Chapter 16, Problem 40E

(a)

To determine

To explain: the corresponding variables can have the overall annual veterinary costs and express them.

Answer to Problem 40E

  $X_1+X_2+Y$

Explanation of Solution

Given:

Given: X shows the dogs

Y shows the cats.

  $\begin{array}{c}{\mu }_x=E\left(X\right)=100\\ {\mu }_y=E\left(Y\right)=120\end{array}$

Formula used:

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

Calculation:

  $\begin{array}{c}SD\left(X\right)=\sqrt{\mathrm{var}\left(X\right)}=30\Rightarrow \mathrm{var}\left(X\right)={30}^2=900\\ SD\left(Y\right)=\sqrt{\mathrm{var}\left(Y\right)}=35\Rightarrow \mathrm{var}\left(Y\right)={35}^2=1225\end{array}$

If X shows the dogs and Y the cats

  $X_1+X_2+Y$

(b)

To determine

To explain: the total cost model. Enter the name, expected value and standard deviation.

Answer to Problem 40E

Mean = $320

Standard deviation = $55

Explanation of Solution

Given:

Given: X shows the dogs

Y shows the cats.

  $\begin{array}{c}{\mu }_x=E\left(X\right)=100\\ {\mu }_y=E\left(Y\right)=120\end{array}$

Formula used:

  $\begin{array}{c}E\left(X_1+X_2+Y\right)=E\left(X_1\right)+E\left(X_2\right)+E\left(Y\right)\\ SD\left(X_1+X_2+Y\right)=\sqrt{\mathrm{var}\left(X_1+X_2+Y\right)}\end{array}$

Calculation:

Properties mean and variance:

  $\begin{array}{c}E\left(aX+b\right)=aE\left(X\right)+b\\ \mathrm{var}\left(aX+b\right)=a^2\mathrm{var}\left(X\right)\\ E\left(X\pm Y\right)=E\left(X\right)\pm E\left(Y\right)\\ \mathrm{var}\left(X\pm Y\right)=\mathrm{var}\left(X\right)\pm \mathrm{var}\left(Y\right)\end{array}$

Using the properties

  $\begin{array}{c}E\left(X_1+X_2+Y\right)=E\left(X_1\right)+E\left(X_2\right)+E\left(Y\right)\\ =100+100+120\\ =\$320\\ SD\left(X_1+X_2+Y\right)=\sqrt{\mathrm{var}\left(X_1+X_2+Y\right)}\\ =\sqrt{\mathrm{var}\left(X_1\right)+\mathrm{var}\left(X_2\right)+\mathrm{var}\left(Y\right)}\\ =\sqrt{900+900+1225}\\ =\sqrt{3025}\\ =\$55\end{array}$

(c)

To determine

To find: the probability of gross expenditures exceeding $400 is to be defined.

Answer to Problem 40E

0.0735 or 7.35%

Explanation of Solution

Given:

Given: X shows the dogs

Y shows the cats.

  $\begin{array}{c}{\mu }_x=E\left(X\right)=100\\ {\mu }_y=E\left(Y\right)=120\end{array}$

Formula used:

  $z=\frac{x-\mu }{\sigma }$

Calculation:

The z-score is

  $\begin{array}{c}z=\frac{x-\mu }{\sigma }\\ =\frac{400-320}{55}\\ =1.45\end{array}$

Find the associating probability

  $\begin{array}{c}P\left(X>\$400\right)=P\left(Z>1.45\right)\\ =1-P\left(Z>1.45\right)\\ =1-0.9265\\ =0.0735\\ =7.35\%\end{array}$