Chapter 4.1, Problem 8E

a.

To determine

Construct the graph of the pdf of Y.

Answer to Problem 8E

The graph of the pdf of Y:

Explanation of Solution

Given info:

The information is based on a professor who must get her first bus near her house and then transfer to a second bus. Here, the waiting time at the bus stop follows the uniform distribution with parameters $A=0$ and $B=5$.

The probability density function of total waiting time Y is given by:

$f\left(y\right)=\{\begin{array}{l}\frac{1}{25}y0\le y<5\\ \frac{2}{5}-\frac{1}{25}y5\le y\le 10\\ 0y<0\text{ or }y>10\end{array}$

Calculation:

From the equation, it can be observed that the values of y lie between 0 and 10.

For $0\le y<5$,

The values of the $f\left(y\right)$ obtained as follows:

For y at 0:

Substitute 0 for y in the equation $f\left(y\right)=\frac{1}{25}y$

$f\left(y\right)=\frac{1}{25}\left(0\right)$

Similarly the remaining points are obtained as follows:

y	$f\left(y\right)$
0	0
1	0.04
2	0.08
3	0.12
4	0.16

Table 1

For $5\le y\le 10$,

The values of the $f\left(y\right)$ obtained as follows:

For y at 5:

Substitute 5 for y in the equation $f\left(y\right)=\frac{2}{5}-\frac{1}{25}y$

$\begin{array}{c}f\left(y\right)=\frac{2}{5}-\frac{1}{25}\left(5\right)\\ =0.4-0.2\\ =0.2\end{array}$

Substitute 6 for y in the equation $f\left(y\right)=\frac{2}{5}-\frac{1}{25}y$

$\begin{array}{c}f\left(y\right)=\frac{2}{5}-\frac{1}{25}\left(6\right)\\ =0.4-0.24\\ =0.16\end{array}$

Similarly the remaining points are obtained as follows:

y	$f\left(y\right)$
5	0.2
6	0.16
7	0.12
8	0.08
9	0.04
10	0

Table 2

Procedure for drawing the density curve of the variable $f\left(y\right)$ for $0<y<10$ otherwise $y=0$ is as follows:

• Let horizontal axis take y values and vertical axis take $f\left(y\right)$ values
• Plot the points obtained from Table 1.

The graph of the pdf of $f\left(y\right)$ is given below:

Figure 1

In the above graph, y takes value in the interval of $\left[0,10\right]$

$f\left(y\right)$ represents the probability density function.

Thus, the graph for the pdf $f\left(y\right)$ for an interval $\left[0,10\right]$ has been obtained.

b.

To determine

Verify that ${\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy=1}$.

Explanation of Solution

Calculation:

Proof:

The total area of the $f\left(y\right)$ can be proved as given below:

$\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}={\displaystyle \underset{0}{\overset{5}{\int }}\frac{1}{25}ydy}+{\displaystyle \underset{5}{\overset{10}{\int }}\left[\frac{2}{5}-\frac{1}{25}y\right]dy}\\ =\frac{1}{25}{\displaystyle \underset{0}{\overset{5}{\int }}ydy}+{\displaystyle \underset{5}{\overset{10}{\int }}\left[\frac{2}{5}-\frac{1}{25}y\right]dy}\\ =\frac{1}{25}{\frac{y^2}{2}]}_0^5+{\frac{2}{5}y-\frac{1}{25}\times \frac{y^2}{2}]}_5^{10}\\ ={\frac{y^2}{50}]}_0^5+{\frac{2}{5}y-\frac{y^2}{50}]}_5^{10}\end{array}$

$\begin{array}{c}=\left(\frac{5^2}{50}-\frac{0^2}{50}\right)+\left\{\left(\frac{2\left(10\right)}{5}-\frac{{10}^2}{50}\right)-\left(\frac{2\left(5\right)}{5}-\frac{5^2}{50}\right)\right\}\\ =\frac{25}{50}+\left\{\left(\frac{20}{5}-\frac{100}{50}\right)-\left(\frac{10}{5}-\frac{25}{50}\right)\right\}\\ =\frac{1}{2}+\left\{\left(4-2\right)-\left(2-\frac{1}{2}\right)\right\}\\ =0.5+\left\{4-2-2+\frac{1}{2}\right\}\end{array}$

$\begin{array}{c}=0.5+0.5\\ =1\end{array}$

Thus, $\underset{\_}{{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy=1}}$.

