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.)

Question

Greyson Phillips

Mathematics

13 November, 03:53

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.)

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Danna Kerr · 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