To elaborate on @robin’s comment let’s be clear that the n-1 arises precisely when computing the standard deviation of a sample of numerical data. This is something that’s typically done on a computer and done by hand only when you’ve got a very small list of numbers. Thus, this is not an issue in problem number 9.
While we’re at it, let’s take a look at problem 9, which states:
Suppose now that we are interested in the proportion of in-state UNCA students who are from outside Western NC this year. Since data on the entire student body is not yet available for this year, we draw a simple random sample of 64 in-state UNCA students and find that 34 of them are from outside Western NC. Use this data to write down a 95% confidence interval for the proportion of in-state UNCA students who from outside Western NC.
In this problem, the confidence interval will have the form
[\hat{p} - z^* \times SE, \hat{p} + z^* \times SE] ,
where hat(p) = 34//64 ~~ 0.53125, z^* = 2 (or z^* = 1.96, if you want to be more precise), and the standard error is
SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{(34/64)(30/64)}{64}} \approx 0.0624.
Thus, the confidence interval is:
[0.53125 - 2\times 0.0624, 0.53125 + 2\times 0.0624] \approx [0.40645 0.65605].