Math 115 Probability & Statistics Section 1.3 HW Section 1.3 Exercises (Solutions) 1.109, 1.110, 1.111, 1.114*, 1.115, 1.119*, 1.122, 1.125, 1.127*, 1.128*, 1.131*, 1.133*, 1.135*, 1.137*, 1.139*, 1.145*, 1.146 - 148. 1.109 Sketch some normal curves. (a) Sketch a normal curve that has mean 10 and standard deviation 3. (b) On the same x axis, sketch a normal curve that has mean 20 and standard deviation 3. (c) How does the normal curve change when the mean is varied but the standard deviation stays the same. Solution (a) It is easiest to draw the curve first, and then mark the scale on the axis. (b) Draw a copy of the first curve, with the peak over 20. (c) The curve has the same shape, but is translated left or right. The curve has the same shape, but is translated left or right. 1.110 The effect of changing the standard deviation. (a) Sketch a normal curve that has mean 10 and standard deviation 3. (b) On the same x axis, sketch a normal curve that has mean 10 and standard deviation 1. (c) How does the normal curve change when the standard deviation is varied but the mean stays the same. Solution (a) As in the previous exercise, draw the curve first, and then mark the scale on the axis. (b) In order to have a standard deviation of 1, the curve should be 1/3 as wide, and three times taller. (c) The curve is centered at the same place (the mean), but its height and width change. Specifically, increasing the standard deviation makes the curve wider and shorter; decreasing the standard deviation makes the curve narrower and taller. (b) Single peak and skewed to the left. 1. (a) Using these values for µ and σ. (b) If the grading policy is to give grades of A to the top 15% of scores based on the normal distribution with mean 70 and standard deviation 10.Math 115 Probability & Statistics Section 1. Sketches will vary. but with two peaks (that is. two strong clusters of observations).111. Solution 1.114 Total scores. what is the cut-off for an A in terms of a standardized score? (c) Which students earned a grade of A on the final exam for this course? .3 HW 1. Here are the total scores of 10 students in an introductory statistics course. Sketch density curves that might describe distributions with the following shapes: (a) Symmetric. standardize the first exam scores of these 10 students.111 Know your density. Previous experience with this course suggests that these scores should come from a distribution that is approximately normal with mean 70 and standard deviation 10. next 40% receive C.2 54 -1. (a) Why is the total area under this curve equal to 1? (b) What proportion of the observations lie below 0. From Table A. (b) The cut-off for an A is the 85th percentile for the N(0. (c) Do you think that this method of assigning grades is a good one? Give reasons for your answer. These cut-offs are based on the N(70. 1. and 85th percentiles for a N(0. Use areas under this density curve to answer the following questions. 55th. (68−70)/10 = −0. (c) The top two students (with scores of 92 and 98) received A’s.5 73 0.0364. Figure 1. (b) To convert to actual scores. If you ask a computer to generate “random numbers” between 0 and 1.3 98 2. this is approximately 1. The grading policy says the cut-offs for the other grades correspond to the following: bottom 5% receive F.35? .Math 115 Probability & Statistics Section 1.3 HW Solution 1. 1) distribution. 1) distribution. software gives 1.6 92 2. the point where one’s grade drops from an A to a B.2. The complete list is given below. (a) For example. Refer to the previous exercise. These cut-offs are the points where one’s grade jumps up. (a) Give the cut-offs for the grades in this course in terms of standardized scores.115. 10) distribution. this is only an issue for a score that falls exactly on the border between two grades. (c) Opinions will vary.5 80 1 70 0 1. 68 -0. take the standardscore cut-off z and compute 10z + 70.6 55 -1. (a) We need the 5th.04.115 Assign more grades. 