the Functions of a Variable Representation Ordered Pairs Graphs of Equations Linear Equations Quadratic Equations Equations with x as the Denominator Exponential and Logarithmic Equations Gradients Gradient of a Straight Line Equation of a Straight Line and Gradient Using Graphs in Business Problems Business Costs Break-even Analysis Economic Ordering Quantities Plotting Time Series Data Page 120 121 121 122 124 125 126 126 127 132 136 139 141 142 145 147 147 151 152 155 160 B. There are many different types of graph. You will immediately realise that since equations are a mathematical expression of the relationship between two variables. which way its graph will slope identify. by knowing how to interpret it. We shall spend some time on this in the unit. From this. One element of this is the need to use graph paper. it is essential that you are very careful when drawing graphs. and thus provide a measure of something that changes with time.120 Introduction to Graphs INTRODUCTION For many problems and issues which fall within the remit of quantitative methods. it is possible to analyse and solve them both mathematically and graphically. whether two lines are parallel describe a shaded region on a graph in terms of the lines bounding it use graphs to solve a variety of problems. indicate a trend in the data from which we might be able to make further predictions. Whenever one measurable quantity changes as a result of another quantity changing. For example. showing graphically what happens to one variable when the other changes. exponential and logarithmic graphs. You will find it very useful to have a supply to hand as you work through this unit. The graph might also be used to approximate a solution to a particular problem. from looking at their equations. since the answers must be read from the graph and hence cannot be more accurate than the thickness of the pencil. from an equation. For this reason. The importance of graphs is that they give a visual picture of the behaviour of the two variables. quadratic (curved line). they are less accurate. many graphs involve time as one of the variables. and calculate its gradient from its graph or its equation decide. and we cover some of the key design aspects here. f(x)) calculate and tabulate the values of y from an equation in x draw graphs from linear. for example. quadratic. The picture may. it is possible to represent equations by means of graphs. Graphical approaches to problem solving are often easier to carry out than mathematical ones. but here we will only be concerned with linear (straight line). © ABE and RRC . However. exponential and logarithmic equations identify the characteristics of a straight line in terms of its equation. it is possible to draw a graph. Objectives When you have completed this study unit you will be able to: explain the construction of graphs calculate functions of x (that is. information can very quickly be found. This study unit introduces the principles and role of graphs in such problems. Line graphs illustrate the relationships between two variables. We have shown the main gridlines here as grey.1: Sample graph framework Height of plant (cm) 6 5 4 3 y-axis 2 1 origin 0 1 2 3 4 5 6 7 8 9 co-ordinate (3. This grid may actually be marked on the paper (as is the case with graph paper. 0–9 on the x-axis and 0–6 on the y-axis). The values of the variables are marked along each axis according to a scale (here. We shall start by considering the framework of the graph. Particular points within the graph are identified by a pair of numbers called co-ordinates which fix a point by reference to the scales along the axes. The area within the axes (known as the plot area) is divided into a grid by reference to the scale on the axes. Here this is the time since planting (which causes the height of the plant to increase). The scale need not be the same on both axes. The x-axis is the horizontal axis and usually displays the independent variable – i. The following figures show the relationship between the height of a plant and the time since it was planted. but the dividing marks do need to be equally spaced along the axis and the scale used must be clear to the reader. Each axis should be titled to show clearly both what it represents and the units of measurement. BASIC PRINCIPLES OF GRAPHS A graph is a diagram illustrating one or more mathematical relationships between two variables. We shall use this to explain some of the basic features of graphs and their construction.e. the variable which changes according to changes in the other variable.3) co-ordinate (6. Not all values need to be marked. The first co-ordinate number always refers to a value on the scale along the x-axis and the second to the scale on the y- © ABE and RRC .2) x-axis Time since planting (days) A graph has two axes along which the values of the variables are shown: The y-axis is the vertical axis and usually displays the dependent variable – i. Framework of a Graph Figure 5.Introduction to Graphs 121 A. dotted lines.e. which is why it is much easier to draw graphs using graph paper) or may be invisible. the variable which causes change in the other variable. Here this is the height of the plant (which changes in accordance with the time since planting). 1) x-axis Drawing a Graph When drawing a graph. These are the values given (or determined) for firstly. Draw the axes and mark off the scales along them. you may be given the following data about the growth of a plant: © ABE and RRC . Determine the range of values which will need to be shown on each of the axes. If you are not using graph paper.2) 3 2 1 2 3 co-ordinate (3. The point at which the two axes meet is called the origin. for the dependent variable. use a ruler to mark the scale and select easy-to-identify measurements for the divisions (such as one cm for each main unit). the framework of the graph will look like that shown in Figure 5. Remember that the scales must be consistent along each axis.0). Finally. since the range of values for the independent variable will determine the range of values needed for the dependent variable on the yaxis. Points on the graph are identified by their co-ordinates.122 Introduction to Graphs axis. number the main divisions along each scale and give the whole graph a title.2: Graph framework including negative scales 3 co-ordinate (2. You should start by considering the x-axis. Give each axis a title.2) 2 1 y-axis 3 2 1 0 1 co-ordinate (1. secondly.2. The aim is usually to draw as large a graph as possible since this will make it easier to plot the co-ordinates accurately. Figure 5. Now you are ready to start plotting the graph itself. If you are using graph paper. In this case. but do not need to be the same on both. the independent variable and. We have shown two points on the framework in Figure 5. This has the co-ordinate reference (0. and also make it easier to read. note that the x and y-axes may extend to include negative numbers. Thus. follow this procedure for constructing the framework. use the thick lines as your main divisions.1 – make sure you understand their referencing. The full graph derived from the previous data on the height of a plant and extended over nine days is given in Figure 5. then the plot line will form a curve.) However. when joined and extended across the whole scale of the x-axis. draw the curve through the points in one movement. you should join them all up. When you are satisfied that the path of the curve is smooth. if your graph is a quadratic graph. Figure 5. (In fact.) Once you have plotted all the given points.1) you count one unit across (along the x-axis) and then count one unit up (along the y-axis). The points should be joined with a smooth curve.7). The easiest way of doing this is to trace through the points with a pencil. you only need to plot two points which. which is somewhere completely different! (Remember that if the x co-ordinate is negative you count to the left and if the y co-ordinate is negative you count downwards.3: Graph of height of plant against time 6 Height of plant (cm) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 Time since planting (days) Now that we have reviewed the basic principles.3. give the complete graph for all values of the independent variable. to draw a linear graph.Introduction to Graphs 123 Height of plant (cm) Day 1 Day 3 Day 5 Day 7 1 2 3 4 The points should be plotted with a small x using a sharp pencil. we shall move on to consider in detail the way in which variables and equations are represented on graphs. but without touching the paper. If your graph is a linear graph. Be careful not to get these in the wrong order or when you come to plot the co-ordinate (7. to find the shape of the curve. © ABE and RRC . To plot the co-ordinate (1. all the points will be in a straight line and you can join them using a ruler.4) you will have plotted the point (4. when x 1. You could start by making the following table: x x2 3 y x2 3 2 4 3 7 1 1 3 4 0 0 3 3 1 1 3 4 2 4 3 7 3 9 3 12 4 16 3 19 Alternatively. If y f(x). For example. as follows: x yx3 0 3 1 4 2 5 3 6 4 7 5 8 Let's look at another example. y 6. you could tabulate just the first and the last lines. y will be equal to 3. x. y 4. when x 3. working the rest in your head as you go along. and so on. if y x 3.124 Introduction to Graphs B. then when x is 0. For convenience. We can also find a single function of x by substituting that value directly into the given equation. so that we can obtain a succession of values for y by giving successive values to x. GRAPHING THE FUNCTIONS OF A VARIABLE The statement that something is a "function of x" simply means that the value of that something depends upon the value of a variable quantity. for example: f(x) x2 2x Replace x by 3 and find the value: f(3) 32 2 3 15 Replace x by 2 and find the value f(2) (2)2 2(2) 0 © ABE and RRC . It is denoted by f(x). y 5. when x 2. Suppose you are given the equation y x2 3 and are asked to find the values of y from x 2 to x 4. y will have a different value for each value of x. as this is a simple problem. we usually tabulate our values for x and y. 4. The first figure is the value of x along the scale of the x (horizontal) axis. x y 3x 2 2. and the second figure is the value of y along the scale of the y (vertical) axis. Representation From the tables we see that as the value of x changes.5 3. One way of showing how these values change is by means of a graph.5 2 2. C is (3. Complete the table for y 3x 2. If f(x) x2 2 find: (a) (b) f(2) f(3) Now check your answers with those given at the end of the unit. 2 ).1 2 ). E is (1.Introduction to Graphs 125 Questions for Practice 1 1.1). in Figure 1 1 5.1). so does the value of y.2). A is the point (2. If f(x) 5x 2 find: (a) (b) f(2) f(3). The values in the table can be expressed as a pair of co-ordinates which may be plotted on a graph.0) and F is (0. © ABE and RRC . 4. For example. D is (5. B is (4.5 1 1. Which one of the following functions could be represented by the following table? yx4 y 2x 1 x y 3 7 y x2 6x 16 y x3 1 5 11 7 15 9 19 11 23 0 0. (1. Quadratic equations – those of the second degree in x and y (i.23) These are then the points which we would plot if we were going to draw the graph of the function in question.15) (9. Thus. such as y 2x2 x 1. As the name implies. (2. (0.126 Introduction to Graphs Figure 5. The table in Question 2 of Questions for Practice 1 could be written as a set of ordered pairs as follows: (3. logarithmic and hyperbolic equations.e. y ½x2 3). Such graphs have a different curved shape – a hyperbola. When dealing with ordered pairs. Linear equations – equations of the first degree in x and y (i.7). © ABE and RRC .4: Co-ordinates for y f(x) 3 B 2 1 E 2 A 6 4 2 0 1 4 D 6 C 2 F Ordered Pairs Another way of describing the varying values of x and y is by means of a set of ordered pairs. (1.4). This is a very useful and helpful way of noting the points we required in the graphical representation of a function. the graph of expressions like these will always be a straight line.1).e. but no higher power. and if the function was x2 3 for the same values of x it would be (1.4). exponential. Such equations always give a curved graph with a characteristic "U" shape. you will normally be told the values of x with which you are concerned. C.11) (7. y 5x 1).3). (0. called a parabola.3). a Hyperbolic equations – we shall look at graphs from the expression y where a is a x constant. GRAPHS OF EQUATIONS In this section we shall consider graphs of linear. containing no higher power of x than the first. containing the square of the independent variable. quadratic.7) (5.5). (2.19) (11. such as y 2x 3. we would write (1. if we had to write down the ordered pairs of the function 2x 1 for integral (whole number) values of x from 1 to 2 inclusive.1). the minimum number of points you ought to plot for a linear graph is three. x 2 and x 2. Linear Equations A linear equation is of the form y mx c. Example 1: Draw the graph of y 2x 3. Suppose we choose x 0.g. (Note that such an equation can be written in different forms. on joining them. When plotted. we have the graph of y 2x 3. If we choose any three values of x we can find the corresponding values of y and this will give us the co-ordinates of three points which lie on the line. So. where m and c are constants. It is even better if we know three points.Introduction to Graphs 127 Exponential and logarithmic equations – an exponential or logarithmic function occurs when the independent variable appears as the exponent or as a logarithm. using the third as a check. e.5: Graph of y 2x – 3 4 2 0 -3 -2 -1 -2 -4 -6 -8 0 1 2 y x 3 © ABE and RRC .) To draw a line we must know two points on it. Figure 5. y b ax. y 2x 3: when x 0: y 0 3 3 when x 2: y (4 3) 1 when x 2: y (4 3) 7 This is better tabulated as: x y 2x 3 0 3 2 1 2 7 We now have three points which we can plot and. equations of both exponentials and logarithms produce non-linear lines. This will always give a straight line. Taking the given equation. This is because the constant c in the general equation y mx c is always the point where the line passes through the y-axis (i. The point at which the line crosses an axis is called the intercept. plot your points and join them up. y rises from 2 to 8 (i. 2).6 you will notice that the line y 2x 2 meets the y-axis at the point where y 2. We have thus found that in the equation y mx c: the constant c the point where the line crosses the y-axis the constant m the slope of the line.6: Graph of y 2x + 2 8 6 4 2 3 0 2 y x -4 -2 0 0 -2 -4 -6 -8 2 4 If you look carefully at Figure 5.e. mark the scale.e. Draw the graph of y 3x 2. for values of x from 3 to 2.6. increases by two).128 Introduction to Graphs Example 2: Try this one yourself. where x 0). 0 and 2. We shall return to these points in some detail later. Then compare your result with Figure 5. (0. Then draw up a table of values – fill in the spaces for yourself to find the co-ordinates of the points to be plotted. © ABE and RRC . increases by four). (2. say 3. 8). Now draw your two axes to cover the range of values for x specified (although it is good practice to extend them beyond this – leaving some spare room at each end of each axis). First choose three values of x. The change in the value of y for a change of one in the value of x is given in the equation y mx c by the constant m. A further fact which emerges is that when x goes from 0 to 2 (i. The y intercept is where the line crosses the y-axis and the x intercept is where the line crosses the x-axis. Figure 5. x y 3x 2 You should have found the points: (3.e. 7). 1. Working these out gives us the following table of values: x y 4 3 1 1 2 1 Now plot these points and join them.Introduction to Graphs 129 Example 3: Suppose we are asked to draw the graph of 3y 2x 1. Note that. Figure 5. we must change the equation to the form y mx c. from x 4 to 2.7: Graph of 3y 2x + 1 3 y 1 x -2 -1 -1 0 1 2 3 4 -3 © ABE and RRC . Before we can work out the table of values in this case. 2.7. as in the title to Figure 5. having divided both sides of it by 3: y 2x 1 3 Now choose some values of x – let's say 4. this is still a graph of the original equation. So we rewrite the equation. 130 Introduction to Graphs Example 4: We are not limited to having just one line plotted on a graph.8: Graphs of y ⅓ x and y 8 – x 10 8 6 4 2 y A x -4 -2 0 0 -2 2 4 6 8 10 You will notice that the lines intersect (i.2) © ABE and RRC .e. Taking values of x between 4 and 10. cross) at A (6. We can draw two or more graphs using the same axes – a very useful method of comparing different equations of the same variables.8. Figure 5. we get the following tables of values: For y 1 3 x x y 3 1 2 10 0 0 4 4 3 1 10 2 For y 8 x x y The graphs are shown in Figure 5. we could draw the graphs of y 1 3 x and y 8 x on the same axes. For example. 3.2) (4. For values of x from 3 to 6. For values of x from 3 to 6. draw on the same axes the graphs of: (i) yx2 (ii) 3y 6 x Identify the point of intersection. 6. from x 1 to x 2.0) (3. for values of x from 4 to 6. be a line parallel to the x-axis at a distance c from it.3) (0. then. from x 4 to x 2.3) (1. and plays a very important part in calculations concerning a straight-line graph.2) Write down the co-ordinates of the points where (i) and (ii) each cut the lines (iii) and (iv). draw on the same axes the graphs of: (i) 7. and those you drew in the Questions for Practice. including 0. draw on the same axes the graphs of: (i) y5 (ii) x 5 (iii) y 5x (iv) y1x 5 Describe briefly the resultant lines. On the same axes. Characteristics of Straight-Line Graphs If you consider the graphs we have drawn as examples. Lines parallel to an axis Consider the case where m 0.0) (1.6) (0. Now check your answers with those given at the end of the unit. Draw the graph of y 2x 3. y 2x (ii) y 2x 3 (iii) y 2x 4 Describe briefly the resultant lines and their intercepts. The general equation for a straight line is y mx c where m and c are constants which may have any numerical value. 4. and draw the line through them: (i) (ii) (iii) (iv) (3. y is always equal to c. This means y c. plot the following pairs of points on squared paper. The value of m is known as the coefficient of x.Introduction to Graphs 131 Questions for Practice 2 1. 5. The graph of y c must. we can start to identify a number of characteristics of straight-line graphs. Draw the graph of y 5x 4. Draw the graph of y 2 x. For values of x from 3 to 6. © ABE and RRC . 2. In the same way.5) (2. Note that y 0 is the x-axis itself. x c is a line parallel to the y-axis at a distance c from it. x 0 is the y-axis. Whatever value x has. if m is kept the same and c has different values. so the graph of y mx is a line through the origin (0. then the equation of the line is y mx. where a. Quadratic Equations A quadratic equation is of the form y ax2 bx c. Thus.9. Intercepts The distances from the origin to the point where the line cuts the axes are called the intercepts on the axes. and the part that is usually asked for is the part that does a "U" turn. y 4 3x) the line will slope down from left to right instead of up. The general shape of a quadratic equation will be as shown in Figure 5. in order to draw a quadratic graph accurately. the intercept on the y-axis is always at c. the lines are parallel. Direction of slope If the coefficient of x is negative (for example. b and c are constants. In y mx c. Parallel lines When the coefficient of x is the same for several lines. © ABE and RRC . The maximum and minimum values of the quadratic equation can be found by careful inspection of the graph. When x 0 it is clear that y 0. if m is negative the line slopes down. we get a set of parallel straight lines. if m is positive the line slopes up from left to right. in y mx c. They are found by putting x 0 and y 0 in the equation of the graph. as it does when the coefficient of x is positive.132 Introduction to Graphs Lines through the origin If c 0. you need as many points as possible – especially around the dip. Thus. Figure 5.9: General shape of a quadratic equation y y x or x where a is negative where a is negative In contrast to linear graphs. This will always give you a curved graph.0). 0)).10. What are the co-ordinates at which the function y x2 is minimised? As ever.e.Introduction to Graphs 133 Example 1: Draw the graph of y x2 taking whole number values of x from x 3 to x 3 inclusive.10: Graph of y x2 10 y 8 6 y x2 4 2 x 0 -3 -2 -1 -2 0 1 2 3 © ABE and RRC . as shown in Figure 5. the co-ordinates where y x2 is minimised are (0. Figure 5. x y x2 3 9 2 4 1 1 0 0 1 1 2 4 3 9 Plot these seven points on the graph and then join the points with a smooth curve. From the graph you will see that function y x2 is minimised where y 0 and x 0 (i. the first step is to draw up a table of values for the given range of x. 11: Graph of y 2x2 18 16 14 12 y y 2x2 A 10 8 6 4 B y 10 y 1.25 the values of x and y where the function y 2x2 is minimised.134 Introduction to Graphs Example 2: Draw the graph of y 2x2 from x 3 to 3. © ABE and RRC .25 C 2 x 0 -3 -2 -1 0 1 2 3 Reading off from this: (a) (b) (c) when y 10.25 and x 2.3 when the function y 2x2 is minimised. there are two values of x. Table of values: x x2 y 2x2 3 9 18 2 4 8 1 1 2 0 0 0 1 1 2 2 4 8 3 9 18 Plot the points and draw the graph as in Figure 5. Then from the graph find: (a) (b) (c) the values of x when y 10 the value of y when x 1.25. at points A and B: x 2. Figure 5.25 when x 1. the value of y is at the point C: y 3. x 0 and y 0.11. we can see that the minimum value of y occurs when x 1. © ABE and RRC . Figure 5.12.Introduction to Graphs 135 Example 3: Draw the graph of y x2 2x 1 from x 2 to 3.12: Graph of y x2 – 2x + 1 10 y 8 6 4 2 x -2 -1 0 0 1 2 3 Reading from the graph. At what value of x does the minimum value of y occur? Table of values: x x2 2x 1 y x2 2x 1 2 4 4 1 9 1 1 2 1 4 0 0 0 1 1 1 1 2 1 0 2 4 4 1 1 3 9 6 1 4 Plot the points and draw the graph as in Figure 5. 5 3 0. Draw the graph of y 4x2 from x 2 to 2. x y 1 x 1 1 2 0. Draw the graph of the profit (£000s) function x2 8x 1 from x 0 to 8. 1 2 x2 from x 4 to 4.25 5 0. where x ('000) represents output. Draw the graph of y From the graph find: (a) (b) 2.33 4 0.13.17 1 is not defined (i.136 Introduction to Graphs Questions for Practice 3 1.2 6 0. since the expression meaning). Draw the graph of y x2 from x 3 to 3.e.000. the direction of which may vary. 3. Find the values of x when y 9. Equations with x as the Denominator a where a is a constant. x They result in a characteristic curve. it has no 0 Note that we have avoided taking x 0. the value of y when x 3. These equations (hyperbolic equations) take the form of y Example 1: Let's look at the case where a 1: y 1 x We'll take x to have values from x 1 to x 6. From the graph find: (a) (b) the output at which profit is maximised the output range at which profits generated amount to at least £8. which gives the table following and the graph in Figure 5. © ABE and RRC . Now check your answers with those given at the end of the unit. 4.2 the values of x when y 7. 8 0. as x x gets bigger.6 0.57 8 0.5. x From the graph.14: Graph of y 4 3 2 1 0 0 2 4 6 8 Reading from the graph.8 4 x 6 0.13: Graph of y 1. This graph illustrates the general shape of graphs representing y Example 2: Draw the graph of y 4 for values of x from x 1 to x 8.2 1 0.33 4 1 5 0. Table of values: x y 4 x 1 4 2 2 3 1.Introduction to Graphs 137 Figure 5. © ABE and RRC . x 1.5 Figure 5. when y 2. There is also no point corresponding to x 0.4 0.5.7. In each case. find the value of x when y 2.67 7 0. y gets smaller.2 0 0 2 4 1 x 6 a . x From your graph.15: Shape of graphs from the equation y a x a x 0 x a x a x © ABE and RRC .5. x a x These are illustrated in Figure 5. so the shape of graphs developed from equations where x is the denominator change: a (i. Now check your answers with those given at the end of the unit. Draw the graph of y 2 from x 1 to x 6. where both the a and x are positive) gives a graph which slopes down from left x to right a gives a graph which slopes down from left to right x a gives a graph which slopes up from right to left x a gives a graph which slopes up from right to left. find the value of x when y 1.e.15. Figure 5.138 Introduction to Graphs Questions for Practice 4 1. Just as we saw that the shape of the quadratic graphs changed according to whether the coefficient of x2 is positive or negative. In the final part of this section we will now plot graphs for exponential and logarithmic equations.Introduction to Graphs 139 Exponential and Logarithmic Equations So far in this unit we plotted and interpreted graphs for linear and quadratic equations (also known as polynomial functions). Plot a graph of this exponential function and from the graph find the value of x when y 10.37 0 1. © ABE and RRC . or otherwise.60 5 148. The mathematics behind exponentials and logarithms was discussed in Study Unit 4. calculate the value of y ex for the values of x from x 2 to x 5. Table of values: x y ex -2 0.16: Graph of equation y ex 160 140 120 100 80 60 40 20 0 -2 -1 0 1 2 3 4 5 Reading from the graph. x 2.1 4 54. Using a scientific calculator.5 (approximately).14 -1 0.00 1 2.39 3 20. when y 10.41 Figure 5.72 2 7. Example 1: Let us begin by plotting the graph of an exponential function. 099 3. Use the table to draw a graph of S against t and from your graph estimate the value of S when t is very large.693 2. Now check your answers with those given at the end of the unit. 20 and 25 (give your answers to 2 decimal places).0 1. Plot a graph of this logarithmic function and from the graph find the value of x when y 0. (a) (b) Construct a table of values for S for t 0. The sales S (in thousands of units) of a product after it has been on the market for t months is given by the equation: S 100(1 e0. when y 0. 10.253 4. Table of values: x ln (x) 0. Questions for Practice 5 1.17: Graph of equation ln(x) 1.5 0. 20 and 25 (give S values to the nearest whole number).5 1 2 3 4 -1 Reading from the graph.5 to x 4.5 1.5 0.5 0.0.00 1. © ABE and RRC . 2. Using a scientific calculator.916 3. 10.5.405 2.0 1. 15.0 0.693 1.140 Introduction to Graphs Example 2: In this example we will plot a graph of the natural logarithmic function. x 1.5 1 0.386 Figure 5.5 0 0 -0.5. Use a scientific calculator to construct a table of values for y log10(x) when x 0. calculate the values of y ln(x) for the values of x from x 0.0 0. 15. Plot these values on a graph.1625t). 18: Gradient of a slope In (i). the gradient of s is 8/2 4 s 8 s s s s 2 2 1 2 2 2 4 2 (i) (ii) (iii) (iv) (v) © ABE and RRC . In the other cases. the gradient of s is 4/2 2 in (v). A horizontal line has a gradient of zero: A vertical line has an infinitely large gradient: In between.5 in (iii). the slope of the line(s) increases as we go from (i) through to (v). the gradient is defined as the height of the right-angled triangle divided by the base. GRADIENTS The gradient of a line is a measure of how much it slopes.18. the gradient of s is 1/2 0. the gradient of s is 2/2 1 in (iv). the gradient increases: or decreases Now. In the five shapes.Introduction to Graphs 141 D. consider the five slopes illustrated in Figure 5. Therefore in (ii). the gradient (s) is zero (no slope). Figure 5. 3) and (4. 8) y 9 8 7 6 5 5 4 3 2 1 3 1 2 3 4 5 x (b) Make this line the hypotenuse of a right-angled triangle by drawing in a horizontal "base" and a vertical "height".20. Example 1: Plot the points (1. as illustrated by Figure 5.142 Introduction to Graphs If a line slopes down from left to right (instead of up).20: Gradient of the line connecting points (1. 2 Gradient of a Straight Line We can easily calculate the gradient of a straight line joining two points from the co-ordinates of those points. as shown in Figure 5. © ABE and RRC . we shall examine the method by reference to plotting the line on a graph.19: Negative gradient 2 s 2 The gradient of the line s is 2 1.8) and find the gradient of the line joining them. then the gradient is negative.3) and (4. This is denoted by introducing a negative sign. The procedure would be as follows: (a) Plot the two points on a graph and join them to give a sloping line.19. Firstly. Figure 5. Figure 5. 4) and (2.3) and find the gradient of the line joining them.21: Gradient of the line connecting points (–1. –3) y 4 3 2 1 7 2 1 1 2 3 x 1 2 3 3 This time we shall work out the lengths of the base and the height from the co-ordinates of the points. height 5 2 13 base 3 As we have seen. base 2 (1) 3 height 4 (3) 7 We can check that this is correct by reference to the plotted triangle on the graph. 3) and 4. Therefore the gradient height 1 7 2 3 3 base © ABE and RRC . join them and construct a right-angled triangle with the line forming the hypotenuse: Figure 5. at step (c). 8) is 1 3 Note that. Plot the points. the gradient of the straight line joining (1. 2 Therefore.Introduction to Graphs 143 (c) Work out the lengths of the base and the height from the graph: base 3 height 5. Note that the slope is negative. we could have worked out the lengths of the base and the height from the co-ordinates of the points: length of the base difference between x co-ordinates 4 1 3 length of the height difference between y co-ordinates 8 3 5 Example 2: Plot the points (1. 4) and (2. the gradient is given by the equation Note that the slope is positive. the gradient would be: 32 1 3 ( 2) 2 ( 1) 21 3 © ABE and RRC .22: Gradient of a straight line y B B (i) (ii) A C C A O x Note that these lengths are directed lengths: AC is positive if the movement from A to C is in the same direction as O to x.1) and (7. take any two points A and B on the line.2) as follows: 2 1 1 75 2 For points (1.144 Introduction to Graphs General formula We can now develop a general formula for the gradient of a straight line by reference to its co-ordinates. say. AC is negative. (0. hence the gradient CB is positive. Therefore. and CB parallel to the y-axis.) Taking the lines in Figure 5. and negative if the movement from A to C is in the opposite direction to O to x.22. We can further see that AC is the difference between the co-ordinates along the x-axis and CB is the difference between the co-ordinates along the y-axis. CB is positive if the movement from point C to B is in the same direction as O to y. we could calculate the gradient where the points are.3). AC In line (ii). and the gradient is negative. AC and CB are both positive.0).2) and (2. (5. To find the gradient of a line. In the same way. (The point at O is the origin. CB is positive. Draw AC parallel to the x-axis. we find the following: In line (i). i. Figure 5.e. The gradient is then given by the ratio CB : AC. Consider a general straight line drawn on a graph. Using this method.23: Proving the equation of a straight line y B D P G O A x OAPB is a rectangle and OAGD is a rectangle. For example: Find the gradient of a line with the equation 6x 3y 19.23. Figure 5. This is shown in Figure 5. D is the point on that line where it crosses the y-axis. OB y. Also. the intercept OD c and the gradient of DP m. and P is another point along the line. given any linear equation. Let: OA x. its equation can be found immediately. given any straight-line graph. its intercept is negative. Note that c is the intercept that the line makes with the y-axis.Introduction to Graphs 145 Equation of a Straight Line and Gradient We can now go back to consider aspects of the equation of a straight line which we have met before. If the line cuts the y-axis below the origin. Then: m PG PG BD DG OA OA BO DO yc OA x yc we get: x Rearranging the equation m mx y c y mx c This is the general equation of any straight line. we can find the gradient of the line (m) immediately. c. We need to put the equation into the general form y mx c: y 6x 19 3 © ABE and RRC . 1) (4.6) (2.7) (2.5) and (2. (a) (b) (c) (d) (e) (f) 2.4) (0.24. (1. Figure 5. c 5.0) (7. the gradient is 6 2 3 Finally. m The intercept.6) and (1.146 Introduction to Graphs Therefore.24: Deriving the equation of a straight line 10 8 6 4 2 0 0 2 4 6 8 The gradient of the line.1) and (5. the equation of this line is: y or 6y 5x 30 5 x5 6 Questions for Practice 6 1.0) and (1. consider the graph in Figure 5.2) and (4.3) and (3.7) Find the gradients of the lines with the following equations: (a) (b) (c) 2y 3x 6 0 7y 4x 7y 1 2x © ABE and RRC . 10 5 5 6 6 Therefore. Find the gradients of the straight lines joining the following points. printing room and warehouse. not least in calculating profit. its fixed costs would include the rent on its offices. Write down the co-ordinates of the points where the line and the curve intersect. but the variable costs rise with the number of books produced. One of these is to divide them into fixed and variable costs. USING GRAPHS IN BUSINESS PROBLEMS In this final part of the unit. These are an important consideration in many aspects of business finance. 