Chapter 6.1, Problem 33E

To determine

To find: the percentages of researchers have a number of cars within 2 standard deviation of the mean

Answer to Problem 33E

93%

Explanation of Solution

Given:

X	0	1	2	3	4	5
P(X)	0.09	0.36	0.35	0.13	0.05	0.02

Formula used:

  $\begin{array}{l}\mu ={\displaystyle \sum x\text{P}\left(x\right)}\\ \sigma =\sqrt{{\displaystyle \sum {\left(x-\mu \right)}^2\text{P}\left(x\right)}}\end{array}$

Calculation:

  $\begin{array}{c}\mu ={\displaystyle \sum x\text{P}\left(x\right)}\\ =0\times 0.09+1\times 0.36+2\times 0.35\\ +3\times 0.13+4\times 0.05+5\times 0.02\\ =1.75\end{array}$

  $\begin{array}{c}\sigma =\sqrt{{\displaystyle \sum {\left(x-\mu \right)}^2\text{P}\left(x\right)}}\\ $ {\sigma }^2={(0-1.75)}^2\times 0.09+{(1-1.75)}^2\times 0.36+{(2-1.75)}^2\times 0.35$$ +{(3-1.75)}^2\times 0.13+{(4-1.75)}^2\times 0.05+{(5-1.75)}^2\times 0.02$\sigma =\sqrt{{\sigma }^2}\\ =\sqrt{1.675}\\ \approx 1.0805\end{array}$

Required the percentage of households have a number of cars within two standard deviation of the mean

  $\begin{array}{c}\mu -2\sigma =1.75-2\left(1.0805\right)\\ =-0.4110\\ \mu -2\sigma =1.75+2\left(1.0805\right)\\ =3.9110\end{array}$

  $\begin{array}{c}\text{P}\left(\mu -2\sigma <X<\mu +2\sigma \right)=\text{P}\left(-0.4110<X<3.9110\right)\\ =\text{P}\left(0\le X\le 3\right)\\ =\text{P}\left(X=0\right)+\text{P}\left(X=1\right)+\text{P}\left(X=2\right)+\text{P}\left(X=3\right)\\ =0.09+0.36+0.35+0.13\\ =0.93\\ =93\%\end{array}$

About 93% of the researchers have a number of cars within 2 standard deviations from the mean.

Hence, the correct option is (c)