Chapter 2, Problem 25SE

a.

To determine

Find the joint probability mass function of X and Y.

Answer to Problem 25SE

The joint probability mass function of X and Y is,

	Y
X	0	1	2
0	0.0667	0.2000	0.0667
1	0.2667	0.2667	0
2	0.1333	0	0

Explanation of Solution

Given info:

A box contains four 75W light bulbs, three 60W light bulbs, and three burned-out light bulbs. Also, two bulbs are selected at random from the box.

Here, X represents the number of 75W bulbs selected, and let Y represent the number of 60W bulbs selected.

Calculation:

From the given information, it can be observed that the two bulbs are selected random. The number of not burned-out light bulbs is 7 $\left(=4+3\right)$ and the number of burned-out light bulbs is 3.

Therefore, the probabilities for finding the joint probability mass function of X and Y is,

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

$\begin{array}{c}p\left(0,0\right)=P\left(X=0\text{ and }Y=0\right)\\ =\left(\frac{3}{10}\right)\left(\frac{2}{9}\right)\\ =\frac{1}{15}\\ =0.0667\end{array}$

For $\left(x=1,y=0\right)$

$\begin{array}{c}p\left(1,0\right)=P\left(X=1\text{ and }Y=0\right)\\ =\left(\frac{4}{10}\right)\left(\frac{3}{9}\right)+\left(\frac{3}{10}\right)\left(\frac{4}{9}\right)\\ =\frac{4}{15}\\ =0.2667\end{array}$

For $\left(x=2,y=0\right)$

$\begin{array}{c}p\left(2,0\right)=P\left(X=2\text{ and }Y=0\right)\\ =\left(\frac{4}{10}\right)\left(\frac{3}{9}\right)\\ =\frac{2}{15}\\ =0.1331\end{array}$

For $\left(x=0,y=1\right)$

$\begin{array}{c}p\left(0,1\right)=P\left(X=0\text{ and }Y=1\right)\\ =\left(\frac{3}{10}\right)\left(\frac{3}{9}\right)+\left(\frac{3}{10}\right)\left(\frac{3}{9}\right)\\ =\frac{3}{15}\\ =0.2000\end{array}$

For $\left(x=1,y=1\right)$

$\begin{array}{c}p\left(1,1\right)=P\left(X=1\text{ and }Y=1\right)\\ =\left(\frac{4}{10}\right)\left(\frac{3}{9}\right)+\left(\frac{3}{10}\right)\left(\frac{4}{9}\right)\\ =\frac{4}{15}\\ =0.2667\end{array}$

For $\left(x=0,y=2\right)$

$\begin{array}{c}p\left(0,2\right)=P\left(X=0\text{ and }Y=2\right)\\ =\left(\frac{3}{10}\right)\left(\frac{2}{9}\right)\\ =\frac{1}{15}\\ =0.0667\end{array}$

$\begin{array}{c}p\left(1,2\right)=P\left(X=1\text{ and }Y=2\right)\\ =\left(\frac{0}{10}\right)\left(\frac{0}{9}\right)\\ =0\end{array}$

$\begin{array}{c}p\left(2,1\right)=P\left(X=2\text{ and }Y=1\right)\\ =\left(\frac{0}{10}\right)\left(\frac{0}{9}\right)\\ =0\end{array}$

$\begin{array}{c}p\left(2,2\right)=P\left(X=2\text{ and }Y=2\right)\\ =\left(\frac{0}{10}\right)\left(\frac{0}{9}\right)\\ =0\end{array}$

By using the above probabilities, the joint probability mass function of X and Y is,

	Y
X	0	1	2
0	0.0667	0.2000	0.0667
1	0.2667	0.2667	0
2	0.1333	0	0

Table 1

b.

To determine

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

Answer to Problem 25SE

The value of ${\mu }_X$ is 0.0534.

Explanation of Solution

Calculation:

The totals of the joint probability mass function of tensile strength (in thousands of pounds/in²) and additive concentration is,

	Y
X	0	1	2	Total
0	0.0667	0.2000	0.0667	0.3334
1	0.2667	0.2667	0	0.5334
2	0.1333	0	0	0.1333
Total	0.4667	0.4667	0.0667	1

Table 1

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

$\begin{array}{c}{\mu }_X={\displaystyle \sum x{p}_X\left(x\right)}\\ =0p_X\left(0\right)+1p_X\left(1\right)+2p_X\left(2\right)\\ =0\left(0.3334\right)+1\left(0.5334\right)+2\left(0.1333\right)\\ =0.8\end{array}$

Thus, the value of ${\mu }_X$ is 0.8.

c.

