Chapter 2.6, Problem 9E

a.

To determine

Find the marginal probability mass function X.

Answer to Problem 9E

The marginal probability mass function X is,

X	$p_x\left(x\right)$
0	0.10
1	0.20
2	0.30
3	0.25
4	0.15

Explanation of Solution

Given info:

In a lot of 500 bolts, X be the number that are downgraded and Y be the number that are scrap.

Calculation:

The marginal probability mass function of X can be written as,

$p_X\left(x\right)=P\left(X=x\right)={\displaystyle \sum _{y}p\left(x,y\right)}$

The joint probability mass function of X and Y their totals are given in the following table:

		Y
X	0	1	2	3	4	Total
0	0.06	0.03	0.01	0.00	0.00	0.10
1	0.06	0.08	0.04	0.02	0.00	0.20
2	0.04	0.05	0.12	0.06	0.03	0.30
3	0.00	0.03	0.07	0.09	0.06	0.25
4	0.00	0.00	0.02	0.06	0.07	0.15
Total	0.16	0.19	0.26	0.23	0.16	1

Table 1

From the Table 1, the marginal probability mass function X is,

X	$p_x\left(x\right)$
0	0.10
1	0.20
2	0.30
3	0.25
4	0.15

Here, it is observed that $p_X\left(x\right)=0\text{ if }x\ne 0,1,2,3\text{ or 4}$.

b.

To determine

Find the marginal probability mass function Y.

Answer to Problem 9E

The marginal probability mass function Y is,

Y	$p_Y\left(y\right)$
0	0.16
1	0.19
2	0.26
3	0.23
4	0.16

Explanation of Solution

Calculation:

The marginal probability mass function of Y can be written as,

$p_Y\left(y\right)=P\left(Y=y\right)={\displaystyle \sum _{x}p\left(x,y\right)}$

From the Table 1, the marginal probability mass function Y is,

Y	$p_Y\left(y\right)$
0	0.16
1	0.19
2	0.26
3	0.23
4	0.16

Here, it is observed that $p_Y\left(y\right)=0\text{ if }y\ne 0,1,2,3\text{ or 4}$.

c.

To determine

Explain whether X and Y are independent.

Answer to Problem 9E

No, X and Y are not independent because the joint probability mass function is not equal to the product of the marginals.

Explanation of Solution

Calculation:

Independent:

If two variables are independent then,

$P\left(X=x\text{ and }Y=y\right)=P\left(X=x\right)P\left(X=y\right)$

For $\left(x=0,y=0\right)$:

$\begin{array}{c}{p}_{X,Y}\left(0,0\right)=p_X\left(0\right)p_Y\left(0\right)\\ 0.06=\left(0.10\right)\left(0.16\right)\\ 0.06\ne 0.016\end{array}$

Thus, it can be concluded that X and Y are not independent.

d.

To determine

Find the value of ${\mu }_X$ and ${\mu }_Y$.

Answer to Problem 9E

The value of ${\mu }_X$ and ${\mu }_Y$ are 2.15 and 2.04.

Explanation of Solution

Calculation:

The value of ${\mu }_X$ can be obtained as follows:

$\begin{array}{c}{\mu }_X={\displaystyle \sum x{p}_X\left(x\right)}\\ =0p_X\left(0\right)+1p_X\left(1\right)+2p_X\left(2\right)+3p_X\left(3\right)+4p_X\left(4\right)\\ =0\left(0.10\right)+1\left(0.20\right)+2\left(0.30\right)+3\left(0.25\right)+4\left(0.15\right)\\ =2.15\end{array}$

Thus, the value of ${\mu }_X$ is 2.15.

The value of ${\mu }_Y$ can be obtained as follows:

$\begin{array}{c}{\mu }_Y={\displaystyle \sum y{p}_Y\left(y\right)}\\ =0p_Y\left(0\right)+1p_Y\left(1\right)+2p_Y\left(2\right)+3p_Y\left(3\right)+4p_Y\left(4\right)\\ =0\left(0.16\right)+1\left(0.19\right)+2\left(0.26\right)+3\left(0.23\right)+4\left(0.16\right)\\ =2.04\end{array}$

Thus, the value of ${\mu }_Y$ is 2.04.

e.

