A sample of 100 elements produced a mean of 48 and a standard deviation of 5. What is the 99% confidence inter?

Sx= S/n^.5 = 5/10 = 0.5





I think you can use the standard normal distribution since n%26gt;100, so Z for P 0.005 and P 0.995 = +/- 2.58





Confidence interval = mean+- Z*Sx = 48 +- 1.49