SOLUTION: A study of 800 homeowners in a certain area showed that the average value of the homes was $82,000, and the standard deviation was $5000. If 50 homes are for sale, find the probab

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Question 1203981: A study of 800 homeowners in a certain area showed that the average value of the homes was
$82,000, and the standard deviation was $5000. If 50 homes are for sale, find the probability
that the mean of the values of these homes is greater than $83,500.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
population mean is 82,000.
population standard deviation is 5000.
sample size is 50.
standard error = standard deviation / sqrt(sample size) = 5000 / sqrt(50) = 707.106781.
sample mean is assumed to be 83,500.
formula to use is z = (x - m) / s
z is the z-score
x is the sample mean
m is the population mean
s is the standard error.
formula becomes z = (83,500 - 82,000) / 707.106781 = 2.121320.
area to the right of that z-score = .016947.
that's the probability of getting a z-score greater than 2.121320.
that's also the probability of getting a sample of 50 house from the same poulation that has a mean greater than 83,500.
here's what that looks like on the calculator at https://davidmlane.com/hyperstat/z_table.html