3. for example by acquiring different premises which may be cheaper. Draw an x-axis from 3 to 5 and a y-axis from 4 to 4. Variable costs are those which vary with output – such as the cost of raw materials used as part of the production process. E. On them draw the graphs of y x 1 and 2y 4 x. Fixed costs are those which the business must always meet. 2 2 4. the fixed costs remain the same. Now check your answers with those given at the end of the unit.5) and (3. Whether they print six or six hundred books a week. (Note that fixed costs are only fixed in the short term – over a long period. This means that. there are many ways of looking at costs. they may change. within that. Business Costs All businesses incur costs in the production and supply of the goods and/or services which they provide. Draw the graph of y 3x for values of x from 3 to 3. 5.3). Find (a) the gradient and (b) the equation of the straight line joining them. We can turn this observation into an algebraic expression: let: x number of units produced (level of production) b variable costs © ABE and RRC . so do the variable costs.) If we consider variable costs. if we take the case of a printing company. we shall consider how we can use the properties of graphs we have been examining to analyse and solve business problems. as the level of production changes. Write down the co-ordinates of the point where the two lines intersect. Thus. Plot the points (6.Introduction to Graphs 147 The next three questions bring together aspects of the whole unit so far. and the variable costs would include paper and ink. we can say that they are a function of the level of production. On the same axes draw the line y 5. which at heart is simply: revenue from sales less costs of production and supply (or cost of sales) The analysis of costs is a subject of its own and. regardless of the number of units produced – such as rent on buildings. They are actual examination questions. So. These do not vary with the level of production.25: Graph of relationship between variable costs and level of production Costs 6 (£000s) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 Variable costs Production (thousands of units) We can work out the gradient of the line on this graph. if we know the level of fixed costs. we can show this on the same graph as in Figure 5. so we can express the relationship of variable costs to units of production by the following equation in the form of y mx c: a or 3a x Now consider fixed costs. One of those co-ordinates must be (0. 1 3 x0 © ABE and RRC . 0). Let us suppose that we do have such values and can produce such a graph (Figure 5.25): Figure 5.26. since when there is no production. If we know that one other point on the graph is (9. then the gradient (which is m) is: m 30 3 1 90 9 3 We also know that the intercept (c) of the line is 0. based on two co-ordinates. 3). no variable costs will be incurred.148 Introduction to Graphs Then: b f(x) This is where we started the unit! We know that we can draw a graph of this relationship if we know some of the values for x and y. Note that the line for total costs is in fact a combination of the lines for variable and fixed costs. © ABE and RRC .26: Graph of fixed costs and level of production Costs 6 (£000s) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 Variable costs Fixed costs Production (thousands of units) We can see that this line is parallel to the x-axis. again on the same graph as in Figure 5. This is represented by the equation: b 2 (where b is fixed costs) Businesses are not only interested in costs as either fixed or variable.27. We could work out a table of values for it (which we shall skip here) and draw a graph. We can say that: total costs variable costs fixed costs If total costs y. Its intercept is at the level of fixed costs and its gradient is that of variable costs. This has been done. in Figure 5. but in their total costs. and using the notation from above. with an intercept at 2.Introduction to Graphs 149 Figure 5. we get the following equation: yab Substituting from the equations established for a and b.26. we get: y 1 3 x2 This is a standard linear equation. Despite this. Which one is more economical? We can develop equations for these charges and then use a graph to show which company would be more economical at different journey lengths: Let: x number of miles y total cost (£) Then. This takes place at 5 miles. A business wishes to use a courier company on a regular basis. The point at which the two lines intersect means that the values of x and y are the same for each line. A charges £7 and B charges £4. Thus at that point.00 per mile. Some of these situations are shown by dotted lines on the graph. which is why we can show the variable and total costs lines as linear. Consider the following situation.00 plus £1. whilst company B charge £2. it is cheaper to use company B. For any distance above this. for company A. A charges £12 and B charges £14 at 2 miles. variable costs do not necessarily rise at a constant rate with increases in production. the cost of using each company is the same. whereas for distances less than five miles. © ABE and RRC .28. it is a useful way of looking at costs.27: Graph of total costs and level of production Costs 6 (£000s) 5 4 3 2 Total costs Variable costs Fixed costs 1 0 0 1 2 3 4 5 6 7 8 9 Production (thousands of units) Note that the chart is only a model. the costs is: y 2x The graph is shown in Figure 5.150 Introduction to Graphs Figure 5. total cost may be defined as : yx5 For company B. In real life. company A is cheaper. For example: at 7 miles.00 per mile but no call out charge. Company A charge a call out charge of £5. 28: Comparison of courier company charges Costs 20 (£) 15 Company B Company A 10 5 0 0 2 4 6 8 10 Journey length (miles) Break-even Analysis We can further develop the costs graph by including revenue on it.Introduction to Graphs 151 Figure 5. Here. © ABE and RRC .29. with both costs and revenue being £3.29: Costs and revenue graph Costs (£000s) Revenue 8 6 Break-even point 4 Total costs 2 0 0 1 2 3 4 5 6 7 8 9 Sales(thousands of units) The point at which the revenue and cost lines intersect is known as the break-even point.000 units. we get Figure 5. revenue exactly equals costs.27. this takes place at sales of 3. At that point.000.) Figure 5. If we assume that each unit sells for £1. Revenue is the total income from sales of the units produced. (Note that we are assuming that all units produced are sold. we can say that y x where y is the income from sales (in £) and x is the number of units sold. Adding the graph of this to the costs graph shown in Figure 5. Questions for Practice 7 1. It has worked out that the following costs for its best selling teapot "Standard" are: fixed costs £20. Now check your answers with those given at the end of the unit. there are also costs involved in holding a stock of unused materials – very large stocks can incur large warehousing costs. how many must Teapots R U Ltd. On the other hand. The fixed cost of producing them is £80. One of the central issues in purchasing materials to be used as part of the production process is how much to order and when. we need to look at the relationships between the usage.667 whilst revenue is only £2. costs are £2.000 units. so buyingin stock on a daily basis might not be appropriate. Economic Ordering Quantities We shall now look at the role of graphs in analysing and helping to solve a purchasing problem.00.000 whilst revenue is £6.00. supply and costs involved.000. However. 2. The purchaser must determine the optimum point. In order to do this. find the number of bicycles needed to be sold for the company to break even. but just review the general approach and the way in which graphs can be used. Give your answer to the nearest 100 bicycles.150 variable costs £5. The break-even point represents the minimum position for the long term survival of a business.000 – there will a loss of £667 at sales of 6.152 Introduction to Graphs Any sales above the break-even point will mean that revenue exceeds costs. We shall not work through the problem in detail. Teapots R U Ltd makes and sells a small range of teapots. and can also mean getting cheaper prices for buying in bulk.000 units. sell each week to break even? Use a fixed and variable costs chart (break-even chart) to find the answer.50 per Standard teapot. If the price of a Standard teapot is £12. costs exceed revenue and there will be a loss. © ABE and RRC . Set against this. By "economic order quantity" we mean the quantity of materials that should be ordered in order to minimise the total cost.00. buying in larger quantities has the effect of spreading the fixed costs of purchase over many items.000 with a variable cost per bicycle of £36. Using a fixed and variable costs chart (break-even chart). For example: at sales of 2. Many producers and most retailers must purchase the materials they need for their business in larger quantities than those needed immediately. A company makes bicycles and sells them for £80. there are costs in the ordering of materials – for example. each order and delivery might incur large administrative and handling costs. costs are £4. and there will be a profit.000 – there will be a profit of £2. At sales levels below the break-even point though. including transportation costs. 800 units. The cost of placing the order is £30.e.) (6 £30) (£2 300) £780. Q days later this will be down to zero. so the average holding is 150. then the stock will last for Q days.600 units over one year and these units are used at a constant rate. (As we place an order for 3. (600 is the average holding on orders of 1. © ABE and RRC . If it buys in lots of 1. i. the average stock held in storage will be 1.180 (36 orders of 100 each. (Six orders of 600 each. There will be 1.600 units and use them evenly through the year. Make 36 orders at 10-day intervals. Make three orders at four-monthly intervals.Introduction to Graphs 153 Suppose we have the following problem: (a) (b) (c) (d) (e) We want to order 3.000Q sprockets in stock immediately after a purchase has been made. Let us consider another problem. when another purchase will be made. But is it really the cheapest? We would need to try many other options before we could be sure.30). It could purchase daily. monthly orders of 300. Make six orders bimonthly. We have several options as to how we could tackle this: There are many other possibilities but these will serve our purpose.630.290. The average stock level is 500Q (see Figure 5. Place a single order for the whole year. The relevant costs for each option above are: (a) £30 for one order storage costs of £2 1.000Q sprockets.) (b) (c) (d) (e) (12 £30) for 12 orders (£2 150) storage £660 (In this case.800 £3. As soon as we order.000 sprockets every day. but considers this to be unacceptable as the cost of raising an order and sending a van to a cash-and-carry to collect the goods has been estimated at £100.) (3 £30) (£2 600) £1.200 each.) We can see that the cheapest option is (b). giving the average holding of 50. The cost of keeping one unit in stock for a year is £2.) (36 £30) (£2 50) £1. We order as soon as existing stock runs out. The graph of stock level is a saw-tooth. Make 12 monthly orders. we receive the whole order instantaneously – unrealistic but let us keep the problem simple. A company called Oily Engineers Ltd uses 1. giving the average holding of 300. the orders are for 300 each. The cost per annum depends on the number of orders placed.154 Introduction to Graphs Figure 5. on average. Each time the amount purchased is doubled. © ABE and RRC . This shown in Figure 5.30: Stock levels over time Stock level 1. then the warehousing costs are doubled as there are.31. i. twice as many items in the warehouse. The graph is shown in Figure 5. a batch every day: cost £100 250 £25.e.5 £6. the cost per annum halves. assume that there are 250 working days each year: (a) (b) (c) Sprockets purchased in batches of 1.000 Sprockets purchased in batches of 2.000.250. a batch every fourth day: cost £100 62.32. For example. as it is a fixed amount per order.000.e. a batch every other day: cost £100 125 £12.500 Sprockets purchased in batches of 4. i.31: Warehousing costs Cost Size of purchased lot The cost of purchasing is not linear. i.000.000Q 0 Q 2Q 3Q Days The warehousing cost is linear – if the amount purchased in each order is doubled. Figure 5.e. This can give us a good visual guide of the trend in the time series data. For example. quarterly or monthly intervals could be examined in an attempt to predict their future behaviour. we can identify the economic order quantity from the minimum cost position on the graph.33. Such sets of values observed at regular intervals over a period of time are called time series. Figure 5. The first step in analysing the data is to plot the observations on a scattergram. The analysis of this data is a complex problem as many variable factors may influence the changes which take place. Plotting Time Series Data Businesses and governments statistically analyse information and data collected at regular intervals over extensive periods of time. with time evenly spaced across the x-axis and measures of the dependent variable forming the y-axis.Introduction to Graphs 155 Figure 5. sales values or unemployment levels recorded at yearly. © ABE and RRC .33: Total costs of purchasing Cost Total cost Minimum cost Storage or warehousing cost Purchasing cost Size of purchased lot Thus. to plan future strategies and policies.32: Cost of purchasing (per annum) Cost Size of purchased lot These costs can be added to the warehouse costs to give a total cost per annum. The graph of this is shown in Figure 5. (Do not confuse this with a histogram. Naturally. say thermometers. © ABE and RRC .34. The only function of these lines is to help your eyes to see the pattern formed by the points.156 Introduction to Graphs Let us consider a factory employing a number of people in producing a particular commodity. as shown in Figure 5. The following table shows the number of days lost through sickness over the last five years.) Note the following characteristics of a historigram: It is usual to join the points by straight lines.34. depending on the scale of the yaxis. which is a type of bar chart. Each year has been broken down into four quarters of three months. A historigram is simpler than other scattergrams since no time value can have more than one corresponding value of the dependent variable.1: Days lost (per quarter) through sickness at a thermometer factory We can plot this information using a particular type of scattergram. Intermediate values of the variables cannot be read from the historigram. The scattergram of a time series is often called a historigram. It is possible to change the scale of the y-axis to highlight smaller changes. Year 2003 Quarter 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Days Lost 30 20 15 35 40 25 18 45 45 30 22 55 50 32 28 60 60 35 30 70 2004 2005 2006 2007 Table 5. Every historigram will look similar to the one in Figure 5. at such a factory during the course of a year some employees will be absent for various reasons. We have assumed that the number of employees at the factory remained constant over the five years. 34: Historigram (scattergram) of days lost (per quarter) through sickness (2003–2007) Lost 70 days 60 50 40 30 20 10 0 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 2003 2004 2005 2006 2007 There are four factors that influence the changes in a time series over time. These are the trend. and irregular or random fluctuations.34.34 that the trend is definitely upwards. we can clearly see that lost days from work (per quarter) have increased over the period. © ABE and RRC . However. You can see in Figure 5. Based on this trend line. in spite of the obvious fluctuations from one quarter to the next. If we were to draw a simple linear trend line through the time series data shown in Figure 5. The trend is the change in general level over the whole time period and is often referred to as the secular trend. then we would get a trend line similar to that shown in Figure 5.Introduction to Graphs 157 Figure 5. A trend can thus be defined as a clear tendency for the time series data to travel in a particular direction in spite of inter-period fluctuations (be they large or small). seasonal variations. you only need to concern yourself with the trend.35. cyclical fluctuations. from approximately 25 days lost per quarter in 2003 to approximately 50 days lost per quarter in 2007. 158 Introduction to Graphs Figure 5. with linear trend line Lost 70 days 60 50 40 30 20 10 0 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 2003 2004 2005 2006 2007 © ABE and RRC .35: Historigram (scattergram) of days lost (per quarter) through sickness (2003–2007). Sales (£000) 320 200 150 400 400 210 180 450 800 400 340 900 950 500 400 980 Plot the time series on a historigram. 2. Using a ruler. draw a linear trend line through the time series plotted in Question 1 and comment on the trend. © ABE and RRC .Introduction to Graphs 159 Questions for Practice 8 The following table shows sales of premium bonds at a large post office over five years: Year and Quarter 2004 1 2 3 4 2005 1 2 3 4 2006 1 2 3 4 2007 1 2 3 4 1. Now check your answers with those given at the end of the unit. Table of values: x y 5x 4 1 9 0 4 2 6 Graph of y 5x – 4 6 4 2 0 -2 -1 -2 -4 -6 -8 -10 0 1 2 y x 3 © ABE and RRC .5 2 4 2. x y 3x 2 2. 0 2 0.160 Introduction to Graphs ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1. (a) (b) (a) (b) f(2) 10 2 8 f(3) 15 2 17 f(2) 4 2 6 f(3) 9 2 11 Questions for Practice 2 1. 3.5 1 1 1. 4.5 5.5 0.5 y 2x 1 is the only function which satisfies all values in the table.5 2. Introduction to Graphs 161 2. Table of values: x y2x 4 6 1 1 6 4 Graph of y 2 – x 6 y 4 2 0 -4 -2 -2 0 2 4 x 6 -4 © ABE and RRC . Table of values: x y 2x 3 4 5 0 3 2 7 Graph of y 2x + 3 8 6 4 2 0 -4 -3 -2 -1 -2 -4 -6 0 1 y x 2 3. cuts (iv) at (1. cuts (iv) at (2. 2 2 cuts (iii) at ( 1 3 . 1 y 5 x is a line through the origin. y 5 is a line parallel to the x-axis and 5 units above it. 7. sloping down from left to right with a gradient of 1 in 5 (i.162 Introduction to Graphs Note that for the next four questions. The lines intersect at (3. cuts (iii) at (1.7.6.1) to 1 decimal place. 1). along 5 and down 1).8). 2.7 and 3. y 5. 0). we do not show the graphs themselves. 4. we get: (a) (b) when x 3.7 © ABE and RRC . x 5 is a line parallel to the y-axis and 5 units to the left of it.e.7) to 1 decimal place.2.1 when y 7. x 3.e. Questions for Practice 3 1. (i) (i) (ii) (ii) 5. Reading off from it. These three graphs are parallel lines: (i) (ii) (iii) passes through the origin. makes an intercept of 4 on the y-axis and 2 on the x-axis. crosses the yaxis at 3 and the x-axis at 1½). 0. if you go along 1 unit you go up 5 units). makes an intercept of 3 on the y-axis and 1½ on the x-axis (i.e. sloping up from left to right with a gradient 5 in 1 (i. Table of values: x x2 y 1 2 4 16 x2 8 3 9 4½ 2 4 2 1 1 ½ 0 0 0 1 1 ½ 2 4 2 3 9 4½ 4 16 8 The graph is shown on the next page. (i) (ii) (iii) (iv) 6.4. 2. 2 3 ) or (1. y 5x is a line through the origin. Introduction to Graphs 163 Graph of y 8 1 2 x2 6 4 2 0 -4 -2 0 2 4 2. Table of values: x y x2 3 9 2 4 1 1 0 0 1 1 2 4 3 9 Graph of y –x2 -3 -2 -1 0 0 1 2 3 -2 -4 -6 -8 -10 © ABE and RRC . when y 9. y 8 or above).e. when x 4. x 1.5 and 1.e. Table of values: x x2 4x2 y 4x2 2 4 16 16 1 1 4 4 0 0 0 0 1 1 4 4 2 4 16 16 Graph of y –4x2 -2 -1 0 0 1 2 -4 -8 -12 -16 Reading from the graph. At an output of between 1.000 units (i.000 and 7. Reading off from it. © ABE and RRC . between x 1and x 7).e.5.164 Introduction to Graphs 3.000 (i.000 units of output. i. we get: (a) (b) Profit is maximised (£17.000) at 4. Table of values: x -x2 8x 1 y x2 8x 1 0 0 0 1 -1 8 2 -4 16 3 -9 24 4 -16 32 5 -25 40 6 -36 48 7 -49 56 8 -64 64 1 1 1 8 1 13 1 16 1 17 1 16 1 13 1 8 1 1 The graph is shown on the next page. 4. profits generated are in excess of £8. 5.5 2 x 5 0.3.4 6 0. Table of values: x y 2 x 1 2 2 1 3 0.5 0 0 1 2 3 4 5 6 When y 1.Introduction to Graphs 165 Graph of y x2 + 8x + 1 18 Profit (£000s) 16 14 12 10 8 6 4 2 0 0 1 2 3 4 5 6 7 8 Output (£000s) Questions for Practice 4 1.5 1 0.33 Graph of y 2 1.67 4 0. x 1. © ABE and RRC . 70 10 1.30 25 1.6 0.2 0 0 5 10 15 20 25 2. © ABE and RRC .2 1 0.1625t) 120 100 80 60 40 20 0 0 5 10 15 20 25 From the graph.00 5 0.1625t ): t S 0 0 5 56 10 80 15 91 20 96 25 98 Graph of S 100(1 e0. when t is very large S will be approximately 100.4 1.40 Graph of y log10(x) 1.8 0.000. Table of values for y log10(x): x y log10(x) 0 0.4 0. Table of values for S 100(1 e0.166 Introduction to Graphs Questions for Practice 5 1.18 20 1.00 15 1. (a) (b) (c) (d) (e) (f) 2.Introduction to Graphs 167 Questions for Practice 6 1. © ABE and RRC . 73 4 2 3 1 2 6 (1) 7 1 23 53 3 7 6 ( 1) 1 23 ( 2) 1 3 5 5 50 ( 4 ) 2 6 6 4 0 ( 4 ) 1 2 7 ( 1) 8 7 ( 2) 9 1 22 40 4 y y y 3x 6 2 4 x 7 2x 1 7 gradient gradient gradient 3 2 4 7 2 7 Tables of values: x yx1 2y 4 x y2 x 2 1 2 1 2 3 4 1 0 5 4 x 2 2 0 2 0 2 4 2 2 0 x 1 2 1 x 3 y2 The graph is shown on the following page. (a) (b) (c) 3. 5 2 6 3 13. 1) 4.5 Graph of y 14 12 10 8 6 3x 2 and y 5 2 3x 2 2 y y5 4 2 0 -3 -2 -1 0 1 2 3 From the graph.8. 5) and (1. 5).5 2 6 1 1. Table of values: x y 3x 2 2 3 13. © ABE and RRC . the lines intersect at point (2. you will see that the line and curve intersect at (1.5 0 0 1 1.168 Introduction to Graphs Graph of y x – 1 and 2y 4 – x 4 3 yx1 2y 4 x 2 1 0 -3 -2 -1 -1 -2 -3 -4 0 1 2 3 4 5 Reading from this.8. Introduction to Graphs 169 5. the equation is: y x 1 or 3y 2x 3 Questions for Practice 7 1. © ABE and RRC .1). 5 4 3 2 1 0 0 1 2 3 4 5 6 (a) (b) Gradient 53 2 3 63 2 3 The line cuts the y-axis at (0. Therefore. Break-even chart: Revenue/ Costs (£000s) 50 40 30 20 10 0 0 1 2 3 4 Sales (000s) The break-even point is at sales of 3.100 Standard teapots. Time series historigram: Sales (£000s) 1000 800 600 400 200 0 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 2004 2005 2006 2007 2.000 per quarter in 2007.000 per quarter in 2003 to approximately £700.800 bicycles (to the nearest 100). Break-even chart Revenue/ Costs (£000s) 250 200 150 100 50 0 0 1 2 3 Sales (000s) The break-even point is at sales of 1. from approximately £250. The historigram clearly shows that sales of premium bonds have increased over the period. Questions for Practice 8 1. Linear trend is shown on the historigram above. © ABE and RRC .170 Introduction to Graphs 2. D. C.171 Study Unit 6 Introduction to Statistics and Data Analysis Contents Introduction A. Answers to Questions for Practice © ABE and RRC . Data for Statistics Preliminary Considerations Types of Data Requirements of Statistical Data Methods of Collecting Data Published Statistics Personal Investigation/Interview Delegated Personal Investigation/Interview Questionnaire Classification and Tabulation of Data Forms of Tabulation Secondary Statistical Tabulation Rules for Tabulation Frequency Distributions Simple Frequency Distributions Grouped Frequency Distributions Choice of Class Interval Cumulative Frequency Distributions Relative Frequency Distributions Page 172 173 173 174 174 175 175 176 176 176 177 179 181 183 185 186 186 188 189 189 191 B. Inferential methods use the available data in order to make general statements about the whole from which that data is only a part. In this study unit we shall review the basis of the statistical method – the collection of facts (data) about a problem or issue – and then consider initial ways of analysing this data so that it has meaning in relation to the problem. For example. we would need to use inferential methods. "Statistics" is a word which is used in a variety of ways and with a variety of meanings. Descriptive methods simply manipulate the available data in order to convey information about it. and with methods of analysing these facts in order to solve or shed light on the problem or issue. Predictions and estimates are typical of the general conclusions which may be made by using inferential methods. we talk of the "statistics" of steel production. These methods are concerned only with the available data. However. In other words. even if that data is only a part of a (possibly much larger) whole. If we determine the average age of the team from that data. Sometimes the subject is called "statistical method". In this course we shall only be concerned with descriptive statistical methods. all professional English football players – and clearly their average age will not necessarily be an accurate estimate of the average age of this larger group. the players are also a part of a much larger group – for example. statistics is concerned with numerical facts about a problem or issue – the facts themselves. namely: The numerical facts themselves – for example. The numerical facts with which statistics are concerned are commonly called data. "statistics" is the title of a subject like "arithmetic" or "chemistry" or "physics". We therefore talk about the collection and analysis of data in respect of a particular problem. The ages of the individual players form the data. If we wanted to draw conclusions about this larger group from the study of the eleven players in the England team. a statistical analysis may be made of the ages of the eleven players in the England football team. and usually for the purpose of simplifying comprehension or appreciation of the facts. in whatever way it is used. The methods used for analysing data can be divided into two categories – descriptive and inferential. Objectives When you have completed this study unit you will be able to: explain the concept of data in statistics distinguish between primary and secondary data explain the differences between quantitative and qualitative variables and between continuous and discrete variables © ABE and RRC . often in the form of summaries or graphical displays. The methods of analysing the facts – in this sense. but. it is always concerned with numerical information. we are applying descriptive methods to tell us something about the eleven players as a whole. Thus. they allow us to infer something about the whole mass of data from a study of just a part of it. There are two particular meanings of the word which concern us.172 Introduction to Statistics and Data Analysis INTRODUCTION This unit is the first of five concerned specifically with statistics. Preliminary Considerations It is important to be aware of these issues as they impact on the data which is to be collected and analysed. It is no use just asking for "output" from several factories – some may give their answers in numbers of items. the more precisely we wish to estimate a value. grouped. The problem as originally put to the statistician is often of a very general type and it needs to be specified precisely before work can begin. and that effort is not wasted by collecting irrelevant data. there are some important points to consider when planning a statistical investigation. or only some? Are we to concern ourselves with our own business only. DATA FOR STATISTICS A statistical investigation involves a number of stages: definition of the problem or issue collection of relevant data classification and analysis of the collected data presentation of the results. and prepare tables from given data explain the concept of a frequency distribution and define the terms class frequency. The degree of precision required in an estimate might affect the amount of data we need to collect. or with others of the same kind? Accuracy of the data To what degree of accuracy is data to be recorded? For example. some in number of inspected batches and so on. Exact definition of the problem This is necessary in order to ensure that nothing important is omitted from the enquiry. In general. some in weight of items. cumulative and relative frequency distributions from given data. class limits and class interval prepare simple. Scope of the enquiry No investigation should be commenced without defining the field to be covered. then the accuracy of the measuring instrument will determine the accuracy of the results.Introduction to Statistics and Data Analysis 173 explain the reasons for classifying and tabulating data. Definition of the units The results must appear in comparable units for any analysis to be valid. the more readings we need to take. If the analysis is going to involve comparisons. So even before the collection of data starts. class boundary. © ABE and RRC . A. then the data must all be in the same units. are ages of individuals to be given to the nearest year or to the nearest month or as the number of completed years? If some of the data is to come from measurements. Are we interested in all departments of our business. for example. there are two methods of adjustment possible. because it is impossible to have other than a whole number of children. However each possible non-numeric value (hair colour say) may be associated with a numeric frequency of occurrence. etc. A discrete variable however can take only certain values occurring at intervals between stated limits. it includes all published statistics. Homogeneity The data must be in properly comparable units. Standardise the data.174 Introduction to Statistics and Data Analysis Types of Data There are a number of different ways in which data may be categorised. determine a relationship between the different units so that all may be expressed in terms of one – in food consumption surveys. for example. "Five houses" means little since five slum hovels are very different from five ancestral mansions. and study them separately. Thus it includes any data which has been subject to the processes of classification or tabulation or which has resulted from the application of statistical methods to primary data. 2. shoe sizes are stated in half-units. then the only possible values are 0. Alternatively. Use units such as "output per person-hour" to compare the output of two factories of very different size. For most (but not all) discrete variables. etc. to which we shall restrict discussion here. if the variable is the number of children per family. and so here we have an example of a discrete variable which can take the values 1. 2½. The desirable qualities of statistical data are as follows. these interval values are the set of integers (whole numbers). Secondary data is any data other than primary data. Requirements of Statistical Data Having decided upon the preliminary matters about the investigation. Quantitative and qualitative data Variables may be either quantitative or qualitative. If the data is found not to be homogeneous. © ABE and RRC . are those for which observations are numerical in nature (for example a person's height or the number of children in a school). in Britain. 1. and has to be classified and processed using appropriate statistical methods in order to reach a solution to the problem. Primary data is therefore raw data. primary data is data which is both original and has been obtained in order to solve the specific problem in hand. Qualitative variables relate to non-numeric observations. such as colour of hair. the statistician must look in more detail at the actual data to be collected. Houses cannot be compared unless they are of a similar size or value. A continuous variable may take any value between two stated limits (which may possibly be minus and plus infinity). Primary and secondary data In its strictest sense. This is because a person's height may (with appropriately accurate equipment) be measured to any minute fraction of a millimetre. For example. Continuous and discrete data Quantitative variables may be either continuous or discrete. However. Quantitative variables. a child may be considered equal to half an adult. is a continuous variable. 2. Height. 