Question

Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let X be the random variable representing the time it takes her to complete one review. Assume X is normally distributed. Let X be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

Give the 95^th Percentile. (Round your answer to two decimal places.)

Answer 1

Given that,

mean = = 16

standard deviation = = 1.2

n = 16

= 16

= / n = 1.2/16=0.3

Using standard normal table,

P(Z < z) = 95%

= P(Z < z) = 0.95  

= P(Z <1.65 ) = 0.95

z = 1.65 Using standard normal table,

Using z-score formula  

= z * +   

= 1.65 *0.3+16

= 16.495

= 16.50