To determine

Find the value of ${\sigma }_X$ and ${\sigma }_Y$.

Answer to Problem 9E

The value of ${\sigma }_X$ and ${\sigma }_Y$ are 1.1948 and 1.3032.

Explanation of Solution

Calculation:

The value of ${\sigma }_{{}_X}^2$ can be obtained as follows:

$\begin{array}{c}{\sigma }_{{}_X}^2={\displaystyle \sum x^2{p}_X\left(x\right)-{\mu }_{{}_X}^2}\\ =0^2{p}_X\left(0\right)+1^2{p}_X\left(1\right)+2^2{p}_X\left(2\right)+3^2{p}_X\left(3\right)+4^2{p}_X\left(4\right)-{2.15}^2\\ =0\left(0.10\right)+1\left(0.20\right)+4\left(0.30\right)+9\left(0.25\right)+16\left(0.15\right)-4.6225\\ =1.4275\end{array}$

Thus, the value of ${\sigma }_{{}_X}^2$ is 1.4275.

The formula for finding ${\sigma }_X$ is,

$\begin{array}{c}{\sigma }_X=\sqrt{{\sigma }_{{}_X}^2}\\ =\sqrt{1.4275}\\ =1.1948\end{array}$

Thus, the value of ${\sigma }_X$ is 1.1948.

The value of ${\sigma }_Y^2$ can be obtained as follows:

$\begin{array}{c}{\sigma }_Y^2={\displaystyle \sum y^2{p}_Y\left(x\right)-{\mu }_{{}_Y}^2}\\ =0^2{p}_Y\left(0\right)+1^2{p}_Y\left(1\right)+2^2{p}_Y\left(2\right)+3^2{p}_Y\left(3\right)+4^2{p}_Y\left(4\right)-{2.04}^2\\ =0\left(0.16\right)+1\left(0.19\right)+4\left(0.26\right)+9\left(0.23\right)+16\left(0.16\right)-4.1616\\ =1.6984\end{array}$

Thus, the value of ${\sigma }_Y^2$ is 1.6984.

The formula for finding ${\sigma }_Y$ is,

$\begin{array}{c}{\sigma }_Y=\sqrt{{\sigma }_{{}_Y}^2}\\ =\sqrt{1.6984}\\ =1.3032\end{array}$

Thus, the value of ${\sigma }_Y$ is 1.3032.

f.

To determine

Find the value of $\text{Cov}\left(X,Y\right)$.

Answer to Problem 9E

The value of $\text{Cov}\left(X,Y\right)$ is 5.44.

Explanation of Solution

Calculation:

The value of $\text{Cov}\left(X,Y\right)$ can be obtained as follows:

