SOLUTION: given the mean of 100 and the standard deviation of 30 for a normally distributed population of observations. Suppose you randomly selected from the population a sample size of 36�

Question 953455: given the mean of 100 and the standard deviation of 30 for a normally distributed population of observations. Suppose you randomly selected from the population a sample size of 36��b. what is the probability that the sample mean will fall above 92? c. what is the probability that the sample mean will fall more than 8 points above the population mean of 100?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
given the mean of 100 and the standard deviation of 30 for a normally distributed population of observations. Suppose you randomly selected from the population a sample size of 36��
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b. what is the probability that the sample mean will fall above 92?
z(92) = (92-100)/[30/sqrt(36)] = -8/5
P(x-bar > 92) = P(z > -8/5) = normalcdf(-8/5,100) = 0.9452
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c. what is the probability that the sample mean will fall more than 8 points above the population mean of 100?
z(108-100)/[10/sqrt(36)] = 8/5
P(x-bar > 108) = normalcdf(8/5,100) = 0.0548
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Cheers,
Stan H.
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