Question

The GRE is an entrance exam that most students are required to take upon entering graduate school. In 2014, the combined scores for the Verbal and Quantitative sections were approximately normally distributed with a mean of 310 and a standard deviation of 12.

What percent of scores were between 286 and 322? Round your answer to the nearest whole number.

Answer 1

82% of the scores will be between 286 and 322.

To find this answer, we need to find the z-scores for 286 and 322. This is done by diving the difference from the mean by the standard deviation.

(286 - 310) / 12 = -2 which gives a percent of 2.28%
(322 - 310) / 12 = 1 which gives a percent of 84.13%

Subtracting those 2 percents will tell us the difference between the two scores.

84.13 - 2.28 = 81.85% or about 82%

Answer 2

82%

Explanation:

Data:

mean, μ = 310

standard deviation, σ = 12

We need to use the standard normal distribution table (see figures attached).

Z is calculated as follows:

Z = (x - μ)/σ

Replacing with x = 286 and x = 322:

Z = (286 - 310) / 12 = -2

Z = (322 - 310) / 12 = 1

So, we need to find:

P(-2 ≤ Z ≤ 1) =

= P(-2 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 1)

From the first table:

P(-2 ≤ Z ≤ 0) = 0.5 - P(Z ≤ -2) = 0.5 - 0.0228 = 0.4772

From the second table:

P(0 ≤ Z ≤ 1) = 0.3413

Then,

P(-2 ≤ Z ≤ 1) = 0.4772 + 0.3413 = 0.8185, which as a percentage is 0.8185*100 = 81.85% ≈ 82%