Arecent survey of 8 social networking sites has a mean of 13.1 million visitors for a specific month. the standard deviation of the sample was 4.1 million. find the 95% confidence interval of the true mean.

For the answer to the question above, the working equation is shown in the image. The z for a 95% confidence level is 1.96. The bar x is 13.1 million, sigma is 4.1 million and n is 8. Substituting,
 13.1 million - (1.96)(4.1 million/sqrt8) <truemean <13.1 million +
(1.96)(4.1 million / sqrt 8)
 
 10.26 million < truemenan < 15.94 million
 I hope my answer helped you.