Chapter 7.2, Problem 66E

(a)

To determine

Calculate the probability distribution of two team’s scores.

Explanation of Solution

The probability distribution of scores is shown in below table.

Table 1

	Home team
Visiting team	0	1	2	3
0	0.14	0.11	0.09	0.10
1	0.12	0.10	0.05	0.02
2	0.09	0.07	0.04	0.01
3	0.03	0.02	0.01	0

Probability distribution of two team’s scoresis shown in below table.

P(T)	Probability distribution
P(T=0)	0.14
P(T=1)	0.23
P(T=2)	0.28
P(T=3)	0.25
P(T=4)	0.08
P(T=5)	0.02
P(T=6)	0

Economics Concept Introduction

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

(b)

To determine

Mean,variance and standard deviation.

Explanation of Solution

Mean of team’s scores is calculated as follows.

$\begin{array}{c}\text{μ}=\text{E}\left(\text{X}\right)={\displaystyle \sum \text{xP(x)}}\\ =\left(\begin{array}{l}\text{0}\left(\text{0}\text{.14}\right)+\text{1}\left(\text{0}\text{.23}\right)+\text{2}\left(\text{0}\text{.28}\right)+3\left(0.25\right)\\ +4\left(0.08\right)+5\left(0.02\right)+6\left(0\right)\end{array}\right)\\ =1.96\end{array}$

Mean value is 0.96.

Variance of team’s scores is calculated as follows.

$\begin{array}{c}{\text{σ}}^{\text{2}}=\text{V}\left(\text{X}\right)={\displaystyle \sum {\text{(x}-\text{μ)}}^{\text{2}}\text{P(x)}}\\ =\left(\begin{array}{c}{\left(\text{0}-1.96\right)}^{\text{2}}\left(\text{0}\text{.14}\right)\text{ + }{\left(\text{1}-1.96\right)}^{\text{2}}\left(\text{0}\text{.23}\right)\text{ + }{\left(2-1.96\right)}^{\text{2}}\left(\text{0}\text{.28}\right)+{\left(3-1.96\right)}^2\left(0.25\right)\\ +{\left(4-1.96\right)}^2\left(0.08\right)+{\left(5-1.96\right)}^2\left(0.02\right)+{\left(6-1.96\right)}^2\left(0\right)\end{array}\right)\\ =\text{ 1}\text{.538}\end{array}$

Variance is 1.538.

Standard deviation (square root of variance) of team’s scores is calculated as follows.

$\begin{array}{c}\text{σ}=\sqrt{{\text{σ}}^{\text{2}}}\\ =\sqrt{1.538}\\ =1.240\end{array}$

Standard deviation is 1.240.

(c)

To determine

Covariance and coefficient of correlation.

Explanation of Solution

Covariance and coefficient of correlation of two team’s score is calculated as follows.

$\begin{array}{c}{\displaystyle \sum _{\text{all}\text{x}}}{\displaystyle \sum _{\text{all}\text{y}}\text{xyP(x,y)}}=\left(\begin{array}{l}\left(0\right)\left(0\right)\left(\text{.14}\right)+\left(0\right)\left(1\right)\left(\text{.12}\right)+\left(0\right)\left(2\right)\left(\text{.09}\right)+\left(0\right)\left(3\right)\left(\text{.03}\right)\\ +\left(1\right)\left(0\right)\left(\text{.11}\right)+\left(1\right)\left(1\right)\left(.10\right)+\left(1\right)\left(2\right)\left(.07\right)+\left(1\right)\left(3\right)\left(.02\right)\\ +\left(2\right)\left(0\right)\left(.09\right)+\left(2\right)\left(1\right)\left(.05\right)+\left(2\right)\left(2\right)\left(.04\right)+\left(2\right)\left(3\right)\left(.01\right)\\ +\left(3\right)\left(0\right)\left(.10\right)+\left(3\right)\left(1\right)\left(.02\right)+\left(3\right)\left(2\right)\left(.01\right)++\left(3\right)\left(3\right)\left(0\right)\end{array}\right)\\ =\text{1}\text{.76}\\ \text{COV}\left(\text{X,Y}\right)={\displaystyle \sum _{\text{all}\text{x}}}{\displaystyle \sum _{\text{all}\text{y}}\text{xyP(x,y)}}-{\text{μ}}_{\text{x}}{\text{μ}}_{\text{y}}\\ =0.74-\left(1.07\right)\left(0.89\right)\\ =-0.212\\ \text{ρ}=\frac{\text{COV(X,Y)}}{{\text{σ}}_{\text{x}}{\text{σ}}_{\text{y}}}\\ =\frac{-0.212}{(1.085)(0.937)}\\ =-0.217\end{array}$