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 Quantitative Reasoning scores are normally distributed with a mean of 152.47 and a standard deviation of 8.93. (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 49 test scores. What is the standard deviation of the sample means? (Round your answer to three decimal places)

(b) What is the probability that 49 randomly selected test scores will have a mean test score that is greater than 150? (Round your answer to four decimal places)

Explanation / Answer

given mean =152.47

standard deviation =8.93

n =49

a) standard deviation or error is given by standard devition / sqrtn = 8.93/sqrt49 = 1.275

b) z = (x^-mean)/ standard deviation = (150-152.47) / 1.275= -2.47/1.275 =-1.973

now looking intp standard ztable for the value

p(-1.973) = 0.0268

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