Chapter 15.3, Problem 6WE

a.

To determine

To calculate: The percent of students who are expected to score above 800 points.

Answer to Problem 6WE

The percent of students who are expected to score above 800 pointsis $0.13\%$ .

Explanation of Solution

Given information:

Mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

Formula used:

The number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Calculation:

Consider the provided information that mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

To compute percent of students who are expected to score above 800 points.

Recall that the number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Here, z is 800, m is 500 and $\sigma$ is 100.

Apply it,

  $\begin{array}{c}x=\frac{800-500}{100}\\ x=\frac{300}{100}\\ x=3\end{array}$

The total area under the curve is 1 and standard normal curve is symmetric about y -axis.

The required percent is the area under the curve of standard normal to the right of $x=3$ that is,

  $\begin{array}{c}0.5-A\left(0\le x\le 3\right)=0.5-0.4987\\ =0.0013\end{array}$

Multiply the result by 100, to obtain the answer in percentage form,

  $\begin{array}{c}0.5-A\left(0\le x\le 3\right)=0.5-0.4987\\ =0.0013\\ =0.13\%\end{array}$

Thus, the percent of students who are expected to score above 800 pointsis $0.13\%$ .

b.

To determine

To calculate: The percent of students who are expected to score less than 400 points.

Answer to Problem 6WE

The percent of students who are expected to score less than 400 pointsis $15.87\%$ .

Explanation of Solution

Given information:

Mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

Formula used:

The number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Calculation:

Consider the provided information that mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

To compute percent of students who are expected to score less than 400 points.

Recall that the number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Here, z is 400, m is 500 and $\sigma$ is 100.

Apply it,

  $\begin{array}{c}x=\frac{400-500}{100}\\ x=\frac{-100}{100}\\ x=-1\end{array}$

The total area under the curve is 1 and standard normal curve is symmetric about y -axis.

The required percent is the area under the curve of standard normal to the left of $x=-1$ that is,

  $\begin{array}{c}0.5-A\left(0\le x\le 1\right)=0.5-0.3413\\ =0.1587\end{array}$

Multiply the result by 100, to obtain the answer in percentage form,

  $\begin{array}{c}0.5-A\left(0\le x\le 1\right)=0.5-0.3413\\ =0.1587\\ =15.87\%\end{array}$

Thus, the percent of students who are expected to score less than 400 pointsis $15.87\%$ .

c.

To determine

To calculate: The percent of students who are expected to score between 700 and 900 points.

Answer to Problem 6WE

The percent of students who are expected to score between 700 and 900 pointsis $2.28\%$ .

Explanation of Solution

Given information:

Mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

Formula used:

The number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Calculation:

Consider the provided information that mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

To compute percent of students who are expected to score between 700 and 900 points.

Recall that the number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Here, z is 700, m is 500 and $\sigma$ is 100.

Apply it,

  $\begin{array}{c}x=\frac{700-500}{100}\\ x=\frac{200}{100}\\ x=2\end{array}$

Andz is 900, m is 500 and $\sigma$ is 100.

Apply it,

  $\begin{array}{c}x=\frac{900-500}{100}\\ x=\frac{400}{100}\\ x=4\end{array}$

The total area under the curve is 1 and standard normal curve is symmetric about y -axis.

The required percent is the area under the curve of standard normal and between $x=2\text{ and }x=4$ that is,

  $\begin{array}{c}0.5-A\left(0\le x\le 2\right)=0.5-0.4772\\ =0.0228\end{array}$

Multiply the result by 100, to obtain the answer in percentage form,

  $\begin{array}{c}0.5-A\left(0\le x\le 2\right)=0.5-0.4772\\ =0.0228\\ =2.28\%\end{array}$

Thus, the percent of students who are expected to score between 700 and 900 pointsis $2.28\%$ .

d.

To determine

To calculate: The percent of students who are expected to score between 800 and 820 points.

Answer to Problem 6WE

The percent of students who are expected to score between 800 and 820 pointsis $0.06\%$ .

Explanation of Solution

Given information:

Mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

Formula used:

The number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Calculation:

Consider the provided information that mean is 500 and standard deviation is 100 for the score that approximate a normal distribution.

To compute percent of students who are expected to score between 800 and 820 points.

Recall that the number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Here, z is 800, m is 500 and $\sigma$ is 100.

Apply it,

  $\begin{array}{c}x=\frac{800-500}{100}\\ x=\frac{300}{100}\\ x=3\end{array}$

And z is 820, m is 500 and $\sigma$ is 100.

Apply it,

  $\begin{array}{c}x=\frac{820-500}{100}\\ x=\frac{320}{100}\\ x=3.2\end{array}$

The total area under the curve is 1 and standard normal curve is symmetric about y -axis.

The required percent is the area under the curve of standard normal and between $x=3\text{ and }x=3.2$ that is,

  $\begin{array}{c}A\left(0\le x\le 3.2\right)-A\left(0\le x\le 3\right)=0.4993-0.4987\\ =0.0006\end{array}$

Multiply the result by 100, to obtain the answer in percentage form,

  $\begin{array}{c}A\left(0\le x\le 3.2\right)-A\left(0\le x\le 3\right)=0.4993-0.4987\\ =0.0006\\ =0.06\%\end{array}$

Thus, the percent of students who are expected to score between 800 and 820 pointsis $0.06\%$ .