For the answer to the question above asking <span>If the sample mean was 89 and the standard deviation was 6.2, construct a 99% confidence interval of the mean score of all students. Assume that the population has a normal distribution</span> z=2.57 cuts off .005 in each tail of the standard normal distribution 99% CI is 89 +/- 2.57(6.2)/sqrt(30)= 89 +/- 2.909, or from 86.09 to 91.91
