Question

Professor smith conducted a class exercise in which students ran a computer program to generate random samples from a population that had a mean of 50 and a standard deviation of 9 mm. each of smith's students took a random sample of size n and calculated the sample mean. smith found that about 68% of the students had sample means between 48.5 and 51.5 mm. what was n? (assume that n is large enough that the central limit theorem is applicable.)

Answer 1

When xbar is 48.5

-1 is (48.5 - 50)/(9/sqrt n)
-9/sqrt n is -1.5
n is (9/1.5²) is 36

when xbar is 51.5
+1 is (51.5 - 50)/(9/sqrt n)
9/sqrt n is1.5
n is (9/1.5)² is 36