Question 1064321
Assume that​ women's heights are normally distributed with a mean given by mu equals 62.4 in, and a standard deviation given by sigma equals 1.8 in. 
​(a) If 1 woman is randomly​ selected, find the probability that her height is less than 63 in. 
​(b) If 30 women are randomly​ selected, find the probability that they have a mean height less than 63 in.
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For samples of size 30 the standard deviation becomes 1.8/sqrt(30).
z(63) = (63-62.4)/(1.8/sqrt(30)) = 1.8257
P(x-bar< 63) = P(z < 1.8257) = normalcdf(-100,1.8257) = 0.9661
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a)The probability is approximately  0.6293
(b) The probability is approximately 0.9661
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Cheers,
Stan H.
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