Chapter 5.1, Problem 20P

a.

To determine

Calculate the mean, variance, and standard deviation for the given random variable W.

Answer to Problem 20P

The mean of the random variable $W=x_1+x_2$ is ${\mu }_W=118.6$ minutes.

The variance of the random variable $W=x_1+x_2$ is ${\sigma }_W^2=298.28$.

The standard deviation of the random variable $W=x_1+x_2$ is ${\sigma }_W\approx 17.27$ minutes.

Explanation of Solution

Calculation:

Let W be the random variable defined as the total time to examine and repair the computer $W=x_1+x_2$ with $a=1\text{ and }b=1$.

Here, $x_1$ is the time to examine computer with mean ${\mu }_1=28.1$ minutes and standard deviation ${\sigma }_1=8.2$ minutes and $x_2$ is the time to repair computer with mean ${\mu }_2=90.5$ minutes and the standard deviation ${\sigma }_2=15.2$ minutes.

The mean of the random variable is calculated as given below:

$\begin{array}{c}{\mu }_W={\mu }_1+{\mu }_2\\ =28.1+90.5\\ =118.6\end{array}$

Thus, the mean of the random variable $W=x_1+x_2$ is ${\mu }_W=118.6$ minutes.

The variance of the random variable is calculated as given below:

$\begin{array}{c}{\sigma }_W^2={\sigma }_1^2+{\sigma }_2^2\\ ={8.2}^2+{15.2}^2\\ =298.28\end{array}$

Thus, the variance of the random variable $W=x_1+x_2$ is ${\sigma }_W^2=298.28$.

The standard deviation of the random variable is calculated as given below:

$\begin{array}{c}{\sigma }_W=\sqrt{{\sigma }_W^2}\\ =\sqrt{298.28}\\ \approx 17.27\end{array}$

Thus, the standard deviation of the random variable $W=x_1+x_2$ is ${\sigma }_W\approx 17.27$ minutes.

b.

To determine

Calculate the mean, variance, and standard deviation for the given random variable W.

Answer to Problem 20P

The mean of the random variable $W=1.5x_1+2.75x_2$ is ${\mu }_W=291.03$.

The variance of the random variable $W=1.5x_1+2.75x_2$ is ${\sigma }_W^2=1,898.53$.

The standard deviation of the random variable $W=1.5x_1+2.75x_2$ is ${\sigma }_W\approx 43.57$.

Explanation of Solution

Calculation:

Let W be the random variable defined as the service charges $W=1.5x_1+2.75x_2$ with $a=1.5andb=2.75$

Here, $x_1$ is the time to examine computer with mean ${\mu }_1=28.1$ minutes and standard deviation ${\sigma }_1=8.2$ minutes and $x_2$ is the time to repair computer with mean ${\mu }_2=90.5$ minutes and standard deviation ${\sigma }_2=15.2$ minutes.

The mean of the random variable is calculated as given below:

$\begin{array}{c}{\mu }_W=1.5{\mu }_1+2.75{\mu }_2\\ =1.5\left(28.1\right)+2.75\left(90.5\right)\\ =291.03\end{array}$

Thus, the mean of the random variable $W=1.5x_1+2.75x_2$ is ${\mu }_W=291.03$.

The variance of the random variable is calculated as given below:

$\begin{array}{c}{\sigma }_W^2={\left(1.50\right)}^2{\sigma }_1^2+{\left(2.75\right)}^2{\sigma }_2^2\\ =2.25{\left(8.2\right)}^2+7.5625{\left(15.2\right)}^2\\ =1,898.53\end{array}$

Thus, the variance of the random variable $W=1.5x_1+2.75x_2$ is ${\sigma }_W^2=1,898.53$.

The standard deviation of the random variable is calculated as given below:

$\begin{array}{c}{\sigma }_W=\sqrt{{\sigma }_W^2}\\ =\sqrt{1,898.53}\\ \approx 43.57\end{array}$

Thus, the standard deviation of the random variable $W=1.5x_1+2.75x_2$ is ${\sigma }_W\approx 43.57$.

c.

To determine

Calculate the mean, variance, and standard deviation for the given random variable L.

Answer to Problem 20P

The mean of the random variable $L=1.5x_1+50$ is ${\mu }_L=92.15$.

The variance of the random variable $L=1.5x_1+50$ is ${\sigma }_L^2=151.29$.

The standard deviation of the random variable $L=1.5x_1+50$ is ${\sigma }_L\approx 12.30$.

Explanation of Solution

Calculation:

Let L be the random variable defined as the charges $L=1.5x_1+50$ with $a=50andb=1.5$.

Here, $x_1$ is the time to examine computer with mean ${\mu }_1=28.1$ minutes and standard deviation ${\sigma }_1=8.2$ minutes and $x_2$ is the time to repair computer with mean ${\mu }_2=90.5$ minutes and standard deviation ${\sigma }_2=15.2$ minutes.

The mean of the random variable is calculated as given below:

$\begin{array}{c}{\mu }_L=50+1.5{\mu }_1\\ =50+1.5\left(28.1\right)\\ =92.15\end{array}$

Thus, the mean of the random variable $L=1.5x_1+50$ is ${\mu }_L=92.15$.

The variance of the random variable is calculated as given below:

$\begin{array}{c}{\sigma }_L^2={\left(1.5\right)}^2{\sigma }_1^2\\ =2.25{\left(8.2\right)}^2\\ =151.29\end{array}$

Thus, the variance of the random variable $L=1.5x_1+50$ is ${\sigma }_L^2=151.29$.

The standard deviation of the random variable is calculated as given below:

$\begin{array}{c}{\sigma }_L=\sqrt{{\sigma }_L^2}\\ =\sqrt{151.29}\\ \approx 12.30\end{array}$

Thus, the standard deviation of the random variable $L=1.5x_1+50$ is ${\sigma }_L\approx 12.30$.