The VP of HR for a large company is interested in the distribution of sick-leave hours for employees at the company. A recent study revealed that the distribution was consistent with a normal model, with a mean of 58 hours per year, and a standard deviation of 14 hours. An office manager of one division believes that during the past year, two of the division's employees have taken excessive sick leave. One took 74 hours and the other used 90 hours. What would you conclude about the division manager's claim, and why?

The z score and % to the left of z (also known as the normal distribution probability after you multiply it by 100) are computed in Excel. A value of 1 is used for the cumulative sample size when computing the normal distribution probability since we are considering an individual sheet within the distribution.

1st Upper End Hours Sick Leave (X)	90		2nd Upper End Hours Sick Leave (X)	74
Mean ()	58	Mean ()	58
Standard Deviation ()	14	Standard Deviation ()	14
z-score = (x-)/	2.29	z-score = (x-)/	1.14
% to the left of z = NORMDIST*100	98.9	% to the left of z = NORMDIST*100	87.3