Suppose that a random sample of size 64 is to be selected from a population with

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 sampling distribution?
x =  
x =  

(b) What is the approximate probability that x will be within 0.3 of the population mean ? (Round your answer to four decimal places.)
P =  

(c) What is the approximate probability that x will differ from by more than 0.6? (Round your answer to four decimal places.)
P =

You may need to use the appropriate table in Appendix A to answer this question.

Explanation / Answer

Mean is 40 and s is 5, standard error for 64 sample size is s/sqrt(N)=5/sqrt(64)=5/8=0.625

z is given as (x-mean)/s

a) Mean is 40 and standard error is 0.625

b) We need P(39.7<x<40.3)=P((39.7-40)/0.625<z<(40.3-40)/0.625))=P(-0.48<z<0.48) or 2*P(z<0.48)-1, from normal distribution table we get 2*0.6844-1=0.3688

c) P(x>40.6)=P(z>(40.6-40)/0.625)=P(z>0.96) or 1-P(z<0.96), from normal table we get 1-0.8315 =0.1685

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