18. We do not expect these times to be normally distributed because the second criterion is not satisfied. The distribution is likely to be left-skewed because most competing swimmers would have times near the world record for this event. The outliers would correspond to people having a bad day.
20.
(a) 68% of the heart rates lie within one standard deviation
of the mean. That is, 68% lie between 70-15 and 70+15, or
55 and 85. The remaining 32% are evenly split below and above
55 and 85; i.e., 16% lie below 55 (and 16% lie above 85).
(b) 95% of the heart rates lie within two standard deviations
of the mean. That is, 95% lie between 70-30 and 70+30, or
40 and 100. The remaining 5% are evenly split below and above
40 and 100; i.e., 2.5% lie below 40 (and 2.5% lie above 100).
(c) As indicated in (a), 16% of the heart rates lie below 55
and 68% lie between 55 and 85. Therefore 68% + 16% = 84% of
the heart rates are below 85.
(d) As indicated in (b), 2.5% of the heart rates lie below 40
and 95% lie between 40 and 100. Therefore 2.5% + 95% = 97.5% of
the heart rates are below 100.
(e) As indicated in (a), 16% of the heart rates lie above 85.
(f) As indicated in (a), 16% of the heart rates are below 55,
and the rest, or 84% are above 55.
(g) As shown in (b), 2.5% of the heart rates lie below 40.
Therefore, the rest, or 97.5%, lie above 40.
(h) As shown in (a), 68% of the heart rates lie between 55
and 85.
You can also do all of these problems with the Table on page 408.
For instance, for problem (h), we want to find out how many people
have z-scores between (55-70)/15 = -1 and (85-70)/15 = +1. Since
a z-score of +1.0 corresponds to the 84.13th percentile and a
z-score of -1.0 corresponds to the 15.87th percentile, the proportion
of the people who lie in between is 84.13-15.87 percent, which is
68%.
32.
(a) From the table, the 85th percentile corresponds to a
standard score of z = 1.0; the data value lies 1 standard
deviation above the mean.
(b) From the table, the 10th percentile corresponds to a
standard score of z = -1.3; the data value lies 1.3 standard
deviations below the mean.
(c) From the table, the 54th percentile corresponds to a
standard score of z = 0.1; the data value lies 0.1 standard
deviations above the mean.
34. About 95% of births occur within 2 standard deviations of the mean; i.e., within 30 days of the due date.
40. The scenario is not possible. If it were true, then 68% of the scores would lie between -15 and 185 (i.e., within 1 standard deviation of the mean) with half of the remaining 32% of the scores being less than -15.
42. From the table, the 95th percentile corresponds to a standard score of z = 1.65. The required GRE score is thus about 1.65 standard deviations above the mean: 497 + 1.65 × 115 = 686.75 or about 687.
"Hivaria and Lovaria":. We will assume for the sake of
concreteness that there are a million Hivarians and a million
Lovarians. (This will not affect the answer, since increasing
the number of Hivarians by some factor and increasing the number
of Lovarians by the same factor, e.g., doubling both populations,
will change the absolute numbers but will not change the proportions
we are asked to compute.)
(a) For a Hivarian, being 6 feet tall or taller means being
1 standard deviation above the mean, so 15.87% of Hivarians
are 6 feet tall or taller;
that is, there are 158,700 such Hivarians.
For a Lovarian, being 6 feet tall or taller means being
2 standard deviations above the mean, so 2.28% of Lovarians
are 6 feet tall or taller;
that is, there are 22,800 such Lovarians.
Hence, among the 158,700+22,800 = 181,500 Varians
who are 6 feet tall or taller,
only 22,800 are from Lovaria;
so that proportion of the 6-feet-taller-or-taller Varians
who come from Lovaria is 22,800/181,500 = 0.1256, or about 13%.
(b) For a Hivarian, being 6 foot 2 or taller means being
1.5 standard deviations above the mean, so 6.68% of Hivarians
are 6 foot 2 or taller;
that is, there are 66,800 such Hivarians.
