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.
Accepted Solution
A:
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