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 3227 Round your answer to the nearest whole number,

Answer 1

82% of scores were between 286 and 322

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 310 and a standard deviation of 12.

This means that
`\mu = 310, \sigma = 12`

What percent of scores were between 286 and 322?

The proportion is the pvalue of Z when X = 322 subtracted by the pvalue of Z when X = 286. So

X = 322

`Z = (X - \mu)/(\sigma)`

`Z = (322 - 310)/(12)`

`Z = 1`

`Z = 1` has a pvalue of 0.8413

X = 286

`Z = (X - \mu)/(\sigma)`

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

`Z = -2`

`Z = -2` has a pvalue of 0.0228

0.8413 - 0.0228 = 0.8185

0.8185*100% = 81.85%

Rounding to the nearest whole number

82% of scores were between 286 and 322