15th. (b) Give the cut-offs in terms of actual total scores. These are given in the table below. and next 30% receive B. Note: The cut-off for an A given in the previous solution is the lowest score that gets an A—that is. Solution 1.34 graphs the density curve for a uniform distribution.2 75 0.8 64 -0. you will get observations from a uniform distribution. next 10% receive D.114. In practice.116 A uniform distribution. each with three points marked on it.34 The density curve of a uniform distribution.116 .Math 115 Probability & Statistics Section 1.116. for Exercise 1. Figure 1. and the quartiles.35 displays three density curves. the median.3 HW (c) What proportion of the observations lie between 0. 1.1.35 and 0.34? What are the quartiles? Solutions (1.118) 1. At which of these points on each curve do the mean and the median fall? .118 Find the mean. What are the mean and the median of the uniform distribution in Figure 1.119 Three density curves.65? Figure 1. Solution . median is B (the right skew pulls the mean to the right).120 Length of pregnancies. (a) Mean is C. (c) Mean A. Draw a density curve for this distribution on which the mean and standard deviation are correctly located.119. median B (the left skew pulls the mean to the left). Solution 1. for Exercise 1.119. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. median A.3 HW Figure 1. 1. (b) Mean A.35 Three density curves.Math 115 Probability & Statistics Section 1. (b) About 16% are longer than 339 days since 339 days or more corresponds to at least one standard deviation above the mean. Bigger animals tend to carry their young longer before birth.24% (100/105) are in the range x ± 2s = 4. (a) 99. How well do these scores satisfy the 68–95–99. It will hold only approximately for actual data.7 rule? To find out.35 to 6. (b) The shortest 2.04.50. The mean and standard deviation are x = 5.125 Acidity of rainwater.7%) horse pregnancies fall in what range of lengths? (b) What percent of horse pregnancies are longer than 339 days? Solution 1. About 67.7 rule to answer the following questions. Use the 68–95–99. The Normal quantile plot in Figure 1.32 (page 66) shows that the acidity (pH) measurements for rainwater samples in Exercise 1. The length of horse pregnancies from conception to birth varies according to a roughly Normal distribution with mean 336 days and standard deviation 3 days.36 are approximately Normal.81 to 7.7 rule.123 Horse pregnancies are longer. calculate the mean and standard deviation s of the observations.5% last? Solution 1.6476) of the pH measurements are in the range x ± s = 4.96.62% (71/105 = 0. Use the 68–95–99. (a) The middle 95% fall within two standard deviations of the mean: 266 ± 2(16).7% of horse pregnancies fall within three standard deviations of the mean: 336 ± 3(3).122 Pregnancies and the 68–95–99.123.89 to 5.4256 and s = 0.7 rule to answer the following questions. All (100%) are in the range x ± 3s = 3. About 95.7 rule is exact for any theoretical Normal distribution.Math 115 Probability & Statistics Section 1. See the sketch of the curve in the solution to Exercise 1.122. . 1.5379. (The 68–95–99. Then calculate the percent of the 105 measurements that fall between − s and + s and compare your result with 68%.125. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. 1. (a) Almost all (99.) Solution 1. or 234 to 298 days.5% of all pregnancies? How long do the longest 2.120. or 327 to 325 days. Do the same for the intervals covering two and three standard deviations on either side of the mean.3 HW 1.5% of pregnancies are shorter than 234 days (more than two standard deviations below the mean). (a) Between what values do the lengths of the middle 95% of all pregnancies fall? (b) How short are the shortest 2. 3 HW 1. (b) Z ≥ −1.0495.8 (b) Z ≥ −1. In each case. find the proportion of observations from a standard Normal distribution that satisfies each of the following statements.0359. (b) Z < 1.9093.9505 − 0.76 < Z < 1.6 Solution 1.9641. sketch a standard Normal curve and shade the area under the curve that is the answer to the question.65: 0. In each case.9452 − 0. (d) −1. Using either Table A or your calculator or software. (a) Z ≤ −1.7269. .76: 0.9505. 