CHAPTER7 CHAPTER CONTENTS 7.1 7.2 7.3 7.4 7.5 7.6 7.7 Continuous Probability Distributions Describing a Continuous Distribution Uniform Continuous Distribution Normal Distribution Standard Normal Distribution Normal Approximations Exponential Distribution Triangular Distribution (Optional) CHAPTER LEARNING OBJECTIVES When you finish this chapter you should be able to LO 7-1 LO 7-2 LO 7-3 LO 7-4 LO 7-5 LO 7-6 LO 7-7 LO 7-8 LO 7-9 Define a continuous random variable. Calculate uniform probabilities. Know the form and parameters of the normal distribution. Find the normal probability for given z or x using tables or Excel. Solve for z or x for a given normal probability using tables or Excel. Use the normal approximation to a binomial or a Poisson. Find the exponential probability for a given x. Solve for x for a given exponential probability. Use the triangular distribution for “what-if” analysis (optional). 254 I n Chapter 6, you learned about probability models and discrete random variables. We will now expand our discussion of probability models to include models that describe continuous random variables. Recall that a discrete random variable usually arises from counting something such as the number of customer arrivals in the next minute. In contrast, a continuous random variable usually arises from measuring something such as the waiting time until the next customer arrives. Unlike a discrete variable, a continuous random variable can have noninteger (decimal) values. Probability for a discrete variable is defined at a point such as P(X 5 3) or as a sum over a series of points such as P(X # 2) 5 P(0) 1 P(1) 1 P(2). But when X is a continuous variable (e.g., waiting time), it does not make sense to speak of probability “at” a particular X value (e.g., X 5 54 seconds) because the values of X are not a set of discrete points. Rather, probabilities are defined as areas under a curve called the probability density function (PDF). Probabilities for a continuous random variable are defined on intervals such as P(53.5 # X # 54.5) or P(X , 54) or P(X $ 53). Figure 7.1 illustrates the differences between discrete and continuous random variables. This chapter explains how to recognize data-generating situations that produce continuous random variables, how to calculate event probabilities, and how to interpret the results. LO 7-1 Define a continuous random variable. FIGURE 7.1 Discrete Variable: Defined at Each Point 0 1 2 3 4 5 Continuous Variable: Defined over an Interval 53 53.5 54 54.5 55 Discrete and Continuous Events 7.1 DESCRIBING A CONTINUOUS DISTRIBUTION PDFs and CDFs A probability distribution can be described either by its probability density function (PDF) or by its cumulative distribution function (CDF). For a continuous random variable, the PDF is an equation that shows the height of the curve f (x) at each possible value of X. Any continuous PDF must be nonnegative and the area under the entire PDF must be 1. 255 256 Applied Statistics in Business and Economics The mean, variance, and shape of the distribution depend on the PDF and its parameters. The CDF is denoted F(x) and shows P(X # x), the cumulative area to the left of a given value of X. The CDF is useful for probabilities, while the PDF reveals the shape of the distribution. There are Excel functions for many common PDFs or CDFs. For example, Figure 7.2 shows a hypothetical PDF and CDF for a distribution of freeway speeds. The random variable miles per hour is a continuous variable that can be expressed with any level of precision we choose. The curves are smooth, with the PDF showing the probability density at points along the X-axis. The CDF shows the cumulative probability of speeds, gradually approaching 1 as X approaches 90. In this illustration, the distribution is symmetric and bell-shaped (normal or Gaussian) with a mean of 75 and a standard deviation of 5. FIGURE 7.2 Freeway Speed Examples Normal PDF .09 .08 .07 .06 .05 .04 .03 .02 .01 .00 60 65 70 75 80 Miles per Hour 85 90 1.00 .90 .80 .70 .60 .50 .40 .30 .20 .10 .00 60 65 70 75 80 Miles per Hour 85 90 Normal CDF Probabilities as Areas With discrete random variables, we take sums of probabilities over groups of points. But continuous probability functions are smooth curves, so the area at any point would be zero. Instead of taking sums of probabilities, we speak of areas under curves. In calculus terms, we would say that P(a , X , b) is the integral of the probability density function f(x) over the interval from a to b. Because P(X 5 a) 5 0 the expression P(a , X , b) is equal to P(a # X # b). Figure 7.3 shows the area under a continuous PDF. The entire area under any PDF must be 1. FIGURE 7.3 Probability as an Area F(x) a f(x) b Expected Value and Variance The mean and variance of a continuous random variable are analogous to E(X ) and Var(X ) for a discrete random variable, except that the integral sign e replaces the summation sign . Integrals are taken over all X-values. The mean is still the balancing point or Chapter 7 Continuous Probability Distributions 257 fulcrum for the entire distribution, and the variance is still a measure of dispersion about the mean. The mean is still the average of all X-values weighted by their probabilities, and the variance is still the weighted average of all squared deviations around the mean. The standard deviation is still the square root of the variance. Continuous Random Variable 1` Discrete Random Variable E(X ) 5 5 Mean E(X ) 5 5 # x f (x) dx 2` 1` Variance Var(X ) 5 2 5 # (x 2 2` )2 f (x) dx Var(X ) 5 2 O O all x x P(x) (7.1) 5 [x 2 ]2 P(x) (7.2) all x Calculus notation is used here for the benefit of those who have studied it. But statistics can be learned without calculus, if you are willing to accept that others have worked out the details by using calculus. If you decide to become an actuary, you will use calculus (so don’t sell your calculus book). However, in this chapter, the means and variances are presented without proof for the distributions that you are most likely to see applied to business situations. 7.1 Flight 202 is departing Los Angeles. Is each random variable discrete (D) or continuous (C)? a. Number of airline passengers traveling with children under age 3. b. Proportion of passengers traveling without checked luggage. c. Weight of a randomly chosen passenger on Flight 202. SECTION EXERCISES 7.2 It is Saturday morning at Starbucks. Is each random variable discrete (D) or continuous (C)? a. Temperature of the coffee served to a randomly chosen customer. b. Number of customers who order only coffee with no food. c. Waiting time before a randomly chosen customer is handed the order. 7.3 Which of the following could not be probability density functions for a continuous random variable? Explain. Hint: Find the area under the function f (x). a. f (x) 5 .25 for 0 # x # 1 b. f (x) 5 .25 for 0 # x # 4 c. f (x) 5 x for 0 # x # 2 7.4 For a continuous PDF, why can’t we sum the probabilities of all x-values to get the total area under the curve? 7.2 UNIFORM CONTINUOUS DISTRIBUTION Characteristics of the Uniform Distribution The uniform continuous distribution is perhaps the simplest model one can imagine. If X is a random variable that is uniformly distributed between a and b, its PDF has constant height, as shown in Figure 7.4. The uniform continuous distribution is sometimes denoted U(a, b) for short. Its mean and standard deviation are shown in Table 7.1. Since the PDF is rectangular, you can easily verify that the area under the curve is 1 by multiplying its base (b 2 a) by its height 1y(b 2 a). Its CDF increases linearly to 1, as shown in Figure 7.4. Since events can easily be shown as rectangular areas, we rarely need to refer to the CDF, whose formula is just P(X # x) 5 (x 2 a)y(b 2 a). The continuous uniform distribution is similar to the discrete uniform distribution if the x values cover a wide range. For example, three-digit lottery numbers ranging from 000 to 999 would closely resemble a continuous uniform with a 5 0 and b 5 999. LO 7-2 Calculate uniform probabilities. 90 . .4 Uniform Distribution Uniform PDF 1 b a 1. X .00 .50 .20 . This situation is illustrated in Figure 7. =a+(b–a)*RAND() Used as a conservative what-if benchmark and in simulation. or 33.3333. 25) 5 (25 2 20)y(30 215) 5 5y15 5 0. 30). d) 5 (d 2 c)y(b 2 a) (area between c and d in a uniform model) For example.80 . Thus.258 Applied Statistics in Business and Economics FIGURE 7.5.60 F(x) . we obtain the mean and standard deviation: a 1 b 15 1 30 5 _____ 5 _______ 5 22. the probability of taking between c and d minutes is (7.40 . the probability that the anesthetic takes between 20 and 25 minutes is P(20 .00 Uniform CDF f(x) 0 a Value of X b a Value of X b TABLE 7.1 Uniform Continuous Distribution Parameters PDF CDF Domain Mean Standard deviation Shape Random data in Excel Comments a 5 lower limit b 5 upper limit 1 f (x) 5 ______ b2a x2a P(X # x) 5 ______ b2a a#x#b a1b ______ 2 ________ (b 2 a) _______ 12 Symmetric with no mode. In short notation. Setting a 5 15 and b 5 30. 2 EXAMPLE 7.3%. we could say that X is U(15.5 minutes 2 2 ________ __________ 2 5 (b 2 a)2 _______ 5 12 (30 2 15) _________ 5 4. the dentist views the time for anesthesia effectiveness as a uniform random variable that takes between 15 minutes and 30 minutes.33 minutes 12 An event probability is simply an interval width expressed as a proportion of the total.1 Anesthesia Effectiveness An oral surgeon injects a painkiller prior to extracting a tooth.3) P(c .30 .10 . X .70 . Given the varying characteristics of patients. If you create random numbers by using =RAND(). The uniform distribution can be useful in business for what-if analysis. This special case is important because Excel’s function =RAND() uses this distribution.5000 and 5 0. This important distribution is discussed in more detail in later chapters on simulation and goodness-of-fit tests.5 7. 500) for U(0.7 For a continuous uniform distribution. denoted U(0. 7. U(0. P(X . you know what their mean and standard deviation should be.Chapter 7 Continuous Probability Distributions 259 FIGURE 7. 50) b.6 Find the mean and standard deviation for each uniform continuous model. is shown in Figure 7. b) is used only when you have no reason to imagine that any X values are more likely than others. 1. Using the formulas for the mean and standard deviation. X . 45) for U(15.030 .5 Uniform Probability P(20 .000 10 15 X 25) 20 25 Minutes 30 35 A special case of the continuous uniform distribution.070 . U(1. U(100. That may sound like a conservative approach. has limits a 5 0 and b 5 1. P(X . you can easily show that this distribution has 5 0.060 . 200) c. 1). X .000) c.6. 10) a. 1) Distribution f(x) 0 0 Value of X 1 Uses of the Uniform Model The uniform model U(a. P(25 .010 . SECTION EXERCISES 7.6 Unit U(0. 10) for U(0. why is P(25 . but don’t want to make any assumptions about the distribution in between.040 . 99) Find each uniform continuous probability and sketch a graph showing it as a shaded area. in situations where you know the “worst” and “best” range. a.2887. But bear in mind that if the data-generating situation has any central tendency at all. the assumption of a uniform distribution would lead to a higher standard deviation than might be appropriate.020 . 45) the same as P(25 # X # 45)? . 65) b. 1 FIGURE 7.050 f(x) . The expected value of a normal random variable is and its variance is 2.DIST(x.70 .50 .” Its CDF has a “lazy-S” shape.INV(area.7 Normal PDF and CDF 1. f(x) does not reach the X-axis beyond 6 3 . x . It approaches. and .000 and 4.60 .00 . . named for German mathematician Karl Gauss (1777–1855). but never reaches.80 .500 pounds.30 .7. Shape PDF in Excel* CDF in Excel* Random data in Excel Symmetric. 1 3 ] includes almost all the area (as you know from the Empirical Rule in Chapter 4).10 . Its importance gives it a major role in our discussion of continuous models. It is often denoted N( .500 pounds to 4.7.40 . . as a practical matter. The normal or Gaussian distribution. A normal probability distribution is defined by two parameters. 1` 2 TABLE 7.90 .DIST(x. mesokurtic. cumulative) and =NORM. ). ).000 pounds? (d) More than 4.14159).000 pounds? 7.2 Normal Distribution *Excel 2010’s new functions =NORM. The domain of a normal random variable is 2` . cumulative) and =NORMINV(area.3 NORMAL DISTRIBUTION Characteristics of the Normal Distribution LO 7-3 Know the form and parameters of the normal distribution. Table 7.8 Assume the weight of a randomly chosen American passenger car is a uniformly distributed random variable ranging from 2. Besides and . . Dev.DIST(x.1) =NORM. .000 pounds? (e) Between 3. x .260 Applied Statistics in Business and Economics 7. . . The normal probability density function f(x) reaches a maximum at and has points of inflection at 6 as shown in the left chart in Figure 7. the interval [ 2 3 . the normal probability density function f (x) depends on the constants e (approximately 2. . Parameters PDF Domain Mean Std.00 1 2 3 3 2 Normal PDF Normal CDF 3 2 1 1 1 2 3 . but is merely asymptotic to it. ) give the same result as =NORMDIST(x.INV(RAND(). . ) 5 population mean 5 population standard deviation 1 x2 2__ _____ 1 ___ f (x) 5 ______ e 2 ( ) s 2 2` . Its single peak and symmetry cause some observers to call it “mound-shaped” or “bell-shaped. (a) What is the mean weight of a randomly chosen vehicle? (b) The standard deviation? (c) What is the probability that a vehicle will weigh less than 3.2 summarizes its main characteristics. However. and bell-shaped. as shown in the right chart in Figure 7. FIGURE 7.71828) and (approximately 3.20 .0) =NORM. 1 . 