# 3.1. Sketches will vary. Use them to confirm that students understand the meaning of (a) symmetric and (b) skewed to the left.

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1 Chapter Solutions.1. Sketches will vary. Use them to confirm that students understand the meaning of (a) symmetric and (b) skewed to the left..2. (a) It is on or above the horizontal axis everywhere, and because it forms a x rectangle, the area beneath the curve is 1. (b) One-third of accidents occur in the first this is a x 1 rectangle, so the proportion is ~. (c) One-tenth of accidents occur next Sue s property: this is a x 0. rectangle, so the proportion is j.~ = 1.5 the obvious balance point of the rectangle. The median is also 1.5 because the distribution is symmetric (so that median = mean) and because half of the area lies to the left and half to the right of (a) Mean is C, median is B (the right skew pulls the mean to the right). (b) Mean B, median B (this distribution is symmetric). (c) Mean A, median B (the left skew pulls the mean to the left). Note: For each of the three curves, the highest point(s) of the curves are marked, but are neither the mean nor the median. Those points are the modes of those distributions; your students might have heard of this other measure of center, but for most of the data considered in this text, the mode is not as usefid as the mean and median..5. Students may at first make mistakes such as draw ing a half-circle instead of the correct bell-shaped curve, or being careless about locating the change-ofcurvature (inflection) point..6. Refer to the sketch in the solution to the previous exercise. (a) About 99.7% of male college students can run a mile in between.89 and 9. minutes three standard deviations above and below the mean: (0.7) =.89 to 9. minutes. (b) About 16% can run a mile in less that 6.7 minutes (one standard deviation below the mean); this is half of the approximately 2% of observations which do not fall within one standard deviation of the mean..7. (a) In 95% of all years, monsoon rain levels are between 688 and 1016 mm two standard deviations above and below the mean: (82) = 688 to 1016 mm. (b) The dryest 2.5% of monsoon rainfalls are less than 688 mm; this is more than 2 standard deviations below the mean Note: This exercise did not ask for a sketch of the Normal curve, but students should be encouraged to make such sketches anyway..8. Eleanor s standardized score is z = , and Gerald s standardized score is z Eleanor s score is higher. mile: to ,

2 Solutions First note that 6 ft is 72 inches. The z-scores are z~, = for women and = 72~% for men. The z-scores tell us that 6 feet is quite tall for a woman, but not at all extraordinary for a man..10. (a) (b) (c) (ci) = ~~.11. Let x be the monsoon rainfall in a given year. (a) x ~ 697 mm corresponds to z < , for which Table A gives = 2.9%. (b) 68 < x < 1022 corresponds to 6S~852 < z < IO2~852 that is, 2.06 < z < This proportion is = = 96.11%..12. Let x be the length of the thorax of a given fly. (a) x ~ 0,9 mm corresponds to z > 1.28, for which Table A gives I = (b) 0.9 < x < 1 corresponds to 1.28 < z < zz This proportion is = (a) Search Table A for 0.20: z 0.8 (a) (software gives 0.816). (b) Search Table A for 0.60: a 0.25 (software: 0.25)..1. The median is the same as the mean: mm. The quartiles are 0.67 standard deviations above and below the mean: ± (0.67)(0.078) and mm..15. (a) Income distributions are typically skewed to the right. In particular, there will be many countries where average income is low to moderate, but in a few countries, average income will be much higher..16. (a) Mean and standard deviation tell you center and spread, which is all you need for a Normal distribution..17. (b) The curve is centered at (b) Estimating a standard deviation is more difficult than estimating the mean, but among the three options, 2 is clearly too small, and 5 is clearly too large..19. (b) 266 ± 2(16) = 2 to 298 days..20. (c) 10 is two standard deviations above the mean, so 2.5% of adults have IQs of 10 or more.

3 80 Chapter The Normal Distributions.21. (a) Corinne s z-score is z = = (c) Table A shows an area of for z = (1,) Table A shows an area of below z = 0.75 (or 0.77 below z = 0.75)..2. (b) Corinne s standard score is z = 118doo = 1.2, for which Table A gives Sketches will vary, but should be some varia tion on the one shown here: the peak at 0 should be tall and skinny: while near 1, the curve should be short and fat? For each distribution, take the mean plus or minus two standard deviations. For mildly obese people, this is 7 ± 2(67) = 29 to 507 minutes. For lean people, this is 526±2(107) = 12 to 70 minutes is two standard deviations below the mean (that is, it has standard score z = 2), so about 2.5% (half of the outer 5%) of adults would have WAIS scores below (a) (b) I = (c) I = (d) = ~-~~Ô* ~.29. (a) Search Table A for 0.80: z 0.8 (software: 0.816). (b) Search Table A for 0.65: z 0.9 (software: 0.85)..0. The proportion of students who could run a mile in 5 minutes or less is about : x <5 corresponds to z < 2.85, for which Table A gives About : The proportion of rainy days with rainfall ph below 5.0 is about : x <5.0 corresponds to z < 0.7, for which Table A gives (a) Less than 2% of runners have heart rates above 10 bpm: for the N(10, 12.5) distribution, x > 10 corresponds to z > 1?_b0 = 2.08, for which Table A gives = = 1.88%. (b) About 50% of nonrunnei-s have heart rates above 10 bpm: for the N(10, 17) distribution, x > 10 corresponds to z > 0.

