126 NORMAL DISTRIBUTIONS AND STANDARD (z) SCORESscore. Unlike z scores, transformed standard scores usually lack (r) signs and (s)points.

126 NORMAL DISTRIBUTIONS AND STANDARD (z) SCORESscore. Unlike z scores, transformed standard scores usually lack (r) signs and (s)points. Transformed standard scores accurately refl ect the relative standing of the original (t) score. Answers on page 505. Finding Proportions 5.11 Scores on the Wechsler Adult Intelligence Scale (WAIS) approximate a normal curve with a mean of 100 and a standard deviation of 15. What proportion of IQ scores are (a) above Kristen’s 125? (b) below 82? (c) within 9 points of the mean? (d) more than 40 points from the mean? 5.12 Suppose that the burning times of electric light bulbs approximate a normal curve with a mean of 1200 hours and a standard deviation of 120 hours. What proportion of lights burn for (a) less than 960 hours? (b) more than 1500 hours? (c) within 50 hours of the mean? (d) between 1300 and 1400 hours? Finding Scores 5.13 IQ scores on the WAIS test approximate a normal curve with a mean of 100 and a standard deviation of 15. What IQ score is identifi ed with (a) the upper 2 percent, that is, 2 percent to the right (and 98 percent to the left)? (b) the lower 10 percent? (c) the upper 60 percent? (d) the middle 95 percent? [Remember, the middle 95 percent straddles the line perpendicular to the mean (or the 50th percentile), with half of 95 percent, or 47.5 percent, above this line and the remaining 47.5 percent below this line.] (e) the middle 99 percent? 5.14 For the normal distribution of burning times of electric light bulbs, with a mean equal to 1200 hours and a standard deviation equal to 120 hours, what burning time is identifi ed with the (a) upper 50 percent? (b) lower 75 percent? Copyright © 2015 John Wiley & Sons, Inc. *5. 15 An investigator polls common cold sufferers, asking them to estimate the number of hours of physical discomfort caused by their most recent colds. Assume that their estimates approximate a normal curve with a mean of 83 hours and a standard deviation of 20 hours. (a) What is the estimated number of hours for the shortest suffering 5 percent? (b) What proportion of sufferers estimate that their colds lasted longer than 48 hours? (c) What proportion suffered for fewer than 61 hours? (d) What is the estimated number of hours suffered by the extreme 1 percent either above or below the mean? (e) What proportion suffered for between 1 and 3 days, that is, between 24 and 72 hours? (f) What is the estimated number of hours suffered by the middle 95 percent? [See the comment about “middle 95 percent” in Question 5.13(d) .] (g) What proportion suffered for between 2 and 4 days? (h) A medical researcher wishes to concentrate on the 20 percent who suf-fered the most. She will work only with those who estimate that they suf-fered for more than hours. (i) Another researcher wishes to compare those who suffered least with those who suffered most. If each group is to consist of only the extreme 3 percent, the mild group will consist of those who suffered for fewer than hours, and the severe group will consist of those who suffered for more than hours. (j) Another survey found that people with colds who took daily doses of vita-min C suffered, on the average, for 61 hours. What proportion of the orig-inal survey (with a mean of 83 hours and a standard deviation of 20 hours) suffered for more than 61 hours? (k) What proportion of the original survey suffered for exactly 61 hours? (Be careful!)