1. has already been mentioned several times. Despite its appearance. =NORM. . . . The normal distribution is always symmetric. even with excellent quality control. Possess a clear center.71 42. Conversely.70 0. The right chart in Figure 7.8 shows the distribution of diameters of golf balls from a manufacturing process that produces normally distributed diameters with a mean diameter of 5 42.015 . a random variable should: • • • • • Be measured on a continuous scale. Have only one peak (unimodal). Because the area under the entire curve must be 1.01 mm. and operations analysis. marketing research.70 42. FIGURE 7.005 .06 mm).72 Golf Ball Diameter (mm) 42. • X 5 cockpit noise level in a Boeing 777 at the captain’s left ear during cruise.. or N(42. To be regarded as a candidate for normality. ) for short.8 you can see that this is not the case.70.8. Be symmetric about the mean (equal tails). but in the left chart in Figure 7.68 42. Normal random variables also can be found in economic and financial data. when X has a large range (e.000 40 50 60 70 80 CPA Exam Scores 90 42. the CPA exam range is about 60 points). For example.69 42. What Is Normal? Many physical measurements in engineering and the sciences resemble normal distributions.01) in short notation.70 mm and a standard deviation 5 0. differing only in the axis scales.73 100 It is a common misconception that f (x) must be smaller than 1.030 . behavioral measurement scales.010 . notice that the horizontal and vertical axis scales differ.045 .01 70 10 42. the left chart in Figure 7.040 .020 . the golf ball diameter range is about 0. 0.8 shows the distribution of scores on the CPA theory exam.g.Chapter 7 Continuous Probability Distributions 261 A normal distribution with mean and standard deviation is sometimes denoted N( . Although the shape of each PDF is the same.67 . When the range is large. when X has a small range (e. 10) in short notation. assumed to be normal with a mean of 5 70 and a standard deviation 5 10. exam scores are discrete (range from 0 to 100) but are often treated as continuous data.40 for the exam scores). For example. Each of these variables would tend toward a certain mean but would exhibit random variation. • X 5 diameter in millimeters of a manufactured steel ball bearing.. For example.035 . the height of f(x) is small (about 0. The normal distribution is especially important as a sampling distribution for estimation and hypothesis testing. Here are some random variables that might be expected to be approximately normally distributed: • X 5 quantity of beverage in a 2-liter bottle of Diet Pepsi. or N(70. Exhibit tapering tails. we often treat a discrete variable as continuous. the height of f(x) is large (about 40 for the golf ball diameters).g.8 All Normal Distributions Look Alike Except for Scaling 45 40 35 30 25 20 15 10 5 0 42. not every bottle of a soft drink will have exactly the same fill (even if the variation is only a few milliliters). The mean and .025 . as shown in the right chart in Figure 7. All normal distributions have the same general shape. . For now.11 State the Empirical Rule for a normal distribution (see Chapter 4).g. More precise tests will be discussed in Chapter 15. mesokurtic. 7. how do they differ? 7.S. denoted N(0. 1` 0 1 Symmetric.DIST(z. Table 7. Shoe sizes of adult males.S. In Chapter 4.262 Applied Statistics in Business and Economics standard deviation depend on the nature of the data-generating process. This important transformation is shown in formula 7.3 Standard Normal Distribution Parameters PDF Domain Mean Standard deviation Shape CDF in Excel* Random data in Excel Comment 5 population mean 5 population standard deviation x2 2 1 ___ f (z) 5 _____ e2z y2 where z 5 ______ 2 2` . our task is to learn more about the normal distribution and its applications. The shape of the distribution is unaffected by the z transformation.4 STANDARD NORMAL DISTRIBUTION Characteristics of the Standard Normal LO 7-3 Know the form and parameters of the normal distribution. Thus.S. 7.9 If all normal distributions have the same shape. Time to process insurance claims for residential fire damage. even though they may share a common shape.S.g.3 summarizes the main characteristics of the standard normal distribution.10 (a) At what x value does f (x) reach a maximum for a normal distribution N(75. and bell-shaped. so we need normal tables or Excel to find areas.. Days from mailing home utility bills to receipt of payment. Precision manufacturing can achieve very small in relation to (e. you saw that a histogram can be used to assess normality in a general way. b. The maximum height of f (z) is at 0 (the mean) and its points of inflection are at 61 (the standard deviation). c. 1).4) x2 z 5 ______ (transformation of each x-value to a z-value) If X is normally distributed N( . steel ball bearing diameter). Hint: Would you expect a single central mode and tapering tails? Would the distribution be roughly symmetric? Would one tail be longer than the other? a.DIST(z) and =NORM. *Excel 2010’s new functions =NORM. (7. Its mean is 0 and its standard deviation is 1. 5)? (b) Does f (x) touch the X-axis at 6 3 ? 7. your driving fuel mileage). each normally distributed random variable may have a different coefficient of variation. TABLE 7.12 Discuss why you would or would not expect each of the following variables to be normally distributed. except that now we are talking about a population distribution instead of sample data. ). SECTION EXERCISES 7.1) =NORM. the standardized variable Z has a standard normal distribution. There are statistical tests to see whether a sample came from a normal population. z .INV(RAND()) There is no simple formula for a normal CDF. we often transform the variable by subtracting the mean and dividing by the standard deviation to produce a standardized variable.. Since there is a different normal distribution for every pair of values of and . for example. d.INV(area) give the same result as =NORMSDIST(z) and =NORMSINV(area). Years of higher education of 30-year-old employed women.4. while other data-generating situations produce relatively large in relation to (e. =NORM. just as in Chapter 4. This row and column are shaded in Table 7.96 . Z .40 .00 . as you can approximately verify by treating it as a triangle (area 5 1⁄2 base 3 height). you select the row for z 5 1. a normal area can be approximated to any desired degree of accuracy using various methods (e.35 .20 .96 .96 # Z # 11. 1. we can use a common scale.30 .96 encloses 95 percent of the area under the normal curve. usually labeled from 23 to 13.10 . z2) is a definite integral of f (z). Z .g.96 . Since a point has no area in a continuous distribution.45 . Z . Z . 0) 1 P(0 . we can find a right-tail area by subtraction. as shown in Figure 7. 0) 5 . Z .60 .96). we omit the equality. As a rule.4.96) 5 .96) 5 . Suppose we want a middle area such as P(21.96 . to calculate P(0 . which shows areas from 0 to z using increments of 0. 1.4750 5 . Since half the total area under the curve lies to the right of the mean.9.96 5 1. 11. Z .9 Standard Normal PDF and CDF . the entire area under the curve is 1.05 . 1. and a lowercase variable name like z or x to denote a particular value of Z or X.00 0 1 2 3 3 2 Standard Normal PDF Standard Normal CDF 2 1 1 0 1 2 3 Normal Areas from Appendix C-1 Tables of normal probabilities have been prepared so you can look up any desired normal area.80 .96) 5 .Chapter 7 Continuous Probability Distributions 263 Notation Use an uppercase variable name like Z or X when speaking in general. we see P(0 . This area is illustrated in Figure 7.25 . Z .96).96) is the same as P(21. z .06). we get P(21.20 .10.9 and the column for 0. Adding these areas. Such tables have many forms. Table 7.4750.06 (since 1. P(Z . Z .90 . covering the area from 0 to f (z) with many narrow rectangles and summing their areas). For example.96) 5 .69 (beyond this range.9500 So the interval 21. Figure 7. 11. At the intersection of the shaded row and column.5000 2 . Z . The probability of an event P(z1 .01 from z 5 0 to z 5 3.10 . so. FIGURE 7.96) 5 P(21..40 .15 . Because the normal distribution is symmetric.4 illustrates Appendix C-1.4750 5 . we also know that P(21. You do not need to worry about this because tables or Excel functions are available.0250. areas are very small).30 . . Z . Although there is no simple integral for f (z). LO 7-4 Find the normal probability for given z or x using tables or Excel.96).4750 1 .90 1 0. 1. Since f (z) is a probability density function.4750.50 .00 3 1. 11. the probability P(21.96 .70 . 1.96 .5000 2 P(0 . Since every transformed normal distribution will look the same.11 illustrates this calculation. 1. we are not interested in the height of the function f (z) but rather in areas under the curve (although Excel will provide either). For example. for simplicity. 0987 . Z .05 .5000 .08 .8 1.4911 0.4893 3.4793 .4772 .4564 .00) 5 2 3 .4732 . or 99. Z .1 2.0398 0.4686 .49988 . 11.4693 .4750 .00 .4599 .03 .264 Applied Statistics in Business and Economics TABLE 7.0675 .4861 .4554 .4699 .4890 .26% P(22.9973.4788 .4821 .4812 .0279 . Thus.4857 .4463 .0517 .0832 .73% .4898 0.49991 . Z .49990 .96 0 .0359 .4 Normal Area from 0 to z (from Appendix C-1) z 0.0239 . 1.4706 .4842 .7 1.49992 .0319 . 12.4719 .4772 5 .04 .49989 3.4750 .4864 .4738 .0438 .4871 .96 would yield a 95 percent area): P(21.49990 .4573 .49990 .4783 .12.96 FIGURE 7.4633 .00) 5 2 3 .4495 .4649 .2 2.1064 .0 .4887 .49984 .0250 0 1.0714 .0000 0.6826.4750 1.4608 . 3.4830 .4901 0.11 Finding Areas by Using Appendix C-1 .07 .4761 .49989 .00 0.9 2.4798 .10 Finding Areas Using Appendix C-1 .4616 .49986 .4591 .3413 5 .0080 . 13.4826 .0871 .0040 .00 .06 .00) 5 2 3 .00) 5 2 3 P(0 .49992 . Z .4803 .0753 .4778 . illustrated in Figure 7.49988 .49992 . or 95.4913 0.4713 .0557 .00 . or 68.09 .4881 .4678 .4505 .4808 .4656 .0910 .1026 .00) 5 2 3 P(0 .4525 .4909 0.44% P(23.49992 FIGURE 7. These are the “k-sigma” intervals mentioned in Chapter 4 and used by statisticians for quick reference to the normal distribution.4625 .4484 .0948 .4756 .4868 .4916 0.4896 0.4641 . Z .49986 .0478 .4750 1.0160 .49987 .01 .4854 .9544.4726 .4834 .9500 .49987 .4671 .4767 .0199 .1141 .49985 .4817 .4838 .0596 .2 .0793 1. Z .96 0 1.4582 .6 1.96 From Appendix C-1 we can see the basis for the Empirical Rule.4846 .49865 5 .4906 0. it is approximately correct to say that a “2-sigma interval” contains 95 percent of the area (actually z 5 1.49991 .02 .4474 .4744 .4904 0.4878 .4545 .4875 .3 .0 2. 2.6 .1103 .00) 5 2 3 P(0 .4850 .0636 .4515 .7 .4884 .49985 .4664 .1 .4452 .0120 .4535 . 9857 .0436 .99989 .9463 .3 22.0274 .9834 .9686 . z 0.9756 .00011 22.5000 0.00008 . This second table corresponds to the way Excel calculates normal areas.0526 .0110 .9821 .9838 .7 1.0132 .5120 .0233 .5239 .00016 .9767 .9808 .1 2.9625 .0367 .0244 .9916 0.0250 5 .2 .0129 .0122 .9 21.00010 .9898 .00014 .1 22.0087 .00012 .0089 .0116 .01 0.9554 .0446 . 21.0202 .0162 . Z .00013 .0154 .9812 .9871 .6103 .99992 .0465 .6026 .9904 .0336 .9793 .0516 .9881 .05 0. LO 7-4 Find the normal probability for given z or x using tables or Excel.96) 2 P(Z .0268 .6 .00015 .0256 .9798 .09 TABLE 7.0250 .0548 .0329 .0125 .5832 .9495 .99991 .9 2.00012 . This calculation is illustrated in Figure 7.9909 .5636 .9693 . 11.00010 .0359 .00 0.9649 . Appendix C-2 shows cumulative normal areas from the left to z.0146 .0192 .9901 .0170 .73% 3 2 1 1 2 0 One-Sigma Interval 3 3 2 1 0 1 2 Two-Sigma Interval 3 3 2 1 0 1 2 Three-Sigma Interval 3 Normal Areas from Appendix C-2 Table 7.9864 .0207 .99988 .9783 .9545 .7 .0 .9750.2 2.0427 .5753 .96) 5 .5 Cumulative Normal Area from Left to z (from Appendix C-2) 23.5160 .0537 .99984 .9582 .9505 .96 encloses 95 percent of the area under the normal curve.9854 .96) 5 . The interval 21.9842 .9884 .03 0.0094 .1 .0495 .9484 .96) 5 P(Z .00010 .9906 .9678 .0096 .00014 .0113 .02 0.0183 .0174 .0217 .5948 .9525 .9826 .9641 .00013 .7 21.00008 .5438 .9875 .0262 .9452 .9726 .0409 .0104 .00009 .9861 .00009 .8 21.9803 .13.0139 .9817 .9515 .3 .0188 .99985 .5675 .0322 .0107 .9744 .9893 3.99992 .0392 .0455 .9913 .0384 .06 0.9719 .5319 .0287 .0 21.6 .99986 .9772 .99990 .0119 .0158 .0136 .9633 .Chapter 7 Continuous Probability Distributions 265 FIGURE 7.0212 .0143 .6 1.0351 . 