Assume that women\'s heights are normally distributed with a mean given by p = 6
ID: 3158966 • Letter: A
Question
Assume that women's heights are normally distributed with a mean given by p = 64.7 in. and a standard deviation given by sigma = 2.2 in. If 1 woman is randomly selected, find the probability that her height is between 64.0 in and 65 0 in. The probability is approximately (Round to four decimal places as needed.) If 7 women are randomly selected, find the probability that they have a mean height between 64.0 in and 65.0 in. The probability is approximately (Round to four decimal places as needed) Why can the central limit theorem be used in part (b). even though the sample size does not exceed 30? The population size is greater than 30. The sample size needs to be less than 30. not greater than 30. The sample is normally distributed. The population is normally distributed.Explanation / Answer
mean = 64.7
standard deviation = 2.2
a) p(64<x<65) = )
For x = 64 , z = (64 - 64.7) / 2.2 = -0.31
and for x = 65, z = (65 - 64.7) / 2.2 = 0.13
Hence P(64 < x < 65) = P(-0.31 < z < 0.13) = [area to the left of z = 0.13] - [area to the left of -0.31]
= 0.5517 - 0.3783 = 0.1734
b)
p(64<x<65) = )
For x = 64 , z = (64 - 64.7) / (2.2/sqrt(7)) = -0.84
and for x = 65, z = (65 - 64.7) / (2.2/sqrt(7) = 0.36
Hence P(64 < x < 65) = P(-0..84 < z < 0.36) = [area to the left of z = 0.36] - [area to the left of -0.84]
= 0.6406 - 0.2005 = 0.4401