What's the percentage using the quantitative mean and SD?

Question

The GRE has a combined verbal and quantitative mean of 1000 and a standard deviation of 200. Scores are distributed normally.

(A) What percentage of test takers score above 600?

(B) What percentage of test takers score below 1400?

(C) Below what score do about 84% of test takers score?

Answer 1

In this kind of problem, you can use the formula for calculating a z-score which is z = (x-μ)/σ. It is also commonly used in normal distribution.

where:

z= is the z score

x = raw score

μ = mean =1000

σ = standard deviation = 200

Take note: Since the mean is 1000, all scores greater 1000 is 50% of the test takers score and all scores less than 1000 is 50% as well. Furthermore, the z score for a raw score of 1000 is z=0.

(A) Therefore, the percentage of test takers score above 600 should be above 50%

z= (600 - 1000)/200 = -2

Using the table or the z calculator:

P(z > -2) = 0.9772 or 97.72%

(B) z = (1400 - 1000)/200 = 2

P(z < 2) = 0.9772 or 97.72%

(C) Using a z calculator, the value of z of the score we are looking for is z=0.994

.994 = (x-1000)/200

198.8 = x - 1000

x = 1000 +198.8 = 1198.8

Therefore, below a score of 1198.8 is 84% of test takers score.