21.99989 3.9713 .5910 .0166 . 11.0485 .99987 .9608 .5517 .9778 .9911 .0418 .99986 .0197 .6141 .99992 .0375 .9599 .5199 .9474 . Using this approach.00008 23.0 2.96 .44% 99.9591 .9890 .0250 and P(Z .12 Normal Areas within k 68.0314 .9896 . we see that P(Z . By subtraction.9656 .0099 .00015 .0294 .5478 . 11.0091 .2 22.0102 .00008 .9706 .9500 The result is identical to that obtained previously.0084 .0179 . 1.5714 .5871 .9850 .9664 .9616 .08 0.9878 .9564 .9535 .9671 .9868 .7 . we get P(21.5359 .5040 .0307 .0344 .0228 .26% 95.6064 .9699 .6 .99991 .5987 .0505 .8 1.5080 .99988 .96 .0150 .9846 .99992 .9887 .9732 .9573 .0301 .99990 .5793 1.9750 .5279 .9750 2 .99987 .0401 .5 illustrates another kind of table.96) 5 .0222 .0239 .0281 . Z .9830 .9761 .07 0.99990 .5398 0.99985 .9788 .04 0.5596 .00011 .0475 .9738 .5557 . 266 Applied Statistics in Business and Economics FIGURE 7.50) b.75 standard deviations faster than the women’s average for her age group.00) d. P(Z . a.18 Bob’s exam score was 2.” It also has the advantage of being more compact (it fits on one page).00) d.17 Find the standard normal area for each of the following. Assuming a normal distribution. 5 percent.28) c.00 . or the middle 50 percent. 90 percent.65) 7.13 Finding Areas by Using Appendix C-2 .9750 .17 standard deviations above the mean. 21. There were 405 women who ran in her age group.96 1.. Z . how many scores were higher than Bob’s? 7. a. a. 1 percent. Further. P(Z .4900 we see that z 5 2. P(0 . 2.96 .00 . Z . either table is equally convenient.g. Z . Z . Z .14. Z . 1.22 .28) b.96) d. 0) d.96) c. P(Z . P(1. 3.50) 7. 1. P(20. Z . etc. showing your reasoning clearly and indicating which table you used.4901. 7. that is as close as we can get to 49 percent.65) 7.96 1. P(Z . marketing). 0) c. Appendix C-2 corresponds to the way Excel calculates normal areas. Sketch the normal curve and shade in the area represented below. 21.33 yields an area of . P(21. 1.15 Find the standard normal area for each of the following. how many women ran faster than Joan? Finding z for a Given Area We can also use the tables to find the z-value that corresponds to a given area. P(21. P(Z . showing your reasoning clearly and indicating which table you used. 21.00 . P(Z 5 0) 7.9500 0 1. P(Z . 0.0250 1. SECTION EXERCISES Note: Use Appendix C-1 or C-2 for these exercises. Since we are often interested in the top 25 percent. Z .6 and Figure 7. 10 percent.96) b.19 Joan’s finishing time for the Bolder Boulder 10K race was 1. P(22.16 Find the standard normal area for each of the following.. For example. which is one reason why it has traditionally been used for statistics exams and in other textbooks (e. P(Z .00) c.65) d. P(21.96 Since Appendix C-1 and Appendix C-2 yield identical results. 2. This is illustrated in Table 7. a. 2.15) b. you should use whichever table is easier for the area you are trying to find. . Z . P(23. When subtraction is required for a right-tail or middle area. P(Z . But Appendix C-2 is easier for left-tail areas and some complex areas.65 . 0. 1.15) b.22 . We can find other important areas in the same way. a.13 Find the standard normal area for each of the following. 2.96 .50 . what z-value defines the top 1 percent of a normal distribution? Since half the area lies above the mean. The exam was taken by 200 students. showing your reasoning clearly and indicating which table you used. Without interpolation. P(Z 5 0) 7. an upper area of 1 percent implies that 49 percent of the area must lie between 0 and z. Searching Appendix C-1 for an area of .14 Find the standard normal area for each of the following. P(2.00) c. P(Z . Appendix C-1 is often easier for “middle areas. 2. 1. Sketch the normal curve and shade in the area represented below. Assuming a normal distribution. 0832 .7 summarizes some important normal areas.4608 .Chapter 7 Continuous Probability Distributions 267 z 0.8 1.2 2.4808 .4803 .49986 .645 1.49986 .0636 .4864 .4535 .4826 . Table 7.0793 1.4884 .49987 .576 Right Tail Area .4901 .01 .49991 .49988 .00 0.4834 .6 .49985 .49989 3.4901 3 2 1 0 1 2 3 TABLE 7.0987 .0000 0.0099 Finding Areas by Using Appendix C-1 .4761 .4744 . without interpolating you may only be able to approximate the actual area.7 1.0714 .4881 .4904 0.4573 .4495 .4591 .4554 .49989 .0040 .4656 .0160 .4812 .0199 .49987 .4778 .49985 .07 .20 Find the associated z-score for each of the following standard normal areas. a.4633 .4706 .4505 .01 .4906 0.0279 . 99 percent.960 2. 7.0438 .6 1.326 2.4671 .9 2.4756 .4901 0.5000 .04 .0675 .90 .7 .49990 .0120 .0948 .4854 .0 .1141 .4868 . Highest 10 percent b.4599 .4545 .4913 0.4664 .4788 .4452 .3 .0319 .4767 .4641 .49991 .4649 .1064 .4719 .4909 0.02 .7 Important Normal Areas 0 z z 0 z z 0.0478 . Note: For each problem below.4525 .4616 .005 Middle Area .4861 .4625 .4850 .4463 . Lowest 50 percent c.4793 .0239 .4887 .0596 .0398 0.4821 .06 .4898 0.675 1.4878 .4772 .49992 . Highest 7 percent SECTION EXERCISES .4857 .4916 TABLE 7.0080 .4686 .4838 .4484 .0 2.08 .49992 .14 .05 .4582 .4732 .2 .0359 .4693 .25 .0753 .1 . these z-values are shown to three decimals (they were obtained from Excel).50 .99 95 percent.4713 .03 .1026 .4699 .4842 .80 .4846 .4890 .4726 . and so forth.4783 .09 .49990 .49984 .4515 . it is convenient to record these important z-values for quick reference.4564 .025 .4750 .4798 .49992 .10 .4896 0.95 .05 .0910 .0871 .4911 0.4738 .49988 .4830 .98 .0557 .6 Normal Area from 0 to z (from Appendix C-1) 0.4817 . For greater accuracy.1 2.4678 .4474 .1103 .0517 .4893 3.4875 .282 1.4871 .49990 .49992 FIGURE 7. 15.43) as P(Z $ 21. Highest 2 percent c. as illustrated in Figure 7. so John is approximately in the 94th percentile. or 92. 7.43) 5 1 – P(Z .9418. Middle 95 percent 7. What percentile is John in? That is. Lowest 7 percent c. Lowest 5 percent b.25 The fastest 10 percent of runners who complete the Nosy Neighbor 5K race win a gift certificate to a local running store. how many standard deviations above the mean does a student have to score to be publicly recognized? 7.23 Find the associated z-score or scores that represent the following standard normal areas.57 7 7 This says that John’s score is 1. Z .24 High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capability Challenge Exam.9418 86 .4% . Middle 90 percent c. The class mean was 75 with a standard deviation of 7. Middle 50 percent a. what is P(X .43 .22 Find the associated z-score or scores that represent the following standard normal areas. 7. Assuming a normal distribution.0764 5 .43) 5 P(21. From Appendix C-2 we get P(X .268 Applied Statistics in Business and Economics 7.9236.43) as P(Z $ 21. 86)? We need first to calculate John’s standardized Z-score: xJohn 2 86 2 75 11 zJohn 5 _________ 5 _______ 5 ___ 5 1.4% Using Appendix C-2 we can calculate P(X $ 65) 5 P(Z $ 21. The table gives a slightly different value from Excel due to rounding.4236 1 . 1. Lowest 6 percent a. how many standard deviations below the mean must a runner’s time be in order to win the gift certificate? Finding Areas by Using Standardized Variables John took an economics exam and scored 86 points.15 Two Equivalent Areas Original Scale P(X 86) Standard Scale P(Z 1.57 54 61 68 75 82 89 96 3 2 1 0 1 2 3 On this exam. or 92. a.57) X . FIGURE 7.5000 5 .57 standard deviations above the mean. 86) 5 P(Z . That means that his score was better than 94 percent of the class.43) 5 1 2 . 0) 1 . 21.57) 5 . Highest 40 percent b. what is the probability that a randomly chosen test-taker would have a score of at least 65? We begin by standardizing: x2 65 2 75 210 z 5 ______ 5 _______ 5 ____ 5 21.9418 Z 1. Assuming a normal distribution.9236. Middle 60 percent b.21 Find the associated z-score for each of the following standard normal areas.43 7 7 Using Appendix C-1 we can calculate P(X $ 65) 5 P(Z $ 21. Students who score in the top 20 percent are recognized publicly for their achievement by the Department of the Treasury.5000 5 . . there is a 92.8 illustrates Excel functions that return left-tail normal areas for a given value of x or z.S.762475 z NORM. . Here. TABLE 7.16 Two Ways to Find an Area Using Appendix C-1 Using Appendix C-2 . you must be careful of syntax.75.25% of the exam-takers 75 and 7.5000 .7. Finding Normal Areas with Excel Excel offers several functions for the normal and standard normal distributions.DIST(x.96. cumulative) NORM. as shown in Figure 7. It is a good idea to sketch a normal curve and shade the desired area to help you visualize the answer you expect.1) NORM.DIST(80.4 percent chance that a student scores 65 or above on this exam. score 80 or less if Area to the left of z in a standard normal.1) 0.43 0 1. These calculations are illustrated in Figure 7. so that you will recognize if you are getting an unreasonable answer from Excel.96.8.S. Excel is more accurate than a table.8 Excel Normal CDF Functions ? ? Syntax: Example: x NORM.4236 .975002 What it does: Area to the left of x for given and . In Table 7. Here.DIST(z.17 Inserting Excel Normal Functions Table 7. 76. note that the cumulative argument is set to 1 (or TRUE) because we want the CDF (left-tail area) rather than the PDF (height of the normal curve).16. however. FIGURE 7. we see that 97.9236 . .50% of the area is to the left of z 1.17.Chapter 7 Continuous Probability Distributions 269 FIGURE 7.DIST(1.1) 0.43 Using either method.0764 1. 2.039) Excel’s P (X .8185.DIST(2. let X be the diameter of a manufactured steel ball bearing whose mean diameter is 5 2. 2.270 Applied Statistics in Business and Economics Excel’s NORM.040 cm and whose standard deviation 5 .042.000 barrels? (c) Less than 239. we could do exactly the same thing by using Appendix C-2: P(2.000 barrels of crude oil per day with a standard deviation of 7.18.000 barrels? (b) Between 232.9% FIGURE 7.19.818595 2.9772 2 ..DIST(2.042) 5 P(X . refinery is normally distributed with a mean of 232. The desired area is approximately 81.977250 . What is the probability that a given steel bearing will have a diameter between 2.2.04. (Data are from McDonalds.042) – P(X .000 and 239. (a) What is the probability of producing at least 232.977250 Using Excel.DIST(2.2.001.042.27 Assume that the number of calories in a McDonald’s Egg McMuffin is a normally distributed random variable with a mean of 290 calories and a standard deviation of 14 calories.039 2.042 cm? We use Excel’s function =NORM. .039) 5 NORM.DIST .2.039.158655 . 2.cumulative) where cumulative is TRUE. 2.042) as in Figure 7. We then obtain the area between by subtraction.com) 7.1) NORM. (b) What is the probability . as illustrated in Figure 7. or 81.DIST(2.DIST(x.04.2.18 Left-Tail Areas Using Excel Normal CDF Excel’s P (X ..001.158655 .1587 5 .001. 2.000 barrels? (e) More than 225.04.042) FIGURE 7.039.040 2.1) 2 NORM. X .13 gram.000 barrels? (d) Less than 245. (a) What is the probability that a particular serving contains fewer than 300 calories? (b) More than 250 calories? (c) Between 275 and 310 calories? Show all work clearly. Of course.28 The weight of a miniature Tootsie Roll is normally distributed with a mean of 3.039 and 2.001. For example. .9 percent.1) .. Since Excel gives left-tail areas.INV functions let us evaluate areas and inverse areas without standardizing.1) 5 . NORM.DIST and NORM. we first calculate P(X ..818595 .039) and P(X .000 barrels? 7.039 . (a) Within what weight range will the middle 95 percent of all miniature Tootsie Rolls fall? Hint: Use the Empirical rule.19 Cumulative Areas from Excel’s NORM.042 SECTION EXERCISES 7. 2.001 cm.26 Daily output of Marathon’s Garyville. Lousiana.30 grams and standard deviation of 0.000 barrels.04. 2. the 99th percentile for exam-takers is a score of 91. .324 students. 95th. Value of z corresponding to a given left-tail area.254 received a merit scholarship to help offset tuition costs their freshman year (although the amount varied per student). for a given area and x.) 7. The amount a student received was N($3.675 20.480.INV(0.282 0.7) 91.75. If a business traveler’s height is N(59100. Of those. 2011.675 21. ( ) Using Table 7.008). 2.7 (or looking up the areas in Excel) we obtain the results shown in Table 7.30 The cabin of a business jet has a cabin height 5 feet 9 inches high.675. we can solve for the value of x or z that corresponds to a given normal area. Here.645 x z x (to nearest integer) 86. What percentage of their bats will exceed the BBCOR standard? (See http://batrollingblog.645)(7) Using the two Excel functions =NORM. etc. The number of patients needing a bed at any point in time is N(19. If the cost of tuition was $4. 2. 10th.97.9. Note that to find a lower tail area (such as the lowest 5 percent).S. ) NORM.75) 0.500. BigBash Inc.31 On January 1.10.INV(area) NORM. A higher BBCOR allows the ball to travel farther when hit. Also. or 63 (rounded) TABLE 7.32 Last year’s freshman class at Big State University totaled 5. or 70 (rounded) 66. TABLE 7.645)(7) x 5 75 1 (1.0.INV() and =NORM. . 90th.282)(7) x 5 75 2 (1. so bat manufacturers want a high BBCOR. or 84 (rounded) 79.52. If 75 and 7.) 