Chapter 5, Problem 5.6P

If a distribution of test scores is normal, with a mean of 78 and a standard deviation of 11, what percentage of the area lies

a. below 60? __________

b. below 70? ________

c. below 80? _____________

d. below 90? ____________

e. between 60 and 65? ____________

f. between 65 and 79? ___________

g. between 70 and 95? ___________

h. between 80 and 90? __________

i. above 99? _______

j. above 89? ____________

k. above 75?_____________

l. above 65?____________

To determine

a)

To find:

The percentage of area below 60.

Answer to Problem 5.6P

Solution:

The percentage of area below 60 is 5.05%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 60,

Substitute 60 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{60-78}{11}\\ & =& \frac{-18}{11}\\ & =& -1.64\end{array}$

The area below the score $-1.64$ is 0.0505.

The percentage of area below the score of $-1.64$ is 5.05%.

Conclusion:

Therefore, the percentage of area below the score of 60 is 5.05%.

To determine

b)

To find:

The percentage of area below 70.

Answer to Problem 5.6P

Solution:

The percentage of area below 70 is 23.27%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 70,

Substitute 70 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{70-78}{11}\\ & =& \frac{-8}{11}\\ & =& -0.73\end{array}$

The area below the score $-0.73$ is 0.2327.

The percentage of area below the score of $-0.73$ is 23.27%.

Conclusion:

Therefore, the percentage of area below the score of 70 is 23.27%.

To determine

c)

To find:

The percentage of area below 80.

Answer to Problem 5.6P

Solution:

The percentage of area below 80 is 57.14%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 80,

Substitute 80 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{80-78}{11}\\ & =& \frac{2}{11}\\ & =& 0.18\end{array}$

The area between the mean and the score $0.18$ is 0.0714.

Use the concept of symmetry, the area below the score 0.18 is,

$0.0714+0.5=0.5714$

The percentage of area below the score of $0.18$ is 57.14%.

Conclusion:

Therefore, the percentage of area below the score of 80 is 57.14%.

To determine

d)

To find:

The percentage of area below 90.

Answer to Problem 5.6P

Solution:

The percentage of area below 90 is 86.21%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 90,

Substitute 90 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{90-78}{11}\\ & =& \frac{12}{11}\\ & =& 1.09\end{array}$

The area between the mean and the score $1.09$ is 0.3621.

Use the concept of symmetry, the area below the score 1.09 is,

$0.3621+0.5=0.8621$

The percentage of area below the score of $1.09$ is 86.21%.

Conclusion:

Therefore, the percentage of area below the score of 90 is 86.21%.

To determine

e)

To find:

The percentage of area between 60 and 65.

Answer to Problem 5.6P

Solution:

The percentage of area between the score 60 and 65 is 6.85%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 60,

Substitute 60 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{60-78}{11}\\ & =& \frac{-18}{11}\\ & =& -1.64\end{array}$

The area between the mean and the score $-1.64$ is 0.4495.

For the score of 65,

Substitute 65 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{c}Z=\frac{65-78}{11}\\ =\frac{-13}{11}\\ =-1.18\end{array}$

The area between the mean and the score $-1.18$ is 0.3810.

Use the concept of symmetry, the area between the score $-1.64$ and $-1.18$ is,

$0.4495-0.3810=0.0685$

The percentage of area between the score $-1.64$ and $-1.18$ is 6.85%.

Conclusion:

Therefore, the percentage of area between the score 60 and 65 is 6.85%.

To determine

f)

To find:

The percentage of area between 65 and 79.

Answer to Problem 5.6P

Solution:

The percentage of area between the score 65 and 79 is 41.69%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 65,

Substitute 65 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{65-78}{11}\\ & =& \frac{-13}{11}\\ & =& -1.18\end{array}$

The area between the mean and the score $-1.18$ is 0.3810.

For the score of 79,

Substitute 79 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{79-78}{11}\\ & =& \frac{1}{11}\\ & =& 0.09\end{array}$

The area between the mean and the score $0.09$ is 0.0359.

Use the concept of symmetry, the area between the score $-1.18$ and $0.09$ is,

$0.3810+0.0359=0.4169$

The percentage of area between the score $-1.18$ and $0.09$ is 41.69%.

