Chapter 5.1, Problem 19P

a.

To determine

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

Answer to Problem 19P

The mean of the random variable $W\text{= }{x}_1-x_2$ is ${\mu }_W=15$.

The variance of the random variable $W\text{= }{x}_1-x_2$ is ${\sigma }_W^2=208$.

The standard deviation of the random variable $W\text{= }{x}_1-x_2$ is ${\sigma }_W\approx 14.4$.

Explanation of Solution

Calculation:

Let W be the random variable defined as the difference between the scores $W\text{= }{x}_1-x_2$ with $a=1andb=-1$.

Here, $x_1$ is the score of Person N with mean ${\mu }_1=115$ and standard deviation ${\sigma }_1=12$ and $x_2$ is the score of Person G with mean ${\mu }_2=100$ and standard deviation ${\sigma }_2=8$.

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

$\begin{array}{c}{\mu }_W={\mu }_1-{\mu }_2\\ =115-100\\ =15\end{array}$

Thus, the mean of the random variable $W\text{= }{x}_1-x_2$ is ${\mu }_W=15$.

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

$\begin{array}{c}{\sigma }_W^2=1^2{\sigma }_1^2+{\left(-1\right)}^2{\sigma }_2^2\\ ={12}^2+8^2\\ =208\end{array}$

Thus, the variance of the random variable $W\text{= }{x}_1-x_2$ is ${\sigma }_W^2=208$.

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

$\begin{array}{c}{\sigma }_W=\sqrt{{\sigma }_W^2}\\ =\sqrt{208}\\ \approx 14.4\end{array}$

Thus, the standard deviation of the random variable $W\text{= }{x}_1-x_2$ is ${\sigma }_W\approx 14.4$.

b.

To determine

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

Answer to Problem 19P

The mean of the random variable $W\text{= }-0.5x_1+0.5x_2$ is ${\mu }_W=107.5$.

The variance of the random variable $W\text{= }-0.5x_1+0.5x_2$ is ${\sigma }_W^2=52$.

The standard deviation of the random variable $W\text{= }-0.5x_1+0.5x_2$ is ${\sigma }_W\approx 7.2$.

Explanation of Solution

Calculation:

Let W be the random variable defined as the average of the scores $W\text{= }-0.5x_1+0.5x_2$ with $a=0.5andb=0.5$.

Here, $x_1$ is the score of Person N with mean ${\mu }_1=115$ and standard deviation ${\sigma }_1=12$ and $x_2$ is the score of Person G with mean ${\mu }_2=100$ and standard deviation ${\sigma }_2=8$.

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

$\begin{array}{c}{\mu }_W=0.5{\mu }_1+0.5{\mu }_2\\ =0.5\left(115\right)-0.5\left(100\right)\\ =107.5\end{array}$

Thus, the mean of the random variable $W=-0.5x_1+0.5x_2$ is ${\mu }_W=107.5$.

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

$\begin{array}{c}{\sigma }_W^2={\left(0.5\right)}^2{\sigma }_1^2+{\left(0.5\right)}^2{\sigma }_2^2\\ =0.25{\left(12\right)}^2+0.25{\left(8\right)}^2\\ =52\end{array}$

Thus, the variance of the random variable $W=-0.5x_1+0.5x_2$ is ${\sigma }_W^2=52$.

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

$\begin{array}{c}{\sigma }_W=\sqrt{{\sigma }_W^2}\\ =\sqrt{52}\\ \approx 7.2\end{array}$

Thus, the standard deviation of the random variable $W=-0.5x_1+0.5x_2$ is ${\sigma }_W\approx 7.2$.

c.

To determine

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

Answer to Problem 19P

The mean of the random variable $L=0.8x_1-2$ is ${\mu }_L=90$.

The variance of the random variable $L=0.8x_1-2$ is ${\sigma }_L^2=92.16$.

The standard deviation of the random variable $L=0.8x_1-2$ is ${\sigma }_L\approx 9.6$.

Explanation of Solution

Calculation:

Let L be the random variable defined as the handicap formula for Person N $L=0.8x_1-2$ with $a=-2andb=0.8$.

Here, $x_1$ is the score of Person N with mean ${\mu }_1=115$ and standard deviation ${\sigma }_1=12$ and $x_2$ is the score of Gary with mean ${\mu }_2=100$ and standard deviation ${\sigma }_2=8$.

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

$\begin{array}{c}{\mu }_L=-2+0.8{\mu }_1\\ =-2+0.8\left(115\right)\\ =90\end{array}$

Thus, the mean of the random variable $L=0.8x_1-2$ is ${\mu }_L=90$.

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

$\begin{array}{c}{\sigma }_L^2={\left(0.8\right)}^2{\sigma }_1^2\\ =0.64{\left(12\right)}^2\\ =92.16\end{array}$

Thus, the variance of the random variable $L=0.8x_1-2$ is ${\sigma }_L^2=92.16$.

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

$\begin{array}{c}{\sigma }_L=\sqrt{{\sigma }_L^2}\\ =\sqrt{92.16}\\ \approx 9.6\end{array}$

Thus, the standard deviation of the random variable $L=0.8x_1-2$ is ${\sigma }_L\approx 9.6$.

d.

To determine

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

Answer to Problem 19P

The mean of the random variable $L=0.95x_2-5$ is ${\mu }_L=90$.

The variance of the random variable $L=0.95x_2-5$ is ${\sigma }_L^2=92.16$.

The standard deviation of the random variable $L=0.95x_2-5$ is ${\sigma }_L\approx 9.6$.

Explanation of Solution

Calculation:

Let L be the random variable defined as the handicap formula for Person G $L=0.95x_2-5$ with $\text{a}=-5andb=0.95$.

Here, $x_1$ is the score of Person N with mean ${\mu }_1=115$ and standard deviation ${\sigma }_1=12$ and $x_2$ is the score of Person G with mean ${\mu }_2=100$ and standard deviation ${\sigma }_2=8$.

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

$\begin{array}{c}{\mu }_L=-5+0.95{\mu }_2\\ =-5+0.95\left(100\right)\\ =90\end{array}$

Thus, the mean of the random variable $L=0.95x_2-5$ is ${\mu }_L=90$.

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

$\begin{array}{c}{\sigma }_L^2={\left(0.95\right)}^2{\sigma }_2^2\\ =0.9025{\left(8\right)}^2\\ =57.76\end{array}$

Thus, the variance of the random variable $L=0.95x_2-5$ is ${\sigma }_L^2=92.16$.

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

$\begin{array}{c}{\sigma }_L=\sqrt{{\sigma }_L^2}\\ =\sqrt{57.76}\\ \approx 7.6\end{array}$

Thus, the standard deviation of the random variable $L=0.95x_2-5$ is ${\sigma }_L\approx 9.6$.