= Solution
{c}
The independent uniform choices from the student list describe <sampling with replacement>. Under the stated independent <normal distribution> model for the heights, the population <variance> is unknown and the given $8.5$ cm is the <sample standard deviation> $s$. The exact <pivotal quantity> is
$$
\frac{\overline X-\mu}{S/\sqrt{100}}\sim t_{99}.
$$
Thus the exact 95% <Student t confidence interval> is
$$
\boxed{171.7\pm t_{99,0.975}\frac{8.5}{10}\ \mathrm{cm}
\ \approx\ [170.01,173.39]\ \mathrm{cm},}
$$
using $t_{99,0.975}\approx1.9842$. The supplied approximation $\Phi(2)=0.975$ instead gives a <normal approximation> for large <sample size>
$$
\boxed{171.7\pm2(0.85)=[170.0,173.4]\ \mathrm{cm}.}
$$
This latter <confidence interval> is approximate even under <normality> because it substitutes the <sample standard deviation> for the unknown population value. There is no <finite population correction> under the specified independent selections. Interpreting the finite student list literally as a discrete population makes <normality> an idealized model; it does not justify changing the stated sampling design.
Back to article page
