Question

Based on the performance of all individuals who tested between July 1, 2013 and June 30, 2016, the GRE Verbal Reasoning scores are normally distributed with a mean of 149.97 and a standard deviation of 8.49. (https://www.ets.org/s/gre/pdf/gre_guide_table1a.pdf).

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(a) For a sample of size 49, state the standard deviation of the sample mean (the "standard error of the mean"). (Round your answer to three decimal places)

(b) Suppose a sample of size 49 is taken. Find the probability that the sample mean GRE Verbal Reasoning scores is more than 152. (Round your answer to three decimal places)

Answer 1

Let X be the random variable denoting the GRE Verbal

Reasoning score.

Thus, X ~ N(149.97, 8.49) i.e. (X - 149.97)/8.49 ~ N(0,1)

(a) For a sample of size 49, the standard deviation of sample

mean = = 1.2129. (Ans).

(b) Let M be the sample mean of a sample of size 49.

Thus, M ~ N(149.97, 1.2129) i.e. (M - 149.97)/1.2129 ~ N(0,1)

The required probability that the verbal reasoning score is

more than 152 = P(M > 152)

= P[(M - 149.97)/1.2129 > (152 - 149.97)/1.2129]

= P[(M - 149.97)/1.2129 > 1.6737]

= 1 - P[(M - 149.97)/1.2129 < 1.6737]

= 1 - (1.6737) = 1 - 0.9529 = 0.0471. (Ans).