Based on the performance of all individuals who tested between July 1, 2013 and

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/pdfi/gre guide tablela.pdf). Show all work. Just the answer, without supporting work, will receive no credit. 12. (a) For a sample of size 64, 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 64 is taken. Find the probability that the sample mean GRE Verbal Reasoning scores is more than 152. (Round your answer to three decimal places)

Explanation / Answer

a)

std deviation of sample mean =population std deviaition/sqrt(n)=8.49/sqrt(64)=1.061

b)

as z score =(Xbar-mean)/std deviation of sample mean

therefore P(sample mean is more than 152)=P(Xbar>152)=P(Z>(152-149.97)/1.061)

=P(Z>1.91)=0.028

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