7.10 Excel Inverse Normal Functions area area Syntax: Example: ? NORM. we must use negative Z-values.29 The pediatrics unit at Carver Hospital has 24 beds. What is the probability that the number of patients needing a bed will exceed the pediatric unit’s bed capacity? 7.S.9 Percentiles for Desired Normal Area x 5 75 1 (1.5).70).Chapter 7 Continuous Probability Distributions 271 that a randomly chosen miniature Tootsie Roll will weigh more than 3. or 87 (rounded) 83.282)(7) x 5 75 1 (0. 25th.675)(7) x 5 75 2 (0.282 21. produces bats whose BBCOR is N(0. how can we find X for a given area? We simply turn the standardizing transformation around: x2 x5 1z solving for x in z 5 ______ (7. 75th.674490 What it does: Value of x for given left-tail area.INV() shown in Table 7.5) LO 7-5 Solve for z or x for a given normal probability using tables or Excel. or 66 (rounded) 63. Percentile 95th (highest 5%) 90th (highest 10%) 75th (highest 25%) 25th (lowest 25%) 10th (lowest 10%) 5th (lowest 5%) z 1. what percentage of the business travelers will have to stoop? 7. what percentage of students did not receive enough to cover their full tuition? Inverse Normal How can we find the various normal percentiles (5th.28 or 91 to the nearest integer.99.49. $478).28.2.200 last year. The maximum allowable BBCOR is 0.456.03.com.675)(7) x 5 75 2 (1. a new standard for baseball bat “liveliness” called BBCOR (Ball-Bat Coefficient of Restitution) was adopted for teams playing under NCAA rules. 1. we might want to solve for or .INV(area.S.INV(0.50 grams? (Data are from a project by MBA student Henry Scussel.73.) known as the inverse normal? That is. the 75th percentile (third quartile) of a standard normal is z 0.645 1. or 80 (rounded) 70.2844 ? NORM. what is the standard deviation? . x2 x 2 75 • Substitute the given information into z 5 ______ to get 21. a. Assuming a normal distribution.28 to satisfy P(Z . Find the weight that corresponds to each event. Middle 50 percent d.5). Find the time for each event. Highest 5 percent c. a.0 pounds or more? (b) What would be the 90th percentile for birth weight? (c) Within what range would the middle 95 percent of birth weights lie? 7. Highest 80 percent b. Lowest 10 percent 7.272 Applied Statistics in Business and Economics For example.9 pounds and a standard deviation of 1. Find the time for each event.33 The time required to verify and fill a common prescription at a neighborhood pharmacy is normally distributed with a mean of 10 minutes and a standard deviation of 3 minutes. If 20 percent of all part-time seasonal employees make more than $13.35 The weight of a McDonald’s cheeseburger is normally distributed with a mean of 114 ounces and a standard deviation of 7 ounces. 7. (a) Find the credit score that defines the upper 5 percent.38 The credit scores of 35-year-olds applying for a mortgage at Ulysses Mortgage Associates are normally distributed with a mean of 600 and a standard deviation of 100. SECTION EXERCISES 7. 2. Highest 10 percent c. Show your work. At a certain mountain.00. if 70 percent of the filings cost less than $171. (b) Seventy-five percent of the customers will have a credit score higher than what value? (c) Within what range would the middle 80 percent of credit scores lie? 7.28 5 ______. Highest 10 percent c.28)(7) 5 66. the hourly rates have a normal distribution with 5 $3.10. Show your work. Middle 95 percent b. The steps to solve the problem are: • Use Appendix C or Excel to find z 5 21.41 The average cost of an IRS Form 1040 tax filing at Thetis Tax Service is $157.04 (or 66 after rounding) Students who score below 66 points on the economics exam will be required to retake the exam. what is the average hourly pay rate at this mountain? 7.37 The weights of newborn babies in Foxboro Hospital are normally distributed with a mean of 6. Lowest 80 percent 7.2. Lowest 50 percent d.16 an hour.34 The time required to cook a pizza at a neighborhood pizza joint is normally distributed with a mean of 12 minutes and a standard deviation of 2 minutes. Lowest 10 percent 7.39 The number of patients needing a bed at any point in time in the pediatrics unit at Carver Hospital is N(19. Find the weight that corresponds to each event. x) 5 . 21. Highest 5 percent c. a.36 The weight of a small Starbucks coffee is a normally distributed random variable with a mean of 360 grams and a standard deviation of 9 grams. 7 • Solve for x to get x 5 75 2 (1.28) 5 .00.40 Vail Resorts pays part-time seasonal employees at ski resorts on an hourly basis. Highest 80 percent b. a. Middle 50 percent d.10. Show your work. Show your work.2 pounds. suppose that John’s economics professor has decided that any student who scores below the 10th percentile must retake the exam. Lowest 80 percent 7. Lowest 50 percent d. What is the score that would require a student to retake the exam? We need to find the value of x that satisfies P(X . The approximate z-score for the 10th percentile is z 5 21. Middle 95 percent b.00. The exam scores are normal with 5 75 and 5 7. Find the middle 50 percent of the number of beds needed (round to the next higher integer since a “bed” is indivisible).28. (a) How unusual is a baby weighing 8. 40) 5 . X. is normal with a mean 5 28 minutes and a standard deviation 5 5 minutes. Approximately 66 percent of the cars will be finished in less than half an hour.DIST(30. 5 Using Excel. This information can now be used to answer questions about normal probabilities. (4) Find the area by using one of the tables or Excel.28.4 5 Using Appendix C-2 or Excel we find that P(Z .40 5 Using Appendix C-2 or Excel we find that P(Z .6554. Worked Problem #2 What is the chance that a randomly selected car will take longer than 40 minutes to complete? • Steps 1 and 2: Draw a picture and shade the area to the right of 40 minutes. EXAMPLE 7. N(28. There is less than a 1 percent chance that a car will take longer than 40 minutes to complete. X .2 Service Times in a Quick Oil Change Shop: Four Worked Problems 5 Using Excel. . the shop’s manager has found that the distribution of service times. that is.4) 5 1 2 P(Z # 2. NORM.0082. To answer these types of questions it is helpful to follow a few basic steps.655422 28 x z 30 ? • Step 3: • Step 4: 30 2 28 z 5 _______ 5 0. 5).28.4) 5 1 2 .5.5. 1 NORM.1) .008198 28 x z 40 ? • Step 3: • Step 4: 40 2 28 z 5 _______ 5 2. 2.DIST(40. 0. (2) Shade in the area that will answer your question. (3) Standardize the random variable. Worked Problem #1 What proportion of cars will be finished in less than half an hour? • Steps 1 and 2: Draw a picture and shade the area to the left of 30 minutes. (1) Draw a picture and label the picture with the information you know.9918 5 .Chapter 7 Continuous Probability Distributions 273 After studying the process of changing oil.1) . 4 minutes or less. P(X . P(80 . 100) .5) 34. x 2 28 1. What must the mean service time be to accomplish this goal? • Steps 1 and 2: Draw a curve and shade the desired area.28 by using the tables or Excel.INV(. 1) d. • Step 3: • Step 4: Find z 5 1. x2 30 2 • Step 4: Substitute into z 5 ______ to get 0. 1000) b.84 5 _______ and solve for 5 5 30 2 0.80) . X .50 . 15) c. X . NORM. Worked Problem #4 The manager wants to be able to service 80 percent of the vehicles within 30 minutes. 5.80).000) for N(6000. NORM. Given 5 min Using Excel.84 (approximately) for an upper tail area of . X .43 Use Excel to find each probability. 450) for N(600.8 minutes to ensure that 80 percent are serviced within 30 minutes. a. The mean service time would have to be 25. 100) b.40776 Draw a picture and shade the desired area.00) for N(0. 2. P(X . 110) for N(100. 1000) 7.500 .84(5) 5 25. 450) for N(600. 7.9.20 (lower tail area of . P(X .80 ? x z . 2.28 5 ______.8.S. 15) c. P(4. so x 5 28 1 5(1. steps 3 and 4 need to be reversed. X .28.000) for N(6000. 28 z x ? ? In this case.42 Use Excel to find each probability. P(X . P(1.90 Using Excel. a. SECTION EXERCISES 7. 1) d.841621 .00) for N(0.4 minutes. P(225 .20 30 min ? • Step 3: Use tables or Excel to find z 5 0.274 Applied Statistics in Business and Economics Worked Problem #3 What service time corresponds to the 90th percentile? • Steps 1 and 2: 5 . 5 Ninety percent of the cars will be finished in 34.INV(.28) 5 34. 110) for N(100. 50. Figure 7. 1 P(32). we can use a normal approximation. EXAMPLE 7. when translating a discrete scale into a continuous scale we must be careful about individual points. 10th percentile d. 99. does this mixture provide an adequate margin of safety? 7. particularly when many terms must be summed. Assume a normal distribution. setting the normal and equal to the binomial mean and standard deviation: 5n __________ LO 7-6 Use the normal approximation to a binomial or a Poisson.50 n = 64. As sample size increases. . 99.9th percentile c.7) 5 n (1 2 FIGURE 7.000 psi. (a) What is the probability of more than an hour’s wait? (b) Less than 20 minutes? (c) At least 10 minutes.50 n = 16.3 Coin Flips . (7.21. continuous. Use Excel to find each probability.5 NORMAL APPROXIMATIONS Normal Approximation to the Binomial We have seen that (unless we are using Excel) binomial probabilities may be difficult to calculate when n is large. it becomes easier to visualize a smooth.45 A study found that the mean waiting time to see a physician at an outpatient clinic was 40 minutes with a standard deviation of 28 minutes.000 pounds per square inch (psi). normal curve. The logic of this approximation is that as n becomes large. 7. bell-shaped curve overlaid on the bars. As a rule of thumb. but due to variability in the mixing process it has a standard deviation of 420 psi. 75th percentile f. . Could the normal approximation be used? With n 5 32 and 5 . it is safe to use the normal approximation to the binomial. What is the probability that a given pour of concrete from this mixture will fail to meet the high-strength criterion? In your judgment. 90th percentile b.99th percentile 7.20 Binomial Approaches Normal as n Increases n = 4.Chapter 7 Continuous Probability Distributions 275 7. 16. The event “more than 17” actually falls halfway between 17 and 18 on a discrete scale.20 illustrates this idea for 4. = . and 64 flips of a fair coin with X defined as the number of heads in n tries. = . 32nd percentile e.50 0 1 2 3 4 Number of Successes 0 2 4 6 8 10 12 14 16 Number of Successes 20 25 30 35 40 Number of Successes 45 What is the probability of more than 17 heads in 32 flips of a fair coin? In binomial terms.6) ) (7.44 The weight of a small Starbucks coffee is a normal random variable with a mean of 360 g and a standard deviation of 9 g. this would be P(X $ 18) 5 P(18) 1 P(19) 1 . as shown in Figure 7. Use Excel to find the weight corresponding to each percentile of weight. = . we clearly meet the requirement that n $ 10 and n(1 2 ) $ 10. However. Instead. the discrete binomial bars become more like a smooth.46 High-strength concrete is supposed to have a compressive strength greater than 6. which would be a tedious sum even if we had a table. when n $ 10 and n(1 2 ) $ 10. A certain type of concrete has a mean compressive strength of 7. a. we could use Appendix C-2 to get P(Z .00 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 If you make a diagram like this.276 Applied Statistics in Business and Economics You don’t need to draw the entire distribution. Since the cutoff point for “more than 17” is halfway between 17 and 18.5) 5 2.53 .10 17. 2. FIGURE 7.22.21 . Z .5) 5 16 __________ _______________ 5 n (1 2 )5 (32)(0.2981.53) 5 . by the symmetry of the normal distribution. .05 .53) 5 .5000 2 .5 2 16 z 5 ______ 5 _________ 5 .53) which.5)(1 2 0. the normal approximation is P(X .2981. . .22 Normal Area for P(Z . is the same as P(Z . 17. . . 14 15 16 17 18 19 20 21 18) 22 23 .5).5000 2 P(0 . The 0.5) .2981 0. The calculations are illustrated in Figure 7.5 is an adjustment called the continuity correction.53) 5 .15 Normal Approximation to P(X P(X . Alternately. . .53 2. you can see the correct cutoff point.82843 We then perform the usual standardizing transformation with the continuitycorrected X-value: x2 17. All you need is a little diagram (ignoring the low and high ends of the scale since they are not relevant) to show the event “more than 17” visually: .82843 From Appendix C-1 we find P(Z . The normal parameters are 5 n 5 (32)(0.2019 5 .53) . FIGURE 7. consider the events in the table below. 14 15 16 17 18 19 20 . (See LearningStats on the OLC for a demonstration of this result.5 Use x 5 18. In a right-skewed binomial (when . what is the probability of fewer than 50 defaults? (b) More than 100? Show your work carefully.000 student loans are made. The cashier scans 800 items. (a) What is the expected number of diversions? (b) What is the probability of at least 400 diversions? (c) Fewer than 450 diversions? Show your work carefully. .50 When confronted with an in-flight medical emergency. 