Question

The Graduate Record Examination (GRE) is a standardized test commonly taken by graduate school applicants in the United States. The total score is comprised of three compo- nents: Quantitative Reasoning, Verbal Reasoning, and Analytical Writing. The first two components are scored from 130 - 170. The mean score for Verbal Reasoning section for all test takers was 151 with a standard deviation of 7, and the mean score for the Quantitative Reasoning was 153 with a standard deviation of 7.67. Suppose that both distributions are nearly normal. (a) A student scores 160 on the Verbal Reasoning section and 157 on the Quantitative Reasoning section. Relative to the scores of other students, which section did the student perform better on? (b) Calculate the student’s percentile scores for the two sections. What percent of test takers per- formed better on the Verbal Reasoning section? (c) Computethescoreofastudentwhoscoredinthe80thpercentileontheQuantitativeReasoning section. (d) Compute the score of a student who scored worse than 70% of the test takers on the Verbal Reasoning section.

Answer 1

Let X be the score for Verbal Reasoning section for any given test taker. X is normally distributed with mean and a standard deviation

Let Y be the score for Quantitative Reasoning section for any given test taker. Y is normally distributed with mean and a standard deviation

a) The z score of 160 on the Verbal Reasoning section is

The z score of 157 on the Quantitative Reasoning section is

ans: 160 on the Verbal Reasoning section is 1.29 standard deviations above the average in Verbal Reasoning section. 157 on the Quantitative Reasoning section is 0.52 standard deviations above the average in Quantitative Reasoning section. Since the student's score in Verbal Reasoning section is farther above the average than the student's score in Quantitative Reasoning section, we can say that relative to the scores of other students, the student has perform better on Verbal Reasoning section.

(b) Calculate the student’s percentile scores for the two sections.

The student’s percentile score in  Verbal Reasoning section is same as the proportion of students who have scored less than the student in Verbal Reasoning section.

The probability that a randomly selected student scores less than 160 on the Verbal Reasoning section is

ans: The student’s percentile score in  Verbal Reasoning section is 90.15%

The student’s percentile score in  Quantitative Reasoning section is same as the proportion of students who have scored less than the student in Quantitative Reasoning section.

The probability that a randomly selected student scores less than 157 on the Quantitative Reasoning section is

ans: The student’s percentile score in  Quantitative Reasoning section is 69.85%

The percent of test takers that performed better on the Verbal Reasoning section is

100- (The percent of test takers that performed worse on the Verbal Reasoning section)

100-90.15 = 9.85%

ans: The percent of test takers that performed better on the Verbal Reasoning section is 9.85%

(c) Compute the score of a student who scored in the 80th percentile on the Quantitative Reasoning section

Let q be the 80th percentile on the Quantitative Reasoning section.

That is, the probability that any given student scored less than q on the Quantitative Reasoning section is 0.80

We need

In terms of the z values, we need

Using the standard normal tables, we get for z=0.84,

P(Z<0.84)=0.80

We need

We can equate the z score of q to 0.84 and get

ans: the score of a student who scored in the 80th percentile on the Quantitative Reasoning section is 159.44

(d) Compute the score of a student who scored worse than 70% of the test takers on the Verbal Reasoning section.

Let q be score of a student who scored worse than 70% of the test takers on the Verbal Reasoning section. This means 70% of the students have scored greater than q on the Verbal Reasoning section

That is, the probability that any given student scored greater than q on the Verbal Reasoning section is 0.70

We need

In terms of the z values, we need

However, we know that P(Z<0)=0.5. We need a probability which is less than 0.5, and hence z must be negative.

which is

Using the z tables, we get for z=0.52, P(Z<0.52)=0.70

Or

We need

We can equate the z score of q to -0.52 and get

ans: the score of a student who scored worse than 70% of the test takers on the Verbal Reasoning section is 147.36