Conclusion:

Therefore, the percentage of area between the score 65 and 79 is 41.69%.

To determine

g)

To find:

The percentage of area between 70 and 95.

Answer to Problem 5.6P

Solution:

The percentage of area between the score 70 and 95 is 70.55%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 70,

Substitute 70 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{70-78}{11}\\ & =& \frac{-8}{11}\\ & =& -0.73\end{array}$

The area between the mean and the score $-0.73$ is 0.2673.

For the score of 95,

Substitute 95 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{95-78}{11}\\ & =& \frac{17}{11}\\ & =& 1.54\end{array}$

The area between the mean and the score $1.54$ is 0.4382.

Use the concept of symmetry, the area between the score $-0.73$ and $1.54$ is,

$0.2673+0.4382=0.7055$

The percentage of area between the score $-0.73$ and $1.54$ is 70.55%.

Conclusion:

Therefore, the percentage of area between the score 70 and 95 is 70.55%.

To determine

h)

To find:

The percentage of area between 80 and 90.

Answer to Problem 5.6P

Solution:

The percentage of area between the score 80 and 90 is 29.07%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 80,

Substitute 80 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{80-78}{11}\\ & =& \frac{2}{11}\\ & =& 0.18\end{array}$

The area between the mean and the score $0.18$ is 0.0714.

For the score of 90,

Substitute 90 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{90-78}{11}\\ & =& \frac{12}{11}\\ & =& 1.09\end{array}$

The area between the mean and the score $1.09$ is 0.3621.

Use the concept of symmetry, the area between the score $0.18$ and $1.09$ is,

$0.3621-0.0714=0.2907$

The percentage of area between the score $0.18$ and $1.09$ is 29.07%.

Conclusion:

Therefore, the percentage of area between the score 80 and 90 is 29.07%.

To determine

To find:

The percentage of area above 99.

Answer to Problem 5.6P

Solution:

The percentage of area above 99 is 2.81%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 99,

Substitute 99 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{99-78}{11}\\ & =& \frac{21}{11}\\ & =& 1.91\end{array}$

The area above the score $1.91$ is 0.0281.

The percentage of area above the score of $1.91$ is 2.81%.

Conclusion:

Therefore, the percentage of area above the score of 99 is 2.81%.

To determine

j)

To find:

The percentage of area above 89.

Answer to Problem 5.6P

Solution:

The percentage of area above 89 is 15.87%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 89,

Substitute 89 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{89-78}{11}\\ & =& \frac{11}{11}\\ & =& 1\end{array}$

The area above the score $1$ is 0.1587.

The percentage of area below the score of $1$ is 15.87%.

Conclusion:

Therefore, the percentage of area above the score of 89 is 15.87%.

To determine

k)

To find:

The percentage of area above 75.

Answer to Problem 5.6P

Solution:

The percentage of area above 75 is 60.64%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 75,

Substitute 75 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{75-78}{11}\\ & =& \frac{-3}{11}\\ & =& -0.27\end{array}$

The area between mean and the score $-0.27$ is 0.1064.

Use the concept of symmetry, the area above the score $-0.27$ is,

$0.1064+0.5=0.6064$

The percentage of area above the score of $-0.27$ is 60.64%.

Conclusion:

Therefore, the percentage of area above the score of 75 is 60.64%.

To determine

l)

To find:

The percentage of area above 65.

Answer to Problem 5.6P

Solution:

The percentage of area above 65 is 88.10%.

Explanation of Solution

Given:

The distribution of the test scores is normal. The mean score of the test is 78 and the standard deviation is 11.

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 78 and standard deviation is 11.

For the score of 65,

Substitute 65 for $X_i$, 78 for $\overline{X}$, and 11 for $s$ in the above mentioned formula,

$\begin{array}{rll}Z& =& \frac{65-78}{11}\\ & =& \frac{-13}{11}\\ & =& -1.18\end{array}$

The area between mean and the score $-1.18$ is 0.3810.

Use the concept of symmetry, the area above the score $-1.18$ is,

$0.3810+0.5=0.8810$

The percentage of area above the score of $-1.18$ is 88.10%.

Conclusion:

Therefore, the percentage of area above the score of 65 is 88.10%.