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Unformatted text preview: Homework Ch56 Solutions Chapter 5 1. a. Using the normal table on page A105, the area between —1.25 and 1.25 is 78.87%. So
the number of scores within 1.25 SDs of the average is 0.7887 x 25 z 20. b. The number of scores that are really within 1.25 SDs of the average is the number of
scores between 50 — 1.25 x 10 = 37.5 and 50 + 1.25 x 10 : 62.5. There are 18 of them. 2. There is something wrong with the computer. Some standard units are more than 6 SDS
away from the average 0. 3. a. In 1967', a score of 600 is 1.22 SDs above the average ((600 — 466)/110]. Using the normal curve, the area to the right of 1.22 is about 11%.
b. In 1994, 600 is 1.61 SDs above the average, so the percent is about 5%. 4. This is similar to problem 3. (a) 20% (b) 12%. 5. The percentage of men with heights between between 66 and T2 inches is exactly equal
to the area between 66 inches and 72 inches under the histogram. This percentage is
approximately equal to the area between 1 and 1 under the normal curve. 6. No. The score 178 is 1 SD above the average. If it is the highest, then no scores are 1
SD above the average, which disagrees with the normal curve. 7. a. 350 has standard units —1.5. The area (under the normal curve) below —l.5 is about
7%. The student was then at the 7th percentile. b. The standard unit corresponding to the 75th percentile is 0.7. So the student need to
score 500 + 0.7 x 100 2 570. 8. (a) True (see Ch5, Section 6) (b) False (C) True (d) True (e) True (f) False [SD is always nonnegative) 9. (a) False (for example 0, 1, 101 has a median of 1 and an average of 34. (b) False (see the above example]. (c) False, the income data on page 88 of the text book have a long righthand tail.
(d) False, the shape of the histograms for both lists could be very different. 10. The histogram for the income has a long right hand tail, thus the average is above the
median. So the area between the average ($35,000) and $150,000 is about 40%. 11. It would look like histogram [i]. For (ii) , many people would be taking a negative
number of corirses. Same with (iii) Chapter 6. 4. You expect to get 36 inches give or take the SD of the three measurem ents 35.96, 36.01,
and 36.03 inches, which is .03 inches or so. 5. a. No. For example, students 2 and 1 0 copied from each other.
b. There is variability in the measurements, even from the same student. 86. (a) (650 + 600)/2 : 625.
(b) More than 125, the scores are more spread out. SS. Too low, because the histogram is above the normal curve in that region. SID. False. This is a cross—sectional study, not longitudinal. The 20—yearolds were born in 19561960 whereas the TUyear—olds in 19061910. There was much more pressure to conform
and be right handed earlier in the century. 811. Since the normal curve is symmetric, the average is the mid—point of the 25th and
75th percentiles, M = 64 inches. The standard units of the 75th percentile is 0.67, therefore the SD is 6—5??? = 2.7 inches. The 90th percentile has standard unit of 1.28, and it is 64 + 1.28 x 2.? m 67.5inches. 812. This is probably due to rounding of incomes. for instance
$12,500 includes $10,000, $11,000, and $12,000 {round number
but the next class 312,500815,000 only includes tw
(recall that in a class interval, the left point , the class interval $10,000 s that people tend to use), 0 round numbers, $13,000 and $14,000
is included but. not the right point). 813. (a) Because heart disease rates go up with age. (1)) The investigators wanted to explain the difference in rates by the exercise on the job,
not any pre—existing factors; it might take time for exercise to have its beneﬁcial effect.
(c) Drivers and conductors will be similar with respect to many factors, such as age,
tion, income, etc... (d) Because this study is observation
(e) Sizes of the uniforms could tell us ' when they were hired, that would sugges
argument that exercise matters. educa ting high risk women from the control group to e in the treatment group, which makes the
the study against screening. ...
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 Spring '08
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