1½. (a) (b) Break down a non-homogeneous group into smaller component groups which are homogeneous. The information you require may not be found in its entirety in one source. nationalised industries. in that the investigator has not been personally responsible for collecting it. the United Nations. it is so easy to be slack about this and to run into trouble. B. For example. the term "accident" may mean quite different things to the injured party. but only to reassemble and reanalyse data which has already been collected by someone else for some other purpose. but parts of it may appear in several different sources. It is sometimes difficult to establish the definitions from published information. Uniformity The circumstances of the data must remain the same throughout the whole investigation. This type of data is sometimes referred to as secondary data (as we have mentioned). Acts of Parliament may. Likewise. it is not much use comparing a firm's profits at two different times if the working capital has changed. Of course.Introduction to Statistics and Data Analysis 175 Completeness Great care must be taken to ensure that no important aspect is omitted from the enquiry. the police and the insurance company! Watch out also. © ABE and RRC . the disadvantage is that you could spend a considerable amount of time looking for information which may not be available. By contrast. and it thus came to the investigator "second-hand". When using this method. it can lead to data being obtained relatively cheaply. Another disadvantage of using data from published sources is that the definitions used for variables and units may not be the same as those you wish to use. The source must be reliable and the information up-to-date. this is one of the advantages of this type of data collection. you must establish what it represents. we come to the question of how to collect the data we require. The methods usually available are as follows: use of published statistics personal investigation/interview delegated personal investigation/interview questionnaire. but. It is no use comparing the average age of employees in an industry at two different times if the age structure has changed markedly. Accurate definition Each term used in an investigation must be carefully defined. when using other people's statistics. chambers of trade and commerce and so on. data which has been collected by the investigator for the particular survey in hand is called primary data. METHODS OF COLLECTING DATA When all the foregoing matters have been considered. for changes in definition. for example. before using the data. We can often make good use of the great amount of statistical data published by governments. alter the definition of an "indictable offence" or of an "unemployed person". Although the search through these may be timeconsuming. it is particularly important to be clear on the definition of terms and units and on the accuracy of the data. Published Statistics Sometimes we may be attempting to solve a problem that does not require us to collect new information. for example. you would try to get both people to do part of each town. However the field that may be covered by personal investigation may be limited. For this reason. Delegated Personal Investigation/Interview When the field to be covered is extensive. any difference that is revealed might be due to the towns being different.176 Introduction to Statistics and Data Analysis Personal Investigation/Interview In this method the investigator collects the data personally. © ABE and RRC . The most convenient way to collect such data is to issue questionnaires to the people concerned. For example. a greater response is likely and queries can be answered. personal biases may tend to cancel out. If there are many investigators. If you do. there is sometimes a danger that the investigator may be tempted to select data that accords with preconceived notions. Additionally. Questionnaire In some enquiries the data consists of information which must be supplied by a large number of people. the task of collecting information may be too great for one person. suppose you were investigating the public attitude to a new drug in two towns. or it might be due to different personal biases on the part of the two investigators. This method is usually cheaper than delegated personal investigation and can cover a wider field. Care in allocating the duties to the investigators can reduce the risks of bias. in the UK Population Census.) The distribution and collection of questionnaires by post suffers from two main drawbacks: The forms are completed by people who may be unaware of some of the requirements and who may place different interpretations on the questions – even the most carefully worded ones! There may be a large number of forms not returned. If the forms are distributed and collected by interviewers. In such a case. (A carefully thought out questionnaire is also often also used in personal or delegated interview in order to reduce the effect of personal bias. The people employed should be properly trained and informed of the purposes of the investigation. The personal investigation method is also useful if a pilot survey is carried out prior to the main survey. asking them to fill in the answers to a set of printed questions. This method has the advantage that the data will be collected in a uniform manner and with the subsequent analysis in mind.) However care must be taken that the interviewers do not lead respondents in any way. it is essential to include a reply-paid envelope to encourage people to respond. (This is the method used. as personal investigation will reveal the problems that are likely to occur. Their instructions must be very carefully prepared to ensure that the results are in accordance with the requirements described in the previous section of this study unit. Then a team of selected and trained investigators or interviewers may be used. Do not put investigator A to explore town X and investigator B to explore town Y. The result is that we end up with completed forms only from a certain kind of person and thus have a biased sample. and these may be mainly by people who are not interested in the subject or who are hostile to the enquiry. If we are required to present to the factory management a statement about the age structure of the workforce (both male and female). T Mr Dee. we need to organise it so that we can extract useful information and then present our results. Very often the data will consist of a mass of figures in no special order. Mr Edwards. J Mrs Green. technical qualifications and so on. The cards are probably kept in alphabetical order of names. For example. L Mr 39 20 22 22 32 30 37 33 45 42 24 27 28 44 39 As it stands. but they will contain a large amount of other data such as wage rates. we need to classify the data in some way. F Miss Gunn. One way would be to classify the individual cases by sex: © ABE and RRC . Let us assume that our raw data about employees consists of 15 cards in alphabetical order. L Mr Black. we may have a card index of the 3.000 cards. sex. J Mr Jones. S Miss Dawson.000 employees in a large factory. A Mr Brown.Introduction to Statistics and Data Analysis 177 C. The age of each employee is stated in years. D. Example: Of course we cannot give an example involving 3. C Mrs Edwards. R Mr Gray. What is needed is to classify the cards according to the age and sex of the employee and then present the results of the classification as a tabulation. is usually known as raw data (i. this data does not allow us to draw any useful conclusions about the group of employees. W Miss Carter. so we will look at a shortened version. involving only a small number. Archer. and no one could possibly gain any idea about the topic from merely looking through the cards as they are.e. J Miss Johnson. it is unprocessed). then the alphabetical arrangement does not help us. CLASSIFICATION AND TABULATION OF DATA Having completed the survey and collected the data. W Mr Jackson. as follows. The data in its original form. type of work. In order to do this. before classification. W Mrs Brown. age. A Mr Dawson.178 Introduction to Statistics and Data Analysis Male Archer. T Mr Dee. L Mr 39 22 30 33 45 27 44 39 Female Black. is counted and the results presented in a table. is interested in the proportion of men or women who die at certain ages. L Mr Carter. J Miss 20 22 32 37 42 24 28 An alternative way of classifying the group would be by age. D Mr Edwards. although the data has now acquired some of the characteristics of useful information. splitting them into three groups as follows: Age between 20 and 29 Black. R Mr Gray. In comparing the height of Frenchmen with the height of Englishmen. S Miss Dee. A Mr Brown. W Mrs Brown. but it is not concerned with the age at which John Smith or Mary Brown. we are concerned with things in groups rather than with individuals. L Mr 39 32 30 37 33 39 Again. J Mr 45 42 44 20 22 22 24 27 28 Age between 30 and 39 Archer. L Mr Brown. W Miss Green. to give another example. We need to transform the data into information about the group. J Mr Jones. C Mrs Gray. In statistics though. W Mr Johnson. W Miss Carter. W Mr Jackson. D Mr Jones. The number of cards in each group. and we can do this by tabulation. we are concerned with Frenchmen in general and Englishmen in general. S Miss Dawson. W Mrs Brown. R Mr Gunn. will die. J Miss Age over 40 Edwards. T Mr Edwards. F Miss Gunn. but not with Marcel and John as individuals. it is still a collection of data about individuals. An insurance company. after classification. C Mrs Edwards. J Mrs Johnson. F Miss Jackson. as individuals. © ABE and RRC . J Mrs Green. when tabulating continuous variables such as a person’s age. days. could not be classified into a single group.2. Forms of Tabulation We can classify the process of tabulation into simple tabulation and complex or matrix tabulation. exactly 40 years old.Introduction to Statistics and Data Analysis 179 Table 6.it is to condense an unwieldy mass of raw data to manageable proportions and then to present the results in a readily understandable form. as this person would fall into both the 30 – 40 and 40 – 50 age groups! So choice of date grouping (often referred to as "class intervals") is an important aspect of the classification and tabulation of data. 2 months and 20 days old. In most cases (i. (a) Simple tabulation This covers only one aspect of a set of figures. then it is possible that Mr Archer and Mr Jones. therefore they should not be included in the age group 40 – 49! Thus. Assume that each card carries the name of the workshop in which the person works. Mr Archer and Mr Jones would now clearly fall into the 30 to less than 40 years age group (i. Having understood this.e. 30 to < 40). age is a continuous variable.1: Employees by age and sex Sex Male 2 4 2 8 Female 4 2 1 7 Age group 20 – 29 30 – 39 40 – 49 Total Total 6 6 3 15 You should look through this example again to make quite sure that you understand what has been done. © ABE and RRC . we could use the age groups 20 to < 30 (i. 20 to less than 30). As we learnt in section A of this unit. 30 to < 40 and 40 to < 50. for example. You should note that we did not use the age groups 20 – 30. but they are not as old as (exactly) 40 years.e.000 employees mentioned earlier. we need to ensure that the data groups chosen are appropriate. This is because an employee that is. The idea is best conveyed by an example. 4 months and 3 days old and Mr Jones is 39 years. 30 – 40 and 40 – 50 years.). 30 – 39 and 40 – 49. may be slightly older than 39 years. A question as to how the workforce is distributed can be answered by sorting the cards and preparing a simple table as in Table 6.1 would Mr Archer and Mr Jones now be classified? Clearly they are both older than (exactly) 39 years. days. for example. Of particular importance to note is the presentation of the original age data. therefore they should not be included in the age group 30 – 39. etc. which is given to the nearest year (and not broken down further into months. If the age of employees had been presented more precisely in the original data. which means it is possible to measure a person’s age broken down into months. where a more precise breakdown of a person’s age is given). Let us assume that Mr Archer is 39 years. Consider the card index of 3. you are now in a position to appreciate the purpose behind classification and tabulation –. In which age group in Table 6. We shall look in more detail at the use of class intervals in section D of this unit when we look "frequency distribution".e. minutes and seconds. This has allowed the use of the age groups 20 – 29. 180 Introduction to Statistics and Data Analysis Table 6.2: Distribution of employees by workshop Workshop A B C D E Total Number employed 600 360 660 840 540 3,000 Another question might have been "What is the wage distribution of the employees?" The answer can be given in another simple table, Table 6.3. Table 6.3: Wage distribution Wages group £40 but less than £60 but less than £80 but less than £100 but less than £120 but less than £140 but less than Total £60 £80 £100 £120 £140 £160 Number of employees 105 510 920 1,015 300 150 3,000 Note that such simple tables do not tell us very much – although it may be enough to answer the initial question at the moment. (b) Complex tabulation This deals with two or more aspects of a problem at the same time. In the problem just studied, it is very likely that the two questions presented in Tables 6.2 and 6.3 would be asked at the same time. We could therefore present the answers in a single, although more complex, table or matrix (as shown in Table 6.4). © ABE and RRC Introduction to Statistics and Data Analysis 181 Table 6.4: Distribution of employees and wages by workshop Wages group (£ per week) Workshop 40 – 59.99* 20 11 19 34 21 105 60 – 79.99 101 52 103 167 87 510 80 – 99.99 202 90 210 303 115 920 100 – 119.99 219 120 200 317 159 1,015 120 – 139.99 29 29 88 18 136 300 140 – 159.99 29 58 40 1 22 150 Total no. employed 600 360 660 840 540 3,000 A B C D E Total * Note: 40 – 59.99 is the same as "40 but less than 60" and similarly for the other columns. This table is much more informative than are the two previous simple tables, although it is more complicated. We could have complicated it further by dividing the groups into male and female employees, or into age groups. Later in this unit, we will look at a list of the rules you should try to follow in compiling statistical tables. At the end of that list you will find a table relating to our 3,000 employees, which you should study as you read the rules. Secondary Statistical Tabulation So far, our tables have merely classified the already available figures – the primary statistics. However, we can go further than this and do some simple calculations to produce other figures – secondary statistics. As an example consider again Table 6.2, and this time calculate how many employees there are on average per workshop. This is obtained by dividing the total (3,000) by the number of workshops (5), and the table appears as follows: Table 6.5: Distribution of employees by workshop Workshop A B C D E Total Average number of employees per workshop: The average is a secondary statistic. Number employed 600 360 660 840 540 3,000 600 © ABE and RRC 182 Introduction to Statistics and Data Analysis For another example, we may take Table 6.3 and calculate the proportion of employees in each wage group, as follows: Table 6.6: Wage distribution Wages group £40.00 £60.00 £80.00 £100.00 £120.00 £140.00 – – – – – – £59.99 £79.99 £99.99 £119.99 £139.99 £159.99 Number of employees 105 510 920 1,015 300 150 3,000 Proportion of employees 0.035 0.170 0.307 0.338 0.100 0.050 1.000 Total The proportions given here are also secondary statistics. In commercial and business statistics it is more usual to use percentages than proportions, and so in Table 6.6 these would be 3.5%, 17%, 30.7%, 33.8%, 10% and 5%. Of course secondary statistics are not confined to simple tables. They are used in complex tables too, as in the example presented in Table 6.7. Table 6.7: Inspection results for a factory product in two successive years Year 1 Machine no. Output 800 600 300 500 2,200 550 No. of rejects 40 30 12 10 92 23 % of rejects 5.0 5.0 4.0 2.0 4.2 4.2 Output 1,000 500 900 400 2,800 700 Year 2 No. of rejects 100 100 45 20 265 66.2 % of rejects 10.0 20.0 5.0 5.0 9.5 9.5 1 2 3 4 Total Average per machine The percentage columns and the average row show secondary statistics. All the other figures are primary statistics. Note carefully that percentages cannot be added or averaged to get the percentage of a total or of an average. You must work out such percentages on the totals or averages themselves. Another danger in the use of percentages has to be watched: you must not forget the size of the original numbers. Take the case of two doctors treating a particular disease. One doctor has only one patient and the doctor cures the patient – 100% success! The other doctor has 100 patients of whom 80 are cured – only 80% success! You can see how very unfair it would be to make a comparison based on the percentages alone. © ABE and RRC Introduction to Statistics and Data Analysis 183 Rules for Tabulation There are no absolute rules for drawing up statistical tables, but there are a few general principles which, if borne in mind, will help you to present your data in the best possible way. Try not to include too many features in any one table (say, not more than four or five) as otherwise it becomes rather clumsy. It is better to use two or more separate tables. Each table should have a clear and concise title to indicate its purpose. It should be very clear what units are being used in the table (tonnes, £, people, £000, etc.). Appropriate data groups (or class intervals) for continuous and discrete data should be chosen to ensure that all data can be included within a single group (or class). Blank spaces and long numbers should be avoided, the latter by a sensible degree of approximation. Columns should be numbered to facilitate reference. Try to have some order to the table, for example using size, time, geographical location or alphabetical order. Figures to be compared or contrasted should be placed as close together as possible. Percentages should be placed near to the numbers on which they are based. Use a ruler to draw the tables neatly – scribbled tables with freehand lines nearly always result in mistakes and are difficult to follow. However, it is useful to draw a rough sketch first so that you can choose the best layout and decide on the widths of the columns. Insert totals where these are meaningful, but avoid "nonsense totals". Ask yourself what the total will tell you before you decide to include it. An example of such a "nonsense total" is given in Table 6.8. Table 6.8: Election results Party A B Total Year 1 Seats 350 120 470 Year 2 Seats 200 270 470 Total Seats 550 390 940 The totals (470) at the foot of the two columns make sense because they tell us the total number of seats being contested, but the totals in the final column (550, 390, 940) are nonsense totals for they tell us nothing of value. If numbers need to be totalled, try to place them in a column rather than along a row for easier computation. If you need to emphasise particular numbers, then underlining, significant spacing or bold type can be used. If data is lacking in a particular instance, then insert an asterisk in the empty space and give the reasons for the lack of data in a footnote. Footnotes can also be used to indicate, for example, the source of secondary data, a change in the way the data has been recorded, or any special circumstances which make the data seem odd. © ABE and RRC 184 Introduction to Statistics and Data Analysis It is not always possible to obey all of these rules on any one occasion, and there may be times when you have a good reason for disregarding some of them. But only do so if the reason is really good – not just to save you the bother of thinking! Now study the layout of the following table (based on our previous example of 3,000 employees) and check through the list of rules to see how they have been applied. Table 6.9: ABC & Co. – Wage structure of labour force Numbers of persons in specified categories Wage group (£ per week) Workshop 40 – 59.99 (2) 20 11 19 34 21 105 3.5 60 – 79.99 (3) 101 52 103 167 87 510 17.0 80 – 99.99 (4) 202 90 210 303 115 920 30.7 100 – 119.99 (5) 219 120 200 317 159 1,015 33.8 120 – 140 139.99 159.99 (6) 29 29 88 18 136 300 10.0 (7) 29 58 40 1 22 150 5.0 Total no. employed (8) 600 360 660 840 540 3,000 100.0 % See note (a) (1) A B C D E Total no. in wage group % See note (b) (9) 20.0 12.0 22.0 28.0 18.0 100.0 – Note (a): Total number employed in each workshop as a percentage of the total workforce. Note (b): Total number in each wage group as a percentage of the total workforce. This table can be called a "twofold" table as the workforce is broken down by wage and workshop. Questions for Practice 1 These questions are typical of those likely to be set in an examination. 1. The human resources/personnel manager of a firm submits the following report on labour and staff to the managing director: "The total number of employees is 136, of whom 62 are male. Of these men, 43 are industrial labourers and the rest are staff. The age distribution of the 43 is 'under 18' (seven), '18–45' (22) and the others are over 45. Our total industrial labour force is 113, of whom 25 are 'under 18' and 19 are 'over 45'. The total number of 'under-18s' on the site is 29 (10 male, 19 female). There are 23 'over-45s', of whom only 4 are male staff." The managing director, quite rightly, thinks that this is gibberish and instructs you to present it to him in a tabular form which he can readily understand. Show the table you would produce. © ABE and RRC Introduction to Statistics and Data Analysis 185 2. Consider the following report. "In the period Year 1 – Year 6 there were 1,775 stoppages of work arising from industrial disputes on construction sites. Of these, 677 arose from pay disputes, 111 from conflict over demarcation, 144 on working conditions and 843 were unofficial walk-outs. 10% of all stoppages lasted only one day. Of these, the corresponding proportions were: 11% for pay disputes, 6% in the case of demarcation and 12½% in the case of working conditions. In the subsequent period Year 7 – Year 12, the number of stoppages was 2,807 higher. But, whereas the number of stoppages arising from pay disputes fell by 351 and those concerning working conditions by 35, demarcation disputes rose by 748. The number of pay disputes resolved within one day decreased by 45 compared with the earlier period, but only by 2 in the case of working conditions. During the later period 52 stoppages over demarcation were settled within the day, but the number of stoppages from all causes which lasted only one day was 64 greater in the later period than in the earlier." You are required to: (a) (b) Tabulate the information in the above passage, making whatever calculations are needed to complete the entries in the table. Write a brief comment on the main features disclosed by the completed table. Now check your answers with those given at the end of the unit. D. FREQUENCY DISTRIBUTIONS A frequency distribution is a tabulation which shows the number of times (i.e. the frequency) each different value occurs. For example, the following figures are the times (in minutes) taken by a shop-floor employee to perform a given repetitive task on 20 specified occasions during the working day: 3.5 3.6 3.4 3.7 3.8 3.8 3.7 3.7 3.8 3.9 3.6 3.7 3.4 3.7 3.8 3.5 3.6 3.5 3.6 3.9 If we now assemble and tabulate these figures, we can obtain a frequency distribution as follows: © ABE and RRC 186 Introduction to Statistics and Data Analysis Table 6. but of groups of values: © ABE and RRC . To illustrate the procedure.7 3.11 shows the procedure for the data in Table 6. In this case. Table 6.8 3.8 3. Table 6.10 after all 20 items have been entered. Then we make use of a grouped frequency distribution.5 3.4 3.9 Total Frequency 2 3 4 5 4 2 20 Simple Frequency Distributions A useful way of preparing a frequency distribution from raw data is to go through the records as they stand and mark off the items by the "tally mark" or "five-bar gate" method.5 3. perhaps.9 Total Tally marks || ||| |||| |||| |||| || Frequency 2 3 4 5 4 2 20 Grouped Frequency Distributions Sometimes the data is so extensive that a simple frequency distribution is too cumbersome and.6 3. uninformative. the "length of time" column consists not of separate values.6 3.10: Time taken for one employee to perform task Length of time (minutes) 3. Then mark the items on your table by means of a tally mark.11: Time taken for one employee to perform task Length of time (minutes) 3.4 3.7 3. First look at the figures to see the highest and lowest values so as to decide the range to be covered and then prepare a blank table. As the original readings were given correct to one decimal place. in practice.5 3.35.6 to 3.6 to 3. years..75. If we had had a more precise stopwatch. when an ungrouped table would not have been of much use..7 group. 220) is called the "total frequency". Table 6.12: Time taken for one employee to perform task Length of time (minutes) 3.13: Age distribution of employees in an office Age group (years) 16 to < 21 21 to < 26 26 to < 31 31 to < 36 36 to < 41 41 to < 46 46 to < 51 51 to < 56 56 to < 60 Total Number of employees 10 17 28 42 38 30 25 20 10 220 The various groups (for example.13) the class boundaries are 16. . 28 in the 26 to < 31 class) is called the "class frequency" and the total number of items (in this example.35 and 3.12). as grouped earlier in this section. Table 6.Introduction to Statistics and Data Analysis 187 Table 6. The number of items in each class (for example. 3.7 3.13 shows a grouped frequency distribution used in a more realistic situation. In the length of time example (see table 6. 21.6 minutes when corrected to one decimal place and it goes in the 3. these would be 16 years and 20 years 364 days © ABE and RRC . 26.55. The term "class boundary" is used to denote the dividing line between adjacent classes.4 to 3. would not have been required for the small amount of data in our example. 3. This needs some explanation. so in the age group example (see Table 6. The term "class limits" is used to stand for the lowest and highest values that can actually occur in a class. the times could have been measured more precisely.9 Total Tally marks |||| |||| |||| |||| | Frequency 5 9 6 20 Grouped frequency distributions are only needed when there is a large number of values and.5 we put times which could in fact be anywhere between 3. A time such as 3.8 to 3.57 minutes would not have been in this group as it equals 3. we assume that is the precision to which they were measured. In the first group of 3.95 minutes. "26 to < 31") are called "classes" and the range of values covered by a class (five years in this example) is called the "class interval".4 to 3. In the age group example. the class boundaries are 3.55 if we had been able to measure them more precisely. 3. As a practical guide.500 11. In such cases. © ABE and RRC .6 and 3.7 minutes for the second class.4 and 3.000 Frequency per £200 interval 50.000 12. the class limits are 3. These intervals let you see how the last column was compiled. 400. and column three shows that.001 – 5. Choice of Class Interval When compiling a frequency distribution you should. if possible. 400. however.000 27.600 5. You should remember that the purpose of compiling a grouped frequency distribution is to make sense of an otherwise troublesome mass of figures. you will find that somewhere between about five and 10 groups will usually be suitable. You should make yourself quite familiar with these terms.201 – 5. 200.000 192. This is so that fair comparisons can be made between one class and another.000 3.500 6.5 minutes for the first class and 3. of persons 50.188 Introduction to Statistics and Data Analysis for the first class.000 40. some grouped frequency distributions bring together the first few classes and the last few classes because there are few instances in each of the individual classes. 800. A superficial look at the original table (first two columns only) might have suggested that the most frequent incomes were at the middle of the scale.000 12. the most frequent incomes are at the bottom end of the scale.000 6.14: Annual income statistics Annual income to the nearest £ 4. this rule has to be broken – for example.200 Total No.200 5.801 – 5. as in the example for wage statistics in Table 6.401 – 7. i.000 55. the top of the table.000 40.400 6. 400. Sometimes. it is as well to make the classes comparable by calculating a column showing "frequency per interval of so much". Table 6.000 – Notice that the intervals in the first column are: 200.e.000 5. It therefore follows that we do not want to have too many groups or we will be little better off. make the length of the class interval equal for all classes.000 23.601 – 6. In the length of time example.001 – 6.000. But this apparent preponderance of the middle class is due solely to the change in the length of the class interval. before using the information. 21 years and 25 years 364 days for the second class and so on.14. because of the appearance of the figure 55. nor do we want too few groups or we will fail to see the significant features of the distribution. in fact. assuming that the ages were measured correct to the nearest day. When the frequency in each class of a frequency distribution is given as a proportion or percentage of the total frequency. © ABE and RRC .8 3.15.4 3. remember that proportions or percentages are useful secondary statistics. but in the number of items above or below a certain value.15 gives us the number of items "less than" a certain amount. A cumulative frequency column may also be formed for a grouped distribution. and as an exercise you should now compile the "more than" cumulative frequency column for Table 6. the number of persons having more than some quantity. Relative Frequency Distributions All the frequency distributions which we have looked at so far in this study unit have had their class frequencies expressed simply as numbers of items.16 for you to study. This can easily be done by doing the cumulative additions from the bottom of the table instead of the top. and it is formed by the successive addition of the separate frequencies.15. The example in Table 6. Table 6.15: Time taken for one employee to perform task Length of time (minutes) 3.7 3. the distribution used in Table 6.0 (or 100%). for example. of course. As an example. However we may wish to know.6 3. However.Introduction to Statistics and Data Analysis 189 Cumulative Frequency Distributions Very often we are not especially interested in the separate class frequencies. we form a cumulative frequency distribution as illustrated in column three of Table 6.9 Frequency 2 3 4 5 4 2 Cumulative frequency 2 5 9 14 18 20 The cumulative frequency tells us the number of items equal to or less than the specified value.5 3. Cumulative relative frequency distributions may be compiled in the same way as ordinary cumulative frequency distributions. the result is known as a "relative frequency distribution" and the separate proportions or percentages are the "relative frequencies". The total relative frequency is. When this is the case. always 1.14 is set out as a relative frequency distribution in Table 6. 200 5.600 5.000 90.000 6.000 This example is in the "less than" form.000 23.000 12.000 – Relative cumulative frequency (%) 26 47 76 88 94 100 – Frequency 50.200 Total Relative frequency (%) 26 21 29 12 6 26 100 Cumulative frequency 50.16: Annual income statistics Annual income to the nearest £ 4.601 – 6. the cumulative frequency and the relative cumulative frequency.401 – 7.000 5. © ABE and RRC .000 192.000 192.801 – 5.001 – 5.201 – 5.190 Introduction to Statistics and Data Analysis Table 6.000 12. the relative frequency. Now check your answers with those given at the end of the unit.000 168. Questions for Practice 2 The following figures show the number of faulty items made on successive days by a machine in an engineering workshop: 2 1 4 2 3 1 2 4 1 5 0 4 5 4 1 1 6 4 1 6 2 1 5 0 5 10 5 1 2 1 4 2 6 6 3 1 3 3 6 3 3 1 5 1 2 2 1 7 1 2 2 4 2 0 2 3 2 4 0 2 11 1 4 1 1 4 5 1 8 1 4 7 2 5 3 3 7 7 5 2 0 3 1 1 5 10 2 3 8 6 1 2 2 2 3 6 3 7 1 4 Tabulate these figures as a frequency distribution.000 145. giving columns containing the frequency. and you should now compile the "more than" form in the same way as you did for the non-relative distribution.000 180.400 6.001 – 6.000 55.000 40. 5 Total 677 111 144 843 1.Introduction to Statistics and Data Analysis 191 ANSWERS TO QUESTIONS FOR PRACTICE Questions for Practice 1 1.775 Year 7 – Year 12 1 day 29. (b) The overall number of stoppages in the later period was more than 2½ times greater than in the earlier period.582 Pay Demarcation Working conditions Unofficial* All * Note that we have given the figures as worked out from the information supplied.53 104. fractional numbers of stoppages cannot occur.97 4. 241. with those concerning working conditions being less likely to last more than one day.66 18 78. XYZ Company: Age of Staff and Labour Industrial Male Female Total (2) 7 22 14 43 (3) 18 47 5 70 (4) 25 69 19 113 Staff Male Female Total (5) 3 12 4 19 (6) 1 3 0 4 (7) 4 15 4 23 Total Male Female Total (8) 10 34 18 62 (9) 19 50 5 74 (10) 29 84 23 136 Age group (1) Under 18 18 – 45 Over 45 TOTAL 2. and you should round off each figure to the nearest whole number. In both periods. The subtotals of the columns will then be 177.288 4.597. the number of unofficial stoppages was the highest category. About 95% of stoppages in the later period lasted more than one day.5 More than 1 day 602.03 241.53 807 93 3. © ABE and RRC . though. Obviously.47 6.47 52 16 144.341.37 177.340. (a) The complete table is as follows: Number of Stoppages of Work Arising from Industrial Disputeson Construction Sites Year 1 – Year 6 Cause 1 day 74. compared with 90% in the previous period. and 4. although stoppages over pay and working conditions both declined.5 Total 326 859 109 3.5 More than 1 day 296.34 126 764.63 1. Demarcation disputes rose in number dramatically.598. 1.143. when they amounted to more than twice the number of all other stoppages put together. This was particularly remarkable in the later period. 