14 15 16 17 18 19 20 . . the normal approximation is less accurate. find the probability that (a) at least 175 will receive the desired item and (b) that fewer than 190 will receive the desired item. But when n is large. Event At least 17 More than 15 Fewer than 19 Relevant Values of X .90 that a vending machine in the Oxnard University Student Center will dispense the desired item when correct change is inserted.48 In a certain store. To be sure you understand the continuity correction. 7. Some textbooks allow the approximation when $ 10.49 The probability is .03 probability that the scanned price in the bar code scanner will not match the advertised price. the rule n $ 10 ensures that the mean 5 n is far enough above 0 to prevent severe truncation. . 7. the normal approximation (. (a) What is the expected number of mismatches? The standard deviation? (b) What is the probability of at least 20 mismatches? (c) What is the probability of more than 30 mismatches? Show your calculations clearly. 7. Since Excel’s function is cumulative to the left. In a left-skewed binomial distribution (when . We sketch a diagram to find the correct cutoff point to approximate a discrete model with a continuous one. SECTION EXERCISES Normal Approximation to the Poisson The normal approximation for the Poisson works best when is fairly large.5. .2983 In this case. . partly because this binomial is roughly symmetric ( is near . If 200 customers try the machine.5 Use x 5 15. ceteris paribus.2983). Explain.50). If the global response center is called. 14 15 16 17 18 19 20 .7017 5 . there is a . we find P(X # 17) with the Excel function =BINOM. the normal approximation improves. Suppose the response center is called 8. . . Normal Cutoff Use x 5 16. .50). . . . .01 in the approximation. To use the normal . The rule of thumb results in an error of less than . regardless of . which is comparable to the rule that the binomial mean must be at least 10. you are reasonably safe in using the normal approximation. the rule that n(1 2 ) $ 10 guards against severe truncation at the upper end of the scale by making sure that the mean is well below n. . .465 times in a given year.50).32.2981) is very close to the binomial probability (.0. pilots and crew can consult staff physicians at a global response center located in Arizona.Chapter 7 Continuous Probability Distributions 277 How accurate is this normal approximation to the binomial P(X $ 18) in our coin flip example? We can check it by using Excel.47 The default rate on government-guaranteed student loans at a certain public four-year institution is 7 percent. there is a 4.) When a binomial distribution is badly skewed ( near 0 or 1).DIST(17.8 percent chance that the flight will be diverted for an immediate landing. . 7. If you can’t find in Appendix B (which only goes up to 5 20). (a) If 1. That is why both rules are needed.5 Note: Use Appendix C-2 for these exercises.1) and then subtract from 1: P(X $ 18) 5 1 2 P(X # 17) 5 1 2 . . 0975 In this case.5 (halfway between 50 and 51): . . consider the process of customers arriving at a Noodles & Company restaurant.0951) comes fairly close to the Poisson result (. but the calculation would be tedious at best.278 Applied Statistics in Business and Economics approximation to the Poisson we set the normal standard deviation: 5 __ and equal to the Poisson mean and (7. . 1. We could count the number of customers who arrive in a randomly selected minute. 50. Of course. The standardized Z-value for the event “more than 50” is P(X . For example.23. or we could . 28 patients per hour arrive in the Foxboro 24-Hour Walk-in Clinic on Friday between 6 p. . SECTION EXERCISES Note: Use Appendix C-2 for these exercises.31) 5 .31) because the normal distribution is symmetric.31) since x2 50. 7.9025 5 .1) and subtracting from 1: P (X $ 51) 5 1 – P(X # 50) 5 1 – .8) (7. (a) Estimate the probability of at least 175 attacks in a given day.m.52 For a large Internet service provider (ISP).42. There are two different variables that could be used to describe this process.48074 Using Appendix C-2 we look up P(Z . which is the same as P(Z . (d) Use Excel to calculate the actual Poisson probabilities.31 6.0975). web virus attacks occur at a mean rate of 150 per day. customer billing inquiries arrive at a mean rate of 42 inquiries per hour at Consumers Energy. Characteristics of the Exponential Distribution In Chapter 6 we introduced the idea of a random process.0951. (a) What is the approximate probability of more than 35 arrivals? (b) What is the approximate probability of fewer than 25 arrivals? (c) Is the normal approximation justified? Show all calculations. We can check the actual Poisson probability by using Excel’s cumulative function =POISSON. illustrated in Figure 7. 46 47 48 49 50 51 52 53 . and midnight. you don’t need the approximation at all.DIST(50. (d) Use Excel to calculate the actual Poisson probabilities. whose terms gradually become negligible (recall that the Poisson has no upper limit). if you have access to Excel. We set 5 5 5 42 ___ __ 5 42 5 6. the normal approximation is simple. How close were your approximations? 7. but the mean 5 42 is too large to use Appendix B.5 2 42 z 5 ______ 5 _________ 1.48074 The continuity-corrected cutoff point for X $ 51 is X 5 50.5) 5 P(Z . (c) Is the normal approximation justified? Show all calculations. (b) Estimate the probability of fewer than 125 attacks. What is the probability of receiving more than 50 calls? Call arrivals presumably follow a Poisson model. The formula would entail an infinite sum P(51) 1 P(52) 1 . .6 EXPONENTIAL DISTRIBUTION LO 7-7 Find the exponential probability for a given x. How close were your approximations? 7. 1. . . the normal approximation (.9) 5 EXAMPLE 7.m.4 Utility Bills On Wednesday between 10 a. and noon. 21. However.51 On average. 11. and is very skewed. Fortunately.1) =–LN(RAND())y Waiting time is exponential when arrivals follow a Poisson model.23 Customer Arrival Process at a Noodles & Company Restaurant Time X time between two customer arrivals X is continuous and has an exponential distribution measure the time between two customer arrivals.Chapter 7 Continuous Probability Distributions 279 Customer Arrivals FIGURE 7. Often 1y is given (mean time between events) rather than . no tables are needed. the CDF is simple.11 Exponential Distribution P (X # x) 5 1 2 e2 event. the focus is on the waiting time until the next Parameter PDF CDF Domain Mean Standard deviation Shape CDF in Excel Random data in Excel Comments 5 mean arrival rate per unit of time or space (same as Poisson mean) f (x) 5 e2 x$0 1y 1y Always right-skewed. In the exponential model. The value of e is approximately 2.24. As you learned in Chapter 6.DIST(x. When the count of customer arrivals has a Poisson distribution. as shown in Figure 7. The exponential probability density function approaches zero as x increases. the distribution of the time between two customer arrivals will have an exponential distribution. detailed in Table 7. the count of customer arrivals is a discrete random variable and typically has a Poisson distribution.71828. . just a calculator FIGURE 7. a continuous variable.24 Exponential Areas Left-Tail Exponential Area Right-Tail Exponential Area Left-tail area: P(X x) 1 e x Right-tail area: P(X x) e x 0 x 0 x . x x TABLE 7. =EXPON. We are usually not interested in the height of the function f (x) but rather in areas under the curve. 50 minute.m.00 1.280 Applied Statistics in Business and Economics that has the ex function key. The probability that 30 seconds or less (0.00 .53 In Santa Theresa.50 minute since is expressed in minutes.10.29% There is about a 33 percent chance of waiting more than 30 seconds before the next call arrives. and midnight on Thursday night. 7.m. x) 5 e2 x x (probability of waiting x or less) (probability of waiting more than x) (7.50) 5 1 2 .50 1. x) since the point x has no area.50 Minutes 2.50) 5 . Note that we must convert 30 seconds to 0.50) SECTION EXERCISES 7. Since x 5 0. Mystery Pizza gets an average of 4.54 Between 11 p.2 events per minute and x 5 0.50 1.5 Customer Waiting Time Between 2 p.2)(0.00 1.00 . The probability of waiting more than x units of time until the next arrival is e2 x. For this reason.50) and P(X .2 telephone orders per hour.50) refer to the same event (unlike. in which a point does have a probability). Find the probability that (a) at least 30 minutes will elapse before the next telephone order.00 2.50 .00 0 .50 Minutes 2. and (c) between 15 and 30 minutes will elapse.11) Recall that P(X # x) is the same as P(X . (b) less than 15 minutes will elapse.1 times a minute. say.00 0 .2)(0. What is the probability that (a) less than 60 seconds will pass before the next alarm? (b) More than 30 seconds? (c) At least 45 seconds? .55 A passenger metal detector at Chicago’s Midway Airport gives an alarm 2. 0.6671 These calculations are illustrated in Figure 7. on Wednesday. and 4 p.00 1.50 1.50 0.3329 5 .50 2. FIGURE 7.25 Exponential Tail Areas for l 5 2.10) (7. P(X $ 0. We have P(X . or # in formula 7.3329. patient insurance inquiries arrive at Blue Choice insurance at a mean rate of 2.3 per day.25.00 2. false alarms are received at the downtown fire station at a mean rate of 0.m.50 minute) will be needed before the next call arrives is P(X # 0. What is the probability of waiting more than 30 seconds for the next call? We set 5 2.50) 2.50) 5 1 2 e2(2. we could use either . EXAMPLE 7.50 is a point that has no area in a continuous model. P(X # x) 5 1 2 e2 P(X .50 1.2 P(X 2.50 .2 calls per minute. while the probability of waiting x units of time or less is 1 2 e2 x. a binomial model. or 33. 0.50 P(X 0. (a) What is the probability that more than 7 days will pass before the next false alarm arrives? (b) Less than 2 days? (c) Explain fully.00 1.50) 5 e2 x 5 e2(2.50 2. and the units of measurement must be the same. 7. 25 so e2 x 5 . we would expect the mean waiting time to be above the median. What is the probability that (a) at least 30 seconds will pass before the next customer walks in.00 1. (b) no more than 15 seconds.2 x 5 0.302585y2. During the summer they need to replace their tank on average every 30 days.50 Minutes 2.2 5 0. we want the 90th percentile for waiting time (the top 10 percent of waiting time) as illustrated in Figure 7.2 calls per minute.12 show that the mean waiting time is 1y 5 1y2. At a randomly chosen moment. We want to find the x-value that defines the upper 10 percent.50 1.00 .2876821y2. take the natural logarithm of both sides.386294y2.00 1.302585 x 5 2.50 . (c) more than 1 minute? Inverse Exponential We can use the exponential area formula in reverse.0466 minutes (or 62.0466 minutes So 90 percent of the calls will arrive within 1.56 The Johnson family uses a propane gas grill for cooking outdoors. (b) no more than 20 days? 7.10.50 5 ln(.90 implies P(X .9 seconds Quartiles for Exponential with 2. or 37.26 Finding x for the Upper 10 Percent Call the unknown time x.10 2 x 5 ln(. .Chapter 7 Continuous Probability Distributions 281 7. which it is.6931472y2.75 5 .2 5 0. It is instructive to note that the median waiting time (18.57 At a certain Noodles & Company restaurant. what is the probability that they can grill out (a) at least 40 days before they need to replace their tank.2)x x x 5 1 2 e2 x 5 . 50 percent. and solve for x: P(X # x) 5 1 2 e2 x 5 .25 5 ln(.26.2 Third Quartile Q3 P (X # x) so e2 x 2 x 2(2. For example.8 per minute.6301 minute.75) 2(2.10 LO 7-8 Solve for x for a given exponential probability.50 . x) 5 .25) 5 21. or 18.10. we set the right-tail area to .8 seconds).50) 5 20.2)x 5 22.12 illustrates similar calculations to find the quartiles (25 percent.9 seconds The calculations in Table 7.2)x x x 5 1 2 e2 x 5 .2 x 5 1. TABLE 7. customers arrive during the lunch hour at a rate of 2.1308 minute.9 seconds) is less than the mean.4545 minute.386294 5 1.50 2. FIGURE 7. We can find any percentile in the same way.2876821 x 5 0. or 27 seconds. 75 percent) of waiting time.6931472 5 0. Since P(X # x) 5 .75 2 x 5 ln(.90 .10) 2(2.2)x 5 20.3151 minute.00 .50 5 .00 2.90 so e2 x 5 .50 1.8 seconds Second Quartile Q2 (median) P (X # x) so e2 x 2 x 2(2. 2. If the mean arrival rate is 2.12 First Quartile Q1 P (X # x) 5 1 2 e2 x 5 . or 7. Since the exponential distribution is highly rightskewed.2 5 0. Table 7.00 . 00 0 Finding x for the Lower 30 Percent .500 failures per hour.04 .20 hour) 5 e2(3)(0.6 Flat-Panel Displays The NexGenCo color flat-panel display in an aircraft cockpit has a mean time between failures (MTBF) of 22. What warranty should be offered in order that not more than 30 percent of the GPS units will fail before the warranty expires? The situation is illustrated in Figure 7. FIGURE 7. We set P(X . P(X .14 . 10.08 . We could work a problem using either hours or minutes.000 hours of flight.500 flight hours.60.30 .02 .356675)y . P(X .60 is the same as P(X . then 5 1y20 5 0.000 flight hours? Since 22.10 .500 hours per failure implies 5 1y22. For example.0 arrivals per hour).70 and solve for x by taking the natural log of both sides of the equation: e2 x 5 .20) 5 e20. EXAMPLE 7. 