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.
<span> (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] </span><span> 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/</span>√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