97 0.61 0.12 0.05 0.02 0.192 Introduction to Statistics and Data Analysis Questions for Practice 2 Faulty items made on successive days by a machine in an engineering workshop Number of faulty items 0 1 2 3 4 5 6 7 8 9 10 11 Total Number of days with these faulty items 5 23 20 13 12 10 7 5 2 0 2 1 100 Relative frequency 0.00 0.05 0.13 0.97 0.23 0.90 0.00 – © ABE and RRC .48 0.28 0.73 0.02 0.10 0.95 0.01 1.20 0.05 0.83 0.07 0.00 Cumulative frequency 5 28 48 61 73 83 90 95 97 97 99 100 – Relative cumulative frequency 0.99 1. D. E.193 Study Unit 7 Graphical Representation of Information Contents Introduction A. Answers to Questions for Practice © ABE and RRC . F. C. Graphical Representation of Frequency Distributions Frequency Dot Diagram Frequency Bar Chart Frequency Polygon Histogram The Ogive Stem and Leaf Diagram Pictograms Limited Form Accurate Form Pie Charts Bar Charts Component Bar Chart Horizontal Bar Chart Scatter Diagrams General Rules for Graphical Presentation Page 194 194 195 195 196 197 200 203 204 204 205 206 207 207 208 209 210 215 B. frequency polygons. and details are likely to be lost. A. However. Graphs and charts are normally much superior to tables (especially lengthy complex tables) for showing general states and trends. Presenting information visually is easy to understand. ogives and stem and leaf diagrams to represent frequency distributions prepare pictograms.g. We then move on to discuss a number of popular graphical presentations that are used in everyday life. pie charts. Objectives When you have completed this study unit you will be able to: use frequency dot diagrams. presentations that are not necessarily used by statisticians (e. We will now examine each of these graphical methods in turn. The following is a diagram that to some extent overcomes this limitation: stem and leaf diagram. those used in business journals and the mass media of newspapers and television). bar charts and scatter diagrams from given data. However. and discuss the circumstances in which each form of presentation is most effective describe the general rules and principles for using graphical representations of data. initially using the data on the times (in minutes) taken by an employee to perform a given task on 20 specified occasions during the working day which we previously considered in Study Unit 6 (Table 6.10). such graphs and charts cannot usually be used for accurate analysis of data.194 Graphical Representation of Information INTRODUCTION In this study unit we shall move on and look at further methods of summarising and presenting data as a means of providing useful information. Diagrams are particularly effective as a way of conveying quite complex information. The methods of presenting frequency distributions graphically are as follows: frequency dot diagram frequency bar chart frequency polygon histogram ogive. and enables broad distributions and trends to be taken in quickly. In the first part of this unit we shall examine the specific graphical forms used for presenting frequency distributions of the type considered in the previous study unit. GRAPHICAL REPRESENTATION OF FREQUENCY DISTRIBUTIONS Tabulated frequency distributions are sometimes more readily understood if represented by a diagram. histograms. The tabulated frequency distribution for this data is as follows: © ABE and RRC . frequency bar charts. diagrams are rarely as accurate as the figures themselves. Whilst the two terms do not mean exactly the same thing. at this level.6 3.7 3.1 would be as shown in Figure 7. they can be used interchangeably.8 3.4 3. (In statistics.5 3.9 Total Frequency 2 3 4 5 4 2 20 Frequency Dot Diagram This is a simple form of graphical representation for the frequency distribution of a discrete variate. a frequency dot diagram of the information in Table 7. Above each value on the variate scale we mark a dot for each occasion on which that value occurs. Do not worry too much about the difference between these two terms.6 3. Thus.7 3.4 3.1: Time taken for one employee to perform task Length of time (minutes) 3.1: Frequency dot diagram to show length of time taken by operator to complete a given task Frequency 3. Loosely.) A horizontal scale is used for the variate and a vertical scale for the frequency. roughly speaking.Graphical Representation of Information 195 Table 7. Figure 7.8 3.5 3.9 Length of time (minutes) © ABE and RRC .1. the term "variate" is used when you might have expected the term "variable". a variate is a quantity that can have any of a set of numerical values. 8 3.4 3. as in Figure 7.1 now becomes a frequency bar chart. When we do this.4 3.3. we have marked in groups with zero frequency at 3.3 3. the result is a frequency polygon. Figure 7.2.5 3. Figure 7. the length of which represents the number of dots which should be there.6 3.9 Length of time (minutes) Frequency Polygon Instead of drawing vertical bars as we do for a frequency bar chart.3 and 4.7 3.5 3. This is done to ensure that the area enclosed by the polygon and the horizontal axis is the same as the area under the corresponding histogram which we shall consider in the next section.6 3.0. The frequency dot diagram in Figure 7.e. we could merely mark the position of the top end of each bar and then join up these points with straight lines.2: Frequency bar chart to show length of time taken by operator to complete a given task Frequency 6 5 4 3 2 1 0 3.9 4 Length of time (minutes) Note that we have added two fictitious classes at each end of the distribution.7 3.3: Frequency polygon to show length of time taken by operator to complete a given task Frequency 6 5 4 3 2 1 0 3.8 3.196 Graphical Representation of Information Frequency Bar Chart We can avoid the business of marking every dot in such a diagram by drawing instead a vertical line. as in Figure 7. © ABE and RRC . i. we shall use an example of a grouped frequency distribution similar to one from Study Unit 6 (Table 6. The information in the table may be shown graphically by the histogram in Figure 7. by drawing the vertical line (or marking its top end) at the centre point of the group. Sometimes.13).2. frequency bar charts and polygons are used with grouped data. Table 7.2: Age distribution of employees in an office Age group (years) 15 to < 20 20 to < 25 25 to < 30 30 to < 35 35 to < 40 40 to < 45 45 to < 50 50 to < 55 55 to < 60 Total Number of employees 10 17 28 42 38 30 25 20 10 220 Figure 7. They are mostly used in those cases where the variate is discrete and where the values are not grouped. as set out in Table 7.4: Histogram showing age distribution of employees in an office Frequency 40 30 20 10 0 15 20 25 30 35 40 45 50 55 60 Age (years) © ABE and RRC .Graphical Representation of Information 197 These three kinds of diagram are all commonly used as a means of making frequency distributions more readily comprehensible.4. Histogram A histogram is the best way of showing a grouped frequency distribution. To illustrate this. 198 Graphical Representation of Information In a histogram. Table 7. Note particularly how it illustrates the fact that the frequency falls off towards the higher age groups – any form of graph which did not reveal this fact would be misleading. for the amended data would look like Figure 7.5: Histogram showing age distribution of employees in an office (amended data) Frequency 40 30 This column is the wrong size 20 10 0 15 20 25 30 35 40 45 50 60 Age (years) © ABE and RRC . had given the last few groups as those in Table 7. which represents the frequency.) Look once again at the histogram of ages given in Figure 7. the frequency in each group is represented by a rectangle and – this is a very important point – it is the area of the rectangle. If however the lengths of the class intervals are not all equal. A wrong form of histogram. using heights instead of areas.3: Age distribution of employees in an office (amended extract) Age group (years) 40 to < 45 45 to < 50 50 to < 60 Number of employees 30 25 30 The last two groups have been lumped together as one. (Do not stop at this point if you have not quite grasped the idea. because it will become clearer as you read on. Now let us imagine that the original table had not used equal class intervals but.3. Figure 7. not its height.5. When the lengths of the class intervals are all equal.4. for some reason or other. then the heights of the rectangles represent the frequencies in the same way as do the areas (this is why the vertical scale has been marked in this diagram). then you must remember that the heights of the rectangles have to be adjusted to give the correct areas. less than about 1% or 2%). In © ABE and RRC . then nothing much is lost by leaving it off the histogram altogether.Graphical Representation of Information 199 Now. In the correct form of the histogram. the height of the last group (50 to < 60) would be halved because the class interval is double all the other class intervals. the last few groups in our age distribution table may have been specified as follows: Table 7. If the last group has a larger frequency than about 1% or 2%. in order not to create a false impression of the extent of the distribution. How would we draw this on a histogram? If the last (open-ended) group has a very small frequency compared with the total frequency (say. in published statistics.6 gives the right impression of the falling off of frequency in the higher age groups. For example. The histogram in Figure 7. then you should try to judge from the general shape of the histogram how many class intervals to spread the last frequency over. Figure 7.4: Age distribution of employees in an office (further amended extract) Age group (years) 40 to < 45 45 to < 50 50 and over Number of employees 30 25 30 A distribution of this kind is called an "open-ended" distribution. this clearly gives an entirely wrong impression of the distribution with respect to the higher age groups. that the last group in a frequency table is not completely specified. The vertical axis has been labelled "Frequency density per 5-year interval" as five years is the "standard" interval on which we have based the heights of our rectangles.6: Amended histogram showing age distribution of employees in an office (amended data) Frequency density per 5-year 40 interval 30 20 10 0 15 20 25 30 35 40 45 50 60 Age (years) Often it happens. 750.000 point is plotted against the upper end of the 5.9 Frequency 2 3 4 5 4 2 Cumulative frequency 2 5 9 14 18 20 Table 7.200 Graphical Representation of Information the example given.4 3.000 192.000 90.001 – 5.200 Total Frequency 50. If we had been plotting a "or more" ogive.000 40. Study these two diagrams so that you are quite sure that you know how to draw them.8 3.5 3.000 180. The Ogive This is the name given to the graph of the cumulative frequency. Ogives for two of the distributions already considered are now given as examples: Figure 7.201 – 5. Whatever procedure is adopted.200 5. It can be drawn in either the "less than" or the "or more" form.6 and Figure 7. 5.000 Cumulative frequency 50.000 12.6).5) and Figure 7.6 3.6: Annual income statistics Annual income to nearest £ 4. There is only one point which you might be tempted to overlook in the case of the grouped distribution – the points are plotted at the ends of the class intervals and not at the centre point.7 is for ungrouped data (as per Table 7.601 – 6. but the "less than" form is the usual one.000 23.000 – © ABE and RRC .000 168.8 is for grouped data (as per Table 7. Look at the example Table 7.000 192.401 – 7.001 – 6.8 and see how the 168.400 6.5: Time taken for one employee to perform task Length of time (minutes) 3. but it is often simpler to assume that an open-ended class has the same length as its neighbour.600 5.801 – 5.000 5.000 145.000 group (or class interval) and not against the midpoint.000 55. the plotting would have been against the lower end of the group (or class interval). Table 7. you would probably spread the last 30 people over two or three class intervals.7 3. it is essential to state clearly what you have done and why.601 – 6.000 6.000 12. 2 3.2 5.2 Annual income (£000s) © ABE and RRC .8 4 Length of time (mins) Figure 7.Graphical Representation of Information 201 Figure 7.8 5.8 7.7: Ogive to show length of time taken by operator to complete a given task Cumulative 20 frequency (less than or equal to) 15 10 5 0 3.6 3.0 6.4 6.8: Ogive to show annual income distribution Less than cumulative frequency (000s) 200 150 100 50 0 4.6 6.4 3. shown in Table 7.000 24.2 5. Figure 7.9.600 5.000 102. In each of Figures 7.4 6.000 40.200 Total Cumulative frequency (more than) 192.000 5.000 6.7: Annual income statistics Annual income to nearest £ 4.000 142. we can compile the data from our example of annual income statistics as a "more than" cumulative frequency.601 – 6.8 7.201 – 5.8 and 7.000 47.001 – 6.9 we have added a fictitious group of zero frequency at one end of the distribution.400 6.200 5.001 – 5.401 – 7.000 23.7.000 12.202 Graphical Representation of Information As an example of an "or more" ogive.000 192.801 – 5.000 – Frequency 50.000 12.000 12.000 55.2 Annual income (£000s) Check that you see how the plotting has been made against the lower end of the group and notice how the ogive has a reversed shape.6 6.0 6. © ABE and RRC .8 5.9: Ogive to show annual income distribution More than cumulative frequency (000s) 200 150 100 50 0 4.000 The ogive for this now appears as shown in Figure 7. Table 7. in addition to providing a visual representation of the data. histogram and ogive that we have already studied in that it has the advantage of preserving the original data. there is no loss of information). © ABE and RRC . frequency bar chart. and you should practise using relative frequencies whenever possible. for the stem value of 5 we have leaf values of 0.e. Next. 79 and 77 are in the same group as both have a stem of 7).e.e. the stem is "7" and the leaf is "9"). add a title to your diagram. But what is necessary is that you separate each number into a stem and a leaf. 7 and 9. Unlike the histogram.Graphical Representation of Information 203 It is common practice to call the cumulative frequency graph a "cumulative frequency polygon" if the points are joined by straight lines. it provides a partial sorting of the original data to allow a visual identification of the distribution pattern of the data. Similar to the histogram. each data value is split into a stem and a leaf. In more advanced statistical work the latter are used almost exclusively. In a second column list the leaves to the right of the appropriate stem values. 57 and 59. frequency polygon. Like the histogram. If you have successfully followed these steps. which correspond to the original values of 50.10. always compile a "less than" cumulative frequency. It is probably easiest to understand the stem and leaf diagram through an example. order them now. for the number 79. the stems range from 5 to 8 as the lowest value is 50 and the highest value is 87) in a column. Since these are two digit numbers. Thus. the "leaves") in a second column to the right of the appropriate "stem" (and again sorted in numerical order). then listing the second digit (i. Then. As you will see.g. list the stems in numerical order (in our example. the tens digit is the stem and the units digit is the leaf (e. The leaf is usually the last digit of the number and the other digits to the left of the leaf form the stem. (NB. and a "cumulative frequency curve" if the points are joined by a smooth curve. Unless you are told otherwise. In a stem and leaf diagram. the "stems") in a column (and sorted in numerical order). It differs from the frequency dot diagram. stem and leaf diagrams can also allow further analysis of the data as it is possible to obtain the original data points from the diagram (i. Stem and Leaf Diagram A stem and leaf diagram is the final method of presenting frequency distributions graphically that we will study in this section. it provides extra detail regarding individual values. The stem and leaf diagram is plotted by listing the first digits (i. you might want to arrange the data in numerical order.) Of course all of these diagrams may be drawn from the original figures or on the basis of relative frequencies. a stem and leaf diagram is a way of grouping your original data into classes. group the numbers with the same stems (e.g. If your leaf values are not in increasing order. so let us construct a stem and leaf diagram using the following data of a student's exam marks: 79 61 69 65 83 80 57 64 73 79 82 50 67 76 77 68 62 70 87 71 65 61 62 80 80 59 70 65 82 77 Before you begin. but is not a required step as ordering can be done later. then you should have finished up with a stem and leaf diagram identical to the one shown in Figure 7.) Finally. 204 Graphical Representation of Information Figure 7. Their use is confined to the simplified presentation of statistical data for the general public. PICTOGRAMS Pictograms are the simplest method of presenting information visually. These diagrams are variously called "pictograms".000 180.11: Number of cruises Years 1-3 (Source: Table 7. as in Figure 7. The data we will use is shown in Table 7. or "picturegrams" – the words all refer to the same thing.10: Stem and leaf diagram to show the distribution of exam marks Stem 5 6 7 8 0 7 9 1 1 2 2 4 5 5 5 7 8 9 0 0 1 3 6 7 7 9 9 0 0 0 2 2 3 7 Key: 50 means 50 marks Leaf B. "ideograms". of passengers carried 100.8: Cruises organised by a shipping line between Year 1 and Year 3 Year 1 2 3 Number of cruises 200 250 300 No. Figure 7.000 Limited Form We could represent the number of cruises by showing pictures of ships of varying size.8) Year 1 200 cruises Year 2 250 cruises Year 3 300 cruises © ABE and RRC .11. Table 7. There are two types and these are illustrated in the following examples.000 140. Pictograms consist of simple pictures which represent quantities.8. These diagrams have no purpose other than generally presenting statistics in a simple way. they can give false impressions of the actual increases. They are not capable of providing real accurate detail. It is difficult to judge what increase has taken place.) Accurate Form If the purpose of the pictogram is to covey anything but the most minimal of information. (Sometimes you will even find pictograms in which the sizes shown are actually wrong in relation to the real increases. so there can be no confusion over size.13 – it is difficult to represent a quantity of less than 10m barrels – so we are left wondering what "[" means. This is usually done by representing specific quantities of the variable with pictures or graphics of some sort – usually related to the subject of the variable.12.13: Imports of crude oil Year 1: Year 2: Key: represents 10m barrels of oil © ABE and RRC . Figure 7. their area on the paper. Figure 7. Consider Figure 7. or the volume of the object they represent.000 passengers Each matchstick man is the same height and represents 20. The reader can become confused as to whether the quantity is represented by the length or height of the pictograms.Graphical Representation of Information 205 Although these diagrams show that the number of cruises has increased each year. So.000 passengers.12: Number of passengers carried Years 1-3 (Source: Table 7. to avoid confusion it is better to try and give it a degree of accuracy in representing the numerical data. we could represent the numbers of passengers carried by using pictures of people as in Figure 7.8) Year 1: Year 2: Year 3 Key: 20. including public lighting The data is illustrated as a pie chart or circular diagram in Figure 7.9) Industrial 843m therms Commercial 437m therms Domestic 1. © ABE and RRC .9: Gas sales in Great Britain in one year (Source: Annual Abstract of Statistics) Uses Domestic Industrial Commercial Public* Total Million therms 1.723 % 51 31 16 2 100 * Central and local government uses.9. are used to show the manner in which various components add up to a total.6 100 Construct the diagram by means of a pair of compasses and a protractor. e.14: Example of a pie chart – gas sales in Great Britain (Source: Table 7. They are popular in computer graphics.383 843 437 60 2. 51% of 360 (c) 51 360 183.g. Convert the percentages into degrees. Figure 7. because inaccurate and roughly drawn diagrams do not convey information effectively. Like pictograms. Table 7. (a) (b) Tabulate the data and calculate the percentages.14.383m therms Public 60m therms There are a number of simple steps to creating a pie chart. they are only used to display very simple information to non-expert readers. known also as circular diagrams. Don't overlook this point. PIE CHARTS Pie charts. The data is taken from the Annual Abstract of Statistics as shown in Table 7. Suppose that we wish to illustrate the sales of gas in Great Britain in a certain year.206 Graphical Representation of Information C. However the idea of a bar chart can be extended beyond the field of frequency distributions.15). but only general ideas which are best studied by means of examples. Figure 7. and we will now illustrate a number of the types of bar chart in common use. These bar charts are preferable to circular diagrams because: they are easily read. Ltd 16 14 12 10 8 6 4 2 0 Year 1 Year 2 Key: Wages Royalties Materials Note that the lengths of the components represent the amounts. D. Pie charts are sometimes used side by side to provide comparisons.15 below. since it is not possible to read this exactly from the diagram itself. Component Bar Chart This first type of bar chart serves the same purpose as a circular diagram and.) You can insert the exact values of the data for each segment if more detail is required. even when there are many components they are more easily drawn it is easier to compare several bars side by side than several circles. We say "illustrate" because there are no rigid and fixed types. is sometimes called a "component bar diagram" (see Figure 7. the frequencies of different values of the variate. don't use a diagram of this kind with more than four or five component parts.Graphical Representation of Information 207 (d) Label the diagram clearly. © ABE and RRC . namely the frequency bar chart. Unless the whole diagram in each case represents exactly the same total amount. BAR CHARTS We have already met one kind of bar chart in the course of our studies of frequency distributions. The bars there were simply thick lines representing. (e) The main use of a pie chart is to show the relationship each component part bears to the whole. and that the components are drawn in the same order so as to facilitate comparison. (A key is illustrated in Figure 7. for that reason. If you have the choice.15: Component bar chart showing cost of production of ZYX Co. but this is not really to be recommended. other diagrams such as bar charts (see next section) are much clearer. using a separate legend or key if necessary. by their length. You can supplement the examples in this study unit by looking at the commercial pages of newspapers and magazines. 50% might have been 1 million. The example in Figure 7. namely. you must be careful. the information is expressed in percentages rather than in actual numbers of visitors. the different branches form the different categories.15 we have a numerical variable. Horizontal Bar Chart A typical case of presentation by a horizontal bar chart is shown in Figure 7. © ABE and RRC . Figure 7. If you compare several percentage component bar charts. whereas in another year it may have been nearer to 4 million and in another to 8 million.16).e. Note how a loss is shown by drawing the bar on the other side of the zero line.16: Horizontal bar chart of visitors arriving in the UK in one year Others 4% European 51% Commonwealth 23% USA 22% This figure is also an example of a percentage component bar chart. whereas in Figure 7.17 is also an example of a multiple or compound bar chart as there is more than one bar for each category. i.208 Graphical Representation of Information Bar charts with vertical bars are sometimes called "column charts". but they will not necessarily represent the same actual quantities – for example. Figure 7.17: Horizontal bar chart for the So & So Company Ltd to show profits made by branches in Year 1 and Year 2 London A London B Key: Newcastle Year 1 Year 1 Bristol 5 0 5 10 15 20 25 Loss (£000s) Profit (£000s) Pie charts and bar charts are especially useful for "categorical" variables as well as for numerical variables.17 shows a categorical variable – i. as they each represent 100%. Each bar chart will be the same length. time. Figure 7.e.17. to distinguish them from those in which the bars are horizontal (see Figure 7. whether this relationship is particularly strong.Graphical Representation of Information 209 E. such as the relationship that may exist between sales revenue and the level of advertising expenditure. it is not immediately obviously whether a relationship exists or not and if so.10 presents sales data for a manufacturing company against its expenditure on advertising over a given period. level of advertising expenditure) on the y-axis (vertical axis). scattergram or scatter plot) differs fundamentally from the other graphical forms of presenting data that we have considered so far.g.10: Advertising expenditure and sales for a manufacturing company Advertising expenditure (£000) 23 25 28 31 31 33 36 37 43 49 20 12 8 19 17 15 Sales revenue (£000) 123 134 150 179 187 191 200 202 203 210 85 62 49 106 91 89 © ABE and RRC . Specifically. Scatter diagrams are plotted with one variable (e. SCATTER DIAGRAMS So far we have learnt about various methods used to present data graphically. Table 7.g. The final graphical method that we shall learn about here is called a scatter diagram. pie charts and bar charts). Table 7. Looking at the data. sales revenue) on the x-axis (the horizontal axis) and the other variable (e. scatter diagrams are the appropriate method of presenting data when you want to examine the relationship between two continuous variables. rather than simply providing a visual summary of one (or more) set of variables (as in the case of frequency distribution graphs. pictograms. It is convention to plot the independent variable on the x-axis and the dependent variable on the y-axis. Scatter diagrams are used as a visual aid to examine potential relationships between two sets of (quantitative) variables. So using our example of sales revenue and the level of advertising expenditure. A scatter diagram (also known as a scattergraph. we are examining whether sales revenue (the dependent variable) is dependent on the level of advertising (the independent variable): does an increase in the level of advertising expenditure lead to an increase in sales revenue? Let us look at an example. where both variables are increasing together. These three types of correlation are shown in Figure 7. is described as positive correlation (both variables are positively correlated with each other). we would expect the efficiency of the machines in this manufacturing company to decline as they get older. suggesting that the efficiency and age of machinery are negatively correlated.210 Graphical Representation of Information However.000). we would not expect any correlation between the cost of raw materials and employee output. For example.19. As the company spends more on advertising. In contrast. If the scatter diagram does not show any relationship between the variables then no correlation exists. This association is known as correlation. © ABE and RRC . it is important to stress that these diagrams cannot prove that one variable causes the other. the relationship between the level of advertising expenditure and sales revenue for this manufacturing company becomes immediately clear.18 confirms that there appears to be a positive relationship. if one variable increases while the other decreases.18. if we plot this data on a scatter diagram. However. the relationship shown in Figure 7. Figure 7. the relationship is described as negative correlation. The scatter diagram presented in Figure 7.18 Scatter diagram of advertising expenditure against sales revenue for a manufacturing company Sales revenue (£000) 250 200 150 100 50 0 0 10 20 30 40 50 60 Advertising expenditure (£000) While scatter diagrams are used to study possible relationships between two variables. its sales revenue increases (although sales revenue does seem to flatten out after advertising reaches £350. they may indicate the existence and strength of an association between two variables. Thus. For example. 19: Scatter diagram examples showing different types of correlation Positive correlation Annual Income (£000s) 50 40 30 20 10 0 0 1 2 3 4 Years of post school education Negative correlation Machinery 250 efficiency (output per 200 day) 150 100 50 0 0 10 20 30 40 Age of machinery (months) No correlation Cost of raw 100 materials (£ per tonne) 80 60 40 20 0 0 20 40 60 80 Employee output (items per day) © ABE and RRC .Graphical Representation of Information 211 Figure 7. Wherever necessary. as this defeats the object of diagrams.212 Graphical Representation of Information F. guidelines should be inserted to facilitate reading. GENERAL RULES FOR GRAPHICAL PRESENTATION There are a number of general rules which must be borne in mind when planning and using the graphical methods covered in this unit. Graphs and charts must be given clear but brief titles. and the scales of values clearly marked. Avoid excessive detail.20. Figure 7. but take care not to distort the graph by overemphasising small variations. or at least by a reference to the source of the data. Diagrams should be accompanied by the original data. The axes of graphs must be clearly labelled. In such a case the graph could be "broken" as shown in Figure 7. Obeying this rule sometimes leads to rather a waste of paper space. Try to include the origins of scales.20: Sales of LMN Factory Sales (£000s) 70 60 50 J F M A M J J A S O N D Months © ABE and RRC . 00 56. The following figures show the number of faulty items made on successive days by a machine in an engineering workshop: 2 1 4 2 3 1 2 4 1 5 0 4 5 4 1 1 6 4 1 6 2 1 5 0 5 10 5 1 2 1 4 2 6 6 3 1 3 3 6 3 3 1 5 1 2 2 1 7 1 2 2 4 2 0 2 3 2 4 0 2 11 1 4 1 1 4 5 1 8 1 4 7 2 5 3 3 7 7 5 2 0 3 1 1 5 10 2 3 8 6 1 2 2 2 3 6 3 7 1 4 Tabulate these figures as a frequency distribution.01 to 46. 1.00 61. a frequency polygon and an ogive: Weekly wage (£) 31.00 66.01 to 56.01 to 51.00 3.00 36.01 to 41.01 to 71.00 46. 2.01 to 66.00 51.01 to 36.Graphical Representation of Information 213 Questions for Practice These questions are typical of those likely to be set in an examination.01 to 61.00 41. No. the relative frequency. Illustrate the following data by means of a histogram. giving columns containing the frequency. of employees 6 8 12 18 25 30 24 14 Construct a stem and leaf diagram to illustrate the age distribution of customers purchasing clothing from a retail store over a one-hour period given the following data on customer ages: 8 38 58 13 40 61 16 41 63 25 44 67 26 47 75 29 49 78 30 51 82 32 54 86 37 55 95 © ABE and RRC . the cumulative frequency and the relative cumulative frequency. Illustrate the distribution by means of a frequency bar chart of the relative figures. © ABE and RRC .26 8. Now check your answers with those given at the end of the unit. Illustrate by means of a bar chart the following data relating to the number of units sold of a particular kind of machine: Year 1 2 3 6.500 The unit price of a commodity and its demand over a period of time are shown in the following table: Price (£ per item) 1.000 5.24 9.20 2.400 6.16 Number of items sold (000) 41 38 34 29 26 30 25 22 19 15 Draw a scatter diagram of the data and comment on the relationship between price and the amount sold.700 5.850 5.400 164.450 Sources of revenue in Ruritania for year ending 31 March 5.762.214 Graphical Representation of Information 4. Construct a circular diagram to illustrate the following figures: Source Income Tax Estate Duties Customs & Excise Miscellaneous Amount (£000) 1.500 6. Home sales Commonwealth sales Foreign sales 6.39 3.500 99.27 6.250 5.30 4.18 10.500 1.600 5.800 5.29 7.863.31 5. of faulty items per day © ABE and RRC .15 0.Graphical Representation of Information 215 ANSWERS TO QUESTIONS FOR PRACTICE 1.05 0.99 1.10 0.90 0.00 Cumulative frequency 5 28 48 61 73 83 90 95 97 97 99 100 – Relative cumulative frequency 0.2 0.00 – Frequency bar chart – faulty items produced by machine in workshop 0.05 0.13 0.95 0.28 0.05 0.97 0.23 0.07 0.00 0.61 0.83 0. Number of faulty items 0 1 2 3 4 5 6 7 8 9 10 11 Total Number of days with these items 5 23 20 13 12 10 7 5 2 0 2 1 100 Relative frequency 0.12 0.48 0.01 1.05 0 0 1 2 3 4 5 6 7 8 9 10 11 No.02 0.1 0.20 0.25 Relative frequency 0.97 0.73 0.02 0. 00 41.00 The ogive follows: Frequency 6 8 12 18 25 30 24 14 Cumulative frequency 6 14 26 44 69 99 123 137 © ABE and RRC .01 – 56.01 – 41.00 46.00 66.01 – 36.00 51.01 – 71.00 56.01 – 51.00 61.216 Graphical Representation of Information 2. 