Question 612409: In april 2004 article on Hear the Issue.com stated that Americans have an average of 2.24 televisions per households. If the standard deviations for the number of televisions in a U.S household is 1.2 and a random sample of 80 American households is selected, the mean of this sample belongs to a sampling disributions.
What is the mean and standard deviation of this sampling distribution?
2. Assume that the population of heights of male college students is approximately normally distributed with the mean of 69.09 inches and standard deviation of 4.71 inches. A random sample of 92 heights is obtained.
A. Find the mean and standard error of the x distribution.
B. Find P(x>68.50)
Help is greatly appreciated.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In april 2004 article on Hear the Issue.com stated that Americans have an average of 2.24 televisions per households. If the standard deviations for the number of televisions in a U.S household is 1.2 and a random sample of 80 American households is selected, the mean of this sample belongs to a sampling disributions.
1. What is the mean and standard deviation of this sampling distribution?
u of x-bar = u of x = 2.24
s of x-bar = (s of x)/sqrt(n) = 1.2/sqrt(80) = 0.1342
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2. Assume that the population of heights of male college students is approximately normally distributed with the mean of 69.09 inches and standard deviation of 4.71 inches. A random sample of 92 heights is obtained.
A. Find the mean and standard error of the x-bar distribution.
mean of x-bar = 68.09
std of x-bar = 4.71/sqrt(92) = 0.4911
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B. Find P(x-bar >68.50)
z(68.5) = (68.5-68.09)/0.4911 = 0.8349
P(x-bar > 68.50) = P(z > 0.8349) = 0.2019
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Cheers,
Stan H.
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