Assume that women\'s heights are normally distributed with a mean given by mu eq
ID: 3229297 • Letter: A
Question
Assume that women's heights are normally distributed with a mean given by mu equals 63.5 in=63.5 in, and a standard deviation given by sigma equals 2.8 in=2.8 in. (a) If 1 woman is randomly selected, find the probability that her height is less than 64in. (b) If 32 women are randomly selected, find the probability that they have a mean height less than 64 in.
(a)
The probability is approximately (Round to four decimal places as needed.)
(b) The probability is approximately (Round to four decimal places as needed.
Explanation / Answer
(a) Z = (x - mu)/sigma = (64-63.5)/2.8 = 0.1786
P(X < 64) = P(Z < 0.1786) = 0.5709
(b) In this case, the desired probability is for the mean of a sample of 32 women, therefore use the central limit theorem. The standard deviation of the sample means is 2.8/sqrt(32) = 0.495
Since n > 30, we use the z-distribution.
Z = (64-63.5)/0.495 = 1.01
Then P(X < 64) = P(Z < 1.01) = 0.8438