Chapter 16, Problem 32E

To determine

To find: the mean and standard deviation of red-light numbers that the commuter is supposed to meet over a five-day work week.

Answer to Problem 32E

A red-light number that get in five days' work

Mean= 11.25 and Standard Deviation = 6.3

Explanation of Solution

Given:

X= number of red	0	1	2	3	4	5
$P\left(X\right)$	0.05	0.25	0.35	0.15	0.15	0.05

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:

Expected number is

  $\begin{array}{c}E\left(X\right)={\displaystyle \sum xP\left(x\right)}\\ =0\left(0.05\right)+1\left(0.25\right)+2\left(0.35\right)+3\left(0.15\right)+4\left(0.15\right)+5\left(0.05\right)\\ =0+0.25+0.70+0.45+0.60+0.25\\ =2.25\end{array}$

Researcher can be supposed about 2.25= 2 lights to hit every day

Finding the variance

$X$	$P\left(X\right)$	$\left(X-\mu \right)$	${\left(X-\mu \right)}^2$	${\left(X-\mu \right)}^2.P\left(X\right)$
0	0.05	-2.25	5.0625	0.25313
1	0.25	-1.25	1.5625	0.39063
2	0.35	-0.25	0.0625	0.02188
3	0.15	0.75	0.5625	0.08438
4	0.15	1.75	3.0625	0.45938
5	0.05	2.75	7.5625	0.37813
				1.5875

The variance is,

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^{2}P\left(x\right)}\\ =1.59\end{array}$

Standard deviation is

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{1.59}\\ =1.26\end{array}$

Standard deviation = 1.26

Suppose Y be the number of red lights will hit in a five-day work

  $\begin{array}{c}Y=5X\\ E\left(Y\right)=E\left(5X\right)\\ =5E\left(X\right)\\ =5\times \left(2.25\right)\\ =11.25\\ Var\left(Y\right)=Var\left(5X\right)\\ =5^2\left(1.26\right)\\ =39.69\\ SD\left(Y\right)=\sqrt{Var\left(Y\right)}\\ =\sqrt{39.69}\\ =6.3\end{array}$