Chapter 7.4, Problem 42E

a.

To determine

Compute the probability distribution of $d_1$.

Answer to Problem 42E

The probability distribution of $d_1$ is:

$d_1$	0	650	2,000
$p\left(d_1\right)$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$

Explanation of Solution

It is given that there are three cities, namely City B, City C, and City D. The meeting is going to be held in one of the three cities. The shortest highway distance from City B to City C passes through City D. The highway distance from City B to City D is 1,350 miles, and the distance from City D to City C is 650 miles.

The random variable $d_1$ represents the driving distance from City C to the place of meeting.

Therefore, the random variable $d_1$ can take values 0, 650, and 2,000.

The probability distribution of $d_1$ is as given below:

$d_1$	0	650	2,000
$p\left(d_1\right)$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$

b.

To determine

Obtain the mean of $d_1$.

Answer to Problem 42E

The mean for the random variable $d_1$ is 883.333.

Explanation of Solution

Calculation:

The mean for the random variable $d_1$ is obtained as given below:

$\begin{array}{c}{\mu }_{d_1}={\displaystyle \sum d_1\cdot p\left(d_1\right)}\\ =\left(0\right)\left(\frac{1}{3}\right)+\left(650\right)\left(\frac{1}{3}\right)+\left(2,000\right)\left(\frac{1}{3}\right)\\ =\frac{2650}{3}\\ =883.333\end{array}$

Thus, the mean for the random variable $d_1$ is 883.333.

c.

To determine

Obtain the standard deviation of $d_1$.

Answer to Problem 42E

The standard deviation for the random variable $d_1$ is 834.5.

Explanation of Solution

Calculation:

The variance for the random variable $d_1$ is obtained as given below:

$\begin{array}{c}{\sigma }_{d_1}^2={\displaystyle \sum {\left(d_1-{\mu }_{d_1}\right)}^{2}p\left(d_1\right)}\\ =\left[{\left(0-833.333\right)}^2\left(\frac{1}{3}\right)+{\left(650-833.333\right)}^2\left(\frac{1}{3}\right)+{\left(2,000-833.333\right)}^2\left(\frac{1}{3}\right)\right]\\ =\left[{\left(-833.333\right)}^2\left(\frac{1}{3}\right)+{\left(-183.333\right)}^2\left(\frac{1}{3}\right)+{\left(-1,166.667\right)}^2\left(\frac{1}{3}\right)\right]\\ =\frac{694,443.89+33,610.99+1,361,111.89}{3}\\ =\frac{2,089,166.77}{3}\\ =696,388.923\end{array}$

Thus, the variance for the random variable $d_1$ is 696,388.923.

The standard deviation for the random variable $d_1$ is obtained as given below:

$\begin{array}{c}{\sigma }_{d_1}=\sqrt{{\sigma }_{d_1}^2}\\ =\sqrt{696,388.923}\\ =834.5\end{array}$

Thus, the standard deviation for the random variable $d_1$ is 834.5.

d.

To determine

Check either the probability distributions of $d_2\text{ and }{d}_3$ are same as the probability distribution of $d_1$.

Answer to Problem 42E

Neither the probability distributions of $d_2\text{ and }{d}_3$ are same as the probability distribution of $d_1$.

Explanation of Solution

Calculation:

The random variable $d_2$ represents the driving distance from City D to the place of meeting.

Therefore, the random variable $d_2$ can take values 0, 650, and 1,350.

The probability distribution of $d_2$ is as given below:

$d_2$	0	650	1,350
$p\left(d_2\right)$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$

The random variable $d_3$ represents the driving distance from City B to the place of meeting.

Therefore, the random variable $d_3$ can take values 0, 650, and 1,350.

The probability distribution of $d_2$ is as given below:

$d_3$	0	1,350	2,000
$p\left(d_3\right)$	$\frac{1}{3}$	$\frac{1}{3}$	$\frac{1}{3}$

Therefore, neither the probability distributions of $d_2\text{ and }{d}_3$ are same as the probability distribution of $d_1$.

e.

To determine

Derive the probability distribution of t.

Answer to Problem 42E

The probability distribution of t is,

t	650	3,350
$p\left(t\right)$	$\frac{2}{3}$	$\frac{1}{3}$

Explanation of Solution

Calculation:

The random variable t defined as $t=d_1+d_2$.

