Assume that women\'s heights are normally distributed with a mean given by mu =
ID: 3151047 • Letter: A
Question
Assume that women's heights are normally distributed with a mean given by mu = 64.5 in, and a standard deviation given by sigma = 1.9 in. Complete parts a and b. 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 5 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.)Explanation / Answer
a.For X=64.0, z=(x-mu)/sigma
=(64-64.5)/1.9
=-0.26
For X=65, z=(65-64.5)/1.9=0.26
Thus P(64<X<65)
=P(X<65)-P(X<64)
=P(z<0.26)-P(z<-0.26)
=0.6026-0.3974
=0.2052
b.If 5 women are randomly selected then using central limit theorem the sampling distribution of sample mean is also normal with mean 64.5 and sigma=1.9/root over 5=0.85
For X=64.0, z=(x-mu)/sigma
=(64-64.5)/0.85
=-0.58
For X=65, z=(65-64.5)/0.85=0.58
Thus P(64<X<65)
=P(X<65)-P(X<64)
=P(z<0.58)-P(z<-0.58)
=0.7190-0.2810
=0.438