Chapter 12.5, Problem 2E

To determine

To calculate: the score for the following percentiles ${99}^{\text{th}}$ , $2^{\text{nd}}$ , ${50}^{\text{th}}$ and ${84}^{\text{th}}$

Answer to Problem 2E

Test score at ${99}^{\text{th}}$ percentile is 49.32

Test score at $2^{\text{nd}}$ percentile is 31.76

Test score at ${50}^{\text{th}}\text{percentile is}$ 0

Test score at ${84}^{\text{th}}$ percentile is 44

Explanation of Solution

Given information:

The scores were normally distributed with a mean of 40 and a standard deviation of $4.$

Calculation:

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

  $\mu =40,\sigma =4$

The area left to z-score 2.33 is $0.99$ so test score at ${99}^{\text{th}}$ percentile is,

  $\begin{array}{cc}z& =\frac{x-\mu }{\sigma }\\ 2.33& =\frac{x-40}{4}\\ x& =49.32\end{array}$

Hence, test score at ${99}^{\text{th}}$ percentile is 49.32

The area left to z-score -2.06 is $0.02$ so test score at $2^{\text{nd}}$ percentile is,

  $z=\frac{x-\mu }{\sigma }$

  $-2.06=\frac{x-40}{4}$

  $x=31.76$

Hence, test score at $2^{\text{nd}}$ percentile is 31.76

  $\begin{array}{l}\text{The area left to z-score }0\text{ is }\mathrm{0.50.}\text{ So test score at }{50}^{\text{th}}\text{percentile is,}\\ z=\frac{x-\mu }{\sigma }\\ 0=\frac{x-40}{4}\\ x=40\end{array}$

Hence, test score at ${50}^{\text{th}}\text{percentile is}$ 0

The area left to z-score 1.00 is approximately $1.00$ so test score at ${84}^{\text{th}}$ percentile is,

  $\begin{array}{cc}z& =\frac{x-\mu }{\sigma }\\ 1.00& =\frac{x-40}{4}\\ x& =44\end{array}$

Hence, test score at ${84}^{\text{th}}$ percentile is 44