Thread: Finding the confidence interval given a population sample

I can't seem to figure out the second half of this question:

Consider the sample 100 91 91 86 90 91 90 94 from a normal population with population mean μ and population variance σ². Find the 95% confidence interval for μ.

I know that μ = μ of sample = 91.625, but i need to find the range

I also know the standard deviation of the sample of the population is σ/sqrt(n)

the answer s 91.63±3.37 but I'm not sure where 3.37 comes from, since the σ the sample population is 4.03/sqrt(8) = 1.425

One of the famous properties of normal distribution: interval μ plus/minus 2σ contains 95% of the values.

Thanks, but I just figured it out.. The formula I should be using for the margin of error is x-bar +/- t(a/2)*(σ/sqrt(n))