5 If 82 F Chapter The Normal Distributions (a) Among men, a score of 750 corresponds to standard score z = 759~ = 1 87, so about.07% score 750 or more. (b) Among women, a score of 750 corresponds to standard score z = 2.28, so about 1.1% score 750 or more... if the distribution is Normal, it must be symetric about its mean and in p~cu1ar, the 10th and 90th percentiles must be equal distances below and above the mean so the mean is 250 points. If 225 points below (above) the mean is the 10th (90th) percentile, this is 1.28 standard deviations below (above) the mean, so the distribution s standard deviation is j~j points..5. (a) About 0.6% of healthy young adults have osteoporosis (the cumulative probability below a standard score of 2.5 is ). (b) About 1% of this population of older women has osteoporosis the BMD level that is 2 5 standard deviations below the young adult mean would standardize to 0 5 for these older women and the cumulative probability for this standard score is (a) There are two somewhat low IQs 72 qualifies as an outlier by the 1.5 x IQR rule, while 7 is on the boundary. However, for a small sample, 7 this stemplot looks reasonably Normal. (b) We compute I and 8 8 s 1.27 and find: 9 7.2% of the scores in the range 1± ls 91.6 to 120.1, and % of the scores in the range I ± 2s 77. to For an exactly Normal distribution, we would expect these proportions to be 68% and 95% (a) The percent of scores above 27 is 11.7%. (b) The percent of scores greater than or equal to 27 is ~ 15.%. (c) For the N(21.2,5.0) distribution, x > 27 corresponds to z > = 12.0% above. = 1.16, for which Table A gives an area of.8. (a) The mean (5.) is almost identical to the median (5.), and the quartiles are similar distances from the median: M = 0.9 while Q~ M = 0.5. This suggests that the distribution is reasonably symmetric. (b) x < 5.05 corresponds to z < 5.O~ , and.x < 5.79 corresponds to z < s.7g;s Table A gives these proportions as and These are quite close to 0.25 and 0.75, which is what we would expect for the quartiles, so they are consistent with the idea that the distribution is close to Normal

6 Solutions 8.9. (a) One possible histogram is shown on the right. (b) The mean isi 0.800, the median is M = 0.8, the standard deviation is s , and the quartiles are Q~ = 0.76 and Q~ = 0.86 (all in millimeters). i and M are quite close, and the distances from the median to each quartile are similar: M Qi = 0.0 and Q~ M = This is what we expect for a symmetric distribution. (c) 0.76 <x < 0.86 con-esponds ~ to < Z < , or u.a z < 0.76, for which Table A gives = the proportion falling between 0.76 and 0.86 (inclusive) observations equal to 0.76.) Length (mm) Of the 9 measurements, is (This includes (a) One possible histogram is on the right. The mean is I mm, and the median is M = mm. (b) The histogram shows a left skew; this makes the mean lower than the median. 25 ~ 20 C I 15 IL 10 Monsoon rainfall (mm).51. See also the solution to Exercise 2.7. Both stemplots suggest that the distributions may be slightly skewed to the right, rather than perfectly symmetric (as a Normal distribu tion would be). Loose Intermediate

7 8 Chapter The Normal Distributions.52. (a) The applet shows an area of between and 1.000, while the rule rounds this to (b) Between and 2.000, the applet reports 0.95 (compared with the rounded 0.95 from the rule). Between.000 and.000, the applet reports (compared with the rounded 0.997) Because the quartiles of any distribution have 50% of observations between them, we seek to place the flags so that the reported area is 0.5. The closest the applet gets is an area of 0.50, between and Thus the quartiles of any Normal distribution are about 0.68 standard deviations above and below the mean. Note: Table A places the quartiles at about ±0.67; other statistical software gives ± L / ~ C L Placing the flags so that the area between them is as close as possible to 0.80, we find that the A/B cutoff is about 1.28 standard deviations above the mean, and the B/C cutoff is about 1.28 standard deviations below the mean. / I~ p p *

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