Assume that women's heights are normally distributed with a mean given by, and a standard deviation given by. μ = 63.5 in σ = 2.8 in (a) if 1 woman is randomly selected, find the probability that her height is less than 64 in. (b) if 33 women are randomly selected, find the probability that they have a mean height less than 64 in.

(a) if 1 woman is randomly selected, find the probability that her height is less than 64 in

using z-score formula:

z-score = (x-mu) / sig

(64-63.5) / 2.8

=0.18

thus

P (x<64) = P (z<0.18) - = 0.5714

B] if 33 women are randomly selected, find the probability that they have a mean height less than 64 in

using the central limit theorem of sample means, we shall have:

2.8 / √33=0.49

since n>30 we use z-distribtuion

z (64) = (64-63.5) / 0.49=1.191

The

P (x_bar<64) = P (x<1.191) = 0.8830