7.43: Let [tex]X[/tex] denote the random variable for height and [tex]\overline X[/tex] for the sample mean. Then if [tex]\mu[/tex] is the mean of
So the probability that the difference between the sample and population means does not exceed 0.5 inch is
[tex]P(|\overline X-\mu|\le0.5) = P\left(\left|\dfrac{\overline X-\mu}{0.25}\right|\le\dfrac{0.5}{0.25}\right) = P(|Z|\le 2) \approx \boxed{0.95}[/tex]
per the empirical or 68/95/99.7 rule.
7.44: For a sample of size n, the sample standard deviation would be [tex]\frac{2.5}{\sqrt n}[/tex]. We want to find n such that
[tex]P(|\overline X-\mu| < 0.4) = 0.95[/tex]
Comparing to the equation from the previous part, this means we would need
[tex]\dfrac{0.4}{\frac{2.5}{\sqrt{n}}} = 0.16\sqrt n = 2 \implies \sqrt n = 12.5 \implies n = 156.25[/tex]
so a sample of at least 157 men would be sufficient.
