Chapter 12.5, Problem 1E

To determine

To calculate: the percentiles for the following scores 21, 15,13,9

Answer to Problem 1E

The data value 21 is at ${99.87}^{th}$ percentile.

The data value 15 is ${50}^{\text{th}}$ percentile

The data value 13 is a $15.87$ percentile.

The data value 9 is a ${0.13}^{th}$ percentile

Explanation of Solution

Given information:

The scores were normally distributed with a mean of 15 and a standard deviation of 2

Calculation:

Let $X$ is a random variable shows the score in geometric test. Here $X$ has normal distribution with following parameters,

  $\mu =15,\sigma =2$

The z-score of $x=21$ is,

  $\begin{array}{cc}z& =\frac{x-\mu }{\sigma }\\ & =\frac{21-15}{2}\\ & =3\end{array}$

So, the area left to score 21 is

  $\begin{array}{l}P(x\le 21)=P(z\le 3)\hfill \\ =0.9987\hfill \end{array}$

Hence, data value 21 is at ${99.87}^{th}$ percentile.

The z-score of $x=15$ is,

  $z=\frac{x-\mu }{\sigma }$

  $=\frac{15-15}{2}$

  $=0$

So, the area left to score 15 is

  $\begin{array}{l}P(x\le 15)=P(z\le 0)\hfill \\ =0.50\hfill \end{array}$

Hence, data value 15 is ${50}^{\text{th}}$ percentile

The z-score of $x=13$ is,

  $z=\frac{x-\mu }{\sigma }$

  $=\frac{13-15}{2}$

  $=-1$

So, the area left to score 13 is

  $P(x\le 13)=P(z\le -1)$

  $=0.1587$

Hence, data value 13 is a $15.87$ percentile.

  $\begin{array}{l}\text{The z-score of }x=9\text{ is,}\\ z=\frac{x-\mu }{\sigma }\\ =\frac{9-15}{2}\\ =-3\end{array}$

So, the area left to score 9 is

  $P(x\le 9)=P(z\le -3)$

  $=0.0013$

Hence, data value 9 is a ${0.13}^{th}$ percentile