Question 1109340: Assume that women's heights are normally distributed with a mean given by mu equals 63.1 in, and a standard deviation given by sigma equals 2.7 in.
(a) If 1 woman is randomly selected, find the probability that her height is less than 64 in.
(b) If 32 women are randomly selected, find the probability that they have a mean height less than 64 in
I have a ti-84 but not sure how to solve
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Assume that women's heights are normally distributed with a mean given by mu equals 63.1 in, and a standard deviation given by sigma equals 2.7 in.
(a) If 1 woman is randomly selected, find the probability that her height is less than 64 in.
z(64) = (64-63.1)/2.7 = 0.9/2.7 = 1/3
P(h < 64) = P(z < 1/3) = normalcdf(-100,1/3) = 0.6306
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(b) If 32 women are randomly selected, find the probability that they have a mean height less than 64 in
z(64) = 0.9/(2.7/sqrt(32)) = 1.8856
P(x-bar < 64) = P(z < 1.8856) = normalcdf(-100,1.8856) = 0.9703
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Cheers,
Stan H.
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