# Math Help - Mean & Standard Deviation

1. ## Mean & Standard Deviation

In a survey of twelve Savannah State Business School graduates, the mean starting salary was $93,000, with a standard deviation of$17,000.

The 95% confidence interval for the average starting salary among all Savannah State graduates is?

2. ## Mean & Standard deviation, confidence interval

I think I have solved it and it is as follows:

R =93000 + (1.96) 17000/3.4641061513775
R = 93000 + (1.96) (4907)
R = 93000 + 9618
R = 102,618

R = 93000 - (1.96) 17000/3.4641061513775
R = 93000 - (1.96) (4907)
R = 93000 - 9618
R = 83,382

So the range of salaries would be $83,382 ;$102,618

3. I like it. Do you still doubt?