To determine

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

Answer to Problem 25SE

The value of ${\mu }_Y$ is 0.6.

Explanation of Solution

Calculation:

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

$\begin{array}{c}{\mu }_Y={\displaystyle \sum y{p}_Y\left(y\right)}\\ =0p_Y\left(0\right)+1p_Y\left(1\right)+2p_Y\left(2\right)\\ =0\left(0.4667\right)+1\left(0.4667\right)+2\left(0.0667\right)\\ =0+0.4667+0.1334\end{array}$

       $\approx 0.6$

Thus, the value of ${\mu }_Y$ is 0.6.

d.

To determine

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

Answer to Problem 25SE

The value of ${\sigma }_X$ is 0.6532.

Explanation of Solution

Calculation:

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

$\begin{array}{c}{\sigma }_{{}_X}^2={\displaystyle \sum x^2{p}_X\left(x\right)-{\mu }_{{}_X}^2}\\ =0^2{p}_X\left(0\right)+1^2{p}_X\left(1\right)+2^2{p}_X\left(2\right)-{0.8}^2\\ =0\left(0.3334\right)+1\left(0.5334\right)+4\left(0.1333\right)-0.64\\ =0+0.5334+0.5332-0.64\end{array}$

       $=0.4266$

Thus, the value of ${\sigma }_{{}_X}^2$ is 0.42664.

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

$\begin{array}{c}{\sigma }_X=\sqrt{{\sigma }_{{}_X}^2}\\ =\sqrt{0.4266}\\ =0.6532\end{array}$

Thus, the value of ${\sigma }_X$ is 0.6532.

e.

To determine

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

Answer to Problem 25SE

The value of ${\sigma }_Y$ is 0.6111.

Explanation of Solution

Calculation:

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

$\begin{array}{c}{\sigma }_Y^2={\displaystyle \sum y^2{p}_Y\left(x\right)-{\mu }_{{}_Y}^2}\\ =0^2{p}_Y\left(0\right)+1^2{p}_Y\left(1\right)+2^2{p}_Y\left(2\right)-{0.6}^2\\ =0^2\left(0.4667\right)+1^2\left(0.4667\right)+2^2\left(0.0667\right)-0.36\\ =0+0.4667+0.2668-0.36\end{array}$

       $=0.3735$

Thus, the value of ${\sigma }_Y^2$ is 0.3735.

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

$\begin{array}{c}{\sigma }_Y=\sqrt{{\sigma }_{{}_Y}^2}\\ =\sqrt{0.3735}\\ =0.6111\end{array}$

Thus, the value of ${\sigma }_Y$ is 0.6111.

e.

To determine

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

Answer to Problem 25SE

The value of $\text{Cov}\left(X,Y\right)$ is –0.2133.

Explanation of Solution

Calculation:

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

$\begin{array}{c}\text{Cov}\left(X,Y\right)={\mu }_{XY}-{\mu }_X{\mu }_Y\\ =\left(\begin{array}{l}\left(0\right)\left(0\right)P\left(X=0\text{ and }Y=0\right)+\left(1\right)\left(0\right)P\left(X=1\text{ and }Y=0\right)+\\ \left(2\right)\left(0\right)P\left(X=2\text{ and }Y=0\right)+\left(0\right)\left(1\right)P\left(X=0\text{ and }Y=1\right)+\\ \left(1\right)\left(1\right)P\left(X=1\text{ and }Y=1\right)+\left(2\right)\left(1\right)P\left(X=2\text{ and }Y=1\right)+\\ \left(0\right)\left(2\right)P\left(X=0\text{ and }Y=2\right)+\left(1\right)\left(2\right)P\left(X=1\text{ and }Y=2\right)+\\ \left(2\right)\left(2\right)P\left(X=2\text{ and }Y=2\right)\end{array}\right)-\left(0.8\right)\left(0.8\right)\\ =0+0+0+0+0+0+0+0+0.2667-0.48\\ =-0.2133\end{array}$

Thus, the value of $\text{Cov}\left(X,Y\right)$ is –0.2133.

g.

To determine

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

Answer to Problem 25SE

The value of ${\rho }_{X,Y}$ is –0.5343.

Explanation of Solution

Calculation:

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

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

Substitute $Cov\left(X,Y\right)=-0.2133$, ${\sigma }_X=0.6532$ and ${\sigma }_Y=0.6111$

$\begin{array}{c}{\rho }_{X,Y}=\frac{-0.2133}{\left(0.6532\right)\left(0.6111\right)}\\ =-\frac{0.2133}{0.3992}\\ =-0.5343\end{array}$

Thus, the value of ${\rho }_{X,Y}$ is –0.5343.