Chapter 7.1, Problem 19E

(a)

To determine

Calculate the mean and standard deviation.

Explanation of Solution

The probability distribution of the random variable X is shown below:

Table 1

X	0	1	2	3
P(X)	0.4	0.3	0.2	0.1

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

$\begin{array}{c}\text{μ}=\text{E}\left(\text{X}\right)={\displaystyle \sum \text{xP}\left(\text{x}\right)}\\ =\left(0\left(\text{0}\text{.4}\right)\text{ + 1}\left(\text{0}\text{.3}\right)\text{ + 2}\left(\text{0}\text{.2}\right)\text{ + 3}\left(\text{0}\text{.1}\right)\right)\\ =1\end{array}$

The mean value is 1.

The standard deviation of probability distribution of X is calculated as follows:

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{X}\right)={\displaystyle \sum {\left(\text{x-μ}\right)}^{\text{2}}\text{P}\left(\text{x}\right)}\\ \text{=}\left\{{\left(0-1\right)}^2\left(\text{0}\text{.4}\right)+{\left(1-1\right)}^2\left(\text{0}\text{.3}\right)+{\left(2-1\right)}^2\left(\text{0}\text{.2}\right)+{\left(3-1\right)}^2\left(0.1\right)\right\}\\ =1\\ \text{σ}=\sqrt{{\text{σ}}^{\text{2}}}\\ =\sqrt{1}\\ =1\end{array}$

The standard deviation is 1.

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 $\text{Y}=3\text{X+2}$ is shown in the below table:

X	0	1	2	3
Y	2	5	8	11
P(Y)	0.4	0.3	0.2	0.1

(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:

$\begin{array}{c}\text{μ}=\text{E}\left(\text{Y}\right)={\displaystyle \sum \text{yP}\left(\text{y}\right)}\\ =\left(2\left(\text{0}\text{.4}\right)\text{+ 5}\left(\text{0}\text{.3}\right)\text{+ 8}\left(\text{0}\text{.2}\right)\text{+11}\left(\text{0}\text{.1}\right)\right)\\ =5\end{array}$

The mean value is 5.

The standard deviation of the probability distribution of Y is calculated as follows:

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{Y}\right)={\displaystyle \sum {\left(\text{y-μ}\right)}^{\text{2}}\text{P}\left(\text{y}\right)}\\ \text{=}\left\{{\left(2-5\right)}^2\left(\text{0}\text{.4}\right)+{\left(5-5\right)}^2\left(\text{0}\text{.3}\right)+{\left(8-5\right)}^2\left(\text{0}\text{.2}\right)+{\left(11-5\right)}^2\left(0.1\right)\right\}\\ =9.0\\ \text{σ}=\sqrt{{\text{σ}}^{\text{2}}}\\ =\sqrt{9.0}\\ =3\end{array}$

The standard deviation is 3.

(d)

To determine

The mean, variance, and standard deviation.

Explanation of Solution

The mean value is calculated as follows:

$\begin{array}{c}\text{μ}=\text{E}\left(\text{3X+2}\right)=3\text{E}\left(\text{X}\right)+2\\ =3\left(1\right)+2\\ =5\end{array}$

The mean value is 5.

Variance and standard deviation are calculated as follows:

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{Y}\right)=\text{V}\left(3\text{X+2}\right)\\ \text{=3E}\left(\text{X}\right)\\ =3^2\text{V}\left(\text{X}\right)\\ =9\left(1\right)\\ =9\\ \text{σ}=\sqrt{{\text{σ}}^{\text{2}}}\\ =\sqrt{9}\\ =3\end{array}$

The variance is 9 and standard deviation is 3. The parameters are identical.