Chapter 16, Problem 26E

(a)

To determine

To find: the mean and standard deviation.

Answer to Problem 26E

Mean = 60

Standard deviation = 12

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

X-20

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

  $\begin{array}{c}E\left(X-20\right)=E\left(X\right)-20\\ =80-20\\ =60\\ \mathrm{var}\left(X-20\right)=\mathrm{var}\left(X\right)\\ ={12}^2\\ =144\\ SD\left(X-20\right)=\sqrt{\mathrm{var}\left(X-20\right)}\\ =\sqrt{144}\\ =12\end{array}$

(b)

To determine

To Calculate:

Answer to Problem 26E

Mean = 6

Standard deviation = 1.5

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

0.5Y

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

  $\begin{array}{c}E\left(0.5Y\right)=0.5E\left(Y\right)\\ =\left(0.5\right)12\\ =6\\ \mathrm{var}\left(0.5Y\right)={0.5}^2\mathrm{var}\left(Y\right)\\ =\left(0.25\right)3^2\\ =2.25\end{array}$

  $\begin{array}{c}SD\left(0.5Y\right)=\sqrt{\mathrm{var}\left(0.5Y\right)}\\ =\sqrt{2.25}\\ =1.5\end{array}$

(c)

To determine

To Calculate:

Answer to Problem 26E

Mean = 92

Standard deviation = 12.37

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

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

  $\begin{array}{c}E\left(X+Y\right)=E\left(X\right)+E\left(Y\right)\\ =80+12\\ =92\\ \mathrm{var}\left(X+Y\right)=\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ ={12}^2+3^2\\ =144+9\\ =153\\ SD\left(X+Y\right)=\sqrt{\mathrm{var}\left(X+Y\right)}\\ =\sqrt{153}\\ =12.37\end{array}$

(d)

To determine

To Calculate:

Answer to Problem 26E

Mean = 68

Standard deviation = 12.37

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

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

  $\begin{array}{c}E\left(X-Y\right)=E\left(X\right)-E\left(Y\right)\\ =80-12\\ =68\\ \mathrm{var}\left(X-Y\right)=\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ ={12}^2+3^2\\ =144+9\\ =153\\ SD\left(X+Y\right)=\sqrt{\mathrm{var}\left(X-Y\right)}\\ =\sqrt{153}\\ =12.37\end{array}$

(e)

To determine

To Calculate:

Answer to Problem 26E

Mean = 24

Standard deviation = 4.242

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

  $Y_1+Y_2$

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

  $X_1+X_2$

  $\begin{array}{c}E\left(Y_1+Y_2\right)=E\left(Y_1\right)+E\left(Y_2\right)\\ =12+12\\ =24\\ \mathrm{var}\left(Y_1+Y_2\right)=\mathrm{var}\left(Y_1\right)+\mathrm{var}\left(Y_2\right)\\ =3^2+3^2\\ =18\\ SD\left(Y_1+Y_2\right)=\sqrt{\mathrm{var}\left(Y_1+Y_2\right)}\\ =\sqrt{18}\\ =4.242\end{array}$