30 No. of employees Combined histogram and frequency polygon 20 10 0 31 36 41 46 51 56 61 66 71 Weekly wage (£) Workings for ogive: Wage Group (£) 31.01 – 61.00 36.01 – 66.01 – 46. Stem and leaf diagram to show the age distribution of retail customers for a clothing store over a one-hour period Stem 0 1 2 3 4 5 6 7 8 9 8 3 6 5 6 9 0 2 7 8 0 1 4 7 9 1 4 5 8 1 3 7 5 8 2 6 5 Leaf Key: 08 means 8 years old © ABE and RRC .Graphical Representation of Information 217 Ogive for wages distribution Cumulative frequency 140 120 100 80 60 40 20 0 31 36 41 46 51 56 61 66 71 Weekly wage (£) 3. Sources of revenue in Ruritania Customs & excise Estate duties Miscellaneous Income tax 5.6% 100.000 8.000 16.000 2.218 Graphical Representation of Information 4.9% 4. The amounts must first be turned into percentages and degrees.3% 2. Shading has been added for clarity.000 4.000 10. draw bar charts on a percentage basis. The bar chart is shown in below. Income tax Estate duties Customs & excise Miscellaneous Total 47.000 0 Year 1 Year 2 Year3 Key: Foreign sales Commonwealth sales Home sales © ABE and RRC .000 12.000 14. of course.0% 172 15 163 9 359 Note: The angles do not add up to 360 exactly because of rounding errors. Bar chart of machine sales 18.2% 45. You can.000 6. Graphical Representation of Information 219 6. The scatter diagram showing the relationship between product price and the number of items sold for a company follows. Sales (000s) 45 40 35 30 25 20 15 10 5 0 0 2 4 6 8 10 12 Price (£ per item) © ABE and RRC . the scatter diagram shows that as price increases demand for the product (measured in terms of number of items sold) falls. This suggests that there is a negative correlation between the price of this particular item and product demand. In general. 220 Graphical Representation of Information © ABE and RRC . The Arithmetic Mean The Mean of a Simple Frequency Distribution The Mean of a Grouped Frequency Distribution Simplified Calculation Calculating the Mean from a Stem and Leaf Diagram Characteristics of the Arithmetic Mean The Mode Mode of a Simple Frequency Distribution Mode of a Grouped Frequency Distribution Calculating the Mode from a Stem and Leaf Diagram Characteristics of the Mode The Median Median of a Simple Frequency Distribution Median of a Grouped Frequency Distribution Calculating the Median from a Stem and Leaf Diagram Page 222 223 224 225 226 230 231 232 232 232 234 235 236 236 237 238 243 B.221 Study Unit 8 Measures of Location Contents Introduction A. Answers to Questions for Practice © ABE and RRC . C. There are three such measures with which you need to be familiar: the arithmetic mean the mode the median. Distributions with different averages indicate that there is a different general level of the variate in the two groups. but we would now like to take the matter further. we might have compiled frequency distributions and graphed the results. in reference to the two kinds of egg. © ABE and RRC . Figure 8. which is a topic in which you would be interested if you worked for an egg marketing company. Drawing the frequency curves has enabled us to make an important general statement concerning the relative egg weights of the two breeds of poultry. but often we wish to compare two or more distributions. although some eggs from Breed A are of less weight than some eggs from Breed B. In everyday life. In this case the distributions overlap and.1. Remember that one of the objects of statistical analysis is to condense unwieldy data so as to make it more readily understood. eggs from Breed A are. we commonly use such a figure when we talk about the "average" value of something or other. by means of a quick revision. heavier than those from Breed B. in general. Consider for example the distribution of the weights of eggs from two different breeds of poultry. The two frequency polygons might well look something like Figure 8. Having weighed a large number of eggs from each breed. We would like to calculate some figure which will serve to indicate the general level of the variate under discussion.222 Measures of Location INTRODUCTION We have looked at frequency distributions in detail in Study Unit 6 and you should.1: Frequency polygon showing weight of eggs from two breeds of hen Frequency Breed B Breed A Weight of eggs Examining these distributions you will see that they look much alike except for one thing – they are located on different parts of the scale. The single value which we use to describe the general level of the variate is called a "measure of location" or a "measure of central tendency". that those from Breed A had a higher average weight than those from Breed B. We might have said. A frequency distribution may be used to give us concise information about its variate. make sure that you have understood them well before proceeding. in our example. written Σ. We don't always want to specify what particular items we are discussing. © ABE and RRC . Thus. Notice that some values occur more than once. Take. but we still add them all. and so decide on the appropriate measure of location to use in common situations. egg weights. including stem and leaf diagrams find the median of grouped and ungrouped data (including stem and leaf diagrams) by calculation and. where appropriate. so make sure you understand the meaning of the expression.1: Height of seven men Man A B C D E F G Total Height (cms) 187 160 163 180 180 168 187 1225 The arithmetic mean of these heights is 1225 7 175 cms. median and mode. wages. usually x. Also. we indicate the sum of a number of values of x by the Greek letter sigma. of seven men: Table 8. THE ARITHMETIC MEAN The arithmetic mean is what we normally think of as the "average" of a set of values. For this reason in general discussion we will use (as you will recall from algebra) some general letter.Measures of Location 223 Objectives When you have completed this study unit you will be able to: calculate the arithmetic mean and mode of grouped and ungrouped distributions. we may write: Σx 1225 We shall be using sigma notation extensively in this unit. It is obtained by adding together all the values and then dividing the total by the number of values involved. At this point we must introduce a little algebra. the following set of values which are the heights. whether it is heights. etc. for example. graphical methods utilise the assumed mean method and the "class interval" method in calculating the arithmetic mean list the relative advantages and disadvantages of the arithmetic mean. A. in centimetres. as practice. the arithmetic can be eased somewhat by forming a frequency distribution. you can see that Σf n and that. complete the (fx) column and calculate the value of the arithmetic mean. the formula becomes: x Σ( fx ) Σf which is 1225/7 175 cms as before. However if we had a much larger number of items the method would save a lot of additions. with only seven items it would not be necessary in practice to use this method. 7 You will often find the arithmetic mean simply referred to as "the mean" when there is no chance of confusion with other means (which we are not concerned with here). when the x's are not all the separate values but only the different ones. Example 2: Consider the following table and. The calculation of the arithmetic mean can then be described by formula: x Σx n or 1 x Σx n The latter expression is customary in statistical work. Of course. Consider the following example. Applying it to the example above. The Mean of a Simple Frequency Distribution When there are many items (that is when n is large). Do this before moving on. of men at this height (f) 1 1 1 2 2 Σf 7 Product (fx) 160 163 168 360 374 Σ(fx) 1225 Indicating the frequency of each value by the letter f.224 Measures of Location We indicate the arithmetic mean by the symbol x (called "x bar") and the number of items by the letter n. we have: x 1 (1225 ) 175 cms. x .2: Height distribution data Height in cms (x) 160 163 168 180 187 Total No. © ABE and RRC . Example 1: Table 8. Measures of Location 225 Table 8. which we will discuss later). We do this by assuming that all the values in any group are equal to the midpoint (or mid-value) of that group.3: Data for calculation of arithmetic mean Value of x 3 4 5 6 7 8 9 10 Total You should have obtained the following answers: the total numbers of items. The procedure is very similar to that for a simple frequency distribution (which is why we stressed the need for revision) and is shown in this example: © ABE and RRC . x 713 7. only the groups in which they lie (unless you have a stem and leaf diagram. Revise it if necessary. It is most important that you do not get muddled about calculating arithmetic means. before going on to the next paragraph. The Mean of a Grouped Frequency Distribution Suppose now that you have a grouped frequency distribution. Σf 100 the total product. Σ(fx) 713 the arithmetic mean. but we can make an approximation sufficiently accurate for most statistical purposes. In this case. How then can we calculate the arithmetic mean? The answer is that we cannot calculate the exact value of x .13 100 Number of items (f) 1 4 7 15 35 24 8 6 Product (fx) 3 16 35 Make sure that you understand everything we have covered so far. you will remember. we do not know the actual individual values. 226 Measures of Location Table 8.4: Data for calculation of arithmetic mean of a grouped frequency distribution Group 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 Total x Σfx 1 1,660 33.2 Σf 50 Mid-value (x) 5 15 25 35 45 55 65 Frequency (f) 3 6 10 16 9 5 1 Σf 50 Product (fx) 15 90 250 560 405 275 65 Σ(fx) 1,660 Provided that Σf is not less than about 50 and the number of groups is not less than about 12, the arithmetic mean thus calculated is sufficiently accurate for all practical purposes. There is one pitfall to be avoided when using this method – if all the groups should not have the same class interval, be sure that you get the correct mid-values! Table 8.5 is part of a larger table with varying class intervals, to illustrate the point: Table 8.5: Data for grouped frequency distribution Group 0 to < 10 10 to < 20 20 to< 40 40 to < 60 60 to < 100 Mid-value (x) 5 15 30 50 80 You will remember that in discussing the drawing of histograms we had to deal with the case where the last group was not exactly specified. The same rules for drawing the histogram apply to the calculation of the arithmetic mean. Simplified Calculation It is possible to simplify the arithmetic involved in calculating an arithmetic mean still further by the following two devices: working from an assumed mean in the middle of one convenient class working in class intervals instead of in the original units. Let us consider the first device. © ABE and RRC Measures of Location 227 (a) Working from an assumed mean in the middle of one convenient class If you go back to our earlier examples, you will discover after some arithmetic that if you add up the differences in value between each reading and the true mean, then these differences add up to zero. Take first the height distribution (Table 8.1) considered at the start of this section: Table 8.6: Height distribution data, x 175 cms Man A B C D E F G Height, x (cms) 187 160 163 180 180 168 187 (x – x ) (cms) 12 15 12 5 5 7 12 Σ(x x) 34 34 0 That is, Σ(x x) 0 . Second, consider the grouped frequency distribution (Tables 8.3, 8.4) given earlier in this section: Table 8.7: Data for grouped frequency distribution, x 33.2 Group 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 Total That is, Σ(x x) 0 . If we take any value other than x and follow the same procedure, the sum of the differences (sometimes called deviations) will not be zero. In our first example, let us assume the mean to be 170 cms and label the assumed mean xo. The differences between each reading and this assumed value are: Mid-value (x) 5 15 25 35 45 55 65 Frequency (f) 3 6 10 16 9 5 1 Σf 50 (x – x ) 28.2 18.2 8.2 1.8 11.8 21.8 31.8 f(x – x ) 84.6 109.2 82 28.8 106.2 109 31.8 Σ(f(x – x )) 275.8 275.8 0 © ABE and RRC 228 Measures of Location Table 8.8: Differences between x and xo for height distribution data Man A B C D E F G Height, x (cms) 187 160 163 180 180 168 187 d = (x – xo) (cms) 17 10 7 10 10 2 17 Σ(x xo) 54 19 35 That is Σ(x – xo) 35 cms or Σd 35 cms. We can make use of this property and use it as a short cut for finding x . We start by choosing some value of x as an assumed mean. We try to choose it near to where we think the true mean ( x ) will lie, and we always choose it as the midpoint of one of the groups when we are dealing with a grouped frequency distribution. In the example just given, the total deviation (d) does not equal zero, so 170 cannot be the true mean. As the total deviation is positive, we must have underestimated in our choice of xo, so the true mean is higher than 170. As there are seven readings, we need to adjust xo upwards by one seventh of the total deviation, i.e. by (35) 7 5. Therefore the true value of x is: 170 (35) 170 5 175 cms. 7 We know this to be the correct answer from our earlier work. Let us now illustrate the short cut method for the grouped frequency distribution. We shall take xo as 35 as this is the mid-value in the centre of the distribution. Table 8.9: Differences between x and xo for grouped frequency distribution data Group 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 Total This time Σfd 90. Mid-value (x) 5 15 25 35 45 55 65 Frequency (f) 3 6 10 16 9 5 1 Σf 50 d = (x – xo) 30 20 10 0 10 20 30 f(x – xo) = fd 90 120 100 0 90 100 30 Σfd 310 220 90 © ABE and RRC Measures of Location 229 This means we must have overestimated xo, as the total deviation (Σfd) is negative. As 1 there are 50 readings altogether, the true mean must be 50 th of (90) lower than 35, i.e. x 35 (90) 35 1.8 33.2 50 which is as we found previously. (b) Working in class intervals instead of in the original units This method of calculation can be used with a grouped frequency distribution to work in units of the class interval instead of in the original units. In the fourth column of Table 8.9, you can see that all the deviations are multiples of 10, so we could have worked in units of 10 throughout and then compensated for this at the end of the calculation. Let us repeat the calculation using this method. The result (with xo 35) is set out in Table 8.10. Note that here the symbol used for the length of the class interval is c, although you may also come across the symbol i used for this purpose. Table 8.10: Differences between x and xo for grouped data using class intervals Group Mid-value (x) 5 15 25 35 45 55 65 Frequency (f) 3 6 10 16 9 5 1 Σf 50 (x – xo) d= x xo c fd 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 Total –30 –20 10 0 10 20 30 Σ 3 2 1 0 1 2 3 x xo 22 31 9 c 9 12 10 0 9 10 3 Remembering that we need to compensate, by multiplying by 10, for working in class interval units in the table: x 35 x 35 (9) 10 50 90 35 1.8 33.2 as before. 50 The general formula for x that applies for all grouped frequency distributions having equal class intervals can be written as: x xo Σfd c n where d x xo c © ABE and RRC 230 Measures of Location As we mentioned at an earlier stage, you have to be very careful if the class intervals are unequal, because you can only use one such interval as your working unit. Table 8.11 shows how to deal with this situation. Table 8.11: Differences between x and xo using unequal class intervals Group Mid-value (x) 5 15 25 35 45 60 d= x xo c Frequency (f) 3 6 10 16 9 6 50 Product (fd) 9 12 10 0 9 15 24 (31) 7 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 70 Total 3 2 1 0 1 2½ The assumed mean is 35 as before, and the working unit is a class interval of 10. Notice how d for the last group is worked out: the midpoint is 60, which is 2½ times 10 above the assumed mean. The required arithmetic mean is, therefore: x 35 7 10 70 35 35 1.4 33.6 50 50 We have reached a slightly different figure from before because of the error introduced by the coarser grouping in the "50 to < 70" region. The method just described is of great importance. By using it correctly, you can often do the calculations for very complicated-looking distributions by using mental arithmetic and pencil and paper. With the advent of electronic calculators, the time saving on calculations of the arithmetic mean is not great, but the method is still preferable because: The numbers involved are smaller and thus you are less likely to make slips in arithmetic. The method can be extended to enable us to find easily the standard deviation of a frequency distribution (which we shall examine in Study Unit 9). Calculating the Mean from a Stem and Leaf Diagram You will recall that a stem and leaf diagram presents frequency distributions of grouped data graphically. However because these diagrams preserve the original data, they can be used for further analysis to calculate the mean relatively straightforwardly, as if the data was ungrouped. Consider the following example of a stem and leaf diagram in Figure 8.2, which shows the distribution of marks awarded to an ice skater in a competition. Note that the stem values represent the units of a mark awarded and the leaf values represent the tenths of a mark awarded (i.e. a mark of 5.8 is expressed in the stem and leaf diagram as 5|8). © ABE and RRC Measures of Location 231 Figure 8.2: Stem and leaf diagram to show the marks awarded to a competition skater Stem 0 1 2 3 4 5 5 9 0 1 1 3 0 5 8 Leaf 0 Note: 0|0 0.0 marks Although the stem and leaf diagram presents grouped data, the individual data is preserved so that the mean can be calculated using the simple formula: x Σx n Σ(0.0 3.5 3.9 4.0 4.1 4.1 4.3 5.0 5.5 5.8) 10 40.2 4.0 10 Accordingly: x x Characteristics of the Arithmetic Mean There are a number of characteristics of the arithmetic mean which you must be aware of: It is not necessary to know the value of every item in a distribution in order to calculate the arithmetic mean. Only the total and the number of items are needed. For example, if you know the total wages bill and the number of employees, you can calculate the arithmetic mean wage without knowing the wages of each person. The arithmetic mean is fully representative because it is based on all, and not just some, of the items in the distribution. One or two extreme values can make the arithmetic mean somewhat unrepresentative by their influence on it. For example, if a millionaire came to live in a country village, the inclusion of his or her income in the arithmetic mean for the village might make the place seem very much better off than it really was! The arithmetic mean is reasonably easy to calculate and to understand. In more advanced statistical work, it has the advantage of being amenable to algebraic manipulation. © ABE and RRC which we will discuss later).3724. This is the name given to the most frequently occurring value. This is not an actually occurring value and doesn't help us to answer our problem. it is not quite so easy to determine the mode (unless you have a stem and leaf diagram. THE MODE The first alternative to the mean which we will discuss is the mode. and do not fall into this trap! For comparison. the mode of the distribution obtained from the survey would be an actual value (perhaps size 8) and it would provide the answer to the problem. Mode of a Simple Frequency Distribution Consider the following frequency distribution: Table 8. A common error is to say that the mode of the distribution in Table 8. Mode of a Grouped Frequency Distribution When the data is given in the form of a grouped frequency distribution. An example is the case of a survey carried out by a clothing store to determine what size of garment to stock in the greatest quantity. Watch out.12 is 39. The mode is 1. What. is the mode of the following distribution shown in Table 8.12 Accident distribution data Number of accidents per day (x) 0 1 2 3 4 Total Number of days with the stated number of accidents (f) 27 39 30 20 7 123 In this case the most frequently occurring value is 1 (it occurred 39 times) and so the mode of this distribution is 1.232 Measures of Location B. is a value of the variate x. not the frequency of that value. The mode is used in those cases where it is essential for the measure of location to be an actually occurring value. like the mean. This is wrong. The average size of garment in demand might turn out to be. However.13? © ABE and RRC . Note that the mode. 9. let us say.52. calculate the arithmetic mean of the distribution – it works out at 1. you might ask. and so the midpoint of the modal group is not a good way of estimating the mode. A number of procedures based on the frequencies in the groups adjacent to the modal group can be used. and would then have appeared as follows (only the middle part is shown. You may be tempted to say that the mode is 75. and we shall now describe one procedure. we will get different modal groups if the grouping is by 15 or by 20 or by any other class interval.14: Revised data for determining the mode of a grouped frequency distribution Group 60 to < 65 65 to < 70 70 to < 75 75 to < 80 80 to < 85 85 to < 90 Frequency 16 22 21 19 12 8 The modal group is now "65 to < 70". but this is not true. The reason is that the modal group depends on the method of grouping. that these procedures are only mathematical devices for finding the most likely position of the mode – it is not possible to calculate an exact and true value in a grouped distribution. this determination of the modal group is usually sufficient.13: Data for determining the mode of a grouped frequency distribution Group 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 70 to < 80 80 to < 90 90 to < 100 100 to < 110 110 to < 120 Frequency 4 6 10 16 24 32 38 40 20 10 5 1 All we can really say is that "70 to < 80" is the modal group (the group with the largest frequency). You should note. however. to illustrate the point): Table 8. nor even a useful approximation in most cases. but you may occasionally be asked for the mode to be determined from a grouped distribution. The distribution could have been set out with class intervals of five instead of 10. which can be chosen quite arbitrarily to suit our convenience. © ABE and RRC .Measures of Location 233 Table 8. In practical work. Likewise. as we have remarked the original data is preserved. This means that these diagrams can be used to calculate the mode relatively straightforwardly as if the data was ungrouped. Let us consider our previous example of a stem and leaf diagram which shows the distribution of marks awarded to an ice skater in a competition (Figure 8. This suggests to us that the mode may be some way towards the lower end of the modal group rather than at the centre. This method can be used when the distribution has equal class intervals. Draw in the diagonals AB and CD as shown in Figure 8. and the one above ("80 to < 90") has a frequency of 20. Remember that the stem values represent the units of a mark awarded and the leaf values represent the tenths of a mark awarded. From the point of intersection draw a vertical line downwards.4. Where this line crosses the horizontal axis is the mode. Figure 8. In our example the mode is just less than 71. Now examine the groups on each side of the modal group.2)).3. previously as Figure 8.3: Using a histogram to determine the most likely value of the mode Class frequency 40 A C D 30 20 B Mode 60 70 80 90 Variate Calculating the Mode from a Stem and Leaf Diagram Although a stem and leaf diagram presents frequency distributions of grouped data graphically.3. A graphical method for estimating the mode is shown in Figure 8. Draw that part of the histogram which covers the modal class and the adjacent classes on either side.234 Measures of Location We saw that the modal group of our original distribution was "70 to < 80". © ABE and RRC . The group below ("60 to < 70") has a frequency of 38. 10. It is not affected by the occurrence of extreme values. 9.e. 7. namely 7 and 10. there can be more than one mode. 10.Measures of Location 235 Figure 8. so the mode is 4. Characteristics of the Mode Some of the characteristics of the mode are worth noting. 6. 8. In our example. For example.4: Stem and leaf diagram to show the marks awarded to a competition skater Stem 0 1 2 3 4 5 5 9 0 1 1 3 0 5 8 Leaf 0 Note: 0|0 0. 10. two data points of value 4. 12. i. It is not unique. since no calculation is required. © ABE and RRC . It is not amenable to the algebraic manipulation needed in advanced statistical work. Unlike the arithmetic mean. 10. The mode is very easy to find with ungrouped distributions. in the set of numbers 7. but only on those near its own value. 13. 8. The mode may not exist.0 marks As the mode is defined as the data value that occurs most often. This set of numbers would be referred to as having a bimodal distribution. all we need to do is look for the leaf (number) that occurs the most often on one stem of the diagram. there are two '1' leaves on the '4' stem (i. 8. 7. 12 each number occurs only once so this distribution has no mode. it is not based on all the items in the distribution. For example.1). there are two modes.1. In ungrouped distributions the mode is an actually occurring value. particularly as they compare with those of the arithmetic mean. in the set of numbers.e. 7. 11. It can only be determined approximately with grouped distributions. Now. But what if there is an even number? Then. 12 The two values in the middle are 8 and 9. because there are five values above and five values below 9. The median however is 9. 13 The total of these eleven numbers is 100 and so the arithmetic mean is therefore 100/11 9. and so when those values are arranged in order of magnitude. the median is taken to be the arithmetic mean of the two values in the middle. 9. 7. as follows: Arrange all the values in order of magnitude and the median is then the middle value. 10. by convention. 10.091 (three decimal places). Median of a Simple Frequency Distribution Of course statistical data is rarely in such small groups and. 9. 7. Note that all the values are to be used – even though some of them may be repeated. then we expect about half of the distribution to have larger values. Table 8. 7. therefore. so the median is 8½. 10. For example. This suggests to us that a possible measure of location might be that value which is such that exactly half (50%) of the distribution has larger values and exactly half has lower values.13). and the other half to have smaller values.236 Measures of Location C. as you have already learned. 11.12. The value which divides the distribution into equal parts is called the median. Look at the following set of values: 6. The total number of values is 123. 8. Table 8. 10. 12. they must all be put separately into the list.15: Accident distribution data Number of accidents per day (x) 0 1 2 3 4 Total Number of days with the stated number of accidents (f) 27 39 30 20 7 123 Cumulative frequency (F) 27 66 96 116 123 © ABE and RRC . it was easy to pick out the middle value because there was an odd number of values. THE MEDIAN The desirable feature of any measure of location is that it should be near the middle of the distribution to which it refers. 10. 7. while the mode is 10 because that is the number which occurs most often (three times). Our first rule for determining the median is. 8. but adds a column setting out the cumulative frequency. let us draw up the distribution again. if a value is near the middle of the distribution. 8.15 repeats the data from Table 8. 8. To see what the value of the 62nd item will be. How then do we find the median if our data is in the form of a distribution? Let us take the example of the frequency distribution of accidents already used in discussing the mode (Table 8. the median will be the 62nd item because that will be the middle item. In the example just dealt with. take the following set of values: 6. we usually deal with frequency distributions. so we obviously need to move into that group by 11 items.5).Measures of Location 237 You can see from the last column that. and because we do not know the exact values of the items in each group. a convenient way to find the median of a grouped distribution is to draw the ogive. as you might expect. But exactly where in the "60 to < 70" group? Well. so we need to move 11/38 of the way into that group. from the cumulative frequency column.89 62. Altogether in our "60 to < 70" group there are 38 items. © ABE and RRC . the first 27 would all be 0's and from then up to the 66th would be 1's.16: Data for determining the median of a grouped frequency distribution Group 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 70 to < 80 80 to < 90 90 to < 100 100 to < 110 110 to < 120 Total Frequency (f) 4 6 10 16 24 32 38 40 20 10 5 1 206 Cumulative frequency (F) 4 10 20 36 60 92 130 170 190 200 205 206 The total frequency is 206 and therefore the median is the 103rd item which. the ogive. Because a grouped frequency distribution nearly always has a large total frequency. In practice. against a cumulative frequency of half the total frequency. Our median is therefore: 60 110 60 2. It therefore follows that the 62nd item would be a 1 and that the median of this distribution is 1. If the ogive is drawn with relative frequencies. Then. is very similar to the solution for an ungrouped distribution – we halve the total frequency and then find. then the median is always read off against 50%. must lie in the "60 to < 70" group. read off the median. if we were to list all the separate values in order. that is 11/38 of 10 above 60. there are 92 items before we get to that group and we need the 103 rd item.89 38 The use of the cumulative frequency distribution will no doubt remind you of its graphical representation. it is not necessary to find the two middle items when the total frequency is even – just halve the total frequency and use the answer (whether it is a whole number or not) for the subsequent calculation. from the cumulative frequency column. The solution to the problem. the corresponding value of the variate. In our example the median would be read against 103 on the cumulative frequency scale (see Figure 8. Median of a Grouped Frequency Distribution The final problem connected with the median is how to find it when our data is in the form of a grouped distribution. Table 8. but the individual data is preserved. As the median is the middle value in the set. Again. remember that the stem values represent the units of a mark awarded and the leaf values represent the tenths of a mark awarded.238 Measures of Location Figure 8. This means that the median can be calculated relatively straightforwardly. we can simply go through the data presented in our stem and leaf diagram and cross off pairs of high and low leaves. If we start with the © ABE and RRC . the stem and leaf diagram presents grouped data. let us consider our previous example of a stem and leaf diagram which shows the distribution of marks awarded to an ice skater in a competition (Figure 8. we can use a stem and leaf diagram to calculate the median value of a distribution of grouped data as the original data is preserved.6).5: Reading the median from an ogive Cumulative frequency 220 200 180 160 140 120 100 80 60 40 20 0 0 20 40 60 80 100 120 103 Median 63 Value of the variate Calculating the Median from a Stem and Leaf Diagram Similarly. Again.6: Stem and leaf diagram to show the marks awarded to a competition skater Stem 0 1 2 3 4 5 5 9 0 1 1 3 0 5 8 Leaf 0 Note: 0|0 0. Figure 8.0 marks As noted. we are left with two middle values. which you should compare with those of the mean and the mode. The median is not very amenable to further algebraic manipulation. for his height will be the median height. as its calculation does not depend on the precise values of the variate in these classes. Thus median is of value when we have open-ended classes at the edges of the distribution. and is readily understood as being the "halfway point". is thus 4. we do not have to measure the height of every single one.Measures of Location 239 rightmost leaf on the bottom stem and the leftmost leaf on the top stem. It is only necessary to stand the men in order of their heights and then only the middle one (number 11) need be measured. whereas the value of the arithmetic mean does.1. namely the two '1' leaves on the '4' stem. we can work our way towards the middle of the values crossing off the pairs until we have only one (or two) leave(s) left that have not been crossed out. It is less affected by extreme values than the mean. © ABE and RRC . are as follows: It is fairly easily obtained in most cases. The average of these values. It can be obtained without actually having all the values. That value (or the average of the two values) is the thus median. Characteristic features of the median. and hence the median. if we want to know the median height of a group of 21 men. For example. The millionaire in the country village might alter considerably the mean income of the village but he or she would have almost no effect at all on the median. In our example. 4. from the grouped distribution in the following table. it is found that the sample contains two brothers who are dwarfs.240 Measures of Location Questions for Practice 1. After this has been done. As part of a health study the heights of several dozen men in a particular location are measured as a sample. © ABE and RRC . Calculate the arithmetic mean consumption of the 100 householders. the average) size of shareholding. calculate the arithmetic mean (i. Using the short-cut methods given in this study unit.e. and why? The rate of effluent discharge from two plants based on different systems was sampled at four-hourly intervals and the results are given below: System A 10 13 10 7 12 8 17 4 10 10 18 18 11 9 6 19 9 12 5 10 12 13 10 12 14 10 11 13 8 16 13 16 9 12 6 9 12 7 15 12 7 6 14 3 13 5 10 4 10 10 12 12 12 11 9 4 11 14 8 9 3. The following table shows the consumption of electricity of 100 householders during a particular week. Consumption of electricity of 100 householders (one week) Consumption (kilowatt hours) 0 – under 10 10 – under 20 20 – under 30 30 – under 40 40 – under 50 50 – under 60 60 – under 70 70 – under 80 80 or over Number of householders 5 8 20 29 20 11 6 1 0 2. What sort of "average" would you calculate from the sample. System B Calculate the mean rate of discharge for each system. 