To ensure adequate customer service, a bus company wants to estimate the average number of passengers on a bus during morning rush hours. A number of bus passengers is a random variable with standard deviation of $9.2$. According to records, the average number of bus passengers at $36$ randomly selected times is $40.5$.
(a) Construct a $90%$ confidence interval for the expectation of the number of bus passengers. (b) Suppose that the true population mean number of bus passengers is $41$. If you are to collect another sample of passenger counts at $36$ randomly selected times, what is the probability that a sample average will be within the interval computed in (a)?
Now for (a) I found that the interval is $37,9778$, $43,0223$.
However I am not sure how to go about (b). I tried to calculate the interval using the mean of $41$ $(38,4778, 43,5223)$ and then calculate how much these intervals overlap, which is about $90%$, but I'm am hesitant that this is the right answer.
Calculate the Z scores of $37.9778$ and $43.0223$ for a mean of $41$ and determine the area which lies between them. Hence the probability of being in that confidence interval. You can look up the Z scores on a Z table to get the p values.