Chapter 13, Problem 30RE

a.

To determine

Find the z -score for the first student’s test score.

Answer to Problem 30RE

  $z\approx \mathrm{2.4.}$

Explanation of Solution

Given information: Two students took different college entrance tests .The scores on the

test that the first student took are normally distributed with a mean of 20 points and a standard deviation of 4.2 points .The scores on the test that the second student took are normally distributed with a mean of 500 points and a standard deviation of 90 points .The students scored 30 and 610 on their tests, respectively.

Calculation:

The z-score is given by: $z=\frac{x-\overline{x}}{\sigma }.$

For the first student:

  $\begin{array}{l}x=30,\overline{x}=20,\sigma =\mathrm{4.2.}\\ So,\\ z=\frac{30-20}{4.2}\approx 2.4\end{array}$

b.

To determine

Find the z -score for the second student’s test score.

Answer to Problem 30RE

  $z\approx \mathrm{1.2.}$

Explanation of Solution

Given information: Two students took different college entrance tests .The scores on the

test that the first student took are normally distributed with a mean of 20 points and a standard deviation of 4.2 points .The scores on the test that the second student took are normally distributed with a mean of 500 points and a standard deviation of 90 points .The students scored 30 and 610 on their tests, respectively.

Calculation:

The z-score is given by: $z=\frac{x-\overline{x}}{\sigma }.$

For the second student:

  $\begin{array}{l}x=610,\overline{x}=500,\sigma =90.\\ So,\\ z=\frac{610-500}{90}\approx 1.2\end{array}$

c.

To determine

Which student scored better on their college entrance test.

Answer to Problem 30RE

The first student has higher score since he/she got 2.4 standard deviations higher than the mean.

Explanation of Solution

Given information: Two students took different college entrance tests .The scores on the

test that the first student took are normally distributed with a mean of 20 points and a standard deviation of 4.2 points .The scores on the test that the second student took are normally distributed with a mean of 500 points and a standard deviation of 90 points .The students scored 30 and 610 on their tests, respectively.

Calculation:

The z-score is given by: $z=\frac{x-\overline{x}}{\sigma }.$

For the first student:

  $\begin{array}{l}x=30,\overline{x}=20,\sigma =\mathrm{4.2.}\\ So,\\ z=\frac{30-20}{4.2}\approx 2.4\end{array}$

For the second student:

  $\begin{array}{l}x=610,\overline{x}=500,\sigma =90.\\ So,\\ z=\frac{610-500}{90}\approx 1.2\end{array}$

The first student has higher score since he/she got 2.4 standard deviations higher than the mean.