Question

The HR manager of a large department store believes the number of resignations per week of casual staff at the store can be approximated by a normal distribution with a mean of 42 resignations per week and variance 51.1 (resignations per week)2 . From a large amount of historical data available on the HR database regarding weekly resignations of casuals, a sample of 52 weeks was selected. What must the value of the sample mean be so that only 15% of all possible sample means (of size 52) are less than this value? Give your answer correct to 2 decimal places.

Answer 1

Solution:

Given:  the number of resignations per week of casual staff at the store can be approximated by a normal distribution with a mean of 42 resignations per week and variance 51.1

That is:  

Thus :

Sample size = n = 52

We have to find the value of the sample mean so that only 15% of all possible sample means (of size 52) are less than this value.

That is we have to find:

Thus find z value such that:

P( Z < z ) = 0.1500

Look in z table for area = 0.1500 or its closest area and find corresponding z value.

Area 0.1492 is closest to 0.1500 and it corresponds to -1.0 and 0.04

thus z = -1.04

Now use following z score formula to find mean value.

Thus sample mean should be so  that only 15% of all possible sample means (of size 52) are less than this value.