06 0. Weight in kg 45 – 49 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 No.18 0.02 0 6.e. The following distribution gives the weight in kilograms of 100 schoolboys. A garage keeps two cars for daily hire. Find the median weight.36 0.Measures of Location 241 Distribution of shareholding Number of shares Less than 50 50 – 99 100 – 199 200 – 299 300 – 399 400 – 499 500 – 599 600 – 699 700 – 799 800 – 1. hired out) per day? On what fraction of days is some demand turned away? Proportion of days 0.38 0. The number of cars asked for in a day has varied as follows: Cars wanted in a day 0 1 2 3 4 5 or more (a) (b) (c) What is the mean daily demand? What is the mean number supplied (i.000 Number of persons with this size of holding 23 55 122 136 100 95 81 63 17 8 5. of schoolboys 2 15 37 31 11 4 100 © ABE and RRC . (b) the median holding and (c) the modal holding. of persons with these shares with these shares 80 142 110 75 52 38 300 – 349 350 – 399 400 – 449 450 – 499 500 – 549 550 – 599 28 15 9 7 3 2 8. © ABE and RRC . No.242 Measures of Location 7. of shares held No. Stem 1 2 3 4 Leaf 8 8 9 0 3 5 5 7 9 1 3 8 2 3 5 Note: 18 18 years old Based on the data presented in this diagram. The age distribution of employees working for a manufacturing company is given in the following stem and leaf diagram. of shares held Under 50 50 – 99 100 – 149 150 – 199 200 – 249 250 – 299 No. The holdings of shares (nominal value £1) in DEF plc are given in the following table. Calculate (a) the arithmetic mean holding. calculate the: (a) (b) (c) mean median mode. Now check your answers with those given at the end of the unit. of persons No. 2. Normally one thinks of the arithmetic mean as the best measure of average.53 30 316 10. The mode would not be so affected.53 30 3. the arithmetic mean of the householders' consumption is 36.630 36.015 900 605 390 75 0 3. System A: total number of readings 30 total of rates of discharge 316 mean rate of discharge System B: total number of readings 30 total of rates of discharge 316 mean rate of discharge 316 10. You can group the data first if you wish.3 Σf 100 i.Measures of Location 243 ANSWERS TO QUESTIONS FOR PRACTICE 1. © ABE and RRC .e. it would be best to use the median. or might not exist if all the heights were different. Consumption of electricity of 100 householders (one week) Consumption (kwH) 0 – under 10 10 – under 20 20 – under 30 30 – under 40 40 – under 50 50 – under 60 60 – under 70 70 – under 80 80 or over Mid-value (x) 5 15 25 35 45 55 65 75 Number of householders (f) 5 8 20 29 20 11 6 1 0 100 x fx 25 120 500 1. In order to give the correct impression of the mean in the sample location.630 Σfx 3. but in this case it would be too much affected by the two extremely low values to be representative.3 kilowatt hours. but with such a small sample (a dozen or so) it would be erratic. but it is probably quicker just to add the numbers as they stand. 18 0.0 74. 5.5 648.0 0.5 249.6 -0.36 0.38 0.5 -122.7 85.06 0. (a) Demand (x1) 0 1 2 3 4 5 or more Frequency (as a proportion) (f) 0.5 149.38 0.3 1.0 5. the most common class interval.5 449.00 © ABE and RRC .0 No.7 99.00 fx1 0 0.5 242.18 0.36 0.0 3.7 2.5 189.0 52.000 Total x 250 d= fd -51.0 2.6 251.0 6.5 549.02 0 1.8 1.00 1 car per day Σf 1.0 1.5 349. the middle of the modal group.08 0 1.0 648.8 -96. Distribution of shareholding Mid-value (x) 25.7 343 to the nearest whole number. Take c 100. the mean shareholding is 343 shares.5 749.0 4. of persons with a holding of x (f) 23 55 122 136 100 95 81 63 17 8 700 x xo c No. and xo 250.5 649.7 100 700 250 92. Therefore.244 Measures of Location 4.00 Mean demand Σfx1 1. of shares Less than 50 50 – 99 100 – 199 200 – 299 300 – 399 400 – 499 500 – 599 600 – 699 700 – 799 800 – 1.5 900. 02 0.18 0.02) Frequency (as a proportion) (f) 0. and in the (50 17) 33rd item within it.46 59.46 60 kg to the nearest kg. Weight in kg 45 – 49 50 – 54 55 – 59 60 – 64 65 – 69 70 – 74 No.38 0.90 Mean supply (c) Σfx 2 0. on eight days in every 100 demand is turned away.36 0.38 0.00 fx2 0 0. This proportion of days is 0. Median weight 55 55 4.90 0. of schoolboys 2 15 37 31 11 4 100 Cumulative frequency 2 17 54 85 96 100 6. Therefore.26 1.90 cars per day Σf 1.08. 33 5 37 © ABE and RRC .Measures of Location 245 (b) Supply (x2) 0 1 2 (0. Half the total frequency is 50 so the median lies in the "55 – 59 kg" group.06 0.52 0.00 Demand is turned away whenever three or more cars are requested in one day.06 0. © ABE and RRC . This is shown on the next page.041 242 799 Cumulative frequency 80 222 332 407 459 497 525 540 549 556 559 561 4 3 2 1 0 1 2 3 4 5 6 7 (a) x 224.5 58.424 50) 224. You may have chosen a different xo.5 474.5 424. on a histogram.5 524. Take c 50.B.5 222) item within it.3 153 shares to the nearest whole number.5 324. Median 99. of persons with these shares 80 142 110 75 52 38 28 15 9 7 3 2 561 x xo c Mid-value 24.5 799 50 561 224.1 110 126 shares to the nearest whole number.5 374.) (b) Half of the total frequency is 280.5 and therefore the median lies in the "100 – 149" group and it is the 58. xo 224.5 d= fd 320 426 220 75 0 38 56 45 36 35 18 14 1.5 574.5 74. the estimated mode 83 shares.246 Measures of Location 7.5 71. together with the surrounding class groups. Reading from this.2 153.5 50 126.5 174.5 224. (N.5 Distribution of the sizes of shareholdings in DEF plc Number of shares held Under 50 50 – 99 100 – 149 150 – 199 200 – 249 250 – 299 300 – 349 350 – 399 400 – 449 450 – 499 500 – 549 550 – 599 Total No. but the result will be the same.5 (1. (c) The modal group is the "50 – 99" group and the estimated mode is given by drawing this group.5 274.5th (280.5 124. In our example.e.e. (a) x Σfx (18 18 19 20 23 25 25 27 29 31 33 38 42 43 45) Σf 15 29. the arithmetic mean age of manufacturing employees is 29. © ABE and RRC . two data points of value 18) and two 5 leaves on the 2 stem (i.Measures of Location 247 Frequency (no of persons) 140 120 100 100 – 50 group 80 Under 50 group 50 Mode 83 100 150 Shareholding 8. the data set is bimodal with the mode age of manufacturing employees being 18 years and 25 years. Thus.e.1 years. (b) (c) The median age of manufacturing employees is 27 years. two data points of value 25).1 i. there are two 8 leaves on the 1 stem (i. 248 Measures of Location © ABE and RRC . Deciles and Percentiles The Quartile Deviation Calculation of the Quartile Deviation Deciles and Percentiles Standard Deviation The Variance and Standard Deviation Alternative Formulae for Standard Deviation and Variance Standard Deviation of a Simple Frequency Distribution Standard Deviation of a Grouped Frequency Distribution Characteristics of the Standard Deviation The Coefficient of Variation Skewness Page 250 251 251 251 253 254 255 255 257 258 259 262 262 263 268 C.249 Study Unit 9 Measures of Dispersion Contents Introduction A. B. Answers to Questions for Practice © ABE and RRC . D. E. Range The Quartile Deviation. This example shows that a mean (or other measure of location) does not by itself tell the whole story. we have studied the various measures of location that are available. the mean of 49 and 51 is 50.5 Factory B 136 92 110 36 102 57 108 81 156 117 995 99. the figures which go to make up a distribution may all be very close to the central value. The measures in common use are: range quartile deviation standard deviation. Factory A achieves its mean production with only very little variation from week to week.1: Output from two factories over 10 weeks Weekly output Factory A 94 100 106 100 90 101 107 98 101 98 995 99. but the mean of 0 and 100 is also 50. but the individual values may be spread about the mean in vastly different ways. When applying statistical methods to practical problems. However. which will be discussed later. is of great importance. they are very different in their weekto-week consistency. As was the case with measures of location. Each has its own particular merits and demerits. there are several different measures of dispersion in use by statisticians. We therefore need to supplement it with a measure of dispersion. whereas Factory B achieves the same mean by erratic ups and downs from week to week. Examine the figures in the following table: Table 9. © ABE and RRC . knowledge of this spread.250 Measures of Dispersion INTRODUCTION In order to get an idea of the general level of values in a frequency distribution. or they may be widely dispersed about it – for example. You can see that these two distributions may have the same mean.5 Week 1 2 3 4 5 6 7 8 9 10 Total Mean output Although the two factories have the same mean output. which we call "dispersion" or "variation". We will discuss each of these in this unit. you must cast your mind back to the method of obtaining the median from the ogive. the quartile deviation. the range is therefore 107 90. i. Consequently. © ABE and RRC . For these reasons it is used a great deal in industrial quality control work.17. deciles and percentiles of grouped data calculate the standard deviation and variance of grouped and ungrouped distributions.e. is the value which divides the total frequency into two halves. you remember. As a result. we can see that the lowest weekly output for Factory A was 90 and the highest was 107. DECILES AND PERCENTILES The Quartile Deviation This measure of dispersion is sometimes called the "semi-interquartile range". For Factory B the range is 156 36 120. as shown in Figure 9. The larger range for Factory B shows that it performs less consistently than Factory A. median and mode in a skewed distribution identify the advantages and disadvantages of the range. and takes no account of all those in between.Measures of Dispersion 251 Objectives When you have completed this study unit you will be able to: calculate the range of a distribution from given data find. by both calculation and graphical methods. In the example just given. To understand it. one or two extreme results can make it quite unrepresentative. THE QUARTILE DEVIATION. A. The advantage of the range as a measure of the dispersion of a distribution is that it is very easy to calculate and its meaning is easy to understand.1. the range is not much used in statistical analysis except in the case just mentioned. quartile deviation and standard deviation as measures of dispersion. B. RANGE The range is the simplest measure of dispersion: it is simply the difference between the largest and the smallest. The values which divide the total frequency into quarters are called quartiles and they can also be found from the ogive. The median. using the appropriate formulae calculate the standard deviation using the "corrected/approximate variance" method calculate the coefficient of variation and explain its use calculate Pearson's first and second coefficient of skewness from the appropriate data explain the nature of skewness and describe the relationship between the mean. Its disadvantage is that it is based on only two of the individual values. e. 100% corresponds to 206. the median and the upper quartile (Q3) respectively.75 © ABE and RRC . the total frequency. i. giving the lower quartile (Q1).5 2 Q3 Q1 2 14. 50% and 75% of the cumulative frequency. It is then easy to read off the values of the variate corresponding to 25%.5 2 29.1: Identifying quartiles from an ogive Cumulative 220 frequency 200 180 160 140 120 100 80 60 Relative 100 cumulative frequency (%) 75 50 25 40 20 0 0 20 40 60 80 100 120 Value of the variate This is the same ogive that we drew in Study Unit 8 when finding the median of a grouped frequency distribution. You will notice that we have added the relative cumulative frequency scale to the right of the graph. Q1 46.252 Measures of Dispersion Figure 9.5 median 63 (as found previously) Q3 76 The difference between the two quartiles is the interquartile range and half of the difference is the semi-interquartile range or quartile deviation: Quartile deviation 76 46. 5 on the cumulative frequency scale (i.e.5 and then read from the ogive the values of the variate corresponding to 51.e.5. 4 We can make the calculations in exactly the same manner as we used for calculating the median.2: Table 9. 154. you can work out: 25% of the total frequency. The end result is the same. Remember that graphical methods are never quite as accurate as calculations. and 4 75% of the total frequency. 206 51.e. We shall again use the same example.2: Data for finding the quartile deviation of a grouped frequency distribution Group Frequency (f) 4 6 10 16 24 32 38 40 20 10 5 1 206 Cumulative frequency (F) 4 10 20 36 60 92 130 170 190 200 205 206 0 to < 10 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 70 to < 80 80 to < 90 90 to < 100 100 to < 110 110 to < 120 Total The lower quartile is the 25% point in the total distribution and the upper quartile the 75% point. and the table of values is reproduced for convenience as Table 9.5 and 154. the left-hand scale).Measures of Dispersion 253 Alternatively. © ABE and RRC . Calculation of the Quartile Deviation The quartile deviation is not difficult to calculate in situations where the graphical method is not acceptable. i. i. As the total frequency is 206: the 25% point the 75% point 206 51½ and 4 3 206 154½. the 51½th item comes in the "40 to < 50" group and will be the (51½ 36) 15½th item within it.. Using the same example as previously.254 Measures of Dispersion Looking at Table 9. Deciles and Percentiles It is sometimes convenient.. P99. particularly when dealing with wages and employment statistics. So. P5 is the value below which 5% of the data lies and P64 is the value below which 64% of the data lies.8 to 1 decimal place.125 Remember that the units of the quartiles and of the median are the same as those of the variate. D2 ..125 46.667 2 Q3 Q1 2 24½ 10 40 15½ 10 24 14. So. the upper quartile will be the 154½th item which is in the "70 to < 80" group and is the (154½ 130) 24½th item within it.125 76. to this extent. Such partition values are deciles and percentiles. The deciles are labelled D1. P2 .458 46. Obviously it is only meaningful to consider such values when we have a large total frequency. For instance.8335 14. but which divide the distribution more finely. From their names you will probably have guessed that the deciles are the values which divide the total frequency into tenths and the percentiles are the values which divide the total frequency into hundredths.2.458 Similarly. The quartile deviation is unaffected by an occasional extreme value. In general. to consider values similar to the quartiles. the second decile (D2) is the value below which 20% of the data lies. D9. However it is not based on the values of all the items in the distribution and. then the quartile deviation should be used as the measure of dispersion. Again we can say that.. © ABE and RRC . the lower quartile 40 40 6. and the sixth decile (D6) is the value below which 60% of the data lies. the upper quartile 70 70 6. The percentiles are labelled P1. it is less representative than the standard deviation (which we discuss later). when a median is the appropriate measure of location. let us calculate as an illustration the third decile D3.458 2 29. for example. The quartile deviation 76. The method follows exactly the same principles as the calculation of the median and quartiles. Measures of Dispersion 255 The total frequency is 206 and D3 is the value below which 30% of the data lies: 30% of 206 is 30 206 61. The standard deviation is defined as the square root of the variance.3.6. We could also have found this result graphically – check that you agree with the calculation by reading D3 from the graph in Figure 9.8th item within it. and is the (61. Therefore 30% of our data lies below 50.8th item lies in the "50 to < 60" group. where n is the total frequency. and so we need to know how to calculate the variance first. The Variance and Standard Deviation We start by finding the deviations from the mean. and then squaring them. So: D3 50 50 18 32 1. Except for the use of the range in statistical quality control and the use of the quartile deviation in wages statistics.1.6 to one decimal place. © ABE and RRC . STANDARD DEVIATION The most important of the measures of dispersion is the standard deviation. which removes the negative signs in a mathematically acceptable fashion. C. You will see that the calculation method enables us to give a more precise answer than is obtainable graphically. A glance at the cumulative frequency column shows that the 61. the standard deviation is used almost exclusively in statistical practice.8th item.8 60) 1. n Consider the following example in Table 9.8 100 We are therefore looking for the value of the 61. squared: variance Σ( x x ) 2 . The variance is then defined as the mean of the deviations from the mean.8 10 32 50. although you may sometimes find "s" or "sd" used.50 10 Σ( x x ) 2 n 22.5 7.5 18.85 Week 1 2 3 4 5 6 7 8 9 10 Total Mean output The mean of the squared deviations is the variance: variance 228.25 2.5 0.5 1.5 6.25 42.50 22.5 Deviation (x x) 5.5 1. © ABE and RRC .25 0.25 228.0 18.0 Variance Deviation squared ( x x)2 30.85 4.5 9.25 2. n In this example.5 1.25 56.25 2. then: SD 22. is: SD Σ( x x ) 2 .78.25 0.25 2.25 90.5 1. again where n is the total frequency.85 The formula for the standard deviation.256 Measures of Dispersion Table 9.3: Output from Factory A over 10 weeks Weekly output (x) 94 100 106 100 90 101 107 98 101 98 995 99.5 0. which is defined as the square root of the variance. The symbol usually used to denote standard deviation is "σ" (theta). then you should use formula (c) – particularly if you are not asked to calculate x . is derived directly from the definition of the standard deviation. © ABE and RRC . there is no reason why you should not use formula (a).Measures of Dispersion 257 Alternative Formulae for Standard Deviation and Variance The formula we have used to calculate the standard deviation is: σ Σ(x x )2 n Formula (a) where: x value of the observations x mean of the observations n number of observations. like formula (a). When you are dealing with data in the form of a simple or grouped frequency distribution. then the formula for calculating the standard deviation needs to be expressed slightly differently: σ Σf ( x x )2 Σf Formula (d) or σ 1 (Σfx )2 Σfx 2 n n . as follows: σ Σx 2 x2 n Formula (b) or Σx 2 Σx σ n n 2 Formula (c) The choice of the formula to use depends on the information that you already have and the information that you are asked to give. If you do not know x . therefore: variance σ2 Σ( x x ) 2 n By expanding the expression Σ(x x )2 we can rewrite the formula for standard deviation in two alternative and sometimes more useful forms. you should aim to keep the arithmetic as simple as possible and the rounding errors as small as possible. then formula (b) is the best to use. As the standard deviation is defined as the square root of the variance. Formula (e) is obtained by using the sigma notation as in formula (c) and this is the one to use in calculations. where n Σf Formula (e) Formula (d). If you already know x but it is not an integer (as in the previous example). In any calculation. If you already know x and it is a small integer. as shown in the following sections. 85 (22.25 2.5 1. as follows: σ Σf ( x x )2 Σf σ 228.25 42.25 Frequency (f) 1 1 2 2 2 1 1 10 fx f(x – x )2 90 94 196 200 202 106 107 995 99.4 where we set out the information for calculating the standard deviation by applying formula (d): Table 9.25 0.5 90.85 The standard deviation is then worked out in the same way as previously.4: Determining the standard deviation of a simple frequency distribution Weekly output (x) 90 94 98 100 101 106 107 Total Mean Deviation (x x) 9.5 1.85 is the variance) σ 4.50 0.258 Measures of Dispersion Standard Deviation of a Simple Frequency Distribution If the data in Table 9. then only the different values would appear in the "x" column and we would have to remember to multiply each result by its frequency.50 Variance 22.25 2.5 6.5 5.50 42.5 0.25 228.25 4. This is shown in Table 9.5 Deviation squared ( x x)2 90.25 30.5 7.78 © ABE and RRC .3 had been given as a frequency distribution (as is often the case).25 56.25 30.50 4.5 10 22.25 56. 231 1 99.5 10 1 228. we could have applied formula (e) and reduced the calculation as follows: Table 9. the standard deviation is calculated as follows.78.231 10 10 1 990.Measures of Dispersion 259 However.236 11.236 11.836 19.100 8.5 10 22.231 99.025 99.85 (the variance) 4.000 20.5: Determining the standard deviation of a simple frequency distribution Weekly output (x) 90 94 98 100 101 106 107 Frequency (f) 1 1 2 2 2 1 1 10 Therefore: n Σf 10 Σfx 995 Σfx2 99.231 Calculating the standard deviation proceeds as follows: σ 1 (Σfx )2 Σfx 2 n n 1 995 2 99. Using formula (e) and applying it to the distribution shown in Table 9.201 11.002. © ABE and RRC . we again assume that all the readings in a given group fall at the midpoint of the group.402 11.231 10 10 fx 90 94 196 200 202 106 107 995 x2 8.836 9.604 10.6.449 99.208 20. Standard Deviation of a Grouped Frequency Distribution When we come to the problem of finding the standard deviation of a grouped frequency distribution.100 8.000 10.449 fx2 8. 625 58.125 12.625 fx2 450 3.33 30 1 6.950 30 30 Mid-value Frequency (x) (f) 15 25 35 45 55 65 75 2 5 8 6 5 3 1 30 fx 30 125 280 270 275 195 75 1.800 12. Table 9. but we can extend the short-cut method which we used for finding the arithmetic mean of a distribution to find the standard deviation as well. In that method we: worked from an assumed mean worked in class intervals applied a correction to the assumed mean.260 Measures of Dispersion Table 9.150 15.250 x2 225 625 1.89 (the variance) 15.866.950 Calculating the standard deviation proceeds as follows: σ 1 (Σfx ) 2 Σfx 2 n n 1 1.500 58.950 1 58.250 Σfx2 58.025 4.67 30 228.083. © ABE and RRC .025 3.562.225 5.7 applies this approach to working out the standard deviation.950 52.125 9.950 30 30 1 1.675 5.225 2.13 The arithmetic for this table is rather tedious even with a calculator.6: Determining the standard deviation of a grouped frequency distribution Class 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 70 to < 80 Total Therefore: n Σf 30 Σfx 1.2502 58. as follows: (a) The approximate variance is obtained from 1 82 82 2.4444 2.Measures of Dispersion 261 Table 9. it is necessary to multiply by the class interval as the last step. (c) (d) The corrected variance is thus: 2.7333 30 30 1 Σfd2 which in our case is equal to: n (b) 1 The correction is Σfd n 2 It is always subtracted from the approximate variance to get the corrected variance.7: Determining the standard deviation of a grouped frequency distribution (short cut) Class 10 to < 20 20 to < 30 30 to < 40 40 to < 50 50 to < 60 60 to < 70 70 to < 80 Total Mid-value Frequency (x) (f) 15 25 35 45 55 65 75 2 5 8 6 5 3 1 30 d = x xo c fd 4 5 0 6 10 9 4 29 9 20 d2 4 1 0 1 4 9 16 fd2 8 5 0 6 20 27 16 82 2 1 0 1 2 3 4 xo 35.2889 The standard deviation is then the class interval times the square root of the corrected variance: SD 2. it will all fall into place. As you are working in class intervals.13 This may seem a little complicated. c 10 The standard deviation is calculated in four steps from this table. but if you work through the example a few times. Remember the following points: Work from an assumed mean at the midpoint of any convenient class. © ABE and RRC .4444.7333 0.513 class intervals 10 15. The correction is always subtracted from the approximate variance. In our case the correction is: (20/30)2 0.2889 1. 000. If the intervals vary too much. To avoid the trouble of converting the dollars to euros (or vice versa) we can calculate the coefficients of variation in each case and thus obtain comparable measures of dispersion. the small firm is more erratic in its earnings than the large firm. we get coefficients of variation (usually indicated in formulae by V or CV) as follows: for the large firm: V for the small firm: V 500 100 1. Characteristics of the Standard Deviation In spite of the apparently complicated method of calculation. The column for d2 may be omitted since fd2 fd multiplied by d. Let us suppose that both standard deviations come to £500.000 This shows that. most important. © ABE and RRC . It gives slightly more emphasis to the larger deviations but does not ignore the smaller ones and. the coefficient of variation is a ratio and thus has no units. It is based on all of the individual items.000.262 Measures of Dispersion The correction factor is the same as that used for the short-cut calculation of the mean.000 500 100 25. Another application of the coefficient of variation comes when we try to compare distributions the data of which are in different units – for example. it can be treated mathematically in more advanced statistics. We call the result the "coefficient of variation". we revert to the basic formula. we can express the standard deviation as a percentage of the mean in each case. D. Now. To overcome this difficulty. while the other may be a small firm with average monthly profits of only £2. The formula for calculating the standard deviation in this way can be written as: Σfd 2 Σfd SD c n n 2 The assumed mean should be chosen from a group with the most common interval and c will be that interval. One of them may be a fairly large business with average monthly profits of £50. it would be unrealistic to say that the two businesses are equally consistent in their monthly profits. but for the SD it has to be squared. Note that although a standard deviation has the same units as the variate.0% 50. but what about the month-by-month variability? We can compare the variability of the two firms by calculating the two standard deviations.0% 2. But do not omit it until you have really grasped the principles involved. the standard deviation is the measure of dispersion used in all but the very simplest of statistical studies. £500 is a much more significant amount in relation to the small firm than it is in relation to the large firm. Applying the idea to the figures which we have just quoted. THE COEFFICIENT OF VARIATION Suppose that we are comparing the profits earned by two businesses. when we try to compare a French business with an American business. relatively speaking. Clearly. the general level of profits is very different in the two cases. So although they have the same standard deviations. SKEWNESS When the items in a distribution are dispersed equally on each side of the mean. Two distributions may have the same mean and the same standard deviation but they may be differently skewed. the median comes first and then the mean – see Figure 9. we say that the distribution is symmetrical. but these two are always in the same order: working outwards from the mode. the mean and the median lie somewhere along the side of the "tail". shows two symmetrical distributions. mode and median all occur at the same point. the greater the distance from the mode to the mean and the median.2: Symmetrical distributions Frequency Variate When the items are not symmetrically dispersed on each side of the mean. Figure 9.Measures of Dispersion 263 E.3 and then look at the same one through from the other side of the paper! Figure 9. If a distribution is symmetrical.3: Skewed distributions Frequency (b) (a) Positively skewed Negatively skewed Variate What does skewness tell us? It tells us that we are to expect a few unusually high values in a positively skewed distribution or a few unusually low values in a negatively skewed distribution. The more skew the distribution. i.2. we say that the distribution is skewed or asymmetric. although the mode is still at the point where the curve is highest.4. Figure 9. right in the middle. © ABE and RRC . while one with a tail to the left is negatively skewed. But in a skew distribution. the mean. This will be obvious if you look at one of the skew distributions in Figure 9. A distribution which has a tail drawn out to the right is said to be positively skewed.e. The value of the coefficient of skewness is between 3 and 3. You will get negative answers from negatively skewed distributions. When you do the calculation. and so we can use the amount of this spread to measure the amount of skewness. as it is also called). Examples of variates with positive skew distributions include size of incomes of a large group of workers. although values below 1 and above 1 are rare and indicate very skewed distributions. and so we can use an equivalent formula for Pearson's second coefficient of skewness: Pearson's second coefficient of skewness 3(mean median) SD You need to be familiar with these formulae as the basis of calculating the skewness of a distribution (or its "coefficient of skewness". length of service in an organisation and age of a workforce.4: Positions of the mean. mode and median in skew distributions Frequency Negatively skewed Positively skewed Median Median Mean Mode Mode Mean Variate For most distributions. the more spread out are these three measures of location. One such example is the age at death for the adult population in the UK. and positive answers for positively skewed distributions. except for those with very long tails. The most usual way of doing this is to calculate Pearson's first coefficient of skewness: Pearson's first coefficient of skewness mean mode SD However as we have seen the mode is not always easy to find. size of households. © ABE and RRC . remember to get the correct sign ( or ) when subtracting the mode or median from the mean. Negative skew distributions occur less frequently.264 Measures of Dispersion Figure 9. the following approximate relationship holds: mean mode 3(mean median) The more skewed the distribution. 615 1.200 4. Each loan is a multiple of £1.000 Calculate: (i) (ii) (iii) 2. (a) (b) Explain the advantages of using the coefficient of variation as a measure of dispersion. The standard deviation of the loan amounts (£). Explain carefully what is measured by the standard deviation.000 – 6.200 No.280 1.000 customers.504 1.360 1. The coefficient of variation of the loan amounts.800 1.600 – 4.800 – 5.400 – 3.400 5. of borrowers 1 6 9 14 17 23 28 35 39 48 © ABE and RRC .400 2. Income of borrower £ Under 1.800 – 2.Measures of Dispersion 265 Questions for Practice 1.012 406 0 10.000 – 3.650 1.200 – 4.000 and the details are given below: Loan Amount (£000) 1 2 3 4 5 6 7 8 9 10 11 or more Number of customers 535 211 427 1.600 6.600 – 7. Calculate the arithmetic mean and standard deviation for the following data using the short-cut method.000 3.800 4. A small finance company has loaned money to 10. The arithmetic mean of the loan amounts (£).000 6.600 3.400 – 6. each shift having a different operator – A.28 2 1 0 3.9 4.000 100.000 5.21 1 0 3 3.000 25. Calculate for the two distributions the median values and explain any difference between these results and those obtained in (a).6 19.1 11.9 21.000 Assets not over (£) 1.26 17 27 0 3.4 1.9 6.3 29.8 25. It is decided to investigate the accuracy of the operators.000 – Percentages Year 1 48. Items are produced to a target dimension of 3.22 6 0 22 3.000 100.5 10.000 3. using your results where relevant.3 17.500. Write a brief report on the data. B and C. Estimate for each year the number of people whose assets are less than (i) £2. The results of a sample of 100 items produced by each operator are as follows: Accuracy of operators Dimension of item (cm) 3.000 25.29 1 0 0 Operator A B C © ABE and RRC .000 10. and (ii) £25. Consider the following table: Percentages of owners with assets covered by estate duty statistics Assets over (£) – 1.140 Total number of owners (000s) You are required to: (a) (b) (c) (d) 4.4 0.266 Measures of Dispersion 3.000 10.000 3. lower and upper quartile values for Year 1 and Year 11.24 21 25 22 3. Derive by graphical methods the median.448 Year 11 18.000. Production is carried out on each of three shifts.25 36 34 4 3.5 0.25 cm on a single machine.2 3.23 10 7 49 3.2 18.27 6 6 0 3.000 5. with brief explanation. Calculate the coefficient of skewness for a distribution with a mean of 10. to 3 decimal places. which operator is: (i) (ii) (b) (c) (d) The most accurate in keeping to the target. Why is it that a measure of the mean level of the dimension obtained by each operator is an inadequate guide to his or her performance? Calculate the semi-interquartile range of the results obtained by operator A. Illustrate your answers to part (a) by comparing the results obtained by each operator as frequency polygons on the same graph.Measures of Dispersion 267 Required: (a) State. 5. The most consistent in his or her results. What does your answer tell you about this frequency distribution? Now check your answers with those given at the end of the unit. median of 11. and standard deviation of 5. © ABE and RRC . 