Based on the performance of all individuals who tested between July 1, 2012 and

Question

Based on the performance of all individuals who tested between July 1, 2012 and June 30, 2015, the GRE Verbal Reasoning scores are normally distributed with a mean of 150 and a standard deviation of 8.45. (https://www.ets.org/s/gre/pdf/gre_guide_table1a.pdf). Show all work. Just the answer, without supporting work, will receive no credit.

(a) Consider all random samples of 36 test scores. What is the standard deviation of the sample means?

(b) What is the probability that 36 randomly selected test scores will have a mean test score that is between 148 and 152?

Explanation / Answer

(a) Standard deviation of the sample mean is given by standard deviation of the popoulation/ sqrt(n)

=8.45/sqrt(36)

=8.45/6

=1.4083

(b) P(148<x<152) = P(-2/1.4083 < x <2/1.4083) = P(-1.4201 < z < 1.4201 ) = 0.8444 [from the z table]

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