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Chapter 2 Review Answers 39. The distribution is skewed to the right since most of the values are 25 minutes or less, but the values stretch up to about 90 minutes. The data are centered roughly around 20 minutes and the range of the distribution is close to 90 minutes. The two largest values appear to be outliers 40. a) b) 75. See next slide 6%. Since 6% (3/50) of the sample was left-handed, our best guess for the population is that 6% is left-handed. For both kinds of cars, the highway MPG is higher than the city MPG. The twoseater cars have a wider spread of MPG than the minicompact cars do, both in the city and on the highway. The minicompact cars get slightly better gas mileage, and this difference is more pronounced on the highway. All four distributions appear to be symmetric. 76. a) a) Cold Neutral Hot Hatched 16 38 75 Not hatched 11 18 29 Total 27 56 104 59.26% in a cold nest. 67.86% in a neutral nest, and 72.12% in a hot nest. Hatching percentage increases with temperature; the cold nest did not prevent hatching, but did make it less likely. 40A (from previous slide) Handed-ness 50 45 40 35 30 25 20 15 10 5 0 R L A Chapter 2 AP Statistics Practice Test 1. E 2. D 3. B 4. B 5. A 6. D 7. C 8. E 9. E 10. C 11. a) Jane’s performance was better. She did more curl-ups than 85% of girls her age. Matt did more curl-ups than less than 50% of boys his age. b) We can confidently say that Jane would have the higher Z-score, since her percentile was so much higher 12. a) His Z-score would be 1. So the proportion below this is about .84, meaning that he is at the 84th percentile b) The Z-score for 20 is -2.55 ((20-22.8)/1.1), and the area below Z=-2.55 is .0054 (using Table A). The Z-score for 26 is 2.91 ((26-22.8)/1.1), and the area below Z=2.91 is .9982 (using Table A). Because we want the area in the tails, we must do 1-.9982, to give us .0018 in the right tail. Then we add the two tails together, to get .0072 (.0054+.0018), or .72%. So less than 1% of soldiers requires custom helmets c) Q1 is 22.062 (Z=-.67) and Q3 is 23.537 (Z=.67). 23.537-22.062 gives us an IQR of 1.474 inches. 13. These data do not appear to be normally distributed. The median and the mean are not close to each other here, and they would be if the distribution were symmetric. Because the mean is much larger than the median, we can conclude that the distribution is probably skewed to the right. This is further supported by the fact that the median is much closer to the minimum value than the maximum value. Chapter Review Exercises 1. a) The only way to obtain a z-score of 0 is if the x-value equals the mean. Thus, the mean is 170 cm. 1=(177.5-170)/Stdev, so the standard deviation equals 177.5-170, or 7.5 cm. b) 2.5 = 𝑥−170 7.5 X=188.75. A height of 188.75 cm has a z-score of 2.5 2. a) Z=1.20. Paul is somewhat taller than average for his age. His height is 1.2 standard deviations above the mean for his age. b) 85% of boys Paul’s age are shorter than Paul 3. a) Approximately the 70th percentile b) Median is about 5, Q1 is about 2.5, and Q3 is about 11. There are outliers according to the 1.5(IQR) rule, because values exceeding 23.75 clearly exist. 4. 5. a) The shape would not change. The new mean would be 13.32 meters, the median would be 12.8 meters, the standard deviation would be 3.81 meters, and the IQR would be 3.81 meters. b) The mean error would be 43.7-42.6=1.1 feet. The standard deviation of the errors would be the same as the standard deviation of the guesses, 12.5 feet, because we have just shifted the distribution Both lines will be to the right of the peak, with the mean (B) being further to the right than the median (A) 6. a) (327,345) b) 339 is one standard deviation above the mean, so 16% of the horse pregnancies last longer than 339 days. This is because 68% are within one standard deviation of the mean, so 32% are more than one standard deviation from the mean. Because we only want those that are one standard deviation ABOVE the mean, it is 16%. 7. a) .0122 (proportion) or 1.22% (percent) b) .9878 c) .0384 d) .9494 8. a) Z=.84 b) Z=.39 9. a) Z=-2.29. Approximately 1% of babies will be identified as having a low birthweight b) Z= -.67 and Z=.67. For Z=-.67, X=3325.63. For Z= .67, X=4010.37. Therefore, the quartiles of the birthweight distribution are 3325.6 and 4010.4 10. If the distribution is normal, it must be symmetric about its mean—and in particular, the 10th and 90th percentiles must be equal distances above and below 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 225/1.28=175.8 points.