Chapter 16, Problem 17E

To determine

To find: the expected number and standard deviation of appearance defects in a new car.

Answer to Problem 17E

0.93

Explanation of Solution

Given:

Let X be a new car's number of defects

The probability model of X is given below

X	0	1	2	3	>3
P(X)	0.61	0.21	0.11	0.07	0

Formula used:

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

Calculation:

The expected number of defects is

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\\ =0\left(0.61\right)+1\left(0.21\right)+2\left(0.11\right)+3\left(0.07\right)\\ =0+0.21+0.22+0.21\\ =0.64\end{array}$

Thus, the expected number of defects is 0.64

$X$	$P\left(X\right)$	$\left(X-\mu \right)$	${\left(X-\mu \right)}^2$	${\left(X-\mu \right)}^2.P\left(X\right)$
0	0.61	-0.64	0.4096	0.249856
1	0.21	0.36	0.1296	0.027216
2	0.11	1.36	1.8496	0.203456
3	0.07	2.36	5.5696	0.389872
				0.8704

The variance is

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\\ =0.8704\end{array}$

Standard deviation is

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{0.8704}\\ =0.93\end{array}$

Therefore, the standard deviation is 0.93