Chapter 2 - 6851F_ch02_27_37 13/09/2002 04*58 PM Page 27 2...

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27 2.1 There are many correct drawings. Here are two possibilities: (a) (b) 2 2.2 (a) The area under the curve is a rectangle with height 1 and width 1. Thus the total area under the curve 5 1 3 1 5 1. (b) 20%. (The region is a rectangle with height 1 and base width 0.2; hence the area is 0.2.) (c) 60%. (d) 50%. (e) or 0.5, the “balance point” of the density curve. 2.3 (a) Total area under curve 5 area of triangle 1 area of 2 rectangles 5 (1)(.4) 1 (.4)(1) 1 (.4)(1) 5 0.2 1 0.4 1 0.4 5 1. (b) 0.2. (c) 0.6. (d) 0.35. (e) The median is the “equal-areas” point. By (d), the area between 0 and 0.2 is 0.35. The area between 0.4 and 0.8 is 0.4. Thus the “equal-areas” point must lie between 0.2 and 0.4. 2.4 (a) Mean C, median B; (b) mean A, median A; (c) mean A, median B. 2.5 The uniform distribution. Each of the 6 bars should have a height of 20. 2.6 1 > 2 Mean 5 1 > 2 66.5 69 71.5 2.5 6851F_ch02_27_37 13/09/2002 04*58 PM Page 27
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28 Chapter 2 2.7 (a) 2.5% (this is 2 standard deviations above the mean). (b) 69 6 5; that is, 64 to 74 inches. (c) 16%. (d) 84%. The area to the left of X 5 71.5 under the N (69, 2.5) curve is 0.84. 2.8 (a) 50%. (b) 2.5%. (c) 110 6 50, or 60 to 160. 2.9 (a) 50. The mean (center of the distribution in this case) is 64.5, so 50% of the area to the left of 64.5 under the N (64.5, 2.5) curve. Parts (b) through (d) use similar reasoning. (b) 2.5 (c) 84 (d) 99.85. 2.10 Answers vary. 2.11 Approximately 0.2 (for the tall one) and 0.5. 2.12 (a) 16%. (b) 84th percentile. The area to the left of X 5 23.9 under the N (22.8, 1.1) curve is 0.84. (c) 68%. 2.13 (a) 266 6 32, or 234 to 298 days. (b) Less than 234 days. (c) More than 298 days (2 standard deviations to the right of the mean). 2.14 (a) By the 68-95-99.7 rule, approximately 16% of the scores lie below m 2 1 s 5 110 2 25 5 85. The TI-83 command shadeNorm ( 2 1000, 85, 110, 25) reports a lower left tail area of .158655. (b) The 84th percentile is the area under the N (110,25) curve to the left of m 1 1 s 5 110 1 25 5 135. The command shadeNorm ( 2 1000, 135, 110, 25) reports an area of .841345. The 97.5 percentile is the area to the left of m 1 2 s 5 110 1 50 5 160. ShadeNorm ( 2 1000, 160, 110, 25) reports an area of .97725. 2.15 (a) (b) 68%: (58.3, 67.9). 95%: (53.5, 72.7). 99.7%: (48.7, 77.5). 2.16 (a) 48.7 53.5 58.3 63.1 67.9 72.7 77.5 kg m 2 3 sm 2 2 2 1 1 1 1 2 1 3 s m 2 1 1 2 (b) 50% of the outcomes are less than 1. (c) Because the distribution is symmetric, the median is 1, Q 1 5 .5, and Q 3 5 1.5. (d) For 0.5 , X , 1.3, the proportion of outcomes is 0.8 (1/2) 5 0.4. 2.17 (a) Outcomes around 25 are more likely. (d) The distribution should be roughly symmetric with center at about 25, single peaked at the center, standard deviation about 3.5, and few or no outliers. The normal density curve should fit this histogram well.
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Chapter 2 - 6851F_ch02_27_37 13/09/2002 04*58 PM Page 27 2...

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