460 9.024 8.650 1.108 6.920 9. The coefficient of variation has no units and thus the relative variation of sets of data expressed in different units can be compared.000 £6.280 1.000 10. (a) The coefficient of variation is the standard deviation expressed as a percentage of the arithmetic mean. (iii) coefficient of variation 2. It enables the relative variation of two sets of data to be compared.000 (i) x 60.376 3.360 1.504 1.259 100% 37.65% 6.960 12. This is particularly useful when dealing. We need to draw up the following table: Loan amount (£000) (x) 1 2 3 4 5 6 7 8 9 10 Number of customers (f) 535 211 427 1.259 thousand pounds Σf £2.250 9.000 © ABE and RRC . a situation which makes absolute comparisons difficult.000 (x – x ) –5 –4 –3 –2 –1 0 1 2 3 4 (x – x )2 25 16 9 4 1 0 1 4 9 16 f(x – x )2 13.615 1.440 8.268 Measures of Dispersion ANSWERS TO QUESTIONS FOR PRACTICE 1.259 to nearest £.1028 2.496 51. with profits expressed in different currencies.375 3.012 406 10.028 (ii) SD Σf ( x x ) 2 5.440 1.843 5. say.060 60.281 5. This is of use when comparing sets of data which have widely differing means.650 0 1.000 (b) fx 535 422 1.280 6.108 4. 300 3.400 5.800 to < 2. © ABE and RRC . It does this by taking every reading into account. of borrowers (f) 1 6 9 14 17 23 28 35 39 48 220 d = x xo c fd 6 30 36 42 34 23 0 35 78 144 86 fd2 36 150 144 126 68 23 0 35 156 432 1.100 5.200 – £1.Measures of Dispersion 269 2. We then add all the squared deviations.500 5.363.200 to < 4.800 so that its mid-value is £1.100 86 600 220 £5.334.400 to < 6. We need to draw up the following table.55 £5.200 Mid-value (x) 1.600 6.1654 SD 5.700 6.100 2.300 6.900 No.600 to 7.3182 – 0.800 to < 5.400 2.200 4.500 * 2.000 to < 6.500.400 to < 3.000 6.1528 5.000 3.700 3. with xo £5.170 86 variance in class interval units squared 220 220 2 5.6 £1.800 4.900 4.100 and c £600 Income of borrower £ Under 1.364 to the nearest £.335 to the nearest £ 1.800 1. find their mean and finally take the square root.55 £5.2727 class interval units £1.600 to < 4.000 to < 3. The standard deviation measures the dispersion or spread of a set of data.170 6 5 4 3 2 1 0 1 2 3 * Assume the first class covers the range £1. We take the deviation of each reading from the mean and square it to remove any negative signs in a mathematically acceptable fashion. x £5.100 + 234.1654 2.600 3. 3 77. we can draw ogives for both years: Ogive to show cumulative percentage of owners with assets covered by estate duty statistics Cumulative percentage frequency 100 Year 1 90 80 Year 11 70 60 50 40 30 20 10 0 0 5 10 15 20 25 Assets (£000s) © ABE and RRC .8 43.0 99.1 95.4 From this.8 Year 11 18.000 10.000 100.9 98.000 Cumulative % frequency (less than) Year 1 48.000 25.9 55.000 5.8 88.270 Measures of Dispersion 3 (a) We need first to draw up the following table Assets over (£) 1.8 77.000 3.7 94.3 99. 7 £2.913. © ABE and RRC .000 50 48.000 £3.500.25 £1.21 These figures are more precise than the results in (a).000 group: median £3.3% of owners had assets less than £25. This is to be expected as graphical methods are approximate and are limited by the size of the graph paper.000 group: median £1.000 1.Measures of Dispersion 271 Reading off the graph at a cumulative percentage frequency of 50 we obtain the median: Year 1: median £1. Therefore the number of those with assets less than £25.183.25 Year 11: The 50% mark is in the £3.000 29.000 Reading off the graph at cumulative percentage frequencies of 25 and 75 we obtain the lower and upper quartile figures: Year 1: lower quartile £500.500.3 £2. we note that in Year 1.9 £2.9 £3. 98.800 Year 11: lower quartile £1.177.5 £1. upper quartile £2. 70% of owners had assets less than £2.000. (c) Reading from the ogive at a value on the horizontal scale corresponding to £2.600 100 37. (b) Year 1: The 50% mark is in the £1.500.5 19. In Year 11 the corresponding figure was 37½%.384 in Year 1 and 18. upper quartile £9. 100 Year 11: From the table of values in (a).000 50 43. Thus.500.000 11. (d) The total number of owners with assets covered by estate duty statistics rose by 692.000 11.115.134.5 1.1 £2.000 £1.000 – £5.500 was: Year 1: 70 18. the number with assets less than £2.9 6.000 in Year 11.21 £4.025. we find that in Year 1.140.000 – £3.000 29.000 115.000 from Year 1 to Year 11.500.000 12.448.100 Year 11: median £4. In Year 11 the corresponding percentage was 95%.025.000 7.000 was 18. it implies that no calculations are to be made. Dimension (cm) (x) 3. It is good practice to work them out. The interquartile range for Year 1 was £500 – £2. there appears to have been some redistribution of wealth in this period.27 3.7% had assets over £25.000.500 – £9. 4. C is the most consistent in his/her results as almost 50% of his/her items are 3.25 3. Whereas the modal group in Year 1 was under £1. it is unlikely to account for a change in percentage as large as this. As a typical value for assets.29 Total fA 1 6 10 21 36 17 6 2 1 100 fB 7 25 34 27 6 1 100 x 3.25 0.000 group. presumably because more owners are covered by the statistics in Year 11.500. 50% had assets less than £4.272 Measures of Dispersion As well as an increase in total wealth. You need to draw up the following table. (a) As this question says "with brief explanation".000 – £3.02 cm.25 cm as his/her items are spread about 3. indicating a greater spread of assets in the middle range in the later year.000 whereas 5% had assets over that amount in Year 11. so details of the calculations are now given. £1. Although inflation might be responsible for lifting some owners to higher groups.22 3.28 3.01 fC 3 22 49 22 4 100 dA = dB 4 3 2 1 0 1 2 3 4 dC 2 1 0 1 2 - fAdA 4 18 20 21 0 17 12 6 4 24 fBdB 14 25 0 27 12 3 3 fCdC 6 22 0 22 8 2 fAdA2 fBdB2 fCdC2 16 54 40 21 0 17 24 18 16 206 28 25 0 27 24 9 113 12 22 0 22 16 72 dA dB dC x 3.23 0.25 cm. So we could make the following observations: (i) (ii) B is most accurate in keeping to the target of 3. we note that in Year 1. in Year 11 it was £1.21 3.26 3. You could back up your assertions with calculations of the mean and standard deviation in each case if time permits.01 © ABE and RRC .800 but in Year 11 it was much larger.23 3.025. but with a narrower spread than A.23 cm and his/her maximum spread either way is 0.24 3.115 but in Year 11.000 was greater in Year 11 than in Year 1. The actual number of people with assets less than £25. In Year 1. Similarly for operators B and C.000 – £10. 1. 50% of the owners had assets less than £1. Note that f A and dA refer to operator A.000 closely followed by the £5. 25 cm.01 2.0003 3.0576 0.0024 = 0.7196 0.25 x B 3.0004 100 100 2 0. but his/her output varies from 3.2 3.Measures of Dispersion 273 x A 3.13 0.01 1.25 cm and C has the smallest standard deviation.01 2.1291 = 0.21 cm to 3.23 0.01 1.3 Frequency polygons to show output of three operators Operator C Operator A Operator B Dimension of item (cm) (c) The mean on its own gives no indication of the spread of the readings about the mean. © ABE and RRC .28 3.25 0.0106 cm SDC 0.2302 cm 100 206 24 100 100 2 SDA 0.25 0.25 x C 3.22 3.0142 cm SDB 113 3 0.0024 3.72 0.00848 cm This confirms our earlier assertion as x B is the mean closest to 3.01 0.06 0.2476 cm 100 (3) 0.2503 cm 100 2 0.29 cm.01 0.24 3.01 3.0002 3.01 72 2 0. The mean dimension produced by A is close to 3.01 0.26 3.23 (24) 0. (b) 60 50 40 30 20 10 0 3.009 100 100 2 0.01 0.01 3.01 3. Cumulative frequency 1 7 17 38 74 91 97 99 100 Less than (cm) 3.235 3. 5.01 cm – i.235 Similarly Q3 3.255 0.275 3. we assume that the one item produced by A in the 3. asymmetric with the peak of the curve being to the right.2388 cm.01 3. in fact. Q1 3.e.01 3. © ABE and RRC . This tells us something about the spread of the middle 50% of the data.255 3.6 standard deviation 5 Thus we can say that the distribution is slightly negatively skewed – i.0168 0.01 0.01 3. 21 21 The semi-interquartile range places.215 cm. it is the value of the 25th item.285 3.274 Measures of Dispersion (d) Assume that each dimension has been measured to the nearest 0.215 3.255 (75 74) 1 0. so that possible wastage when items are made too far from the target dimension could be estimated. i.295 The lower quartile Q1 is the value below which 25% of the data lies. It could be of use to the manufacturer for predicting the variation in A's production.008 cm to three decimal 2 2 (25 17) 8 0.e.235 3.21 cm category is. 17 17 Q3 Q1 0. Coefficient of skewness 3(mean median) 3(10 11) 0.2556 cm.245 3. less than 3.225 3.e.265 3. Answers to Questions for Practice Appendix © ABE and RRC . G. E.275 Study Unit 10 Probability Contents Introduction A. Estimating Probabilities Theoretical Probabilities Empirical Probabilities Subjective Probabilities Types of Event The Two Laws of Probability Addition Law for Mutually Exclusive Events Addition Law for Non-Mutually Exclusive Events Multiplication Law for Independent Events Multiplication Law for Dependent Events Listing Equally Likely Outcomes Conditional Probability Tree Diagrams Sample Space Expected Value The Normal Distribution Properties of the Normal Distribution Using Tables of the Normal Distribution Further Probability Calculations Page 276 277 277 278 279 279 280 280 282 283 285 286 286 288 294 295 298 300 301 302 305 311 B. D. H. C. F. © ABE and RRC . but it will help you understand the more practical applications. the significance of the normal frequency distribution explain the properties of the normal distribution in terms of the location of observations. Just as there is a scale of temperature. Initially. independent. from appropriate data. empirical and subjective probability explain the probability terms mutually exclusive. Some things are more probable than others – for example. contingency tables and sample spaces to solve probability problems calculate. The normal distribution has a number of properties which can be used to work out the probability of particular event happening. in simple terms. We shall then move on to consider the application of probability to real situations. but it may not be a very definite one. and the multiplication law for dependent and independent events construct and use tree diagrams.276 Probability INTRODUCTION Probability is one of those ideas about which we all have some notion. you will note. the mean and the standard deviation use normal distribution tables to find probabilities that observations within specified limits will occur. dependent. snow is more likely to fall in winter than in summer. It is the branch of mathematics which deals specifically with matters of uncertainty. some uncertainty in these matters. Most things in real life are uncertain to some degree or other. the expected value of an outcome describe. non-independent. because some things are hotter than others. Objectives When you have completed this study unit you will be able to: explain the meaning of a priori. so there is a scale of probability. but will confine ourselves to the task of grasping the idea generally and seeing how it can be used. we will not spend our time trying to get an exact definition. it is necessary to start with simple things like cointossing and die-throwing. or a healthy person has more chance of surviving an attack of influenza than an unhealthy person. non-mutually exclusive. This is a particular aspect of frequency distributions which we considered previously. and it is for this reason that the theory of probability is of great practical value. There is. and complementary explain what is meant by conditional probability solve probability problems using the addition law for mutually exclusive and nonmutually exclusive events. For the purpose of learning the theory. Other words which convey much the same idea as probability are "chance" and "likelihood". This may seem a bit remote from business and industrial life. This study unit also introduces the concept of the normal distribution. The probability that it will snow next January. The probability that the throw of a die will show a six. events which are more likely not to occur have probabilities which are less than 0. Either the order will be delivered or it will not.Probability 277 A. and any event which is certain to occur has a probability of 1. Theoretical Probabilities Sometimes probabilities can be specified by considering the physical aspects of the situation. There is no reason to favour either side as a coin is symmetrical. The probability of a six is one in six or 1 6 The probabilities are as follows: or 0. Statements like this are trying to put probabilities or chances on uncertain events. Remember that events which are more likely to occur than not have probabilities which are greater than 0. For a reliable car it should be considerably greater than 0. the probability that a light bulb will fail sooner or later is 1. Accordingly. The probability that your car will not break down on your next journey. which we call P(H). For example. is: P(H) 0. This probability is quite low. The probability of heads is one in two or 0. Similarly. even though when Friday arrives there are only two outcomes.5. the probability of heads. The probability that sales for your company will reach record levels next year. You can answer this one yourself. consider the tossing of a coin. The probabilities get closer to 0 as the events get more unlikely. Try to estimate probabilities for the following events. ESTIMATING PROBABILITIES Suppose someone tells you: "there is a 50-50 chance that we will be able to deliver your order on Friday". the probability of all outcomes for a particular event does not exceed 1: the total probability rule is thus: ΣP 1 (for all outcomes). the probability that the sun will not rise tomorrow is 0. Any event which is impossible has a probability of 0. The events are: (a) (b) (c) (d) (e) (a) (b) (c) (d) (e) The probability that a coin will fall heads when tossed. For uncertain events.1. Therefore.5. and the more certain they are the closer the probabilities are to 1. This statement means something intuitively. This depends on how frequently your car is serviced. It is somewhere between 0 and 0.5 © ABE and RRC .5. Probability is measured on a scale between 0 and 1: the probability limits are thus: 0 ≤ P(A) ≤ 1 where P(A) is the probability of an event A occurring. For example. The 50-50 chance mentioned previously is equivalent to a probability of 0. What is the probability that it will fall heads? There are two sides to a coin.5.5. the probability of occurrence is somewhere between 0 and 1.167. all outcomes for this particular event are not equally likely.05 Sometimes it is not possible to set up an experiment to calculate an empirical probability. The probability of getting a one is P(1) 0. Thus. empirically). Empirical Probabilities Often it is not possible to give a theoretical probability of an event.e. Therefore. P(6). we say that the outcomes are equally likely to occur. assuming it is not weighted in favour of any of the sides. For example.278 Probability Another example is throwing a die.167. then: probability of failure. all outcomes are equally likely. If we pick one bead out of the box at random (blindfold and with the box well shaken up). then the probability of that event occurring (called a success) is given by: P(E) h n number of possible ways of E occurring total number of possible outcomes In the preceding examples. The probability of getting a heads is P(H) 0. whereas the probability of getting a white bead is P(R) 0. For example.167 As a third and final example. is: P(6) 1 6 0. while others are not equally likely. what is the probability that an item on a production line will fail a quality control test? This question can be answered either by measuring the probability in a test situation or by relying on previous results (i. If someone throws a die. the chances of getting a six are no different to getting any other number.77. there is no reason to favour one side rather than another. In general terms. the chances of getting a red bead or a black bead are not the same.23. i. imagine a box containing 100 beads of which 23 are black and 77 white. when there are the same chances for more than one outcome.167. we introduced the concept that some outcomes are equally likely. if 5 items actually fail the test P(F) 5 100 no. Again. all outcomes are equally likely. which can be assessed from our prior knowledge of a situation. so the probability is: P(B) 23 100 0. the probability of getting a two is P(2) 0. If 100 items are taken from the production line and tested. For example: If someone tosses a coin.5. In the example about the box containing 100 beads. A die has six sides. of items which fail total no. the probability of getting a black bead is P(B) 0. what is the probability that we will draw a black bead? We have 23 chances out of 100. we can say that if an event E can happen in h ways out of a total of n possible ways.e.5 and the probability of getting a tails is P(T) 0. P(F) So. of items tested 0. the probability of a six showing uppermost. But not all outcomes are equally likely to occur. what are your chances of passing a particular examination? You cannot sit a series © ABE and RRC . In probability. i. the chances of getting a heads or tails are the same.e. are also called a priori probabilities. of which 23 are black and 77 white. i. etc.23 Probabilities of this kind.e. and similarly they know at what price to sell them prior to a downturn in the market. and failed only one. on the throw of the coin. © ABE and RRC . since there is one card. (b) Non-mutually-exclusive events These are events which can occur together. On the throw of a die. the occurrence of any one of them as the top face automatically excludes any of the others. For example. the queen of hearts.92 Subjective Probabilities A subjective probability describes an individual's personal judgement or instinct about how likely a particular event is to occur. then we would all be doing it! But they are not. It is not based on any precise theoretical or empirical foundation.e. all the sides of a die are mutually exclusive. you might estimate: probability of passing. which is both a heart and a queen and so satisfies both criteria for success. a six excludes all other possibilities. Successful investors are those who know which stocks to buy at a relatively low price. a rare event has a subjective probability close to 0. Thus. TYPES OF EVENT In probability. In fact. Like all probabilities. Mutually exclusive events If two events A and B are mutually exclusive then the occurrence of one event precludes the possibility of the other occurring. P(pass) 11 12 0. investors on the stock market may base many of their decisions to buy and sell shares on their instinct. If you have taken 12 examinations in the past. the two sides of a coin are mutually exclusive since. but is often a reasonable assessment by a knowledgeable person. they are based on instinct: investors attach (subjective) probabilities to the likelihood that share prices will move in a particular direction and they make their decisions accordingly. For example. in a pack of playing cards hearts and queens are non-mutually exclusive. If these investment decisions were based on a precise theoretical or empirical foundation. For example. In business. B. Thus: P(A and B) 0. a person's subjective probability of an event describes his or her degree of belief in the outcome of that event. Previous results must be used. we can identify five types of event: (a) mutually exclusive non-mutually exclusive independent dependent or non-independent complementary.Probability 279 of examinations to answer this. given their knowledge of the business environment in which they operate. heads automatically rules out the possibility of tails. a subjective probability is conventionally expressed on a scale from 0 to 1. on their instinct. a very common event has a subjective probability close to 1. managers will often make strategic decisions based on the subjective probability of the outcome of an event – i. We can express this as: P(drive to work|car starts) again where the vertical line is a shorthand way of writing "given that". For example. the tossing of a coin in no way affects the result of the next toss of the coin. and the event where either a one. C. namely the addition law and the multiplication law. four or five is uppermost by A (or not A). Addition Law for Mutually Exclusive Events Consider again the example of throwing a die. each toss has an independent outcome. i. i. The probability of her being able to drive to work given that the car starts is a conditional probability.e. if we throw a die and denote the event where a six is uppermost by A. we can always find P(A) by subtracting P( A ) from 1. (e) Complementary events An event either occurs or it does not occur.280 Probability (c) Independent events These are events which are not mutually exclusive and where the occurrence of one event does not affect the occurrence of the other. as it is often easier to find the probability of an event not occurring than to find the probability that it does occur. they are mutually exclusive with a total probability of 1. then A and A are complementary. the probability of a car owner being able to drive to work in her car is dependent on her being able to start the car. We can express this as: P(A) P(A|B) and/or P(B) P(B|A) where the vertical line is a shorthand way of writing "given that". For example.e. Using this formula. Thus: P(A) P( A ) 1 This relationship between complementary events is useful. THE TWO LAWS OF PROBABILITY In this section we will define the two laws of probability. You will remember that: P(6) Similarly: P(1) P(2) P(3) P(4) P(5) 1 6 1 6 1 6 1 6 1 6 1 6 © ABE and RRC . three. we are certain that one or other of these situations holds. two. For example. (d) Dependent or non-independent events These are situations where the outcome of one event is dependent on another event. and discuss the situations when each can be used. . Five of these are red.07 0. But also. N are mutually exclusive events. The sum of the probabilities of a complete list of mutually exclusive events will always be 1.05 What is the probability that it will be blue? Probability that ball is blue P(B) 7 100 0.07 What is the probability that it will be red or blue? P(R or B) P(R) P(B) 0.05 0. or N) P(A) P(B) . B . then it is certain that one of these outcomes will occur. This is a complete list of mutually exclusive possibilities. this can also be stated as P(1) P(2) P(3) P(4) P(5) P(6) 1 Again this is a general rule.Probability 281 What is the chance of getting 1. when the die is tossed there must be one number showing afterwards. from the equations shown here you can see that: P(1) P(2) P(3) This illustrates that P(1 or 2 or 3) P(1) P(2) P(3) This result is a general one and it is called the addition law of probabilities for mutually exclusive events. Therefore: P(R) P(B) P(W) 1 Then the probability of one ball being white is given as: P(W) 1 P(R) P(B) 1 0.5 0. P(N) where A. If all possible mutually exclusive events are listed.07 0.12 This result uses the addition law for mutually exclusive events since a ball cannot be both blue and red. It is stated more generally as: P(A or B or . seven are blue and the rest are white.5. What is the probability that it will be red? Probability that ball is red P(R) 5 100 1 6 1 6 1 6 0.. For example. What is the probability that it will be white? The ball must be either red or blue or white. 2 or 3? From the symmetry of the die you can see that P(1 or 2 or 3) 0. One ball is to be drawn at random from the urn..05 0.. It is used to calculate the probability of one of any group of mutually exclusive events. P(1 or 2 or 3 or 4 or 5 or 6) 1 Using the addition law for mutually exclusive events.. Example: An urn contains 100 coloured balls.88 © ABE and RRC .. of aces 4 no. but the probability of the events occurring together must be deducted: P(A or B or both) P(A) P(B) P(A and B) Example 1: If one card is drawn from a pack of 52 playing cards. However. Thus. the event being both A and B). in pack 52 no. the conditions are not met by both events occurring. in pack 52 no. of aces of spades 1 no.282 Probability Addition Law for Non-Mutually Exclusive Events Events which are non-mutually exclusive are. of spades 13 no. of aces of diamonds 1 no. We also need to take account of the probability of the card being the ace of spades (i. of spades 13 no. we need to deduct that probability from one of the other probabilities. Here. or (b) that it is either a spade or the ace of diamonds? (a) Let event B be "the card is a spade" and event A be "the card is an ace". in pack 52 no. so the probability is given by: P(spade or ace of diamonds) P(A) P(B) P(A) P(B) no. what is the probability (a) that it is either a spade or an ace.e. capable of occurring together. We need to determine the probability of the event being either a spade or an ace. To determine the probability of one of the events occurring. we can still use the addition law. in pack 52 P(spade or ace of diamonds) 1 13 14 7 52 52 52 26 © ABE and RRC . This is given by: P(spade or ace [or both]) P(A or B) P(A) P(B) P(A and B) P(A) P(B) no. by definition. this probability will be included in the probability of it being a spade and in the probability of it being an ace. in pack 52 P(A and B) P(spade or ace) (b) 4 13 1 16 4 52 52 52 52 13 Let event B be "the card is a spade" and event A be "the card is the ace of diamonds". but only every tenth item is checked.9 So.2 0.. If you toss a coin once and then again a second time. To work out the probability of two independent events both happening.2 probability that it is acceptable. however. This can be stated as: P(A and B) P(A) P(B) if A and B are independent events.5 even if all the previous tests have resulted in heads. 9 10 of the customers buy at least one kind of bread. The results of any third or subsequent test are also independent of any previous results. Again this result is true for any number of independent events. So: P(A and B . Consider the possibility that an individual item is both defective and not checked.Probability 283 Example 2: At a local shop 50% of customers buy unwrapped bread and 60% buy wrapped bread.5 0.. This item could be defective or acceptable. These two possibilities are mutually exclusive and represent a complete list of alternatives. P(N) © ABE and RRC . An item is either checked or it is not. Let: T represent those customers buying unwrapped bread and W represent those customers buying wrapped bread. This is shown as: probability that it is checked.1 probability that it is not checked. P(A) 0.8 Now consider another facet of these items. you use the multiplication law for independent events. There are also other kinds of independent events. and N) P(A) P(B) .6 0. These two events can obviously both occur together so they are not mutually exclusive. There is a system for checking them. independent. What proportion of customers buy at least one kind of bread if 20% buy both wrapped and unwrapped. Multiplication Law for Independent Events Consider an item on a production line. P(C) 0. They are. Assume that: probability that it is defective. The probability of heads on any test is 0. the outcome of the second test is independent of the results of the first one.. Then: P(buy at least one kind of bread) P(buy wrapped or unwrapped or both) P(T or W) P(T) P(W) P(T and W) 0. P(D) 0.. whether an item is defective or acceptable does not affect the probability of it being tested.9 Again these two possibilities are mutually exclusive and they represent a complete list of alternatives. That is to say. P(N) 0. 1 0.1 0.0001 Example 2: A card is drawn at random from a well shuffled pack of playing cards. since the suit and the number of a card are independent: P(H and 3) P(H) P(3) © ABE and RRC .2 0.02 1.02 0. what is the probability that both items are defective? For the item from the first batch: probability that it is defective: P(D1) For the item taken from the second batch: probability that it is defective: P(D2) 50 0.9 0.000 items of which 50 are damaged.1 probability that it is not checked: P(N) 0.02 P(A and N) 0.72 P(A and C) 0.8 0.2 probability that it is acceptable: P(A) 0.8 0. we get: P(D and N) 0.08 Example 1: A machine produces two batches of items. P(D and C) 0.9 0.005 0.2 0.18 The probabilities of the other combinations of independent events can also be calculated. The second batch contains 10.000 items of which 20 are damaged. If one item is taken from each batch.9 Using the multiplication law to calculate the probability that an item is both defective and not checked.284 Probability Consider the last example: for any item probability that it is defective: P(D) 0. What is the probability that the card is a heart? What is the probability that the card is a three? What is the probability that the card is the three of hearts? Probability of a heart: P(H) Probability of a three: P(3) 13 1 52 4 4 1 52 13 1 1 1 4 13 52 Probability of the three of hearts.8 probability that it is checked: P(C) 0.000 Since these two probabilities are independent: P(D1 and D2) P(D1) P(D2) 0.000 20 0.005 10. The first batch contains 1. P(A and B) 3 2 7% 10 9 This means that if this experiment were repeated 100 times. What is the probability of selecting a defective marble followed by another defective marble? Probability that the first marble selected is defective: P(A) 3 10 2 9 Probability that the second marble selected is defective: P(B) Thus.e. What is the probability of one or more sixes in these three throws? Probability of no six in first throw 5 6 5 6 Similarly. thus: 52 P(D and 8) 4 1 1 52 4 52 © ABE and RRC . P(D) ) . probability of no six in second or third throw The result of each throw is independent. Example 1: For example. P(8) ) in the pack and that a quarter of the pack 4 4 are diamonds (i.e. Example 2: Select one card at random from a deck of cards and find the probability that the card is an 8 and a diamond. suppose there are 10 marbles in a bag. in the long run 7 experiments would result in defective marbles on both the first and second selections. Two marbles are to be selected one after the other without replacement.Probability 285 Example 3: A die is thrown three times. and 3 are defective. 1 We know that there are four '8' cards (i. so: probability of no six in all three throws 5 5 5 125 6 6 6 216 Since no sixes and one or more sixes are mutually exclusive and cover all possibilities: probability of one or more sixes 1 125 91 216 216 Multiplication Law for Dependent Events To work out the probability of two dependent events both happening. you use the multiplication law for dependent events. This can be stated as: P(A and B) P(A) P(B|A) where: A and B are dependent events and the vertical line is a shorthand way of writing "given that". the probability that there will be one heads and one tails (heads-tails. the probability of different faces is equal to the probability of the same faces – in both cases 0. For example.e.286 Probability Listing Equally Likely Outcomes Another matter about which you must be careful is the listing of equally likely outcomes. we noted earlier the case where a car owner's ability to drive to work in her car (event A) is dependent on her being able to start the car (event B).) In this example. Suppose 80 of the candidates pass and 50 of them have had 10 or more lessons. the conditional probability of event A occurring. (In this case "not both heads" can result in three different ways. 45 of them passed. This is a case of the addition law at work: the probability of heads-tails (¼) plus the probability of tails-heads (¼). both events A and B may or may not occur). For example. there is no necessary link between the two events – and there is no conditional probability. we can list the possible results of tossing two coins: First Coin Heads Tails Heads Tails Second Coin Heads Heads Tails Tails There are four equally likely outcomes. or tailsheads) is 0. the conditional probability of event B occurring. that there are only two outcomes (both heads or not both heads) – you must list all the possible outcomes. so the probability of this result will be higher than "both heads". There is self evidently a causal link between the two events. for example. n(B) Similarly. given that A also occurs. CONDITIONAL PROBABILITY The probability of an event A occurring where that event is conditional upon event B having also occurred is expressed as: P(A|B) The events A and B must be causally linked in some way for A to be conditional upon B. D. n(A) One approach to determining probability in such circumstances is to construct a table of the possible outcomes – sometimes called a contingency table. Do not make the mistake of saying.5. given that B also occurs. Of those who have had 10 or more lessons. is calculated by the following formula: P(A|B) n(A and B) where n the number of possible outcomes. Be sure that you list all of them. Putting it another way. However. Consider the case of 100 candidates attempting their driving test for the first time. is calculated by the following formula: P(B|A) n(A and B) where n the number of possible outcomes. In a case involving equally likely outcomes for both events (i. if we consider the case of her being able to drive to work given that she is married. What is the probability of passing at the first attempt if you have had only 6 lessons? © ABE and RRC .5. 