Chapter 6.1, Problem 19P

Expand Your Knowledge: Linear Functions and Combinations of Independent Random Variables: Golf How can we compute the mean and standard deviation of new random variables created by a linear function of one random variable or a linear combination of two independent random variables? The following discussion shows us.

Norb and Gary are entered in a local golf tournament. Both have played the local course many times. Their scores are random variables with the following means and standard deviations.

Norb. $x_1:{\mu }_1=115;{\sigma }_1=12\text{ Gary, }{x}_2:{\mu }_2=100;=8$

In the tournament. Norb and Gary are not playing together, and we will assume their scores vary independently of each other.

(a) The difference between their scores is $w=x_1-x_2$ Compute the mean, variance, and standard deviation for the random variable IV.

(b) The average of their scores is $w=0.5x_1+0.5x_2$ Compute the mean, variance, and standard deviation for the random variable IV.

(c) The tournament rules have a special handicap system for each player. For Norb. the handicap formula is $w=0.5x_1+0.5x_2$. Compute the mean, variance, and standard deviation for the random variable L.

(d) For Gary, the handicap formula is $L=0.95x_2-5$. Compute the mean, variance, and standard deviation for the random variable L.

To determine

The mean, variance and standard deviation for the provided linear combination W.

Answer to Problem 19P

Solution: The mean $\left({\mu }_w\right)$ is 15, the variance $\left({\sigma }_w{}^2\right)$ is 208 and the standard deviation $\left({\sigma }_w\right)$ is 14.4.

Explanation of Solution

Given: The provided mean and standard deviation of random variables $x_1$ and $x_2$ are as follow,

$\begin{array}{c}{x}_1:{\mu }_1=115,{\sigma }_1=12\\ x_2:{\mu }_2=100,{\sigma }_2=8\end{array}$

The difference between the scores is,

$W=x_1-x_2$

Calculation: The general formula for the linear combination of random variables is as follows,

$W=a{x}_1+b{x}_2$

Now, comparing the provided linear combination of scores with the above formula as:

$\begin{array}{c}a=1\\ b=-1\end{array}$

Now, the combined mean of the provided score can be obtained as:

$\begin{array}{c}{\mu }_w=a{\mu }_1+b{\mu }_2\\ =1\left(115\right)+\left(-1\right)\left(100\right)\\ =115-100\\ =15\end{array}$

Now, the combined variance can be obtained as:

$\begin{array}{c}{\sigma }_w{}^2=a^2{\sigma }_1{}^2+b^2{\sigma }_2{}^2\\ =1^2{\left(12\right)}^2+{\left(-1\right)}^2{\left(8\right)}^2\\ =144+64\\ =208\end{array}$

The standard deviation can be obtained as:

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

Hence, the mean is 15, the variance is 208 and the standard deviation is 14.4.

To determine

The mean, variance and standard deviation for the provided linear combination W.

Answer to Problem 19P

Solution: The mean $\left({\mu }_w\right)$ is 107.5, the variance $\left({\sigma }_w{}^2\right)$ is 52 and the standard deviation $\left({\sigma }_w\right)$ is 7.2.

Explanation of Solution

Given: The provided mean and standard deviation of random variables $x_1$ and $x_2$ are as follow,

$\begin{array}{c}{x}_1:{\mu }_1=115,{\sigma }_1=12\\ x_2:{\mu }_2=100,{\sigma }_2=8\end{array}$

The average of the scores is,

$W=0.5x_1+0.5x_2$

Calculation: The general formula for the linear combination of random variables is as follows,

$W=a{x}_1+b{x}_2$

Now, comparing the provided linear combination of scores with the above formula as:

$\begin{array}{c}a=0.5\\ b=0.5\end{array}$

Now, the combined mean of the provided score can be obtained as:

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

Now, the combined variance can be obtained as:

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

The standard deviation can be obtained as:

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

Hence, the mean is 107.50, the variance is 52 and the standard deviation is 7.2.

To determine

The mean, variance and standard deviation for the provided linear function L.

Answer to Problem 19P

Solution: The mean $\left({\mu }_L\right)$ is 90, the variance $\left({\sigma }_L{}^2\right)$ is 92.16 and the standard deviation $\left({\sigma }_L\right)$ is 9.6.

Explanation of Solution

Given: The provided mean and standard deviation of random variable $x_1$ is as follows,

$x_1:{\mu }_1=115,{\sigma }_1=12$

For Norb, the handicap formula is as follows,

$L=0.8x_1-2$

Calculation: The general formula for the linear function of a random variable is as follows:

$L=a+bx$

Now, comparing the provided linear handicap formula with the above linear formula as:

$\begin{array}{c}a=-2\\ b=0.8\end{array}$

Now, the mean of the provided linear function can be obtained as:

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

Now, the variance can be obtained as:

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

The standard deviation can be obtained as:

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

Hence, the mean is 90, the variance is 92.16 and the standard deviation is 9.6.

To determine

The mean, variance and standard deviation for the provided linear function L.

Answer to Problem 19P

Solution: The mean $\left({\mu }_L\right)$ is 90, the variance $\left({\sigma }_L{}^2\right)$ is 57.76 and the standard deviation $\left({\sigma }_L\right)$ is 7.6.

Explanation of Solution

Given: The provided mean and standard deviation of random variable $x_2$ is as follows,

$x_2:{\mu }_2=100,{\sigma }_2=8$

For Gary, the handicap formula is as follows,

$L=0.95x_2-5$

Calculation: The general formula for the linear function of a random variable is as follows:

$L=a+bx$

Now, comparing the provided linear handicap formula with the above linear formula as:

$\begin{array}{c}a=-5\\ b=0.95\end{array}$

Now, the mean of the provided linear function can be obtained as:

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

Now, the variance can be obtained as:

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

The standard deviation can be obtained as:

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

Hence, the mean is 90, the variance is 57.76 and the standard deviation is 7.6.