Answer: 
OR
(15.102, 15.498)
Step-by-step explanation:
The formula to find the confidence interval
is given by :-

, where n is the sample size
= sample standard deviation.
= Sample mean
= Two tailed z-value for significance level of
.
Given : Confidence level = 95% = 0.95
Significance level = 

sample size : n= 250 , which is extremely large ( than n=30) .
So we assume sample standard deviation is the population standard deviation.
thus , 
By standard normal distribution table ,
Two tailed z-value for Significance level of 0.05 :

Then, the 95% confidence interval for the mean credit hours taken by a student each quarter :-

Hence, the mean credit hours taken by a student each quarter using a 95% confidence interval. =(15.102, 15.498)