0.27 . we calculate: P(X . we note that if 30 percent fail before the warranty expires.7 Warranty Period A manufacturer of GPS navigation receivers for boats knows that their mean life under typical maritime conditions is 7 years.27.000) 5 1 2 e2 x 5 1 2 e2(1y22. This assumes that failures follow the Poisson model. 70 percent will fail afterward. x) 5 e2 x 5 . if the mean time between patient arrivals in an emergency room is 20 minutes. we might be given 1y instead of .05)(12) 5 e20. To solve this problem.356675 x 5 (0.70 5 10 15 Years 20 25 Let x be the length of the warranty. MTBE 5 1y 5 mean time between events (units of time per event) 1yMTBE 5 5 mean events per unit of time (events per unit of time) For example.06 .500)(10.05 arrival per minute (or 5 3.000) 5 1 2 e20. as long as we are careful to make sure that x and are expressed in the same units when we calculate e2 x.4444 5 1 2 . x) 5 1 2 P(X # x) 5 1 2 0.12 .70) 2 x 5 20. What is the probability of a failure within the next 10.88 percent chance of failure within the next 10. 12 minutes) 5 e2(0. In other words.16 .70.30 5 .6412 5 .70 2 x 5 ln(. That is.282 Applied Statistics in Business and Economics Mean Time between Events Exponential waiting times are often described in terms of the mean time between events (MTBE) rather than in terms of Poisson arrivals per unit of time. EXAMPLE 7.3588 There is a 35. (b) Find the first quartile of waiting time before the next alarm. . x$0 Variable Type Discrete Continuous The exponential model may also remind you of the geometric model. 2. the models are different because the geometric model tells us the number of discrete events until the next success. So we plug in 5 1y7 5 0. (a) Find the median waiting time until the next telephone order.Chapter 7 Continuous Probability Distributions 283 But in this case. the firm would offer a 30-month warranty.58 The time it takes a ski patroller to respond to an accident call has an exponential distribution with an average equal to 5 minutes.13 Model Poisson Exponential Relation between Exponential and Poisson Models Random Variable X 5 number of arrivals per unit of time X 5 waiting time until next arrival Parameter (mean arrivals) 5 _____________ (unit of time) (mean arrivals) 5______________ (unit of time) Domain x 5 0. Both models depend solely on the parameter 5 mean arrival rate per unit of time.m. warranty periods are a policy tool used by business to balance costs of expected claims against the competitive need to offer contract protection to consumers. In general.DIST(x. 7. . Using Excel The Excel function =EXPON. SECTION EXERCISES .13. automobiles typically outlast their warranty period (although competitive pressures have recently led to warranties of 5 years or more.1) will return the left-tail area P(X # x). These two closely related distributions are summarized in Table 7. Every situation with Poisson arrivals over time is associated with an exponential waiting time. and midnight on Thursday night. while the exponential model tells the continuous waiting time until the next arrival of an event. Seven years mean time between failures is the same as saying 5 1y7 failures per year.142857) 5 2. It may seem paradoxical that such a short warranty would be offered for something that lasts 7 years. However. A few long-lived GPS units will pull up the mean. they are similar.5 time a minute. In spirit.2 telephone orders per hour. we are not given but rather its reciprocal MTBF 5 1y . Mystery Pizza gets an average of 4.356675)y(0.1428571 to finish solving for x: x 5 (0. 1. This is typical of electronic equipment. (c) Find the 30th percentile waiting time until the next alarm. The “1” indicates a cumulative area. .497 years Thus. the right tail is very long. (a) In what time will 90 percent of all ski accident calls be responded to? (b) If the ski patrol would like to be able to respond to 90 percent of the accident calls within 10 minutes. If you enter 0 instead of 1. even though it may result in a loss on a few warranties). 7. which helps explain why your laptop computer may have only a 1-year warranty when we know that laptops often last for many years. (c) What is the upper 10 percent of waiting time until the next telephone order? Show all calculations clearly. (a) Find the median waiting time until the next alarm. TABLE 7.60 A passenger metal detector at Chicago’s Midway Airport gives an alarm 0. you will get the height of the PDF instead of the left-tail area for the CDF. Similarly. Show all calculations clearly.Lambda. what does the average response time need to be? 7. (b) Find the upper quartile of waiting time before the next telephone order. However. which describes the number of items that must be sampled until the first success.59 Between 11 p. (a 1 c)y2.m. it has a mode or “peak. it is a simple distribution. Its X values must lie within the interval [a. Negatively skewed if b . useful in business what-if analysis. the triangular does not go on forever. (c) Find the upper quartile. since its X values are confined by a and c. c) or T(min.14 shows the characteristics of the triangular distribution.62 The mean life of a certain computer hard disk in continual use is 8 years. Practical model.28. (b) Explain why the median is not equal to the mean. But unlike the uniform. (a) Find the median wait for pizza order arrivals. TABLE 7.28 Triangular PDFs Skewed Left Symmetric Skewed Right a b c a b c a b c . 7. the mean time between arrival of telephone pizza orders is 20 minutes. (a 1 c)y2.284 Applied Statistics in Business and Economics 7. and 4 a. But unlike the normal.14 Triangular Distribution Parameters PDF LO 7-9 Use the triangular distribution for “what-if” analysis (optional). c]. It can be symmetric or skewed. as you can see in Figure 7. 2 2 2 FIGURE 7. (a) How long a warranty should be offered if the vendor wants to ensure that not more than 10 percent of the hard disks will fail within the warranty period? (b) Not more than 20 percent? 7. Visually. mode. A symmetric triangular is the sum of two identically distributed uniform variates.61 Between 2 a. at an all-night pizza parlor. max).m. CDF Domain Mean Standard deviation Shape Comments a 5 lower limit b 5 mode c 5 upper limit 2(x 2 a) f (x) 5 ____________ for a # x # b (b 2 a)(c 2 a) 2(c 2 x) f (x) 5 ____________ for b # x # c (c 2 a)(c 2 b) (x 2 a)2 P (X # x) 5 ____________ for a # x # b (b 2 a)(c 2 a) (c 2 x)2 P (X # x) 5 1 2 ____________ for b # x # c (c 2 a)(c 2 b) a#x#c a1b1c _________ __________________________ 3 a 1 b 1 c 2 ab 2 ac 2 bc __________________________ 18 Positively skewed if b .” The peak is reminiscent of a normal. The triangular distribution is sometimes denoted T(a.7 TRIANGULAR DISTRIBUTION (OPTIONAL) Characteristics of the Triangular Distribution Table 7. b. which also has a single maximum. with 20 minutes as the most likely time.45. 0.1667 15 20 25 30 Special Case: Symmetric Triangular An interesting special case is a symmetric triangular distribution centered at 0.8 Anesthetic Effectiveness Using Triangular Distribution 5 5 a 1 b 1 c 2 ab 2 ac 2 bc _________________________ 18 __________________________________________ 2 2 2 15 1 20 1 30 2 (15)(20) 2 (15)(30) 2 (20)(30) __________________________________________ 18 2 2 2 5 3. the probability that the anesthetic takes less than 25 minutes is (30 2 25)2 P(X # 25) 5 1 2 _________________ 5 . Why is it different? Because the triangular. 0.8333 (30 2 15)(30 2 20) Basically. we are finding the small triangle’s area (1⁄2 base 3 height) and then subtracting from 1. b 5 20. Assuming a uniform distribution may seem conservative. 12. we obtain a 1 b 1 c 15 1 20 1 30 5 _________ 5 ____________ 5 21. 1). 1).29.30 compares these two distributions. but it could lead to patients sitting around longer waiting to be sure the anesthetic has taken effect.12 minutes Using the cumulative distribution function or CDF.45. If you set c 5 2.13) For example.12) (7.7 minutes 3 3 _________________________ EXAMPLE 7. we can calculate the probability of taking less than x minutes: (x 2 a)2 P(X # x) 5 _____________ for a # x # b (b 2 a)(c 2 a) (c 2 x)2 P(X # x) 5 1 2 ____________ (c 2 a)(c 2 b) for b # x # c (7.8333 .45. Yet over much of the range. making it more likely that a patient will be fully anesthetized within 25 minutes.45 # X # 12.45) always has values within the range 22.6667. has more probability on the low end. Unlike the normal N(0. assuming a uniform distribution with parameters a 5 15 and b 5 30 would yield P(X # 25) 5 .Chapter 7 Continuous Probability Distributions 285 An oral surgeon injects a painkiller prior to extracting a tooth. the triangular distribution T(22. Only experience could tell us which model is more realistic. Figure 7. In contrast. FIGURE 7.45.. with mode 20.g. This situation is illustrated in Figure 7. the dentist views the time for anesthesia effectiveness as a triangular random variable that takes between 15 minutes and 30 minutes. whose lower limit is identical to its upper limit except for sign (e. Given the varying characteristics of patients. and random samples from . the distributions are alike. from 2c to 1c) with mode 0 (halfway between 2c and 1c). the distribution T(22.45) closely resembles a standard normal distribution N(0. and c 5 30. 12.29 Triangular P(X 25) . Setting a 5 15. As shown in Figure 7. The triangular distribution is often used in simulation modeling software such as Arena. (d) Sketch the distribution and shade the area for the event in part (c). Bean has a distribution that is T(0. denoted N( . rather than sums.286 Applied Statistics in Business and Economics FIGURE 7. (c) Find the probability that an order will be less than $25.31. b. denoted U(a. (b) Find the standard deviation. but tables or Excel functions are available to find an area under the curve for given z-values or to find z-values that give a specified area (the “inverse normal”). 1) distribution. with mean 0 and standard deviation 1. Uses of the Triangular The triangular distribution is a way of thinking about variation that corresponds rather well to what-if analysis in business. a sales volume). 105). has two parameters a and b that enclose the range. an interest rate. (a) Find the mean. 25.45 T(22. It is a simple what-if model with applications in simulation. The cumulative distribution function (CDF) shows the area under the PDF to the left of X. It is more versatile than a normal because it can be skewed in either direction.45 0 2. (d) Sketch the distribution and shade the area for the event in part (c). In Chapter 18. The uniform continuous distribution. so a normal wouldn’t be much help. If the analyst can anticipate the range (a to c) and most likely value (b).45 *(RAND()+RAND()-1).64 Suppose that the distribution of oil prices ($/bbl) is forecast to be T(50. Its finite range and simple form are more understandable than a normal distribution. The area under the entire PDF is 1. 1). The mean E(X ) and variance Var(X ) are integrals. such distributions will be skewed. Yet it has some of the nice properties of a normal. 0.L. Many times. (c) Find the probability that the price will be greater than $75. we apply the transformation z 5 (x 2 )y to get a new random variable that follows a standard normal distribution. It serves as a benchmark. There is no simple formula for normal areas. ). The exponential distribution describes waiting time until . is symmetric and bell-shaped. the price of a raw material. as for a discrete random variable. denoted N(0. The triangular model is especially handy for what-if analysis when the business case depends on predicting a stochastic variable (e. 65. 12. It is easy to generate symmetric triangular random data in Excel by summing two U(0. a normal approximation for a binomial or Poisson probability is acceptable when the mean is at least 10. we will explore what-if analysis using the triangular T(a. (b) Find the standard deviation. such as a distinct mode.30 Symmetric Triangular Is Approximately Normal 2.g. b). SECTION EXERCISES 7.45. c) model in simulations. The normal distribution.45) are surprisingly similar to samples from a normal N(0. Because there is a different normal distribution for every possible and . CHAPTER SUMMARY The probability density function (PDF) of a continuous random variable is a smooth curve. it will be possible to calculate probabilities of various outcomes. and probabilities are areas under the curve. (a) Find the mean.. It has two parameters. 7. It is not surprising that business analysts are attracted to the triangular model. approaching 1 as X increases. the mean and standard deviation .63 Suppose that the distribution of order sizes (in dollars) at L. 1) random variables using the function =2. 75). c) has three parameters (a and c enclose the range. 