posted by alex .
The heights of 18-year-old men are approximately normally distributed with mean 68 inches and standard deviation 3 inches. What is the probability that the average height of a sample of ten 18-year-old men will be less than 70 inches? Round your answer to four decimal places.
Are you looking for the average or the individual score? This would be for the latter. For the former, you need to have the sample size to find the SE (see following post).
Z = (score - mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Sorry, it was not your post. Here it is.
98% interval = mean ± SE
SE (Standard error) = SD/√(n-1)
Z = (score - mean)/SE