Chapter 16, Problem 9E

(a)

To determine

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

Answer to Problem 9E

7

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 distribution of random variable

X	10	20	30
P(X = x)	0.3	0.5	0.2

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\\ =10\left(0.3\right)+20\left(0.5\right)+30\left(0.2\right)\\ =3+10+6\\ =19\end{array}$

Expected value is 19

Finding the variance of X is,

  $\begin{array}{c}\text{Var}\left(X\right)={\displaystyle \sum {(x-\mu )}^{2}P\left(x\right)}\\ ={\left(10-19\right)}^2\left(0.3\right)+{\left(20-19\right)}^2\left(0.5\right)+{\left(30-19\right)}^2\left(0.2\right)\\ =81\left(0.3\right)+\left(0.5\right)+121\left(0.1\right)\\ =49\end{array}$

The standard deviation of X is,

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{49}\\ =7\end{array}$

Therefore, the standard deviation of X is 7

(b)

To determine

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

Answer to Problem 9E

1.89

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 distribution of random variable

X	2	4	6	8
P(X = x)	0.3	0.4	0.2	0.1

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xp\left(x\right)}\\ =2\left(0.3\right)+4\left(0.4\right)+6\left(0.2\right)+8\left(0.1\right)\\ =0.6+1.6+1.2+0.8\\ =4.2\end{array}$

Expected value is 4.2

Finding the variance of X

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\\ ={\left(2-4.2\right)}^2\left(0.3\right)+{\left(4-4.2\right)}^2\left(0.4\right)+{\left(6-4.2\right)}^2\left(0.2\right)+{\left(8-4.2\right)}^2\left(0.1\right)\\ =4.84\left(0.3\right)+0.04\left(0.4\right)+3.24\left(0.2\right)+14.44\left(0.1\right)\\ =3.56\end{array}$

The standard deviation is,

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{3.56}\\ =1.89\end{array}$

Therefore, the standard deviation of X is 1.89