Chapter 6.3, Problem 35PS

Solution

a)

To find: The z-score for Lisa’s test score.

The z-score is 2.38.

Given information:

  $\begin{array}{l}\text{Lisa:}\\ \text{Mean}\left(\mu \right)=20\\ \text{Standard deviation}\left(\sigma \right)=4.2\\ \text{Ann:}\\ \text{Mean}\left(\mu \right)=500\\ \text{Standard deviation}\left(\sigma \right)=90\end{array}$

Formula used:

The formula of the z-score is:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ \text{where}\\ x=\text{Row score}\\ \mu \text{=Population mean}\\ \sigma \text{=Population standard deviation}\end{array}$

Calculation:

Lisa’s score is 30.

The z-score for Lisa is calculated as:

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{30-20}{4.2}\\ =2.38\end{array}$

The z -score for Lisa is 2.38.

b)

To find: The z-score for Ann’s test score.

The sample size is 4147.

Calculation:

Ann’s score is 610.

The z-score for Ann is calculated as:

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{610-500}{90}\\ =1.2222\end{array}$

The z -score for Ann is 1.2222

c)

To find: The student that has performed better in the entrance test.

Lisa has performed better.

From the above calculations, it is clear that the z-score of Lisa is higher as compared to that of Ann.

Thus, Lisa has performed better in the entrance test.