Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in , and a standard deviation given by sigma equals 2.6 in. (a) if 1 woman is randomly selected, find the probability that her height is less than 63 in. (b) if 44 women are randomly selected, find the probability that they have a mean height less than 63 in.

If 1 woman is randomly selected, the probability that her height is less than 63 in will be:

The z-score is given by:

z = (x-mu) / sig

z = (63-62.5) / 2.6

z=0.1923

Thus:

P (x<63) = P (z<0.1923) = 0.5753

(b) if 44 women are randomly selected, find the probability that they have a mean height less than 63 in.

In this scenario the desired probability is for the mean of a sample of 44 women, therefore we use the central limit theorem. The standard deviation of the sample means is:

2.6/sqrt44=0.392

since n>30, use the z-distribution

z = (63-62.5) / 0.392

z=1.28

Then

P (x-bar <63) = P (z<1.28) = 0.8997