Question

In: Statistics and Probability

Scores on the GRE. A college senior who took the Graduate Record Examination exam scored 580...

Scores on the GRE. A college senior who took the Graduate Record Examination exam scored 580 on the Verbal Reasoning section and 650 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section was 459 with a standard deviation of 114, and the mean score for the Quantitative Reasoning was 432 with a standard deviation of 148. Suppose that both distributions are nearly normal. Round calculated answers to 4 decimal places unless directed otherwise.

1.Write down the short-hand for these two normal distributions.

The Verbal Reasoning section has a distribution N( , )

The Quantitative Reasoning section has a distribution of N( , )

2. What is her Z score on the Verbal Reasoning section?

3. What is her Z score on the Quantitative Reasoning section?

4. Relative to others, which section did she do better on?

A. She did the same on both sections
B. Verbal Reasoning
C. Quantitative Reasoning

5. What is her percentile score on the Verbal Reasoning section? Round to nearest whole number.

6. What is her percentile score on the Quantitative Reasoning section? Round to nearest whole number.

7. What percent of the test takers did better than she did on the Verbal Reasoning section?  %

8. What percent of the test takers did better than she did on the Quantitative Reasoning section?  %

9. What is the score of a student who scored in the 71th71th percentile on the Quantitative Reasoning section? Round to the nearest integer.

10. What is the score of a student who scored worse than 74% of the test takers in the Verbal Reasoning section? Round to the nearest integer.

Solutions

Expert Solution

  1. The verbal reasoning section has a distribution of N (459, 1142).

The Quantitative reasoning section has a distribution of N (432, 1482)

2.

The formulation of Z score comes from normal distribution. If we consider the observed values coming from a population which follows normal distribution with mean µ and variance σ2 and if we denote the observed values as Xi then these Xi’s can be considered as random samples from normal population having mean µ and variance σ2.

Then, using standardization technique, we know, (Xi - µ)/σ follows standard normal distribution with mean 0 and variance 1, which is denoted as N (0,1)
For each value of Xi, we get a value of (Xi - µ)/σ. These values are actually denoted as Z-scores.

Here we are assuming that both distributions are normal and she scored 580 on the verbal reasoning section.

Hence, Z score in the verbal reasoning section will be (580-459) / 114 = 1.0614 (up to 4 decimal places)

3. She got 650 on the quantitative reasoning section.

Hence her Z score in this section will be (650-432) / 148 = 1.47297 ( up to 5 decimal places)

4. In case of quantitative reasoning she has better Z score.

Positive Z value denotes Xi - µ >0 which implies Xi> µ, that is the value is above mean.

Again, the i-th Z score, Zi= (Xi - µ)/σ implies

                        Xi - µ = σ Zi

The above expression gives that the value Xi is Zi times away from the mean as the LHS of above equation gives the difference from mean.

Z score is zero implies that the observation is equal to mean. Z score is -2 implies that the observation is 2 times away from mean and it is below the mean.

Hence, better Z score implies better result.

So, she did better on the quantitative section.

5 . Percentile score is percentage of students who got less than or equal to what she have got in the particular section.

This means we have to find the probability that P(X<580)

= P (Z< 1.0614 )

= 0.85633 ( up to 5 decimal places) (Using standard normal table, as z follows standard normal distribution)

Hence percentile score in the verbal section is 85.633% ~ 86%

6. we have to find the probability that P(Y<650)

=P(Y<1.47297)

=0.92962

So, in quantitative section her percentile score is 92.963% ~ 93%

7. (100-86) = 14% (almost) students did better than she did in the verbal section.

8. (100-93)% = 7% students did better than what she did in the quantitative section.


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