$\begin{array}{c}\text{Cov}\left(X,Y\right)={\mu }_{XY}-{\mu }_X{\mu }_Y\\ =\left(\begin{array}{c}\left(0\right)\left(0\right)p_{X,Y}\left(0,0\right)+\left(0\right)\left(1\right)p_{X,Y}\left(0,1\right)+\left(0\right)\left(2\right)p_{X,Y}\left(0,2\right)+\\ \left(0\right)\left(3\right)p_{X,Y}\left(0,3\right)+\left(0\right)\left(4\right)p_{X,Y}\left(0,4\right)+\left(1\right)\left(0\right)p_{X,Y}\left(1,0\right)+\\ \left(1\right)\left(1\right)p_{X,Y}\left(1,1\right)+\left(1\right)\left(2\right)p_{X,Y}\left(1,2\right)+\left(1\right)\left(3\right)p_{X,Y}\left(1,3\right)+\\ \left(1\right)\left(4\right)p_{X,Y}\left(1,4\right)+\left(2\right)\left(0\right)p_{X,Y}\left(2,0\right)+\left(2\right)\left(1\right)p_{X,Y}\left(2,1\right)+\\ \left(2\right)\left(2\right)p_{X,Y}\left(2,2\right)+\left(2\right)\left(3\right)p_{X,Y}\left(2,3\right)+\left(2\right)\left(4\right)p_{X,Y}\left(2,4\right)\\ \left(3\right)\left(0\right)p_{X,Y}\left(3,0\right)+\left(3\right)\left(1\right)p_{X,Y}\left(3,1\right)+\left(3\right)\left(2\right)p_{X,Y}\left(3,2\right)+\\ \left(3\right)\left(3\right)p_{X,Y}\left(3,3\right)+\left(3\right)\left(4\right)p_{X,Y}\left(3,4\right)+\left(4\right)\left(0\right)p_{X,Y}\left(4,0\right)+\\ \left(4\right)\left(1\right)p_{X,Y}\left(4,1\right)+\left(4\right)\left(2\right)p_{X,Y}\left(4,2\right)+\left(4\right)\left(3\right)p_{X,Y}\left(4,3\right)+\\ \left(4\right)\left(4\right)p_{X,Y}\left(4,4\right)\end{array}\right)-\left(2.15\right)\left(2.04\right)\\ =\left(\begin{array}{l}\left(0\right)\left(0\right)\left(0.06\right)+\left(0\right)\left(1\right)\left(0.03\right)+\left(0\right)\left(2\right)\left(0.01\right)+\left(0\right)\left(3\right)\left(0.00\right)+\\ \left(0\right)\left(4\right)\left(0.00\right)+\left(1\right)\left(0\right)\left(0.06\right)+\left(1\right)\left(1\right)\left(0.06\right)+\left(1\right)\left(2\right)\left(0.04\right)+\\ \left(1\right)\left(3\right)\left(0.02\right)+\left(1\right)\left(4\right)\left(0.00\right)+\left(2\right)\left(0\right)\left(0.04\right)+\left(2\right)\left(1\right)\left(0.05\right)+\\ \left(2\right)\left(2\right)\left(0.12\right)+\left(2\right)\left(3\right)\left(0.06\right)+\left(2\right)\left(4\right)\left(0.03\right)+\left(3\right)\left(0\right)\left(0.00\right)+\\ \left(3\right)\left(1\right)\left(0.03\right)+\left(3\right)\left(2\right)\left(0.07\right)+\left(3\right)\left(3\right)\left(0.09\right)+\left(3\right)\left(4\right)\left(0.06\right)+\\ \left(4\right)\left(0\right)\left(0.00\right)+\left(4\right)\left(1\right)\left(0.00\right)+\left(4\right)\left(2\right)\left(0.02\right)+\left(4\right)\left(3\right)\left(0.06\right)+\\ \left(4\right)\left(4\right)\left(0.07\right)\end{array}\right)-4.386\\ =5.44-4.386\\ =1.054\end{array}$

Thus, the value of $\text{Cov}\left(X,Y\right)$ is 1.054.

g.

To determine

Find the value of ${\rho }_{X,Y}$.

Answer to Problem 9E

The value of ${\rho }_{X,Y}$ is 0.6769.

Explanation of Solution

Calculation:

The formula for finding the ${\rho }_{X,Y}$ is,

${\rho }_{X,Y}=\frac{\text{Cov}\left(\text{X,Y}\right)}{{\sigma }_X{\sigma }_Y}$

Substitute $\text{Cov}\left(\text{X,Y}\right)=1.0540$, ${\sigma }_X=1.1948$ and ${\sigma }_Y=0.6769$

$\begin{array}{c}{\rho }_{X,Y}=\frac{1.0540}{\left(1.1948\right)\left(1.3032\right)}\\ =\frac{1.0540}{1.5571}\\ =0.6769\end{array}$

Thus, the value of ${\rho }_{X,Y}$ is 0.6769.