Question

In: Statistics and Probability

It was found that an average of 4 weeks job leave was spent by staff at...

It was found that an average of 4 weeks job leave was spent by staff at QR investment during the 2017 operation. Assume that 4 weeks is the population mean and 2.5 weeks is the population variance for a sample of 27 staff members.

a) What is the probability that the mean leave is less or equal to 2.5 weeks?

b) Show the sampling distribution of the mean?

Expert Solution

The average number of weeks of job leave is 4. Thus, the assumed population mean, = 4 weeks

The variance of number of weeks of job leave is 2.5. Thus, the assumed population variance, ²= 2.5 weeks

The sample size, N is 27

Since the population variance is given, we can use that to arrive at the standard error. The standard error is / sqrt(N)= sqrt(2.5/ 27)= 0.3042903 weeks

a)To find the Probability that the mean leave is less or equal to 2.5 weeks(>=2.5)

We must find the z- score of 2.5.

Z score of 2.5= (- )/ = (2.5-4)/0.3042903 = -1.5/ 0.3042903= -4.929503

The required probability is the area under the z score of -4.929503= 0.0011

Thus the probability that the staff mean leave weeks will be 2.5 or less than 2.5 is 0.11% or 0.0011

b. Show the sampling distribution of the mean

According to the central limit theorem (CLT), if samples of enough sizes, and enough number of samples are taken from the population, then the distribution of their means always approximate to a normal distribution. The mean of this sampling distribution of the sample means is the mean of the population. That is, the mean of this distribution will be 4 weeks and the standard deviation is the standard error, that is 0.3042903 weeks

orchestra answered 2 years ago

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