Chapter 16, Problem 25E

(a)

To determine

To find: the mean and standard deviation.

Answer to Problem 25E

Mean = 30

Standard deviation = 6

Explanation of Solution

Given:

	Mean	SD
X	10	2
Y	20	5

3X

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Mean and standard deviation is

  $\begin{array}{c}E\left(3X\right)=3E\left(X\right)\\ =3.10\\ =30\\ \mathrm{var}\left(3X\right)=9\mathrm{var}\left(x\right)\\ ={9.2}^2\\ =36\\ SD\left(3X\right)=\sqrt{\mathrm{var}\left(3X\right)}\\ =\sqrt{36}\\ =6\end{array}$

(b)

To determine

To Calculate:

Answer to Problem 25E

Mean = 26

Standard deviation = 5

Explanation of Solution

Given:

	Mean	SD
X	10	2
Y	20	5

Y+6

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Mean and standard deviation is

  $\begin{array}{c}E\left(Y+6\right)=E\left(Y\right)+6\\ =20+6\\ =26\\ \mathrm{var}\left(Y+6\right)=\mathrm{var}\left(Y\right)\\ =25\\ SD\left(Y+6\right)=\sqrt{\mathrm{var}\left(Y+6\right)}\\ =\sqrt{25}\\ =5\end{array}$

(c)

To determine

To Calculate:

Answer to Problem 25E

Mean = 30

Standard deviation = 5.385

Explanation of Solution

Given:

	Mean	SD
X	10	2
Y	20	5

X+Y

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Mean and standard deviation is

  $\begin{array}{c}E\left(X+Y\right)=E\left(X\right)+E\left(Y\right)\\ =10+20\\ =30\\ \mathrm{var}\left(X+Y\right)=\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ =2^2+5^2\\ =29\end{array}$

  $\begin{array}{c}SD\left(X+Y\right)=\sqrt{\mathrm{var}\left(X+Y\right)}\\ =\sqrt{29}\\ =5.385\end{array}$

(d)

To determine

To Calculate:

Answer to Problem 25E

Mean = -10

Standard deviation = 5.385

Explanation of Solution

Given:

	Mean	SD
X	10	2
Y	20	5

X-Y

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Mean and standard deviation is

  $\begin{array}{c}E\left(X-Y\right)=E\left(X\right)-E\left(Y\right)\\ =10-20\\ =-10\end{array}$

  $\begin{array}{c}\mathrm{var}\left(X-Y\right)=\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ =2^2+5^2\\ =29\\ SD\left(X-Y\right)=\sqrt{\mathrm{var}\left(X-Y\right)}\\ =\sqrt{29}\\ =5.385\end{array}$

(e)

To determine

To Calculate:

Answer to Problem 25E

Mean = 20

Standard deviation = 2.828

Explanation of Solution

Given:

	Mean	SD
X	10	2
Y	20	5

  $X_1+X_2$

Formula used:

  $\begin{array}{l}E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}\\ \text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\end{array}$

Calculation:

Mean and standard deviation of $X_1+X_2$

  $\begin{array}{c}E\left(X_1+X_2\right)=E\left(X_1\right)+E\left(X_2\right)\\ =10+10\\ =20\\ \mathrm{var}\left(X_1+X_2\right)=\mathrm{var}\left(X_1\right)+\mathrm{var}\left(X_2\right)\\ =2^2+2^2\\ =8\\ SD\left(X_1+X_2\right)=\sqrt{\left(X_1+X_2\right)}\\ =\sqrt{8}\\ =2.828\end{array}$