257 KEY TERMS Commonly Used Formulas in Continuous Distributions x2a Uniform CDF: P(X # x) 5 ______ for a # x # b b2a x2 ______ for 2 . It is strongly right-skewed and is used to predict warranty claims or to schedule facilities. Triangular a. FIGURE 7. b. 1. 260 probability density function. b . The triangular distribution T(a.31 Normal . 1. 256 inverse normal. c (a 1 b 1 c)y3 (a2 1 b2 1 c 2 2 ab 2 ac 2 bc)y18 continuity correction. 276 continuous random variable. 255 cumulative distribution function. 282 normal distribution. 285 triangular distribution. x Set OK if 10 2 Relationships among Three Models 2 Set n n (1 ) OK if n 10 and n (1 ) 10 Binomial n. 271 mean time between events. b. and b is the mode). Useful as reference benchmark.15 compares these five models. Table 7. 255 standard normal distribution. 279 Gaussian distribution. Comparison of Models Parameters Mean (a 1 b)y2 Variance (b 2 a) y12 2 2 Characteristics Always symmetric Symmetric. … OK if n 20 and . Right-tail area is e2 x for waiting times. It may be symmetric or skewed in either direction. x 0. Its one parameter is (the mean arrival rate) and its right tail area is e2 x (the probability of waiting at least x time units for the next arrival). 262 symmetric triangular distribution.Chapter 7 Continuous Probability Distributions 287 the next Poisson arrival.05 TABLE 7. 255 exponential distribution.15 Model Uniform Normal Standard normal Exponential a. … . Special case of the normal with z 5 (x 2 )y 0 1y 1 1y 2 Always skewed right. n Set n x Poisson 0. 260 integral. . It is easy to visualize and is a useful model for what-if simulation. x . 1 Standard Normal Random Variable: z5 __________ Normal Approximation to Binomial: Normal Approximation to Poisson: 5n 5 5 5 2 x n (1 2 ) for n $ 10 and n(1 2 ) $ 10 __ for for x $ 0 $ 10 Exponential CDF: P(X # x) 5 1 2 e . 284 uniform continuous distribution. Useful for what-if business modeling. d. 5. skewness). and (d) explain why all normal distributions are alike despite having different and . (a) What is the mean swipe rate? (b) What is the standard deviation of the swipe rate? (c) What are the quartiles? (d) What percentage of subway riders must reswipe the card because they were outside the acceptable range? (Data are from The New York Times.66 Which of the following could be probability density functions for a continuous random variable? Explain.288 Applied Statistics in Business and Economics CHAPTER REVIEW 1. (a) What is the mean age of an applicant? (b) The standard deviation? (c) What is the probability that an applicant will be over 45? (d) Over 55? (e) Between 30 and 60? 7.. (a) Why does a point have zero probability in a continuous distribution? (b) Why are probabilities areas under curves in a continuous distribution? 2. (b) The annual income for a randomly chosen Major League Baseball player. 11. (a) Explain the difference between Appendix C-1 and Appendix C-2. (a) What features of a stochastic process might lead you to anticipate a normal distribution? (b) Give two examples of random variables that might be considered normal. f (x) 5 . and (d) explain when it would be used. (c) The annual hurricane losses suffered by homeowners in Florida. or else the card must be re-swiped through the card reader. You will use them often. (a) What is the transformation to standardize a normal random variable? (b) Why do we standardize a variable to find normal areas? (c) How does a standard normal distribution differ from any other normal distribution. (c) describe its shape. and (d) explain when it would be used. (b) PDF. Research shows that actual swipe rates by subway riders are uniformly distributed between 5 and 50 inches per second.70 Why might the following not be normally distributed? (a) The time it takes you to drive to the airport. List the standard normal z-values for several common areas (tail and/or middle). (b) indicate what the parameters represent.65 Which of the following is a continuous random variable? a. Hint: Would you expect a single central mode and tapering tails? Would the distribution be roughly symmetric? Would one tail be longer than the other? a. b. (c) describe its shape in a general way (e. Write an example of each of the four normal functions in Excel and tell what each function does. For the normal distribution: (a) tell how many parameters it has. Weight of contents of 16-ounce boxes of elbow macaroni. Problems with * are harder or based on optional material.50 for 0 # x # 2 b. p. f (x) 5 2 2 x for 0 # x # 2 c. 7. . c. 2004. (b) indicate what the parameters represent. Size of automobile collision damage claims. 4.68 Passengers using New York’s MetroCard system must swipe the card at a rate between 10 and 40 inches per second.g. 23.67 Applicants for a night caretaker position are uniformly distributed in age between 25 and 65. Diameters of randomly chosen circulated quarters. July 18. 10. Census short form. (c) Which table do you expect to use. CHAPTER EXERCISES Note: Show your work clearly. Number of Honda Civics sold in a given day at a car dealership.S. and (d) explain when it would be used. f(x) 5 .5x for 0 # x # 2 7. (b) indicate what the parameters represent. a. Define (a) parameter. c. and how is it similar? 7. 3. (c) describe its shape. When does the normal give an acceptable approximation (a) to a binomial and (b) to a Poisson? (c) Why might you never need these approximations? (d) When might you need them? 12. For the triangular distribution: (a) tell how many parameters it has. Amount of gasoline used for a 200-mile trip in a Honda Civic. (c) describe its shape. For the uniform distribution: (a) tell how many parameters it has.69 Discuss why you would or would not expect each of the following variables to be normally distributed. 9. For the exponential distribution: (a) tell how many parameters it has. 7. Time for households to complete the U. and (c) CDF. and why? (d) Why not always use Excel? 8.) 7. 7. b. Distance driven on a particular Thursday by the owner of a Honda Civic. 6. (b) indicate what the parameters represent. (b) List advantages of each type of table. 90 ppm? 7. Middle 80 percent 7. (a) Find the mean temperature.975 ounces and 2. At a given moment. Lowest 50 percent c. Find the value(s) of X for each event. Show your work.76 The length of a Colorado brook trout is normally distributed.73 The weekly demand for Baked Lay’s potato chips at a certain Subway sandwich shop is a random variable with mean 450 and standard deviation 80. the Li-ion Hitachi stick driver can drive 207 drywall screws on a single charge. the water temperature is a uniformly distributed random variable U(99. The manufacturer’s recommended correct inflation range is 28 psi to 32 psi. (a) What is the probability that a randomly chosen sample of glass will break at less than 579 MPa? (b) More than 590 MPa? (c) Less than 600 MPa? 7. (c) Tom’s score was 63.35.81 Procyon Manufacturing produces tennis balls. a.75 The amounts spent by customers at a Noodles & Company restaurant during lunch are normally distributed with a mean equal to $7.77 The caffeine content of a cup of home-brewed coffee is a normally distributed random variable with a mean of 115 mg with a standard deviation of 20 mg. (c) Find the 75th percentile for water temperature.74 The weekly demand for Papa Chubby’s pizzas on a Friday night is a random variable with mean 235 and standard deviation 10.72 Chlorine concentration in a municipal water supply is a uniformly distributed random variable that ranges between 0. Regulation tennis balls are required to have a weight between 1. (a) What is the probability that a brook trout’s length exceeds the mean? (b) Exceeds the mean by at least 1 standard deviation? (c) Exceeds the mean by at least 2 standard deviations? (d) Is within 2 standard deviations? 7. the time it takes a randomly chosen kernel of popcorn to pop is normally distributed with a mean of 140 seconds and a standard deviation of 25 seconds. 7. (b) Phyllis’s score was 93. . (b) Find the standard deviation of the temperature. Lowest 25 percent d. Middle 80 percent b. what time would you allow? (d) If you wanted 99 percent to pop? 7.78 The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 579 MPa with a standard deviation of 14 MPa. (a) What amount is the first quartile? (b) The second quartile? (c) The 90th percentile? 7. What is the probability that a cup of coffee will have less caffeine than a very strong cup of tea? 7.DIST function.S. If the standard deviation is 14 screws.00 and a standard deviation equal to $0. on average. Highest 25 percent d.79 Tire pressure in a certain car is a normally distributed random variable with mean 30 psi (pounds per square inch) and standard deviation 2 psi.85 ppm? (e) Will be between 0. (a) Bob’s score was 82.80 ppm and 0.71 Scores on a certain accounting exam were normally distributed with a mean of 75 and a standard deviation of 7. Their manufacturing process has a mean ball weight of 2. 90th percentile b.98 ppm.83 Tests show that. Find the percentile for each individual using Excel’s =NORM. What percentage of the kernels will fail to pop if the popcorn is cooked for (a) 2 minutes? (b) Three minutes? (c) If you wanted 95 percent of the kernels to pop. What proportion of Procyon’s production will fail to meet these specifications? 7.74 ppm and 0.107). Show your work. a. Highest 50 percent c.80 In a certain microwave oven on the high power setting. to warn of dangerously low tire pressure. (a) What is the mean chlorine concentration? (b) The standard deviation? (c) What is the probability that the chlorine concentration will exceed 0. How often would such an alarm be triggered? 7. (a) What is the probability that a randomly chosen cup of home-brewed coffee will have more than 130 mg of caffeine? (b) Less than 100 mg? (c) A very strong cup of tea has a caffeine content of 91 mg. (a) What is the probability that the tire’s inflation is within the recommended range? (b) What is the probability that the tire is underinflated? *(c) A company has developed a microchip that will warn when a tire is 25 percent below the recommended mean.095 ounces.80 ppm on a given day? (d) Will be under 0. Find the value(s) of X for each event. 7. The heater kicks in when the temperature falls to 998F and shuts off when the temperature reaches 1078F. 5th percentile 7. find the probability that Bob can finish his job without recharging.03 ounce.Chapter 7 Continuous Probability Distributions 289 7. Water temperature then falls slowly until the heater kicks in again.035 ounces with a standard deviation of 0.82 Shower temperature at the Oxnard Health Club showers is regulated automatically. Hint: Assume a normal distribution and treat the data as continuous. A motorist’s tire is inspected at random. Bob needs to drive 230 drywall screws. 84 The time it takes to give a man a shampoo and haircut is normally distributed with mean 22 minutes and standard deviation 3 minutes. what is the probability that the box will contain less than the advertised weight of 453 g? 7. and B 5 3 minutes. Customers are scheduled every 30 minutes. what is the probability that all three will be finished within their allotted half-hour times? 7.905 MW (megawatts) and a standard deviation of 355 MW.87 Demand for residential electricity at 6:00 p. 7.89 A statistics exam was given.50. while method B has a mean of 32 minutes and a standard deviation of 2 minutes.m. (a) Which route should he choose if he must be at the airport in 54 minutes to pick up his spouse? (b) Sixty minutes? (c) Sixty-six minutes? Explain carefully.85 The length of a time-out during a televised professional football game is normally distributed with a mean of 84 seconds and a standard deviation of 10 seconds.92 The amount of fill in a half-liter (500 ml) soft drink bottle is normally distributed. a. the utility can supply a maximum of 5. “Two machines fill 2-liter soft drink bottles by using a similar process. B 5 60 minutes.5 inches and a standard deviation of 1.995 ml and 5 3 ml.990 ml and 5 5 ml while Machine B has 5 1. Machine A has 5 1. X is in pounds and Y is in kilograms). If the network runs consecutive commercials totaling 90 seconds. 7.5). Calculate the percentile for each of the following four students. “If we see a standardized z-value beyond 63. The variables cannot both be normally distributed since they have different standard deviations. (a) Where should the mean be set to ensure a 95 percent probability that a half-liter bottle will not be underfilled? (b) A 99 percent probability? (c) A 99. What is the probability that the utility will have to purchase electricity from other utilities or allow brownouts? 7.200 MW at that time.g.94 Times for a surgical procedure are normally distributed. 