Question

Suppose GRE Quantitative scores are normally distributed with a mean of 587587 and a standard deviation of 152152. A university plans to offer tutoring jobs to students whose scores are in the top 14%14%. What is the minimum score required for the job offer? Round your answer to the nearest whole number, if necessary.

Answer 1

Answer:

The minimum score required for the job offer is 751.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

$\mu = 587, \sigma = 152$

What is the minimum score required for the job offer?

Top 14%, so the minimum score is the 100-14 = 86th percentile, which is X when Z has a pvalue of 0.86. So X when Z = 1.08.

Then

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

$1.08 = \frac{X - 587}{152}$

$X - 587 = 1.08*152$

$X = 751.16$

Rounding to the nearest whole number:

The minimum score required for the job offer is 751.