Question

Suppose GRE Verbal scores are normally distributed with a mean of 461 and a standard deviation of 118. A university plans to recruit students whose scores are in the top 4%. What is the minimum score required for recruitment? Round your answer to the nearest whole number, if necessary

Answer 1

Answer:

The minimum score required for recruitment is 668.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 461 \sigma = 118$

Top 4%

A university plans to recruit students whose scores are in the top 4%. What is the minimum score required for recruitment?

Value of X when Z has a pvalue of 1-0.04 = 0.96. So it is X when Z = 1.75.

$Z = \frac{X - \mu}{\sigma}$

$1.75 = \frac{X - 461}{118}$

$X - 461 = 1.75*118$

$X = 667.5$

Rounded to the nearest whole number, 668

The minimum score required for recruitment is 668.