1.126 Find some proportions. (a) Z > 1. Using either Table A or your calculator or software.2236 = 0.65 (c) Z > −0.0548. sketch a standard Normal curve and shade the area representing the proportion. Using values from Table A: (a) Z > 1.8: 0.6 (d) −1.6: 0.8 < Z < 1.8 (c) Z > 1.65: 0. Using values from Table A: (a) Z ≤ −1.6: 0.76 (d) −0.65: 0. (c) Z > 1.8: 0. (c) Z > −0.127.126.0359 = 0.8 < Z < 1.65 Solution 1.7764. find the proportion of observations from a standard Normal distribution for each of the following events.Math 115 Probability & Statistics Section 1.76 < Z <1. (d) −0.65 (b) Z < 1.127 Find more proportions. (b) 40% of the observations fall above z. report the value of z that comes closest to satisfying the condition. Solution 1. (This is the 22nd percentile of the standard Normal distribution. .) In each case.7722. (a) 22% of the observations fall below z.Math 115 Probability & Statistics Section 1.2533 (the 60th percentile of the standard Normal distribution). sketch a standard Normal curve with your value of z marked on the axis.) (b) 40% of the observations fall above 0.3 HW 1.128 Find some values of z. (If you use Table A.128. Find the value z of a standard Normal variable Z that satisfies each of the following conditions. (a) 22% of the observations fall below −0. 129.141 are based on this information. the ACT and the SAT.65 (that is. so about 2. The variable Z has a standard Normal distribution.1257 is the 55th percentile). (b) If z = 0. Exercises 1.45. Solution 1.130 Find some values of z.Math 115 Probability & Statistics Section 1.132 to 1.” requires a WAIS score of 130 or higher for membership.3 HW 1. The scale of scores is set separately for each age group and is approximately Normal with mean 100 and standard deviation 15. 0.45 (0.1257.5% (half of the outer 5%) of adults would have WAIS scores below 70. 70 is two standard deviations below the mean (that is. 1. (b) Find the number z such that the event Z > z has proportion 0. ACT scores are reported on a scale from 1 to 36. The scale of scores is set separately for each age group and is approximately Normal with mean 100 and standard deviation 15.3853 is the 65th percentile of the standard Normal distribution). The SAT scores are approximately Normal with mean µ = 1509 and standard deviation σ = 321.130. . applying for Social Security disability benefits.131 High IQ scores. for example. (a) z = 0. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test.3853 has cumulative proportion 0.129 Find more values of z. People with WAIS scores below 70 are considered mentally retarded when.5 and standard deviation σ = 5. it has standard score z = −2). What percent of adults are retarded by this criterion? Solution 1. SAT scores are reported on a scale from 600 to 2400.4. What percent of adults would qualify for membership? There are two major tests of readiness for college. which calls itself “the high IQ society. then Z > z has proportion 0. The organization MENSA. 1. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test.65. The distribution of ACT scores are approximately Normal with mean µ = 21. (a) Find the number z that has cumulative proportion 0. Solution 1.3889.5% of adults would score at least 130.5234. while Jermaine’s score corresponds to z = (29−21. Maria’s score standardizes to z = (30−21. who has the higher score? Report the z-scores for both students. who has the higher score? Report the z-scores for both students. Maria scores 30 on the ACT. Jermaine scores 29 on the ACT. 130 is two standard deviations above the mean (that is. Emily scores 1020 on the SAT. 1.135 Find the SAT equivalent. while Emily’s score corresponds to z = (1020−1509)/321 = −1. ACT scores are reported as whole numbers.5)/5.4 =1.3 HW Solution 1. Jacob’s score is higher. 1.Math 115 Probability & Statistics Section 1. Jacob scores 16 on the ACT. Assuming that both tests measure the same thing.133.5)/5. Assuming that both tests measure the same thing. so an equivalent ACT score is 21. so an equivalent SAT score is 1509 + 1.134.132.7788.9688.4 = 31. it has standard score z = 2).5)/5. Assuming that both tests measure the same thing. Assuming that both tests measure the same thing. . Solution 1. Jermaine’s score is higher. what score on the ACT is equivalent to Jose’s SAT score? Solution 1.5 + 1. (Of course.131.) 