Chapter 7, Problem 164CE

(a)

To determine

Calculate the mean and standard deviation of violent television program viewers.

Explanation of Solution

The probability distribution of the number of viewers of violent and non-violent television program shows is shown below:

Table 1

Viewers of violent shows
X	0	1	2	3	4	5
P(X)	0.36	0.22	0.20	0.09	0.08	0.05
Viewers of nonviolent shows
X	0	1	2	3	4	5
P(X)	0.15	0.18	0.23	0.26	0.10	0.08

The mean of number of correct answers among viewers of violent television programs is calculated as follows:

$\begin{array}{c}\text{μ}=\text{E}\left(\text{X}\right)={\displaystyle \sum \text{xP}\left(\text{x}\right)}\\ =\left(0\left(\text{0}\text{.36}\right)\text{ + 1}\left(\text{0}\text{.22}\right)\text{ + 2}\left(\text{0}\text{.20}\right)\text{ + 3}\left(\text{0}\text{.09}\right)\text{ + 4}\left(\text{0}\text{.08}\right)+5\left(0.05\right)\right)\\ =1.46\end{array}$

The mean value is 1.46.

The standard deviation (square root of variance) of the number of correct answers among viewers of violent television programs is calculated as follows:

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{X}\right)={\displaystyle \sum {\left(\text{x-μ}\right)}^{\text{2}}\text{P}\left(\text{x}\right)}\\ \text{=}\left\{\begin{array}{l}{\left(\text{0}-1.\text{46}\right)}^2\left(\text{0}\text{.36}\right)+{\left(\text{1}-1.\text{46}\right)}^2\left(\text{0}\text{.22}\right)+{\left(\text{2}-1.\text{46}\right)}^2\left(\text{0}\text{.20}\right)\\ +{\left(\text{3}-1.\text{46}\right)}^2\left(\text{0}\text{.09}\right)+{\left(\text{4}-1.\text{46}\right)}^2\left(\text{0}\text{.08}\right)+{\left(5-1.\text{46}\right)}^2\left(\text{0}\text{.05}\right)\end{array}\right\}\\ =2.23\\ \text{σ}=\sqrt{{\text{σ}}^{\text{2}}}\\ =\sqrt{2.23}\\ =1.49\end{array}$

The standard deviation is 1.49.

Economics Concept Introduction

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

(b)

To determine

Calculate the mean and standard deviation of non-violent television program viewers.

Explanation of Solution

The mean of number of correct answers among viewers of non-violent television programs is calculated as follows:

$\begin{array}{c}\text{μ}=\text{E}\left(\text{X}\right)={\displaystyle \sum \text{xP}\left(\text{x}\right)}\\ =\left(0\left(\text{0}\text{.15}\right)\text{ + 1}\left(\text{0}\text{.18}\right)\text{ + 2}\left(\text{0}\text{.23}\right)\text{ + 3}\left(\text{0}\text{.26}\right)\text{ + 4}\left(\text{0}\text{.10}\right)+5\left(0.08\right)\right)\\ =2.22\end{array}$

The mean value is 2.22.

The standard deviation (square root of variance) of the number of correct answers among viewers of non-violent television programs is calculated as follows:

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{X}\right)={\displaystyle \sum {\left(\text{x-μ}\right)}^{\text{2}}\text{P}\left(\text{x}\right)}\\ \text{=}\left\{\begin{array}{l}{\left(\text{0}-2.\text{22}\right)}^2\left(\text{0}\text{.15}\right)+{\left(\text{1}-2.\text{22}\right)}^2\left(\text{0}\text{.18}\right)+{\left(\text{2}-2.\text{22}\right)}^2\left(\text{0}\text{.23}\right)\\ +{\left(\text{3}-2.\text{22}\right)}^2\left(\text{0}\text{.26}\right)+{\left(\text{4}-2.\text{22}\right)}^2\left(\text{0}\text{.10}\right)+{\left(5-2.\text{22}\right)}^2\left(\text{0}\text{.08}\right)\end{array}\right\}\\ =2.11\\ \text{σ}=\sqrt{{\text{σ}}^{\text{2}}}\\ =\sqrt{2.11}\\ =1.45\end{array}$

The standard deviation is 1.45.