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.

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.

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 $155.8625$.

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.

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 $170.3875$