Question

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. (a) What are the mean and standard deviation of the x sampling distribution? μx = 1 40 Correct: Your answer is correct. σx = 2 .625 Correct: Your answer is correct. (b) What is the approximate probability that x will be within 0.2 of the population mean μ? (Round your answer to four decimal places.) P = 3 .2510 Correct: Your answer is correct. (c) What is the approximate probability that x will differ from μ by more than 0.6? (Round your answer to four decimal places.) P =

Answer 1

Solution:

Given:

Mean =

Standard Deviation =

Sample size = n = 64

Part d) We have to find the approximate probability that x will differ from μ by more than 0.6

Since sample size n = 64 is large , so using Central limit theorem, sampling distribution of sample mean follows approximate Normal with mean of sample means = and standard deviation of sample mean is:

We have to find:

That is:

Dividing both terms by .

Look in z table for z = -0.9 and 0.06 as well as for z = 0.9 and 0.06 and find area.

P( Z <-0.96) = 0.1685

P( Z < 0.96)= 0.8315

Thus we get:

Thus the approximate probability that x will differ from μ by more than 0.6 is   0.3370