Question 653012
for a normal population mean of 70 and population standard deviation of 20, what is the probability of obtaining a sample mean equal to or greater than 75 for a random sample of n= 100?
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standard deviation for sample means of size 100 is 20/sqrt(100) = 2
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z(75) = (75-70)/2 = 5/2 
P(x-bar >= 75) = P(z >= 5/2) = normalcdf(5/2,100) = 0.006
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for a normal population mean of 70 and population standard deviation of 20, what is the probability of obtaining a sample mean equal to or greater than 75 for a random sample of n= 50?
std = 20/sqrt(50) = 2.83
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z(75) = (75-70)/2.83 = 1.77
P(x-bar >= 75) = P(z >= 1.77) = normalcdf(1.77,100) = 0.038
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Cheers,
Stan H.