Chapter 16, Problem 28E

(a)

To determine

To find: the mean and standard deviation.

Answer to Problem 28E

Mean = 140

Standard deviation = 6

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

2Y+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

  $\begin{array}{c}E\left(2Y-20\right)=2E\left(Y\right)-20\\ =2.80-20\\ =160-20\\ =140\\ \mathrm{var}\left(2Y-20\right)=2^2\mathrm{var}\left(Y\right)\\ ={4.3}^2\\ =36\\ SD\left(2Y-20\right)=\sqrt{\mathrm{var}\left(2Y-20\right)}\\ =\sqrt{36}\\ =6\end{array}$

(b)

To determine

To Calculate:

Answer to Problem 28E

Mean = 240

Standard deviation = 36

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

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

  $\begin{array}{c}E\left(3X\right)=3E\left(X\right)\\ =3.80\\ =240\\ \mathrm{var}\left(3X\right)=3^2\mathrm{var}\left(X\right)\\ ={9.12}^2\\ =9.144\\ =1296\\ SD\left(3X\right)=\sqrt{\mathrm{var}\left(3X\right)}\\ =\sqrt{1296}\\ =36\end{array}$

(c)

To determine

To Calculate:

Answer to Problem 28E

Mean = 32

Standard deviation = 4.243

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

0.25X+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

  $\begin{array}{c}E\left(0.25X+Y\right)=0.25E\left(X\right)+E\left(Y\right)\\ =\left(0.25\right)80+12\\ =20+12\\ =32\\ \mathrm{var}\left(0.25X+Y\right)={0.25}^2\mathrm{var}\left(X\right)+\mathrm{var}(Y)\\ =\left(0.0625\right){12}^2+3^2\\ =\left(0.0625\right)144+9\\ =18\\ SD\left(0.25X+Y\right)=\sqrt{\mathrm{var}\left(0.25X+Y\right)}\\ =\sqrt{18}\\ =4.243\end{array}$

(d)

To determine

To Calculate:

Answer to Problem 28E

Mean = 20

Standard deviation = 19.21

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

X-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

  $\begin{array}{c}E\left(X-5Y\right)=E\left(X\right)-E\left(Y\right)\\ =80-5.12\\ =80-60\\ =20\\ \mathrm{var}\left(X-5Y\right)=\mathrm{var}\left(X\right)+5^2\mathrm{var}(Y)\\ ={12}^2+{25.3}^2\\ =144+25.9\\ =144+225\\ =369\\ SD\left(X-5Y\right)=\sqrt{\mathrm{var}\left(X-5Y\right)}\\ =\sqrt{369}\\ =19.21\end{array}$

(e)

To determine

To Calculate:

Answer to Problem 28E

Mean = 36

Standard deviation = 5.196

Explanation of Solution

Given:

	Mean	SD
X	80	12
Y	12	3

  $X_1+X_2+X_3$

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 $X_1+X_2$

  $\begin{array}{c}E\left(X_1+X_2+X_3\right)=E\left(X_1\right)+E\left(X_2\right)+E\left(X_3\right)\\ =12+12+12\\ =36\\ \mathrm{var}\left(X_1+X_2\right)=\mathrm{var}\left(X_1\right)+\mathrm{var}\left(X_2\right)+\mathrm{var}\left(X_3\right)\\ =3^2+3^2+3^2\\ =27\\ SD\left(X_1+X_2+X_3\right)=\sqrt{\mathrm{var}\left(X_1+X_2+X_3\right)}\\ =\sqrt{27}\\ =5.196\end{array}$