Chapter 16, Problem 10E

(a)

To determine

To find: the standard deviations of the random variables in part (a).

Answer to Problem 10E

0.7488

Explanation of Solution

Formula used:

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

Calculation:

Probability model of random variable

X	0	1	2
P(X = x)	0.2	0.4	0.4

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\\ =0\left(0.2\right)+1\left(0.4\right)+2\left(0.4\right)\\ =0.4+0.8\\ =1.2\end{array}$

Expected value is 1.2

Finding the variance of X

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\\ ={\left(0-1.2\right)}^2\left(0.2\right)+{\left(1-1.2\right)}^2\left(0.4\right)+{\left(2-1.2\right)}^2\left(0.4\right)\\ =1.44\left(0.2\right)+0.04\left(0.4\right)+0.64\left(0.4\right)\\ =0.56\end{array}$

The standard deviation is,

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{0.56}\\ =0.748\end{array}$

Therefore, the standard deviation is 0.7488

(b)

To determine

To find: the standard deviations of the random variables in part (b).

Answer to Problem 10E

87.18

Explanation of Solution

Formula used:

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

Calculation:

X	100	200	300	400
P(X = x)	0.1	0.2	0.5	0.2

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\\ =100\left(0.1\right)+200\left(0.2\right)+300\left(0.5\right)+400\left(0.2\right)\\ =10+40+150+80\\ =280\end{array}$

Expected value is 280

Finding the variance of X is

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\\ ={\left(100-280\right)}^2\left(0.1\right)+{\left(200-280\right)}^2\left(0.2\right)+{\left(300-280\right)}^2\left(0.5\right)+{\left(400-280\right)}^2\left(0.2\right)\\ =32400\left(0.1\right)+6400\left(0.2\right)+400\left(0.5\right)+14400\left(0.2\right)\\ =7600\end{array}$

The standard deviation is,

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{7600}\\ =87.18\end{array}$

Therefore, the standard deviation of X is 87.18