140 is the threshold for high blood pressure.90 Are the following statements true or false? Explain your reasoning. (a) If Jim’s systolic blood pressure is taken at a randomly chosen moment. A 5 6 minutes. “If X and Y are two normally distributed random variables measured in different units (e.2 inches.88 Jim’s systolic blood pressure is a random variable with a mean of 145 mmHg and a standard deviation of 20 mmHg. what is the probability that it will be 135 or less? (b) 175 or more? (c) Between 125 and 165? (d) Discuss the implications of variability for physicians who are trying to identify patients with high blood pressure.93 The length of a certain kind of Colorado brook trout is normally distributed with a mean of 12. For Jim’s age group. There are two methods.91 John can take either of two routes (A or B) to LAX airport.9 percent probability? Explain. (a) Which procedure is preferred if the procedure must be completed within 28 minutes? (b) Thirty-eight minutes? (c) Thirty-six minutes? Explain your reasoning fully. (a) What is the probability that a male customer will take longer than the allotted time? *(b) If three male customers are scheduled sequentially on the half-hour. then it is not meaningful to compare the standardized z-values. Frieda’s z-score was 2. Mary’s z-score was 0.290 Applied Statistics in Business and Economics 7.62. At midday on a typical Wednesday the travel time on either route is normally distributed with parameters A 5 54 minutes.” *7.86 If the weight (in grams) of cereal in a box of Lucky Charms is N(470. d. The mean is adjustable. 7. (a) What is the probability that both exceed the mean? (b) Neither exceeds the mean? (c) One is above the mean and one is below? (d) Both are equal to the mean? . What minimum size limit should the Department of Natural Resources set if it wishes to allow people to keep 80 percent of the trout they catch? 7.” b. c.79. The process has a standard deviation of 5 ml.95 The length of a brook trout is normally distributed. 7. a. 7. on the first Monday in October in Santa Theresa County is normally distributed with a mean of 4. Zak’s z-score was 1.” c. Due to scheduled maintenance and unexpected system failures in a generating station. b. the variable cannot be normally distributed.. what is the probability that play will resume before the commercials are over? 7.48. Method A has a mean of 28 minutes and a standard deviation of 4 minutes. Two brook trout are caught. John’s z-score was 21. (a) What is the probability that a customer will be serviced in less than 3 minutes? (b) Why is your answer more than 50 percent? Shouldn’t exactly half the area be below the mean? 7.06.108 The distribution of scores on a statistics exam is T(50. 1) and copy the formula into 10 cells. *7.000 hours. (c) Find the probability that a condo price will be greater than $750K.000 hours but not more than 80. what is the probability of at least 50 twin births? (b) Fewer than 35? 7. (b) Find the standard deviation.109 The distribution of beach condominium prices in Santa Theresa ($ thousands) is T(500. (a) What is the probability that at least 25 cars will not start? (b) More than 40? 7.) 7. (a) If 200 seeds are planted. PROJECTS AND DISCUSSION 7.101 On a cold morning the probability is . occurring at a mean rate of 0.105 Suppose the average time to service a Noodles & Company customer at a certain restaurant is 3 minutes and the service time follows an exponential distribution. what is the probability that fewer than 60 false alarms are received? EXPONENTIAL DISTRIBUTION 7. (b) Find the mean and standard deviation of your sample of 10 random data values. (d) Sketch the distribution and shade the area for the event in part (c).80. (a) In 2. Assume that 1. Of the next 1. what is the probability that fewer than 150 will germinate? (b) That at least 150 will germinate? 7. the probability of a twin birth is . (c) What is the probability that the price will exceed 400? *7.100).asp.000 hours? (Product specifications are from www. . The MTBF for a single axis sensor is 400.000 hours. Are you satisfied that the random data have the desired mean and standard deviation? (c) Press F9 to generate 10 more data values and repeat question (b). Each question has four choices. (b) Find the standard deviation. (a) Find the mean.000 cars that pass through a particular neighborhood.000 board feet) of Douglas fir from western Washington and Oregon varies according to a triangular distribution T(300.Chapter 7 Continuous Probability Distributions 291 APPROXIMATIONS 7.02.000 hours? (b) Less than 50. (Product specifications are from www. In a sample of 100 rental cars. (a) What minimum score should be required to reduce the chance of passing by random guessing to 5 percent? (b) To 1 percent? (c) Find the quartiles for a guesser. (a) Find the mean.000 live deliveries.000 hours? (c) At least 50.107 The price (dollars per 1. false alarms are received at a mean rate of 0.99 A multiple-choice exam has 100 questions.) TRIANGULAR DISTRIBUTION *7.500 cars are started each cold morning. (c) Find the probability that a score will be less than 75. (a) Find the probability that a sensor lasts at least 30 years.com.100 The probability that a certain kind of flower seed will germinate is . The sensors use a one-piece. the probability that a rental car is from Hertz is 25 percent. 350. what are the first and third quartiles for the number of car accidents in this neighborhood? 7. and aerospace and defense applications. what is the probability that the DVD writer will last more than 100. (d) Sketch the distribution and shade the area for the event in part (c). assuming continuous operation. 700.102 At a certain fire station.1 claim per month. 2. In a year. (a) Find the mean. (b) Find the standard deviation. com/techsupp_A. micromachined inertial sensing element to measure angular rotational velocity or linear acceleration. commercial/ industrial. 7. 490). (b) Would you be surprised if a sensor has failed within the first 3 years? Explain.96 Among live deliveries.97 Nationwide.98 The probability of being in a car accident when driving more than 10 miles over the speed limit in a residential neighborhood is .systron. (a) What is the probability that the dealership will wait at least 6 months until the next claim? (b) At least a year? (c) At least 2 years? (d) At least 6 months but not more than 1 year? 7. 95). 60.106 Systron Donner Inertial manufactures inertial subsystems for automotive.hp.02 that a given car will not start in the small town of Eureka. (a) Assuming continuous operation. what is the probability that fewer than 20 are from Hertz? 7.104 Automobile warranty claims for engine mount failure in a Troppo Malo 2000 SE are rare at a certain dealership.2 per day.103 The HP dvd1040i 20X Multiformat DVD Writer has an MTBF of 70.110 (a) Write an Excel formula to generate a random normal deviate from N(0. 05. A Primer on Statistical Distributions.111 (a) Write an Excel formula to generate a random normal deviate from N(4000. March 19. Nicholas Hastings. 200) and copy the formula into 100 cells. N. (b) P(S | G). P(B) 5 . Nevzorov. 191. are A and B independent events? Explain. Does it appear normal? 7. b. If A and B are mutually exclusive events. and 189. 10A.112 On a police sergeant’s examination. Statistical Distributions. 2. B. The number of permutations of 5 things taken 2 at a time is 20. B) 5 . 171. classical. (b) Do you think there was sufficient reason to question these four exam scores? What assumptions are you making? (Data are from Detroit Free Press.mhhe. then the odds against event A’s occurrence are 19 to 1. CHAPTER 7 Online Learning Resources The Online Learning Center (OLC) at www. or you may decide to download the ones that sound interesting.25. and V. If P(A) 5 . c. and Brian Peacock. This led to allegations of irregularity in the exam. 2011. 4th ed. and P(A 4. Which type of probability (empirical. Are you satisfied that the random data have the desired mean and standard deviation? (c) Make a histogram of your sample. on the test. a. respectively. (b) Find the mean and standard deviation of your sample of 100 random data values. find (a) P(H R G H Col Total 10 20 30 S 50 50 100 T ). On a given Friday. the historical mean score was 80 with a standard deviation of 20.292 Applied Statistics in Business and Economics 7.30. then P(A B) 5 0.7%. Topic Calculations Normal approximations Random data Tables LearningStats Demonstrations Normal Areas Probability Calculator Evaluating Rules of Thumb Why the Rule of 10? Random Continuous Data Visualizing Random Normal Data Table C—Normal Probabilities Key: = Excel = Word EXAM REVIEW QUESTIONS FOR CHAPTERS 5–7 1.com/doane4e has several LearningStats demonstrations to help you understand continuous probability distributions. Your chance of going to Disney World next year is 10%. The chance of rolling a 3 on two dice is 1y8.. subjective) is each of the following? a. For the following contingency table. (a) Convert these four officers’ scores to standardized z-values. 1999. 2003. Wiley.) RELATED READING Balakrishnan.70. Forbes. c. b. Wiley. Your instructor may assign one or more of them. Catherine. If P(A) 5 . p. Merran Evans. Four officers who were alleged to be cronies of the police chief scored 195. Which statement is false? Explain. the probability that Flight 277 to Chicago is on time is 23. (c) P(S) T 30 40 70 Row Total 90 110 200 3. . b.60. we add up the probabilities at each point. $5. Assuming independent arrivals with a mean of 2. Which statement is incorrect? Explain. a. The uniform distribution has two parameters. =NORM. (b) the first quartile of waiting times.000) with probabilities . c. and . (c) between 1 and 2 minutes. In the previous problem.700 7. (b) the lowest 10 percent of speeds. find the probability that in a given minute there will be (a) exactly 2 arrivals. If a random experiment whose success probability is .20 is repeated 8 times. find the probability that the speed of a randomly chosen vehicle (a) exceeds 78 mph. find (a) the 95th percentile of waiting times (i. (b) more than 3 successes. c. 60)? a. c. (c) is less than 70 mph.000. (b) more than 30 seconds. 12. Which of the following Excel formulas would be a correct way to calculate P(X . find the probability of (a) exactly 3 successes. The number of dimes older than 10 years in a random sample of 8 dimes. Which statement is false? Explain. binomial. Which statement is false? Explain. We use the geometric distribution to find probabilities of arrivals per unit of time. The triangular cannot be skewed left or right. 450) given that X is N(500. . (c) the mean time between arrivals.Chapter 7 Continuous Probability Distributions 293 5. c. sampling is done without replacement.2 arrivals per minute. If freeway speeds are normally distributed with a mean of 5 70 mph and 5 7 mph. The last digit of a randomly chosen student’s Social Security number. 60) c.300 c. find the expected value. 13. The triangular always has a single mode. b. a. c. Poisson) is most nearly appropriate to describe each situation (assuming you knew the relevant parameters)? a. When no tables are available. Which statement is true for a normal distribution? Why not the others? a. (c) fewer than 4 arrivals. 16.619. A uniform PDF is constant for all values within the interval a # X # b. To find probabilities in a continuous distribution. 18. The binomial distribution assumes dependent random trials. In the hypergeometric distribution. The shape of the PDF is always symmetric regardless of and . In a random experiment with 50 independent trials with constant probability of success . =1–NORM. The Poisson distribution has two parameters. 500. In the previous problem..5 minutes. The mean of the triangular is (a 1 b 1 c)y3. (b) is between 65 and 75 mph. 6. The mean of the uniform distribution is always (a 1 b)y2. If the payoff of a risky investment has three possible outcomes ($1. areas may be found by a simple formula. 60.5 arrivals per minute. 10. $1. $2. Which statement is correct concerning the normal approximation? Why not the others? a. calculate (a) the 95th percentile of vehicle speeds (i. (b) at least 3 arrivals. a.10 respectively. (c) the highest 25 percent of speeds (3rd quartile).e. (c) at most 2 successes.DIST(450. The shape of the CDF resembles a bell-shaped curve. 15. $2.DIST(450. 11.000. 14. Which statement is true? Why not the others? a. 95 percent below).DIST(450. If arrivals follow a Poisson distribution with mean 1. Normal approximations are needed since Excel lacks discrete probability functions. 500.30.500 b. 95 percent below). =NORM. a. b.e. find the mean and standard deviation of the number of successes. $1.21) has mean 13 and standard deviation 4.S. find the probability that the waiting time until the next arrival will be (a) less than 1. 1) b. The normal Poisson approximation is acceptable when $ 10. The number of hospital patients admitted during a given minute on Tuesday morning. 20. 19. 0) 17. A uniform continuous model U(5.30. (d) Which probability distribution did you use and why? 9. c. c. 60. . b. Which probability distribution (uniform. b. The normal binomial approximation is better when n is small and is large. (d) Which probability distribution did you use and why? 8.. b. b.