Chapter 5, Problem 5.1P

Scores on a quiz were normally distributed and had a mean of 10 and standard deviation of 3. For each of the following scores, find the $Z$ score and the percentage of area above and below the score.

$X_i$	$Z$ score	% Area Above	% Area Below
5
6
7
8
9
11
12
14
15
16
18

To determine

To find:

The $Z$ score, percentage of area above and below the score.

Answer to Problem 5.1P

Solution:

The $Z$ score and the percentage of area above and below the score is given in the table below,

$X_i$	$Z$ score	% Area Above	% Area Below
5	$-1.67$	95.25	4.75
6	$-1.33$	90.82	9.18
7	$-1$	84.13	15.87
8	$-0.67$	74.86	25.14
9	$-0.33$	62.93	37.07
11	$0.33$	37.07	62.93
12	$0.67$	25.14	74.86
14	$1.33$	9.18	90.82
15	$1.67$	4.75	95.25
16	$2$	2.28	97.72
18	$2.67$	0.38	99.62

Explanation of Solution

Given:

The scores of the quiz are normally distributed with a mean of 10 and standard deviation of 3.

The scores of the quiz are-

$5,6,7,8,9,11,12,14,15,16,18$

Description:

The normal curve is symmetrical and its mean is equal to the median. The area below and above the mean is 50% or 0.05.

Formula used:

Let the data values be denoted by $X_i$.

The formula to calculate the $Z$ score is given by,

$Z=\frac{X_i-\overline{X}}{s}$

Where, $\overline{X}$ is the mean and $s$ is the standard deviation of the distribution.

Calculation:

Given that the mean is 10 and standard deviation is 3.

For the score of 5,

Substitute 5 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{5-10}{3}\\ & =& \frac{-5}{3}\\ & =& -1.67\end{array}$

From the normal distribution table the area below the score of $-1.67$ is $0.0475$.

The area between the mean and the score $-1.67$ is 0.4525.

Use the concept of symmetry, the area above the score of $-1.67$ is, $0.4525+0.5=0.9525$

The percentage of area below and above the score of $-1.67$ is 4.75% and 95.25% respectively.

For the score of 6,

Substitute 6 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{6-10}{3}\\ & =& \frac{-4}{3}\\ & =& -1.33\end{array}$

From the normal distribution table the area below the score of $-1.33$ is $0.0918$.

The area between the mean and the score $-1.33$ is 0.4082.

Use the concept of symmetry, the area above the score of $-1.33$ is, $0.4082+0.5=0.9082$

The percentage of area below and above the score of $-1.33$ is 9.18% and 90.82%.

For the score of 7,

Substitute 7 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{7-10}{3}\\ & =& \frac{-3}{3}\\ & =& -1\end{array}$

From the normal distribution table the area below the score of $-1$ is $0.1587$.

The area between the mean and the score $-1$ is 0.3413.

Use the concept of symmetry, the area above the score of $-1$ is, $0.3413+0.5=0.8413$

The percentage of area below and above the score of $-1$ is 15.87% and 84.13%.

For the score of 8,

Substitute 8 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{8-10}{3}\\ & =& \frac{-2}{3}\\ & =& -0.67\end{array}$

From the normal distribution table the area below the score of $-0.67$ is $0.2514$.

The area between the mean and the score $-0.67$ is 0.2486.

Use the concept of symmetry, the area above the score of $-0.67$ is, $0.2486+0.5=0.7486$

The percentage of area below and above the score of $-0.67$ is 25.14% and 74.86%.

For the score of 9,

Substitute 9 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{9-10}{3}\\ & =& \frac{-1}{3}\\ & =& -0.33\end{array}$

From the normal distribution table the area below the score of $-0.33$ is $0.3707$.

The area between the mean and the score $-0.33$ is 0.1293.

Use the concept of symmetry, the area above the score of $-0.33$ is, $0.1293+0.5=0.6293$

The percentage of area below and above the score of $-1$ is 37.07% and 62.93%.

For the score of 11,

Substitute 11 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{11-10}{3}\\ & =& \frac{1}{3}\\ & =& 0.33\end{array}$

From the normal distribution table the area above the score of $0.33$ is $0.3707$.

The area between the mean and the score $0.33$ is 0.1293.

Use the concept of symmetry, the area below the score of $0.33$ is, $0.1293+0.5=0.6293$

The percentage of area above and below the score of $0.33$ is 37.07% and 62.93%.

For the score of 12,

Substitute 12 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{12-10}{3}\\ & =& \frac{2}{3}\\ & =& 0.67\end{array}$

From the normal distribution table the area above the score of $0.67$ is $0.2514$.

The area between the mean and the score $0.67$ is 0.2486.

Use the concept of symmetry, the area below the score of $0.67$ is, $0.2486+0.5=0.7486$.

The percentage of area above and below the score of $0.67$ is 25.14% and 74.86%.

For the score of 14,

Substitute 14 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{14-10}{3}\\ & =& \frac{4}{3}\\ & =& 1.33\end{array}$

From the normal distribution table the area above the score of $1.33$ is $0.0918$.

The area between the mean and the score $1.33$ is 0.4082.

Use the concept of symmetry, the area below the score of $1.33$ is, $0.4082+0.5=0.9082$.

The percentage of area above and below the score of $1.33$ is 9.18% and 90.82%.

For the score of 15,

Substitute 15 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{15-10}{3}\\ & =& \frac{5}{3}\\ & =& 1.67\end{array}$

From the normal distribution table the area above the score of $1.67$ is $0.0475$.

The area between the mean and the score $1.67$ is 0.4525.

Use the concept of symmetry, the area below the score of $1.67$ is, $0.4525+0.5=0.9525$

The percentage of area above and below the score of $1.67$ is 4.75% and 95.25% respectively.

For the score of 16,

Substitute 16 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{16-10}{3}\\ & =& \frac{6}{3}\\ & =& 2\end{array}$

From the normal distribution table the area above the score of $2$ is $0.0228$.

The area between the mean and the score $2$ is 0.4772.

Use the concept of symmetry, the area below the score of $2$ is, $0.4772+0.5=0.9772$.

The percentage of area above and below the score of $2$ is 2.28% and 97.72% respectively.

For the score of 18,

Substitute 18 for $X_i$, 10 for $\overline{X}$, and 3 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{18-10}{3}\\ & =& \frac{8}{3}\\ & =& 2.67\end{array}$

From the normal distribution table the area above the score of $2.67$ is $0.0038$.

The area between the mean and the score $2.67$ is 0.4962.

Use the concept of symmetry, the area below the score of $2.67$ is, $0.4962+0.5=0.9962$.

The percentage of area above and below the score of $2.67$ is 0.38% and 99.62% respectively.

Conclusion:

Therefore, the $Z$ score and the percentage of area above and below the score is given in the table below,

$X_i$	$Z$ score	% Area Above	% Area Below
5	$-1.67$	95.25	4.75
6	$-1.33$	90.82	9.18
7	$-1$	84.13	15.87
8	$-0.67$	74.86	25.14
9	$-0.33$	62.93	37.07
11	$0.33$	37.07	62.93
12	$0.67$	25.14	74.86
14	$1.33$	9.18	90.82
15	$1.67$	4.75	95.25
16	$2$	2.28	97.72
18	$2.67$	0.38	99.62