Chapter 7.1, Problem 18E

(a)

To determine

Calculate the mean and variance.

Explanation of Solution

The probability distribution of random variable X is shown below:

Table 1

X	-2	5	7	8
P(X)	0.59	0.15	0.25	0.01

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

$\begin{array}{c}\text{μ}=\text{E}\left(\text{X}\right)={\displaystyle \sum \text{xP}\left(\text{x}\right)}\\ =\left(-2\left(\text{0}\text{.59}\right)\text{ + 5}\left(\text{0}\text{.15}\right)\text{ + 7}\left(\text{0}\text{.25}\right)\text{ + 8}\left(\text{0}\text{.01}\right)\right)\\ =1.40\end{array}$

The mean value is 1.40.

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

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{X}\right)={\displaystyle \sum {\left(\text{x-μ}\right)}^{\text{2}}\text{P}\left(\text{x}\right)}\\ \text{=}\left\{{\left(-2-1.4\right)}^2\left(\text{0}\text{.59}\right)+{\left(5-1.4\right)}^2\left(\text{0}\text{.15}\right)+{\left(7-1.4\right)}^2\left(\text{0}\text{.25}\right)+{\left(8-1.4\right)}^2\left(0.1\right)\right\}\\ =17.04\end{array}$

The variance is 17.04.

Economics Concept Introduction

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

(b)

To determine

The probability distribution of Y.

Explanation of Solution

The probability distribution of Y where Y is equal to 5X is shown in the below table:

X	-2	5	7	8
Y	-10	25	35	40
P(Y)	0.59	0.15	0.25	0.01

(c)

To determine

The mean and variance of Y.

Explanation of Solution

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

$\begin{array}{c}\text{μ}=\text{E}\left(\text{Y}\right)={\displaystyle \sum \text{yP}\left(\text{y}\right)}\\ =\left(-10\left(\text{0}\text{.59}\right)\text{ + 25}\left(\text{0}\text{.15}\right)\text{ + 35}\left(\text{0}\text{.25}\right)\text{ + 40}\left(\text{0}\text{.01}\right)\right)\\ =7\end{array}$

The mean value is 7.

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

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{Y}\right)={\displaystyle \sum {\left(\text{y-μ}\right)}^{\text{2}}\text{P}\left(\text{y}\right)}\\ \text{=}\left\{{\left(-10-7\right)}^2\left(\text{0}\text{.59}\right)+{\left(25-7\right)}^2\left(\text{0}\text{.15}\right)+{\left(35-7\right)}^2\left(\text{0}\text{.25}\right)+{\left(40-7\right)}^2\left(0.1\right)\right\}\\ =426\end{array}$

The variance is 426.

(d)

To determine

The expected value and variance of Y.

Explanation of Solution

The expected value of Y from Parameter X is calculated as follows:

$\begin{array}{c}\text{E}\left(\text{Y}\right)=\text{E}\left(\text{5X}\right)=\text{5E}\left(\text{X}\right)\\ =\text{5}\left(\text{1}\text{.4}\right)\\ =\text{7}\end{array}$

The expected value is 7.

The variance of Y from Parameter X is calculated as follows:

$\begin{array}{l}\text{V}\left(\text{Y}\right)=\text{V}\left(\text{5X}\right)=\text{5V}\left(\text{X}\right)\\ =\text{25}\left(\text{17}\text{.04}\right)\\ =\text{426}\end{array}$

The variance is 426.