Question 974249: Assume that the heights of women are normally distributed with a mean of 63.5 inches and a standard deviation of 2.8 inches.
A. If 1 woman is randomly selected, find the probability that her height is less than 64 in.
B. If 44 women are randomly selected, find the probability that they have a mean height less than 64 in.
So far I got part A.
64-63.5/2.8
=0.18
The probability is approximately 0.5714 (0.1, .08) on the "Positive z Scores"
"Standard Normal (z) Distribution: Cumulative Area from the Left"
For part B I am stuck on
In this case, the desired probability is for the mean of a sample of 44 women, therefore use the central limit theorem.
The probability is approx.?
2.8/sqrt44
Please help me and show the work so that I understand…I am not just looking for an answer! Thank you
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Assume that the heights of women are normally distributed with a mean of 63.5 inches and a standard deviation of 2.8 inches.
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A. If 1 woman is randomly selected, find the probability that her height is less than 64 in.
So far I got part A.
(64-63.5)/2.8 = 0.18
P(x < 64) = P(z< 0.18) = is approximately 0.5714
You are correct
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B. If 44 women are randomly selected, find the probability that they have a mean height less than 64 in.
For part B I am stuck on
In this case, the desired probability is for the mean of a sample of 44 women, therefore use the central limit theorem.
The standard deviation of the sample means is 2.8/sqrt44 = 0.42
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Since n > 30, use the z-distribution.
z(64) = (64-63.5)/0.42 = 1.19
Then P(x-bar < 64) = P(z < 1.19) = 0.88
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Hope that helps.
Cheers,
Stan H.
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