Chapter 16, Problem 31E

(a)

To determine

To find: the number of calls should the stop expect per 8 hour.

Answer to Problem 31E

2

Explanation of Solution

Given:

Repair calls	0	1	2	3
Probability	0.1	0.3	0.4	0.2

Formula used:

  $E\left(X\right)={\displaystyle \sum xP\left(X=x\right)}$

Calculation:

  $\begin{array}{c}\mu =E\left(X\right)\\ ={\displaystyle \sum xP\left(x\right)}\\ =0\left(0.1\right)+1\left(0.3\right)+2\left(0.4\right)+3\left(0.2\right)\\ =0.1+0.3+0.8+0.6\\ =1.7\end{array}$

The shop could expect about 1.7~2 calls per hour.

(b)

To determine

To find: the standard deviation.

Answer to Problem 31E

7.2

Explanation of Solution

Given:

Repair calls	0	1	2	3
Probability	0.1	0.3	0.4	0.2

Formula used:

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

Calculation:

calculate the variance.

  $\begin{array}{c}\mathrm{var}\left(X\right)={\displaystyle \sum {\left(x-\mu \right)}^2}P\left(x\right)\\ ={\left(0-1.7\right)}^2\left(0.1\right)+{\left(1-1.7\right)}^2\left(0.3\right)+{\left(2-1.7\right)}^2\left(0.4\right)+{\left(3-1.7\right)}^2\left(0.2\right)\\ =\left(2.89\right)\left(0.1\right)+\left(0.49\right)\left(0.3\right)+\left(0.09\right)\left(0.4\right)+{\left(1.69\right)}^2\left(0.2\right)\\ =0.81\end{array}$

Standard deviation is

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{0.81}\\ =0.9\end{array}$

The mean number of repairs call he could expect during hour is

Expected value	1.7 calls
Standard deviation	0.9 calls

Suppose Y be the number of repair calls the appliance shop can expect.

There is $Y=8X$

Expected value of Y is

  $\begin{array}{c}E\left(Y\right)=E\left(8X\right)\\ =8E\left(X\right)\\ =8\left(1.7\right)\\ =13.6\end{array}$

Finding the variance first

  $\begin{array}{c}\mathrm{var}\left(Y\right)=\mathrm{var}\left(8X\right)\\ =8^2\mathrm{var}\left(X\right)\\ =64{\left(0.9\right)}^2\\ =51.84\end{array}$

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}\\ =\sqrt{51.84}\\ =7.2\end{array}$

During 8-hour day the mean number of repairs call he can predict is 13.6 with standard deviation of 7.2