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CHAPTER 8 SOLUTIONS AND MINI-PROJECT NOTES
CHAPTER 8
BELL-SHAPED CURVES AND OTHER SHAPES
EXERCISE SOLUTIONS
8.1
a. 16%
b. 97.5%
c. 80%
8.2
a. 1 − 0.9 = 0.1, or 10%
b. 1 − 0.4 = 0.6, or 60%
c. 1 − 0.99 = 0.01, or 1%
8.3
a. −0.67
b. 0.67
c. −0.13
d. 2.05
8.4
a. 2.05 (same as 98% below).
b. 0.00 (same as 50% below).
c. −0.67 (same as 25% below).
d. 1.28 (same as 90% below).
8.5
a. 84% − 16% = 68%.
b. 96% − 10% = 86%.
c. 84% − 50% = 34%.
8.6
a. The standardized score is (116 − 100)/16 = 1.00; by Table 8.1 this is the 84th percentile.
b. Both pictures are identical bell-shaped curves. The top picture should be centered at 100 and have a line
at 116, with 84% of the area below the line at 116 and 16% of the area above it. The bottom picture should
look identical, except it is centered at 0 and has the line at 1 instead of 116.
8.7
The picture should show a bell-shaped curve centered on 100. The interval from 90 to 110 should
encompass 68% of the area, from 80 to 120 should encompass 95% of the area, and 70 to 130 should
encompass 99.7% of the area.
8.8
a. The standardized score is (450 − 497)/115, or about −0.41, which is the 34th percentile.
b. The standardized score is (92 − 100)/16 = −0.5, which is the 31st percentile.
c. The standardized score is (68 − 65)/2.5 = 1.2. This is approximately the 88.5th percentile.
8.9
a. By Table 8.1 you need a standardized score of at least 2.05, which corresponds to an IQ of 100 +
(2.05)(16) = 132.8.
b. Again you need a standardized score of 2.05, or a GRE score of at least 497 + (2.05)(115) = 732.75.
8.10
The 97th percentile corresponds to a standardized score of 1.88 or a cholesterol level of 190 + (1.88)(10) =
208.8.
8.11
The percentage of values within one standard deviation of the mean are the percentage of values between
+1 and −1, 84% - 16% = 68%. The percentage of values within two standard deviations of the mean are
those between +2 and −2, or about 97.5% - 2.5% = 95%. The percentage within three standard deviations
is about 99.87% − 0.13%, or 99.7%.
8.12
a. .67 to +.67
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CHAPTER 8 SOLUTIONS AND MINI-PROJECT NOTES
b. The range from .67 to +.67 covers 1.34 standard deviations.
c. The interquartile range covers 1.34 standard deviations, so 2IQR is 2.68 standard deviations. The
standardized score of 2.68 corresponds to about the 0.4 percentile and +2.68 corresponds to about the 99.6
percentile.
8.13
There are numerous examples that could be used here. Make sure the horizontal axis covers the range of
possible values, and that the height of the curve represents reasonable relative frequencies for those values.
8.14
They should accept students with a standardized score of 0.52 or higher, corresponding to a GRE score of
497 + (0.52)(115) = 556.8 or higher.
8.15
a. 68% of scores are between 84 and 116, 95% are between 68 and 132, and 99.7% are between 52 and
148.
b. The picture should be centered on 100 and have the intervals specified in part (a) marked.
8.16
The mean is 51 and the standard deviation is 5, so the standardized score corresponding to 36 boys is
(36−51)/5 = −3.0, which is at the 0.13th percentile. So yes, it is unusual, and 99.87% of the time there
would be more than 36 boys out of 100 births.
8.17
Notice that 50% is two standard deviations above the mean, or a standardized score of 2.00, which is about
the 97.75th percentile. So about 2.25%, or between 2% and 2.5% of such surveys, should show a majority
favoring the candidate.
8.18
You want your allowed travel time to be at the 90th percentile, which corresponds to a standardized score
of 1.28 or travel time of 15 + (1.28)(2) = 17.56 minutes.
8.19
The answer depends on the individual.
8.20
a. The standardized score for 98.6 degrees is (98.6 − 98.2)/0.6 = 0.67, so about 75% or 0.75 of the
population have temperatures below 98.6 degrees.
b. No, that’s the average. Normal temperatures cover a wide range.
8.21
For bell-shaped data, the Empirical Rule tells us that if we had an infinite sample, 99.7% of the data would
fall within 3 standard deviations of the mean, and 95% would fall within 2 standard deviations of the mean.
For a finite data set, depending on how large it is, it thus makes sense that all of the data will fall within 2 or
3 standard deviations on either side of the mean. Thus, the entire range of the data will be 4 to 6 standard
deviations.
8.22
a. 75, in the middle of the range.
b. The range is 50, so the standard deviation is probably about 50/4 = 12.5 to 50/6 = 8.33.
c. The mean is 75, so your score is 5 points above the mean. The standard deviation is around 10, so your
standardized score is around 5/10 = .5.
d. Using Table 8.1, about 31%.
8.23
a. Range = 600 (800 – 200).
b. The standard deviation is 115, so the range of 600 is 600/115 = 5.2 standard deviations. Yes, the result
said that the range should be about 4 to 6 standard deviations, and in this case it is 5.2.
8.24
The mean is much higher than the median, so they must be skewed or have large outliers.
8.25
a. See the figure on the next page.
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CHAPTER 8 SOLUTIONS AND MINI-PROJECT NOTES
b. z 
490  520  30

 0.5 . It should be drawn halfway between 460 and 520 on the picture
60
60
above.
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