The values of random variable t can be obtained as follows:

Place of meeting	$d_1$	$d_2$	$t=d_1+d_2$	Probability
City C	0	650	650	$\frac{1}{3}$
City D	650	0	650	$\frac{1}{3}$
City B	2,000	1,350	3,350	$\frac{1}{3}$

The probability distribution of t is as given below:

t	650	3,350
$p\left(t\right)$	$\frac{2}{3}$	$\frac{1}{3}$

f.

To determine

Check whether the below statements are true or false.

1. i. $E\left(t\right)=E\left(d_1\right)+E\left(d_2\right)$
2. ii. ${\sigma }_t^2={\sigma }_{d_1}^2+{\sigma }_{d_2}^2$

Answer to Problem 42E

1. i. True.
2. ii. False.

Explanation of Solution

Calculation:

From Part (b), $E\left(d_1\right)=883.333$ and from Part (c), ${\sigma }_{d_1}^2=696,388.923$.

The mean for the random variable $d_2$ is obtained as given below:

$\begin{array}{c}{\mu }_{d_2}=E\left(d_2\right)\\ ={\displaystyle \sum d_2\cdot p\left(d_2\right)}\\ =\left(0\right)\left(\frac{1}{3}\right)+\left(650\right)\left(\frac{1}{3}\right)+\left(1,350\right)\left(\frac{1}{3}\right)\\ =\frac{2000}{3}\\ =666.666\end{array}$

Thus, $E\left(d_2\right)=666.666$.

The variance for the random variable $d_2$ is obtained as given below:

$\begin{array}{c}{\sigma }_{d_2}^2={\displaystyle \sum {\left(d_2-{\mu }_{d_2}\right)}^{2}p\left(d_2\right)}\\ =\left[{\left(0-666.666\right)}^2\left(\frac{1}{3}\right)+{\left(650-666.666\right)}^2\left(\frac{1}{3}\right)+{\left(1,350-666.666\right)}^2\left(\frac{1}{3}\right)\right]\\ =\left[{\left(-666.666\right)}^2\left(\frac{1}{3}\right)+{\left(-16.666\right)}^2\left(\frac{1}{3}\right)+{\left(683.334\right)}^2\left(\frac{1}{3}\right)\right]\\ =\frac{444,443.555+277.755+466,945.355}{3}\\ =\frac{911,666.665}{3}\\ =303,888.888\end{array}$

Thus, ${\sigma }_{d_2}^2=303,888.888$

The mean for the random variable t is obtained as given below:

$\begin{array}{c}{\mu }_t=E\left(t\right)\\ ={\displaystyle \sum t\cdot p\left(t\right)}\\ =\left(650\right)\left(\frac{2}{3}\right)+\left(3,350\right)\left(\frac{1}{3}\right)\\ =\frac{4,650}{3}\\ =1,550\end{array}$

Thus, $E\left(t\right)=1,550$.

The variance for the random variable t is obtained as given below:

$\begin{array}{c}{\sigma }_t^2={\displaystyle \sum {\left(t-{\mu }_t\right)}^{2}p\left(t\right)}\\ =\left[{\left(650-1,550\right)}^2\left(\frac{2}{3}\right)+{\left(3,350-1,550\right)}^2\left(\frac{1}{3}\right)\right]\\ =\left[{\left(-900\right)}^2\left(\frac{2}{3}\right)+{\left(1800\right)}^2\left(\frac{1}{3}\right)\right]\\ =\frac{1,620,000+3,240,000}{3}\\ =\frac{4,860,000}{3}\\ =1,620,000\end{array}$

Thus, ${\sigma }_t^2=1,620,000$

i.

From the above results, it can be obtained as follows:

$E\left(t\right)=1,550$.

$\begin{array}{c}E\left(d_1\right)+E\left(d_2\right)=883.333+666.666\\ =1,549.999\\ \approx 1550\end{array}$

Therefore, $E\left(t\right)=E\left(d_1\right)+E\left(d_2\right)$.

Therefore, the given statement is true.

ii.

From the above results, it can be obtained as follows:

${\sigma }_t^2=1,620,000$

$\begin{array}{c}{\sigma }_{d_1}^2+{\sigma }_{d_2}^2=696,388.923+303,888.888\\ =1,000,277.811\end{array}$

Therefore, ${\sigma }_t^2\ne {\sigma }_{d_1}^2+{\sigma }_{d_2}^2$

Therefore, the given statement is false.