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 is the mean of the xbar sampling distribution? =40

What is the standard deviation of the xbar sampling distribution (to 3 decimal places)? =0.625

For parts b & c round to 4 decimal places:

(b) What is the probability that xbar will be within 0.5 of the population mean μ ?

(c) What is the probability that xbar will differ from μ by more than 0.7 ?

I did part a but I dont know how to do b and c. What formula would I use if doing it on Excel

Answer 1

Solution-b:

Xbar follows mean with mu=40 and standard deviation=sigma/sqrt(n)=5/sqrt(64)=0.625

P(40-0.5<xbar<40+0.5)

P(39.5<Xbar<40.5)

P(X<40.5)-P(X<39.5)

Excel formula is

P(X<40.5)=NORMDIST(40.5,40,0.625,TRUE)

P(X<39.5)==NORMDIST(39.5,40,0.625,TRUE)

=NORMDIST(40.5,40,0.625,TRUE)-=NORMDIST(39.5,40,0.625,TRUE)

=0.788145-0.576289

=0.576289

0.5763

Solution-c:

P(39.3<Xbar<40.7)

=NORMDIST(40.7,40,0.625,TRUE)-=NORMDIST(39.3,40,0.625,TRUE)

=0.868643-0.131357

=0.737286

0.7373