Chapter 7.2, Problem 56E

(a)

To determine

The marginal probability distribution of refrigerator.

Explanation of Solution

The probability distribution of refrigerator and stoves sold daily is shown below:

Table 1

Stoves	Refrigerator
0	0	2	3
1	0.08	0.14	0.12
2	0.09	0.17	0.13
	0.05	0.18	0.04

The marginal probability of refrigerator sold daily is shown below:

Table 2

Refrigerator X	P(x)
0	0.22
1	0.49
2	0.29

Economics Concept Introduction

Probability distribution: The probability distribution shows the probabilities of incidence of different likely outcomes in a test.

(b)

To determine

The marginal probability distribution of stoves.

Explanation of Solution

The marginal probability of stoves sold daily is shown below:

Table 3

Stoves Y	P(y)
0	0.34
1	0.49
2	0.27

(c)

To determine

The means and variance of refrigerator.

Explanation of Solution

The mean value of the probability distribution of refrigerator is calculated as follows:

$\begin{array}{c}\text{μ}=\text{E}\left(\text{X}\right)={\displaystyle \sum \text{xP}\left(\text{x}\right)}\\ =\left(0\left(\text{0}\text{.22}\right)\text{+ 1}\left(\text{0}\text{.49}\right)\text{+ 2}\left(\text{0}\text{.29}\right)\right)\\ =1.07\end{array}$

The mean value is 1.07.

The variance of the probability distribution of refrigerator is calculated as follows:

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{X}\right)={\displaystyle \sum {\left(\text{y-μ}\right)}^{\text{2}}\text{P}\left(\text{x}\right)}\\ \text{=}\left\{{\left(0-1.07\right)}^2\left(\text{0}\text{.22}\right)+{\left(1-1.07\right)}^2\left(0.49\right)+{\left(2-1.07\right)}^2\left(\text{0}\text{.29}\right)\right\}\\ =0.505\end{array}$

The variance is 0.505.

(d)

To determine

The mean and variance of stoves.

Explanation of Solution

The mean value of the probability distribution of stoves is calculated as follows:

$\begin{array}{c}\text{μ}=\text{E}\left(\text{Y}\right)={\displaystyle \sum \text{yP}\left(\text{y}\right)}\\ =\left(0\left(\text{0}\text{.34}\right)\text{+ 1}\left(\text{0}\text{.39}\right)\text{+ 2}\left(\text{0}\text{.27}\right)\right)\\ =0.93\end{array}$

The mean value is 0.93.

The variance of the probability distribution of stoves is calculated as follows:

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{X}\right)={\displaystyle \sum {\left(\text{y-μ}\right)}^{\text{2}}\text{P}\left(\text{x}\right)}\\ \text{=}\left\{{\left(0-0.93\right)}^2\left(\text{0}\text{.34}\right)+{\left(1-0.93\right)}^2\left(0.39\right)+{\left(2-0.93\right)}^2\left(\text{0}\text{.27}\right)\right\}\\ =0.605\end{array}$

The variance is 0.605.

(e)

To determine

Calculate covariance and the coefficient of correlation.

Explanation of Solution

Covariance and the coefficient of correlation are calculated as follows:

$\begin{array}{c}{\displaystyle \sum _{\text{all}\text{x}}}{\displaystyle \sum _{\text{all}\text{y}}\text{xyP(x,y)}}=\left(\begin{array}{c}\left(0\right)\left(0\right)\left(0.08\right)+\left(0\right)\left(1\right)\left(0.09\right)+\left(0\right)\left(2\right)\left(0.05\right)\\ +\left(1\right)\left(0\right)\left(0.14\right)+\left(1\right)\left(1\right)\left(0.17\right)+\left(1\right)\left(2\right)\left(0.18\right)\\ +\left(2\right)\left(0\right)\left(0.12\right)+\left(2\right)\left(1\right)\left(0.13\right) +\left(2\right)\left(2\right)\left(0.04\right)\end{array}\right)\\ =0.95\\ \text{COV}\left(\text{X, Y}\right)\text{ =}{\displaystyle \sum _{\text{all}\text{x}}}{\displaystyle \sum _{\text{all}\text{y}}\text{xyP(x,y)}}-{\text{μ}}_{\text{x}}{\text{μ}}_{\text{y}}\\ =95-\left(1.07\right)\left(0.93\right)\\ =-0.045\\ {\text{σ}}_{\text{x}}=\sqrt{{\text{σ}}_{\text{x}}^{\text{2}}}\text{=}\sqrt{\text{.505}}=0.711\\ {\text{σ}}_{\text{y}}\text{=}\sqrt{{\text{σ}}_{\text{y}}^{\text{2}}}\text{=}\sqrt{\text{.605}}=0.778\\ \text{ρ}=\frac{\text{COV(X,Y)}}{{\text{σ}}_{\text{x}}{\text{σ}}_{\text{y}}}\\ =\frac{-.045}{(.711)(.778)}\\ =-0.81\end{array}$