2. n(A and B) 35 0. that the total number of outcomes in which event B is "had 10 or more lessons" is 50. We have entered in the information that we know: that there are a total of 100 possible outcomes.Probability 287 We can construct a table which identifies the possible outcomes as follows in Table 10. Thus. Table 10. Make sure you understand the construction of this table: We have listed the possible outcomes of event A along the top – these outcomes are either to have passed or failed the test. or less than 10 lessons. Thus. We have listed the possible outcomes of event B along the side – these are either to have had 10 or more lessons. Table 10. the columns will show the number of outcomes for either possibility. that the total number of outcomes in which event A was "passed" is 80.7 n(B) 50 Thus the probability is: P(A|B) © ABE and RRC . the rows will show the number of outcomes for either possibility. Therefore: the total number of outcomes of B 50 the total number of outcomes of both A and B 35.1. We can complete the rest of the table by subtraction as in Table 10. and that the total number of outcomes in which event A was "passed" and event B was "had 10 or more lessons" is 45.2: Contingency table Passed 10 plus lessons Less than 10 lessons Total 45 35 80 Failed 5 15 20 Total 50 50 100 We can now work out the total outcomes applicable to the question "what is the probability of passing at the first attempt if you have had only 6 lessons?" Remember that event A is "passed" and event B is "having had less than 10 lessons".1: A partially completed contingency table Passed 10 plus lessons Less than 10 lessons Total 80 100 45 Failed Total 50 We will call event A "passed" and event B "having had less than 10 lessons". then read off the total number of outcomes for event B and for event A and B together. By mutually exclusive we mean. In the following examples.288 Probability We could consider the question: "what is the probability of failing at the first attempt if you have had more than 10 lessons?". may be regarded as a sequence of similar simple experiments. The calculation is: P(A|B) 5 n(A and B) 0. the probabilities are: P(R) P(W) 5 8 3 8 © ABE and RRC . Furthermore.e. we shall consider the implications of conditional probability – the probability of one event occurring if another event has also occurred. If two are selected at random without replacement. Here event A is "failed" and event B is "having had more than 10 lessons". and then apply the formula. For example. that if one of the outcomes of the compound experiment occurs then the others cannot. Example 1: The concept can be illustrated using the example of a bag containing five red and three white billiard balls. as before. one with more than one component part. TREE DIAGRAMS A compound experiment. i. E. and the tossing of four coins can be thought of as tossing one after the other. what is the probability that one of each colour is drawn? Let: R indicate a red ball being selected W indicate a white ball being selected When the first ball is selected. A tree diagram enables us to construct an exhaustive list of mutually exclusive outcomes of a compound experiment. By exhaustive. we mean that every possible outcome is considered.1 n(B) 50 Questions for Practice 1 See if you can work out the answer to the following questions using the preceding table: "What is the probability of failing at the first attempt if you have only had 8 lessons?" Remember to be clear about which is event A and event B. Now check your answers with those given at the end of the unit. the rolling of two dice can be considered as the rolling of one followed by the rolling of the other. a tree diagram gives us a pictorial representation of probability. e. namely RR. At the start we take a ball from the bag. So we mark in the two possibilities at the end of each of the two branches of our existing tree diagram. This ball is either red or white so we draw two branches labelled R and W. WR and WW. The probabilities for the selection of the second ball are then: P(R) 4 3 . Thus: If the first ball selected was red. We enter on these second branches the conditional probabilities associated with them. We then consider drawing a second ball from the bag. there will be five red and two white balls left. The probabilities for the selection of the second ball are then: P(R) 5 2 .1: Example 1 – tree diagram 1st Ball 2nd Ball R W R 5 3 8 7 4 7 Start W 3 8 R W 2 7 5 7 The diagram is constructed as follows. there are only seven balls left and the mix of red and white balls will be determined by the selection of the first ball – i. © ABE and RRC . Whether we draw a red or a white ball the first time. and P(W) 7 7 This process of working out the different conditional probabilities can be very complicated and it is easier to represent the situation by means of a tree diagram as in Figure 10. and P(W) 7 7 If the first ball selected was white. Each complete branch from start to tip represents one possible outcome of the compound experiment and each of the branches is mutually exclusive. Figure 10. Thus. on the uppermost branch in the diagram we must insert the probability of obtaining a second red ball given that the first was red. we can still draw a red or a white ball the second time. there will be four red and three white balls left.Probability 289 When the second ball is selected. We then also write on the branch the probability of the outcome of this simple experiment being along that branch. We can see that there are four different mutually exclusive outcomes possible. the probability of a red or a white ball being selected is conditional upon the previous event. RW. corresponding to the two possibilities.1. working from left to right. The probabilities at each stage are shown alongside the branches of the tree. let us build up a tree diagram. The blue ball is not drawn.290 Probability To obtain the probability of a particular outcome of the compound experiment occurring. A red then a white are drawn.2: Example 2 – tree diagram of first event R1 3 6 Start W1 2 6 B1 1 6 © ABE and RRC . We thus obtain the probabilities shown in Table 10. Probability 5 4 20 8 7 56 5 3 15 8 7 56 3 5 15 8 7 56 3 2 6 8 7 56 1 To solve this problem. Stage 1 will be to consider the situation when the first ball is drawn: Figure 10. A red and a white are drawn.3: Example 1 – outcomes from compound (conditional) events Outcome RR RW WR WW Total Example 2: A bag contains three red balls. Table 10. Find the probability that: (a) (b) (c) (d) Both white balls are drawn. two white balls and one blue ball. The sum of the probabilities should add up to 1. we multiply the probabilities along the different sections of the branch. as we know one or other of these mutually exclusive outcomes is certain to happen. Two balls are drawn at random (without replacement).3. using the general multiplication law for probabilities. whatever colour is chosen first.Probability 291 We have given the first ball drawn a subscript of 1 – for example. We shall give the second ball drawn a subscript of 2. Figure 10. note there was only one blue ball in the bag. the probability of it being red. white or blue is conditional on the outcome of the first event. of balls 6 2 6 1 6 The possibilities in stage 2 may now be mapped on the tree diagram. The probabilities in the first stage will be: P(R1) P(W 1) (P(B1) no. so if we picked a blue ball first then we can have only a red or a white second ball. Thus.3: Example 2 – tree diagram of first and second events 1st Ball 2nd Ball 2 R2 5 W2 2 B2 R1 3 6 5 1 3 5 5 R2 Start W1 2 6 W2 1 B2 5 1 3 B1 1 6 5 5 R2 W2 2 5 We can now list all the possible outcomes. © ABE and RRC . there are only five balls left as there is no replacement. of red balls 3 total no. Also. red first R1. with their associated probabilities. When the second ball is drawn. 5.e. to a maximum of four.4. We can assume that any one child is equally likely to be a boy or a girl.292 Probability Table 10. they do not have any more children. The tree diagram will be as in Figure 10. Draw a tree diagram to find the possible family size and calculate the probability that they have a son. i. (a) Probability that both white balls are drawn: P(W 1W 2) (b) 1 15 1 1 1 1 2 5 5 5 15 3 Probability the blue ball is not drawn: P(R1R2) P(R1W 2) P(W1R2) P(W 1W 2) (c) Probability that a red then a white are drawn: P(R1W 2) 1 5 1 1 2 5 5 5 (d) Probability that a red and a white are drawn: P(R1W 2) P(W 1R2) Example 3: A couple decide to continue having children. Note that once they have produced a son.4. © ABE and RRC . P(B) P(G) 0. until they have a son.4: Example 2 – outcomes from compound (conditional) events Outcome R1R2 R1W 2 R1B2 W1R2 W1W 2 W1B2 B1R 2 B1W 2 Total Probability 3 2 1 6 5 5 3 2 1 6 5 5 3 1 1 6 5 10 2 3 1 6 5 5 2 1 1 6 5 15 2 1 1 6 5 15 1 3 1 6 5 10 1 2 1 6 5 15 1 It is possible to read off the probabilities we require from Table 10. Table 10.4: Example 3 – Tree diagram 1st Child B ½ 2nd Child Start ½ G B ½ 3rd Child ½ G B ½ 4th Child ½ G B ½ ½ G The table of probabilities is as follows.5: Example 3 – probabilities of outcomes Possible Families 1 Boy 1 Girl. 1 Boy 2 Girls. 1 Boy 3 Girls.Probability 293 Figure 10. 1 Boy 4 Girls Total Probability 1 2 1 1 4 2 1 1 8 2 1 1 16 2 1 1 2 16 4 4 3 2 1 The probability that the family will have a son before their fourth child is therefore: 1 1 1 1 15 2 4 8 16 16 © ABE and RRC . the list of all possible different samples? The sample space is: AB AC AD AE In this example the order of the sample. C. each of A. If we have a fair coin. we call this a sample space.e. We have seen the value of tree diagrams in the last section. B. the probability of obtaining a head or a tail at any toss is ½ and all the outcomes in the sample space are equally likely. and now we consider another method of drawing the possible outcomes of an event. For instance.294 Probability F. What is the sample space. of outcomes Alternatively. E must have the same chance of being chosen. C. to have some diagrammatic form of representation. is not relevant as we would still interview both A and B. i. for example. Let us consider another example. we might be interested in a particular section of it.e. It helps. SAMPLE SPACE You need a clear head to perform probability calculations successfully. then. T H H means that on the first toss we obtained a tail. of outcomes in sample space BC BD BE CD CE DE 3 8 no. There are eight outcomes in all. Having written down our sample space. of outcomes in sample space with one head total no. Looking back at the sample space. D.e. we could write: P(event A) 3 8 © ABE and RRC . When we toss a coin three times and note the outcome. D. we are interested in the event of obtaining exactly one head.e. i. and on the third a head. we are performing a statistical experiment. If we make a list of all the possible outcomes of our experiment. i. Suppose we have five people – A. so we can now deduce that: Probability of obtaining exactly one head in 3 tosses of a fair coin no. of outcomes in A total no. on the second a head. E – and we wish to select a random sample of two for interview. B. The sample space in the coin-tossing experiment is: HHH THH HTH HHT HTT THT TTH TTT where. whether we choose A followed by B or B followed by A. in our first example we might want to see in how many of the outcomes we obtained only one head: i. we see that there are three outcomes making up this event. 5. 6 The total number of possible outcomes is 6. Table 10.4 0. it can categorise the possible revenue from the new outlet into categories of high. the expected value E of variable x with associated probabilities P(x) is calculated by the following formula: E(x) ΣxP(x) where x is the value of the outcome. The expected value is an average for the revenue. 2. Example: Suppose a company is considering opening a new retail outlet in a town where it currently has no branches. On the basis of information from its other branches in similar towns. EXPECTED VALUE Probabilities may also be used to assist the decision-making process in business situations.5 0. © ABE and RRC .Probability 295 Now consider the example of an experiment involving the rolling of one fair die. It is the sum of each value for the outcome multiplied by its probability. for example. Before a new undertaking is started.6: Probabilities of likely revenue returns Likely Revenue High Medium Low Probability 0. Thus. medium or low. You will note that the probabilities sum to 1. These figures can then be used to calculate an expected value for the income from the new undertaking. and it is now easy to read off the probabilities from the sample space: (a) (b) (c) a score greater than 3 – three possibilities: P an odd number – three possibilities: P 3 1 6 2 1 6 3 1 6 2 The sample space is a listing of all the possible outcomes as follows: both a score greater than 3 and an odd number – one possibility: P G. 4. weighted by the probabilities. On the basis of past experience it may also be possible to attribute probabilities to each outcome.6 shows the probabilities that have been estimated for each category. What is the probability of obtaining: (a) (b) (c) a score greater than 3 an odd number both a score greater than 3 and an odd number 1. 3. Table 10.1 Using these figures it is possible to calculate the expected value for the revenue from the new branch. it may be possible to list or categorise all the different possible outcomes. 296 Probability In our example, if we give values to the outcomes of each category (i.e. figures for the high, medium and low revenues) and then multiply each by its probability and sum them, we get an average revenue return for new branches. This expected revenue can then be compared with the outlay on the new branch in order to assess the level of profitability. Table 10.7: Expected value of revenue returns Category High Medium Low Revenue (£000) 5,000 3,000 1,000 Probability 0.4 0.5 0.1 Prob. Rev. (£000) 2,000 1,500 100 Total: £3,600 The concept of expected value is used to incorporate the consequences of uncertainty into the decision-making process. Questions for Practice 2 1. 2. If three coins are thrown, what is the probability that all three will show tails? (a) (i) (ii) (b) What is the probability of getting a total of six points in a simultaneous throw of two dice? What is the probability of getting at least one six from a simultaneous throw of two dice? Out of a production batch of 10,000 items, 600 are found to be of the wrong dimensions. Of the 600 which do not meet the specifications for measurement, 200 are too small. What is the probability that, if one item is taken at random from the 10,000: (i) (ii) (iii) (iv) It will be too small? It will be too large? Both are too large? One is too small and one too large? If two items are chosen at random from the 10,000, what is the probability that: 3. (a) Calculate, when three dice are thrown together, the probability of obtaining: (i) (ii) (iii) three sixes; one six; at least one six. (b) Calculate the probability of drawing in four successive draws the ace and king of hearts and diamonds suits from a pack of 52 playing cards. © ABE and RRC Probability 297 (c) Calculate the probability that three marbles drawn at random from a bag containing 24 marbles, six of each of four colours (red, blue, green and white), will be: (i) (ii) all red; all of any one of the four colours. 4. An analysis of 72 firms which had introduced flexitime working showed that 54 had used some form of clocking in system whilst the rest relied on employees completing their own time sheets. It was found that for 60 of the firms output per employee rose and in all other firms it fell. Output fell in only 3 of the firms using employee-completed forms. Construct a contingency table and then: (a) (b) Calculate the probability of output per employee falling if an automatic clocking in system is introduced. Determine under which system it is more probable that output per employee will rise. 5. The probability that machine A will be performing a useful function in five years' time is 0.25, while the probability that machine B will still be operating usefully at the end of the same period is 0.33. Find the probability that, in five years' time: (a) (b) (c) (d) Both machines will be performing a useful function. Neither will be operating. Only machine B will be operating. At least one of the machines will be operating. 6. Three machines, A, B and C, produce respectively 60%, 30% and 10% of the total production of a factory. The percentages of defective production for each machine are respectively 2%, 4% and 6%. (a) (b) If an item is selected at random, find the probability that the item is defective. If an item is selected at random and found to be defective, what is the probability that the item was produced by machine A? 7. Using a sample space diagram, calculate the probability of getting, with two throws of a die: (a) (b) A total score of less than 9. A total score of 9. Now check your answers with those given at the end of the unit. © ABE and RRC 298 Probability H. THE NORMAL DISTRIBUTION Earlier in the course, we considered various graphical ways of representing a frequency distribution. We considered a frequency dot diagram, a bar chart, a polygon and a frequency histogram. For example, a typical histogram will take the form shown in Figure 10.5. You can immediately see from this diagram that the values in the centre are much more likely to occur than those at either extreme. Figure 10.5: Typical frequency histogram Frequency 40 30 20 10 0 80 90 100 110 120 130 140 150 Value Consider now a continuous variable in which you have been able to make a very large number of observations. You could compile a frequency distribution and then draw a frequency bar chart with a very large number of bars, or a histogram with a very large number of narrow groups. Your diagrams might look something like those in Figure 10.6. Figure 10.6: Frequency bar chart and histogram © ABE and RRC Probability 299 If you now imagine that these diagrams relate to a relative frequency distribution and that a smooth curve is drawn through the tops of the bars or rectangles, you will arrive at the idea of a frequency curve. We can then abstract the concept of a frequency distribution by illustrating it without any actual figures as a frequency curve which is just a graph of frequency against the variate (variable): Figure 10.7: Frequency curve Frequency Variate The "normal" or "Gaussian" distribution was discovered in the early eighteenth century and is probably the most important distribution in the whole of statistical theory. This distribution seems to represent accurately the random variation shown by natural phenomena – for example: heights of adult men from one race weights of a species of animal distribution of IQ levels in children of a certain age weights of items packaged by a particular packing machine life expectancy of light bulbs. The typical shape of the normal distribution is shown in Figure 10.8. You will see that it has a central peak (i.e. it is unimodal) and that it is symmetrical about this centre. The mean of this distribution is shown as "m" on the diagram, and is located at the centre. The standard deviation, which is usually denoted by "σ", is also shown. Figure 10.8: The normal distribution Frequency m Variate © ABE and RRC 300 Probability Properties of the Normal Distribution The normal distribution has a number of interesting properties. These allow us to carry out certain calculations on it. For normal distributions, we find that: 68% of the observations are within 1 standard deviation of the mean and 95% are within 2 standard deviations of the mean. We can illustrate these properties on the curve itself: Figure 10.9: Properties of the normal distribution s.d. s.d. Shaded area 68% m s.d. s.d. s.d. s.d. Shaded area 95% m These figures can be expressed as probabilities. For example, if an observation x comes from a normal distribution with mean m and standard deviation σ, the probability that x is between (m σ) and (m σ) is: P(m σ < x < m σ) 0.68 Also P(m 2σ < x < m 2σ) 0.95 © ABE and RRC Probability 301 Using Tables of the Normal Distribution Because the normal distribution is perfectly symmetrical, it is possible to calculate the probability of an observation being within any range, not just (m σ) to (m σ) and (m 2σ) to (m 2σ). These calculations are complex, but fortunately, there is a short cut available. There are tables which set out these probabilities. We provide one such table in the appendix to this study unit. It relates to the area in the tail of the normal distribution. Note that the mean, here, is denoted by the symbol "μ" (the Greek letter mu). Thus, the first column in the table represents the position of an observation, x, in relation to the mean (x μ), expressed in terms of the standard deviation by dividing it by σ. Figure 10.10: Proportion of distribution in an area under the normal distribution curve m 2σ mσ m mσ m 2σ The area under a section of the curve represents the proportion of observations of that size in relation to the total number of observations. For example, the shaded area shown in Figure 10.10 represents the chance of an observation being greater than m 2σ. The vertical line which defines this area is at m 2σ. Looking up the value 2 in the table gives the proportion of the normal distribution which lies outside of that value: P(x > m 2σ) 0.02275 which is just over 2%. Similarly, P(x > m 1σ) is found by looking up the value 1 in the table. This gives: P(x > m 1σ) 0.1587 which is nearly 16%. You can extract any value from P(x m) to P(x > m 3σ) from this table. This means that you can find the area in the tail of the normal distribution wherever the vertical line is drawn on the diagram. Negative distances from the mean are not shown in the tables. However, since the distribution is symmetrical, it is easy to calculate these. P(x < m 5σ) P(x > m 5σ) So P(x > m 5σ) 1 P(x < m 5σ) This is illustrated in Figure 10.11. © ABE and RRC P(m 0.5σ © ABE and RRC .7σ and m 1. This represents the probability that x is between m 0.7σ < x < m 1.11: Using the symmetry to determine the area under the normal curve Area A is read from tables.5σ: i. Area B Area A B m 2σ mσ m mσ A m 2σ C Area C 1 Area B. m 2σ mσ m mσ m 2σ Further Probability Calculations It is possible to calculate the probability of an observation being in the shaded area shown in Figure 10.7σ m 1.e. using values from the tables.5σ) Figure 10.302 Probability Figure 10.12.12: Probability of an observation being in a particular area m 0. This is because Area C and Area B are mutually exclusive and cover the whole range. 5σ< x < m σ) 1 0.0668 0. Therefore we need to find P(m 0.2420 since the distribution is symmetrical.00621 from the tables.Probability 303 First find P(x > m 1.7σ) 0.3085 or P(69 < x < 72) 0.2420 0. Now find the probability that an item will weigh between 69 and 72 grams.5σ) 0.5σ) 0.1587 0.5σ) 0. and 72 grams is 1 standard deviation above the mean.5 standard deviations below the mean.12 is then: 1 0. this is 2.3085 So P(m 0.5 standard deviations below the mean. Let x be the weight of an individual item.7σ) or P(x > m 0.7σ < x < m 1.5σ) P(x > m 2. find the probability that an item will weigh 65 grams or less. Assuming that the items come from a normal distribution.0668 from the table.5328 © ABE and RRC .5σ) P(x > m 0.5σ) 0.6912 Example: A production line produces items with a mean weight of 70 grams and a standard deviation of 2 grams.1587 P(x < m 0. 65 grams is 5 grams below the mean. Then find P(x < m 0. The proportion of the area under the curve which is shaded in Figure 10. 69 grams is 0.5σ < x < m σ): P(x > m σ) 0. Therefore: P(x < 65) P(x < m 2.6912 Hence P(m 0. Since the standard deviation is 2 grams. The probability that it will last for less than 290 hours. What is the probability that overtime will be worked? What is the probability that the workforce neither gets a bonus nor works overtime? Now check your answers with those given at the end of the unit. The value below which only 1 in 500 batteries will fall. On any one day: (a) (b) (c) If more than 2. the employees get a bonus. A factory which makes batteries tests some of them by running them continuously. What is the probability that a bonus will be earned? When less than 1.800 items a day on average.000 items are produced. A machine produces 1. The probability that it will last for between 310 and 330 hours.304 Probability Questions for Practice 3 1. © ABE and RRC . They are found to have a mean life of 320 hours and a standard deviation of 20 hours. overtime is available. The shape of the distribution of items produced in one day is very similar to a normal distribution with a standard deviation of 80 items. The probability that a battery will last for more than 380 hours.700 items are produced. estimate: (a) (b) (c) (d) 2. By assuming that this distribution is normal. there are a total of 36 possible combinations.Probability 305 ANSWERS TO QUESTIONS Questions for Practice 1 Event A is "failed" and event B is "having less than 10 lessons".5 for each coin.02 9. one too small) 0.e. (a) (i) 1 2 1 1 22 1 8 Six points can be obtained from the following combinations: 1st die 1 2 3 4 5 2nd die 5 4 3 2 1 Since each die can fall with any of its faces uppermost. Therefore: P(both too large) P(1st too large) P(2nd too large) 0.04 10. The calculation then is: P(A|B) n( A and B) 15 0.3 n(B) 50 Questions for Practice 2 1. 6 6 36.04 399 9. Therefore: P(one too large.999 0.001596 (iv) The first item can be either too large or too small. Therefore: P(six points) (ii) P(no six) 5 36 5 5 25 6 6 36 11 36 P(at least one six) 1 P(no six) (b) (i) (ii) (iii) P(too small) P(too large) 200 0.04 0. The probability of tails is 0.999 © ABE and RRC . The events are independent. i. Therefore: P(3T) 2. if the first item is too large.999 9. it will be removed from the calculation of the second item.000 Note that.02 10.001594 199 399 0.000 400 0. 4 or 5 125 91 5 1 1 216 216 6 3 (b) Required probability 4 3 2 1 52 51 50 49 6 24 5 23 0. figs. figs.03953 to 4 sig. (c) (i) Probability that first is red probability that second is red probability that third is red 4 22 probability that all three are red (ii) 6 5 4 0.009881 to 4 sig. 2.306 Probability 3. Number of ways the die showing six can be chosen 3 1 probability of six on one die 6 5 probability of not obtaining six on the other two dice 6 2 Therefore: 25 1 5 P(one six) 3 72 6 6 2 (iii) Probability of obtaining at least one six 1 probability of not obtaining any sixes 1 probability that all three dice show 1. 24 23 22 © ABE and RRC . Therefore: 1 1 P(three sixes) 216 6 3 (ii) One die must show a six and the others must not show a six. Any one of the three dice could be the one that shows a six.000003694 to 4 significant figures. 24 23 22 Probability that the three will be all of any one of the four colours probability that all three are red probability that all three are blue probability that all three are green probability that all three are white 4 6 5 4 0. 3. (a) (i) Each die must show a six. We can draw a tree diagram to map the situation as follows: Joint probabilities 1 3 1 x 1 1 4 3 12 B A 1 B 4 2 3 1 x2 3 4 3 12 3 A 4 1 B 3 3 4 x 1 3 3 12 B 2 3 3 4 x2 3 6 12 Total 1 © ABE and RRC . Let: A Machine A performing usefully. The contingency table will be as follows: Output up Clocking in system Employee completion system Total (a) 45 15 60 Output down 9 3 12 Total 54 18 72 The probability of output per employee falling (event A) given that an automatic clocking in system is used (event B) is: P(A|B) 9 1 n( A and B) n(B) 54 6 (b) The probability of output per employee rising (event A) given that an automatic clocking in system is used (event B) is: P(A|B) n( A and B) 45 5 n(B) 54 6 The probability of output per employee rising (event A) given that an employee completion system is used (event B) is: P(A|B) n( A and B) 15 1 n(B) 18 6 The probabilities are the same. and B Machine B not performing usefully.Probability 307 4. and A Machine A not performing usefully B Machine B performing usefully. 5. 94 0.000 (a) (b) P(D) 0.60 x 0.98 A 0.006 D 0.012 0.94 0.012 0.10 x 0.02 0.012 D 0.10 x 0.588 0.96 C 0.012 D 0.10 0.06 D 0. Joint probabilities 0.06 0.60 0.98 0.094 Total 1.30 x 0.30 x 0.04 B 0.04 0.02 D 0. The tree diagram for the situation is as follows.012 0.4 0.03 P(defective that came from machine A) 0.60 x 0.30 D 0. Probability of both machines operating is Probability of neither operating is Probability of only B operating is 6 12 3 12 1 12 or or 1 2 1 4 1 12 3 2 12 12 1 2 Probability of at least one machine operating is Let: A produced on machine A B produced on machine B C produced on machine C D defective D not defective.308 Probability (a) (b) (c) (d) 6.006 0.03 probability of a defective from A probability of defective © ABE and RRC .288 0.96 0. Therefore: P(total score of 9) 4 1 0.Probability 309 7.00135 from the tables. P(observation < m 0. Therefore: P(total score less than 9) 26 13 0. 290 hours is 30 hours or 1.3085 0. (a) (b) (c) 380 hours is 60 hours or 3 standard deviations above the mean.5σ) 2 0.383 (d) Expressing 1 in 500 as a proportion: 1 in 500 1 0.3085 0.1915.3085 0.5σ) 0.3085 from the tables Similarly. Therefore: P(observation < m 1. The sample space diagram sets out all the possible total scores obtainable when two dice are thrown.5σ) 0. P(m 0.5 standard deviations above the mean.3085 Therefore: P(m 0.722 36 18 (b) 4 outcomes give a score of 9.5σ < observation < m 0.5σ) P(observation > m 1.0668 310 hours is 0.002 500 © ABE and RRC .5σ) P(310 < observation < 330) 1 0.5σ 0.383 Alternatively.5σ) 0. using the symmetry of the normal distribution: P(m < observation < m 0.5 standard deviations below the mean.5σ < observation < m 0.5σ) 0.111 36 9 Questions for Practice 3 1. There are 36 equally likely outcomes: 1st die: 2nd die 1 2 3 4 5 6 (a) 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10 6 7 8 9 10 11 7 8 9 10 11 12 1 2 3 4 5 6 26 outcomes give scores less than 9. Hence.5 standard deviations below the mean and 330 hours is 0. P(observation > m 0. Therefore: P(observation > m 3σ) 0.1915 0. 5 standard deviations from the mean 80 Therefore.5 standard deviations from the mean 80 Therefore. Mean.000 items being produced is: P(x > m 2.000 items is: (2.700 items is: (1.25σ) 0. This is when: x μ 2. σ 80 (a) In a normal distribution.4 Therefore 1 in 500 batteries will have a life of less than 262. m 1.700 1.5σ) 0. the nearest we can get is 0. 2.6 262.002 in the body of the tables.00199.000 1.800) 1.88 standard deviations below the mean. 2.88 σ Therefore. and looking up the value 0.88 20 320 57.4 hours.800 Standard deviation.00621 0. 1 in 500 batteries will be more than 2.88819 © ABE and RRC .800) 2.310 Probability Using the tables in reverse.700 items being produced is: P(x < m 1.1056 (c) Probability of no bonus and no overtime: 1 0.00621 (b) In a normal distribution.1056 0. Answer: 320 2. 1. the probability of 1.25σ) P(x > m 1. the probability of 2. 4761 .00676 .0262 .4960 .0367 .00368 .3372 .0505 .2643 .01 .05 .02222 .1056 .01500 .4483 .02169 .3859 .7 0.5000 .01618 .0336 .04 .00587 .4052 .3594 .0708 .00621 .00205 .01255 .0287 .00639 .01463 .00820 .1292 .01130 .1977 .3707 .2389 .9 2.01191 .4168 .1660 .3015 .2236 .00187 .5 0.00272 .0951 .4522 .0392 .00539 .0934 .02068 .1314 .2090 .2611 .3783 .4721 .3483 .1 2.0351 .2483 .0427 .3897 .0885 .4681 .00554 .01831 .01017 .3121 .0516 .1894 .00307 .2119 .5 1.0764 .4013 .2 1.9 1.Probability 311 Appendix: Areas in the Tail of the Normal Distribution The table of values of the standard normal distribution set out below provides a means of determining the probability of an observation (x) lying within specified standard deviations () of the mean of the distribution (μ).1075 .1446 .2206 .01743 .2033 .0735 .6 0.2912 .0314 .0 2.0537 .1379 .1093 .0301 .4247 .00990 .2843 .2420 .4 1.00149 .0985 .00135 .1401 .1922 .09 .00402 .3336 .00317 .9 3.7 2.00570 .0918 .2327 .3085 .1814 .00212 .2810 .00755 .0968 .3557 .4364 .2946 .0694 .00939 .0418 .1949 .2148 .0322 .0294 .00169 .00264 .0455 .4801 .0233 .1210 .0606 .00734 .00193 .1020 .01539 .1736 .01321 .3228 .1515 .00842 .00159 .0274 .00914 .8 0.4920 .0384 .00256 .0307 .00466 .07 .0465 .00347 .1492 .1230 .2266 .1762 .4286 .01160 .4602 .01700 .00326 .0401 .01659 .2177 .00139 0.8 2.1539 .2578 .0359 .0559 .00248 .01101 .3520 .1151 .4880 .2 0.0643 .1335 .2709 .3300 .01923 .1562 .00289 .02 .1 0.0475 .3264 .2 2.4 2.00336 .3050 .5 2.2358 .3669 .01876 .01044 .00144 .4207 .01390 .4404 .μ) σ .0268 .01786 .2743 .00695 .0244 .0409 .4443 .0618 .4840 .3745 .3409 .3 1.00391 .00427 .00164 .3821 .0721 .02018 .0 1.00714 .6 2.0239 .1587 .03 .2546 .0655 .0571 .1131 .3 2.00154 .0485 .00440 .00508 .00181 .00415 .2451 .0 © ABE and RRC .1112 .0901 .1611 .0838 .0630 .1423 .0446 .2776 .0668 .0250 .2877 .4129 .1357 .0548 .0436 .01426 .4325 .0526 .02118 .3446 .00494 .3 0.1003 .08 .01072 .0329 .2676 .00233 .00357 .1271 .01578 .0808 .1190 .4090 .00280 .2514 .2981 .0681 .1635 .02275 .00776 .01355 .1251 .0778 .01287 .0749 .0495 .3156 .3192 .0281 .1711 .2296 .0256 .1685 .06 .00453 .6 1.1867 .0869 .1038 .1469 .0793 .00240 .0375 .00964 .01222 .00523 .00 .4562 .0823 .2061 .0582 .1 1.2005 .00175 .01970 .00798 .00866 .0344 .3874 .00889 .00298 .00379 .3936 .0594 .00199 .00604 .3632 . (x .0 0.00480 .4 0.1841 .0853 .7 1.8 1.4641 .00219 .00226 .1170 .1788 .00657 . 312 Probability © ABE and RRC .