Question

# Joshua is a personnel manager in a large corporation. Each month he must review 19 of...

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.

• 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 _____.

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.

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