Confidence level Statistics question

I couldn't figure out this question, so if you would help me, I would appreciate it. I would like to know how to do it..

Q: A random variable X follows a normal distribution with standard deviation 16. A random sample of 23 individuals is selected from the population, and a confidence interval is calculated to be (90.406, 107.594). The confidence level for this interval is ___ %.

Claire

The midpoint of (90.406, 107.594) will ALWAYS be the mean ...

midpoint = mean = (90.406 + 107.594) / 2 = 99.000

margin of error = endpoint - mean = 107.594 - 99 = 8.594

But, margin of error also = (z*)(standard deviation) / sqrt N

margin of error = 8.594 = (z*)(16) / sqrt 23, now solve for z*

z* = 2.576 , now use a standard normal table to find P( z > 2.576)

P( z > 2.576) = P( z < -2.576) = 0.005

But, you have 2-tails, so 0.005 x 2 = 0.01 or CL = 1 - 0.01 = 0.99

The confidence level for this interval is 99 %.

hope that helps

Margin of error = z-score for the degree of confidence*standard deviation/sqrt sample size
Margin of error = (upper limit - lower limit)/2 = (107.594 - 90.406)/2 = 8.594
8.594 = z-score*16/sqrt 23
z-score = 8.594*sqrt 23/16 = 2.58 (approximately)
The degree of confidence whose z-score is 2.58 is 99%