Find the margin of error and the 90 percent confidence interval for Julias average running time.

Consider that the population of interest is the total number of times that the subject has been jogging since she started exercising, which means the 95 times that were sampled represent the sample values.  The person is interested in estimating what might be the true average number of minutes that Julia has been exercising. 

Personally, I would approach this question using two procedures, first, using the t-score as the distribution of interest, and second using z-score from the stadndard normal distribution. 

Reason;-  some text use normal distribution even when the population standard deviation is not given, other tutors would use the t-distribution no matter how large the sample size is. 

Confidence interval for the mean number of minutes run the subject 

using t-distribution;-  x_bar +/- t-score * standard deviation/sqrt(n)

t-score can be obtained using an excel function;- =T.INV.2T(0.1,94) = 1.661

You can always obtain any score in the tandard normal table using an Excel Function. 

confidence interval;- 5.60 +/- 1.661 * 1.80/sqrt(95) = 5.60 +/- 0.307. so that the confidence interval using the t-distribution is (5.29, 5.91)

In case you are to use the standard normal distribution instead, the z-score will change to 1.645 (This is the z-score for the 90% confidence interval).  And the 90% confidence interval using the normal distribution becomes, 5.60 +/- 1.645 * 1.80/sqrt(95)  = (5.30, 5.90). 

NB: Observe that the confidence interval is wider when the t-distribution is used, and the reason is because the t-distribution always tails further than the normal distribution.