1.1. Jose scores 2080 on the SAT.4 = 1. Tonya’s score standardizes to z = (1820−1509)/321 = 0.4 = −1. Jose’s score standardizes to z = (2080−1509)/321 = 1. so this would presumably be a score of 31.135.132 Compare an SAT score with an ACT score.5741.7788 × 5. Tonya scores 1820 on the SAT. what score on the SAT is equivalent to Maria’s ACT score? Solution 1. 1.133 Make another comparison. Jacob’s score standardizes to z = (16−21.5741 × 321 = 2014.134 Find the ACT equivalent.0185. so about 2. 140. What is her percentile? Solution 1.5 ± 0.3228. 1. What are the quartiles of the distribution of ACT scores? Solution 1. The quartiles of a Normal distribution are ±0. 1.Math 115 Probability & Statistics Section 1.3 percentile.136. Maria’s score standardizes to z = (2090−1509)/321 = 1. they are 21.137.8416.25 and 0. so for ACT scores.75.139.9 to 25.8416 × 321 =1239 on the SAT.138. for which Table A gives 0.5)/5. 1239 and below: The bottom 20% corresponds to a standard score of z = −0. Her score is the 96. which in turn corresponds to a score of 1509 + 1. The quartiles of any distribution are the values with cumulative proportions 0.4 = 17. 1920 and above: The top 10% corresponds to a standard score of z = 1.5 percentile.6745 × 5.4 = −0. What is his percentile? Solution 1.2816 × 321 = 1920 on the SAT. for which Table A gives 0. Reports on a student’s ACT or SAT usually give the percentile as well as the actual score.136 Find an SAT percentile. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than this one.1.140 Find the ACT quartiles. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than this one. Jacob scores 19 on the ACT. 1.3 HW 1. 1. Maria scores 2090 on the SAT. His score is the 32.137 Find an ACT percentile.138 How high is the top 10%? What SAT scores make up the top 10% of all scores? Solution 1. .4630. Reports on a student’s ACT or SAT usually give the percentile as well as the actual score. Jacob’s score standardizes to z = (19−21.6745 standard deviations from the mean.2816. .9649. which in turn corresponds to a score of 1509 − 0.139 How low is the bottom 20%? What SAT scores make up the bottom 20% of all scores? Solution 1.81. 15.95%).9677.4412.8416×321 = 1779.6.20.85% (software: 51. The quintiles of any distribution are the values with cumulative proportions 0. women aged 20 and over have a mean HDL of 55 mg/dL with a standard deviation of 15.5 mg/dl: (a) 40 mg/dl standardizes to z = (40−55)/15.2533×321 = 1590. What proportion are in this category? Solution 1.99%).Math 115 Probability & Statistics Section 1. neither very good or very bad.95% (software: 45. HDL cholesterol levels for men have a mean of 46 mg/dL with a standard deviation of 13.88%). Assume that the distribution is Normal.60. 1. Using Table A. 0. 1.40.0294. Using Table A. 1509−0.8416 × 321 = 1239. (a) What percent of women have low values of HDL (40 mg/dL or less)? (b) HDL levels of 60 mg/dL are believed to protect people from heart disease. For a Normal distribution with mean 55 mg/dl and standard deviation 15. 33% of men fall below this level (software: 32.80. (b) 60 mg/dl standardizes to z = (60−46)/13.143. so about 52% of men fall in the intermediate range. 37. people over the age of 20 years should have at least 40 mg/dL of HDL cholesterol. The quintiles of the SAT score distribution are 1509 − 0.36 U. 0. 1509+0.5 = −0. Solution 1.142 Do you have enough “good cholesterol?” High-density lipoprotein (HDL) is sometimes called the “good cholesterol” because low values are associated with a higher risk of heart disease. (b) 60 mg/dl standardizes to z = (60−55)/15.S. What percent of women have protective levels of HDL? (c) Women with more than 40 mg/dL but less than 60 mg/dL of HDL are in the intermediate range. 