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posted by Anonymous .
the average number of pushups a United States Marine does daily is 300, with a standard deviation of 50. A random sample of 36 marines is selected. What is the probability that the sample mean is at most 320 pushups?
With a mean of 300, you are looking for a sample using a Standard Error of the Mean (SE). (The SE functions the exact same way for a distribution of means as the SD does for a distribution of raw scores.)
SE = SD/(sq.rt.of N) = 50/6
Once you can determine how many SE above the mean the value of 320 is, use the "Area Under the Normal Distribution" table in the back of your statistics book to find the probability.
I hope this helps. Thanks for asking.
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