Chapter 16, Problem 27E

(a)

To determine

To find: the mean and standard deviation.

Answer to Problem 27E

Mean = 96

Standard deviation = 9.6

Explanation of Solution

Given:

	Mean	SD
X	120	12
Y	300	16

0.8Y

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

  $\begin{array}{c}E\left(0.8X\right)=0.8E\left(X\right)\\ =\left(0.8\right)120\\ =96\\ \mathrm{var}\left(0.8X\right)={0.8}^2\mathrm{var}\left(X\right)\\ =\left(0.64\right){12}^2\\ =92.16\\ SD\left(0.8X\right)=\sqrt{\mathrm{var}\left(0.8X\right)}\\ =\sqrt{92.16}\\ =9.6\end{array}$

(b)

To determine

To Calculate:

Answer to Problem 27E

Mean = 140

Standard deviation = 16.97

Explanation of Solution

Given:

	Mean	SD
X	120	12
Y	300	16

2X-100

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

  $\begin{array}{c}E\left(2X-100\right)=2E\left(X\right)-100\\ =2.120-100\\ =240-100\\ =140\\ \mathrm{var}\left(2X-100\right)=2^2\mathrm{var}\left(X\right)\\ ={2.12}^2\\ =2.144\\ =288\end{array}$

  $\begin{array}{c}SD\left(2X-100\right)=\sqrt{\mathrm{var}\left(2X-100\right)}\\ =\sqrt{288}\\ =16.97\end{array}$

(c)

To determine

To Calculate:

Answer to Problem 27E

Mean = 720

Standard deviation = 34.176

Explanation of Solution

Given:

	Mean	SD
X	120	12
Y	300	16

X+2Y

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

  $\begin{array}{c}E\left(X+2Y\right)=E\left(X\right)+E\left(2Y\right)\\ =E\left(X\right)+2E\left(Y\right)\\ =120+2.300\\ =720\\ \mathrm{var}\left(X+2Y\right)=\mathrm{var}\left(X\right)+4\mathrm{var}\left(Y\right)\\ ={12}^2+{4.16}^2\\ =1168\\ SD\left(X+2Y\right)=\sqrt{\mathrm{var}\left(X+2Y\right)}\\ =\sqrt{1168}\\ =34.176\end{array}$

(d)

To determine

To Calculate:

Answer to Problem 27E

Mean = 60

Standard deviation = 39.395

Explanation of Solution

Given:

	Mean	SD
X	120	12
Y	300	16

3X-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

  $\begin{array}{c}E\left(3X-Y\right)=3E\left(X\right)-E\left(Y\right)\\ =3.120-300\\ =360-300\\ =60\\ \mathrm{var}\left(3X-Y\right)=3^2\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ ={9.12}^2+{16}^2\\ =9.144+256\\ =1552\\ SD\left(3X-Y\right)=\sqrt{\mathrm{var}\left(3X-Y\right)}\\ =\sqrt{1552}\\ =39.395\end{array}$

(e)

To determine

To Calculate:

Answer to Problem 27E

Mean = 600

Standard deviation = 22.63

Explanation of Solution

Given:

	Mean	SD
X	120	12
Y	300	16

  $Y_1+Y_2$

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 $X_1+X_2$

  $\begin{array}{c}E\left(Y_1+Y_2\right)=E\left(Y_1\right)+E\left(Y_2\right)\\ =300+300\\ =360-300\\ =600\\ \mathrm{var}\left(Y_1+Y_2\right)=\mathrm{var}\left(Y_1\right)+\mathrm{var}\left(Y_2\right)\\ ={16}^2+{16}^2\\ =256+256\\ =512\\ SD\left(Y_1+Y_2\right)=\sqrt{\mathrm{var}\left(Y_1+Y_2\right)}\\ =\sqrt{512}\\ =22.63\end{array}$