For a Lovarian, being 6 foot 2 or taller means being
3 standard deviations above the mean, so 0.13% of Lovarians
are 6 foot 2 or taller;
that is, there are 1300 such Lovarians.
Hence, among the 66,800+1300 = 68,100 Varians
who are 6 foot 2 or taller,
only 1300 are from Lovaria;
so that proportion of the 6-foot-2-or-taller Varians
who come from Lovaria is 1300/68,100 = 0.019, or about 1.9%.
(c)
The argument does not make sense. It is conceivable that
Africans and non-Africans have the same mean height, but
that Africans manifest greater variation in height. In that
case, we would find that Africans predominate among extremely
tall people, even though Africans are on average no taller than
non-Africans. (Note that, in this scenario, Africans would
also predominate among extremely short people.)
Section 6D:
22. We are told that the probability that the mean lengths of stay in ICU's differ as much as they do (assuming that the null hypothesis holds) is 1 in 10,000, or 0.0001. Since this is well below both 0.01 and 0.05, it is significant at both those levels. Thus we can be reasonably confident that the use of seat belts with children significantly reduces the length of time they would be in an ICU after a severe accident.
28. For a poll of 1012 people, the margin of error is approximately 1/sqrt(1012) = 0.0314 = 3.14%. So the 95% confidence interval is 76-3.14% to 76+3.14%, or 72.86% to 79.14%. This means that we can be 95% confident that the true proportion of drivers who engage in at least one distracting activity is between roughly 73% and 79%.
34. Null hypothesis: Vitamin C in tablets =
500 mg. Alternative hypothesis: Vitamin C in tablets <
500 mg.
Rejecting the null hypothesis means that there is
evidence that the tablets contain less than 500 mg. Failing
to reject the null hypothesis means that there is insufficient
evidence to conclude that the tablets contain less than 500 mg.
36. Null hypothesis: Water usage = 1675 gallons
per month. Alternative hypothesis: Water usage exceeds
1675 gallons per month.
Rejecting the null hypothesis means
that there is evidence that water usage exceeds 1675 gallons
per month. Failing to reject the null hypothesis means that there
is insufficient evidence to conclude that water usage exceeds
1675 gallons per month.
40. Null hypothesis: Proportion of supporting
voters = 0.5. Alternative hypothesis: Proportion of
supporting voters exceeds 0.5.
Rejecting the null hypothesis
means that there is evidence that the proportion of supporting
voters exceeds 0.5. Failing to reject the null hypothesis means
that there is insufficient evidence to conclude that the proportion
of supporting voters exceeds 0.5.
42. Null hypothesis: Mean gas mileage = 21.4 mpg.
Alternative hypothesis: Mean gas mileage exceeds 21.4 mpg.
Rejecting the null hypothesis means that there is evidence that
gas mileage exceeds 21.4 mpg.
Failing to reject the null hypothesis means
that there is insufficient evidence to conclude that
gas mileage exceeds 21.4 mpg.
44. Null hypothesis: Mean ownership = 7.5 years.
Alternative hypothesis: Mean ownership exceeds 7.5 years.
Rejecting the null hypothesis means that there is evidence that
mean ownership exceeds 7.5 years.
Failing to reject the null hypothesis means
that there is insufficient evidence to conclude that
mean ownership exceeds 7.5 years.
46. For a sample of 1400 people, the margin of error is approximately 1/sqrt(1400) = 0.0267 = 2.67%. So the 95% confidence interval is 80-2.27% to 80+2.26%, or 77.73% to 82.67%. This means that we can be 95% confident that the true proportion of student athletes in this country who have been subjected to some form of hazing is between about 77.3% and 82.7%.
48. For a sample of n people, the margin of error is roughly 1/sqrt(n), and for a sample of 100n people the margin of error is 1/sqrt(100n) = 1/10sqrt(n). So a sample size that is one hundred times larger will have a margin of error one tenth as large.
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