SOLUTION: It is known that the time it takes to mark assignments are normally distributed with a population standard deviation of 3.5 minutes. If a random sample of 81 assignments should be

Question 1195910: It is known that the time it takes to mark assignments are normally distributed with a population
standard deviation of 3.5 minutes. If a random sample of 81 assignments should be marked, what is the probability that the average time to mark these assignments will differ by no more than 1.2 minutes in either direction from the unknown population mean marking time?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
population standard deviation = 3.5
sample standard error = 3.5 / sqrt(81) = .3889
z = (x-m)/.3889
(x-m) is plus or minus 1.2
on the low side, you get z = - 1.2/.3889 = -3.0856
on the high side, you get z = 1.2/.3889 = +3.0856
the probability of getting a z-score between those two limits is equal to .998.
the mean can be anything as long as the standard deviation is plus or minus 1.2 from that.
for example, take a mean of 50 and take a mean of 500.
if the mean is 50, than you want the probability of getting a raw score between 48.8 and 51.2.
if the mean is 500, then you want the probability of getting a raw score between 498.8 and 501.2.
here are the results using the z-score calculator at https://davidmlane.com/hyperstat/z_table.html

as you can see, the probability is .998 in both cases.

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