Question

Joshua is a personnel manager in a large corporation. Each month he must review 19 of the employees. From past experience, he has found that the reviews take him approximately 5 hours each to do with a population standard deviation of 0.7 hours. Let ? be the random variable representing the time it takes him to complete one review. Assume X is normally distributed. Let ?⎯⎯⎯⎯⎯ be the random variable representing the mean time to complete the 19 reviews. Assume that the 19 reviews represent a random set of reviews.

a. Find the 99th percentile for the sampling distribution of sample means, rounded to two decimal places.

b. Interpret your results. The blanks represent your numerical answer in part a.

Select your answer from one of the following options.

• a.99% of employees will have a review that takes less than _______ hours to complete.
• b.99% of employees will have a review that takes more than _______ hours to complete.
• c.99% of samples of 19 employees will have a sample mean less than ______.
• d.99% of samples of 19 employess will have a sample mean more than _____.

Answer 1

Solution,

Given that,

mean = = 5 hours

standard deviation = = 0.7 hours

n = 19

= = 5

= / n = 0.7 / 19 = 0.16

Using standard normal table,

P(Z < z) = 99%

= P(Z < z ) = 0.99

= P(Z < 2.33 ) = 0.99

z = 2.33

Using z-score formula  

= z * +

= 2.33 * 0.16 + 5

= 5.37 hours

c.99% of samples of 19 employees will have a sample mean less than 5.37 hours.