Q. 28 - Elementary Statistics

Question

28. Verbal GRE Scores. The Graduate Record Examination (GRE) is a standardized test that students usually take before entering graduate school. According to the document GRE Guide to the Use of Scores, a publication of the Educational Testing Service, the scores on the verbal portion of the GRE have mean $150$points and standard deviation $8.75$ points. Assuming that these scores are (approximately) normally distributed,

a. obtain and interpret the quartiles.

b. find and interpret the 99th percentile.

Step-by-Step Solution

Answer

a. The variable has quartiles of $144.1375,150,$ and $155.8625$, and $25\%$of the observations are less than $144.1375,50\%$ of the observations are less than $150$ , and $75\%$ of the observations are less than $155.8625$.

b. There are $99\%$ of the observations are smaller than $170.3875.$

Part (a) Step 1: Given Information

On the verbal portion of the GRE, the mean score is $150$ points, the standard deviation is $8.75$points, and the scores are normally distributed.

Determine the quartiles and interpret it.

Part (a) Step 2: Explanation

Applying Formula:

The basic $z-$ scores formula for a sample is:

$$\begin{gathered} X=\mu +z\sigma \\ X-\mu =z \\ \sigma z=\frac{X-\mu }{\sigma } \end{gathered}$$

Determine the $z-$ scores corresponding to the percent $0.25,0.50$ and $0.75$ which are:

$-0.67,0,0.67$

Compute the corresponding value:

$\begin{array}{rl}& x=\mu +z\sigma =150+(-0.67)\times 8.75=144.1375\\ & x=\mu +z\sigma =150+\left(0\right)\times 8.75=150\\ & x=\mu +z\sigma =150+\left(0.67\right)\times 8.75=155.8625\end{array}$

In other words, $25\%$ of the observations are smaller than $144.1375, 50\%$ of the observations are smaller than $150$, and $75\%$ of the observations are smaller than $155.8625$.

Part (b) Step 1: Given Information

The verbal section of the GRE has a mean of $150$ points and a standard deviation of $8.75$ points, with the points distributed regularly.

Part (b) Step 2: Explanation

Applying Formula:

The basic $z-$scores formula for a sample is:

$\begin{array}{rl}& X=\mu +z\sigma \\ & X-\mu =z\sigma \\ & z=\frac{X-\mu }{\sigma }\end{array}$

Computation:

Determine the $z-$scores corresponding to the percent $0.99$ which is $2.33$.

Compute the corresponding value:

$x=\mu +z\sigma =150+(2.33)\times 8.75=170.3875$

Observation:$99\%$of the observations are smaller than $170.3875$