16. and 1509+0.2533×321 = 1428.15 (c) Subtract the answers from (a) and (b) from 100%: Table A gives 51.5 = 0.66%).142. For a Normal distribution with mean 46 mg/dl and standard deviation 13. According to the American Heart Association.3 HW 1. Using Table A. so about 46% of women fall in the intermediate range. .60% of women fall below this level (software: 16. and 0.5 mg/dL. What are the quintiles of the distribution of SAT scores? Solution 1.141.6 = 1.45 (c) Subtract the answers from (a) and (b) from 100%: Table A gives 45.6 = −0.6 mg/dl: (a) 40 mg/dl standardizes to z = (40−46)/13.3226.141 Find the SAT quintiles.143 Men and HDL cholesterol. Using Table A. Answer the questions given in the previous exercise for the population of men. σ) distribution. (a) The quartiles for a standard Normal distribution are ±0. (b) About 54. write an equation that gives the quartiles of the N(µ.21%.6745σ.= 255.25.75.146 Quartiles for Normal distributions.84. subtracting the answer from part (a) leaves about 54.47. . (c) For human pregnancies. (c) About 279 days or longer: Searching Table A for 0.6745.44.625 < z < 0. The area to the left of 0. which corresponds to x > 266 + 0.25 is 0.146. (a) What percent of pregnancies last less than 240 days (that’s about 8 months)? (b) What percent of pregnancies last between 240 and 270 days (roughly between 8 months and 9 months)? (c) How long do the longest 20% of pregnancies last? Solution 1.63 and 5. (a) About 5.6745σ and Q3 = µ + 0.145 Length of pregnancies.26% for −1.6745 × 16 .80 leads to z > 0.Math 115 Probability & Statistics Section 1. Apply your result from (b): what are the quartiles of the distribution of lengths of human pregnancies? Solution 1.8416 gives x > 279.2 and Q3 = 266 + 0. Software (or averaging the two table values) gives 5.2%: x < 240 corresponds to z < −1.67455 × 16 = 276. (a) What are the quartiles of the standard Normal distribution? (b) Using your numerical values from (a).62.625. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days.8 days.16% for −1.7%: 240 < x < 270 corresponds to −1. σ) distribution in terms of µ and σ. (c) The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Q1 = 266 − 0.84(16) = 279. Table A gives 5.25 and 0.) 1.5987. (Using the software value z > 0.145.7%. (b) For a N(µ. Q1 = µ − 0. The quartiles of any distribution are the values with cumulative proportions 0.3 HW 1. 5 × IQR = 2. σ) distribution.147. 1.Math 115 Probability & Statistics Section 1. 0.6745σ.5 × IQR rule is the same for any Normal distribution. Q1 = µ − 0.0235σ.…. and above Q3 + 1.3 HW 1.6745σ.2816. . (a) What is the value of the IQR for the standard Normal distribution? (b) There is a constant c such that IQR = cσ for any Normal distribution N(µ. and IQR = 1. we found that for a N(µ.149 Deciles of Normal distributions. What is this percent? Solution 1. Therefore. What is the value of c? Solution 1.6745. σ).5 × IQR = µ + 2. 1. Continue your work from the previous exercise. 20th.12 ounces and standard deviation 0.3490.148 Outliers for Normal distributions.698σ. the quartiles are Q1 = µ − 0. The percentage outside of this range is 2 × 0.5 × IQR = µ − 2. the first and last deciles are µ − 1.698σ. The first and last deciles are the 10th and 90th percentiles. 90th percentiles.31 ounces.15) distribution.6745σ and Q3 = µ + 0. (a) As the quartiles for a standard Normal distribution are ±0. respectively.3490: For a N(µ. The deciles of any distribution are the 10th.147 IQR for Normal distributions.2816σ = 9. (a) What are the first and last deciles of the standard Normal distribution? (b) The weights of 9-ounce potato chip bags are approximately Normal with mean 9.15 ounce.148. Q3 = µ + 0.6745σ. Continue your work from the previous two exercises. σ) distribution. What are the first and last deciles of this distribution? Solution 1.2816σ = 8.70%. we have IQR = 1.149. and the suspected outliers are below Q1 − 1.12. (a) The first and last deciles for a standard Normal distribution are ±1. The percent of the observations that are suspected outliers according to the 1. In the previous two exercises. (b) For a N(9.3490σ.0035 = 0.93 and µ + 1. 1. (b) c = 1. The interquartile range IQR is the distance between the first and third quartiles of a distribution.