Hence the proof.

c.

To determine

Find the probability that the total waiting time is at most 3 min.

Answer to Problem 8E

The probability that the total waiting time is at most 3 min is 0.18.

Explanation of Solution

Calculation:

The probability that the total waiting time is at most 3 min is obtained as shown below:

$\begin{array}{c}P\left(Y\le 3\right)={\displaystyle \underset{0}{\overset{3}{\int }}\frac{1}{25}ydy}\\ =\frac{1}{25}{\displaystyle \underset{0}{\overset{3}{\int }}ydy}\\ =\frac{1}{25}{\frac{y^2}{2}]}_0^3\\ ={\frac{y^2}{50}]}_0^3\end{array}$

$\begin{array}{c}=\left(\frac{3^2}{50}-\frac{0^2}{50}\right)\\ =\frac{9}{50}-0\\ =0.18\end{array}$

Thus, the probability that the total waiting time is at most 3 min is 0.18.

d.

To determine

Find the probability that the total waiting time is at most 8 min.

Answer to Problem 8E

The probability that the total waiting time is at most 8 min is 0.92.

Explanation of Solution

Calculation:

The probability that the total waiting time is at most 8 min is obtained as shown below:

$\begin{array}{c}P\left(Y\le 8\right)={\displaystyle \underset{0}{\overset{8}{\int }}f\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{5}{\int }}\frac{1}{25}ydy}+{\displaystyle \underset{5}{\overset{8}{\int }}\left[\frac{2}{5}-\frac{1}{25}y\right]dy}\\ =\frac{1}{25}{\displaystyle \underset{0}{\overset{5}{\int }}ydy}+{\displaystyle \underset{5}{\overset{8}{\int }}\left[\frac{2}{5}-\frac{1}{25}y\right]dy}\\ =\frac{1}{25}{\frac{y^2}{2}]}_0^5+{\frac{2}{5}y-\frac{1}{25}\times \frac{y^2}{2}]}_5^8\end{array}$

$\begin{array}{c}={\frac{y^2}{50}]}_0^5+{\frac{2}{5}y-\frac{y^2}{50}]}_5^8\\ =\left(\frac{5^2}{50}-\frac{0^2}{50}\right)+\left\{\left(\frac{2\left(8\right)}{5}-\frac{8^2}{50}\right)-\left(\frac{2\left(5\right)}{5}-\frac{5^2}{50}\right)\right\}\\ =\frac{1}{2}+\left\{\left(\frac{16}{5}-\frac{64}{50}\right)-\left(\frac{10}{5}-\frac{25}{50}\right)\right\}\\ =0.5+\left\{\left(3.2-1.28\right)-\left(2-\frac{1}{2}\right)\right\}\end{array}$

$\begin{array}{c}=0.5+\left\{1.92-2+\frac{1}{2}\right\}\\ =0.5+0.42\\ =0.92\end{array}$

Thus, the probability that the total waiting time is at most 8 min is 0.92.

e.

To determine

Find the probability that the total waiting time between 3 and 8 min.

Answer to Problem 8E

The probability that the total waiting time between 3 and 8 min is 0.74.

Explanation of Solution

Calculation:

The probability that the total waiting time between 3 and 8 min is obtained as shown below:

$P\left(3\le Y\le 8\right)=P\left(Y\le 8\right)-P\left(Y\le 3\right)$

From the previous part (c), $P\left(Y\le 3\right)=0.18$

From the previous part (d), $P\left(Y\le 8\right)=0.92$

The probability of the event is given below:

$\begin{array}{c}P\left(3\le Y\le 8\right)=0.92-0.18\\ =0.74\end{array}$

Thus, the probability that the total waiting time between 3 and 8 min is 0.74.

f.

To determine

Find the probability that the total waiting time is either less than 2 min or more than 6 min.

Answer to Problem 8E

The probability that the total waiting time is either less than 2 min or more than 6 min is 0.4.

Explanation of Solution

Calculation:

The probability that the total waiting time is either less than 2 min or more than 6 min is obtained as shown below:

$\begin{array}{c}P\left(Y<2\text{ or }Y>6\right)={\displaystyle \underset{0}{\overset{2}{\int }}\frac{1}{25}ydy}+{\displaystyle \underset{6}{\overset{10}{\int }}\left[\frac{2}{5}-\frac{1}{25}y\right]dy}\\ =\frac{1}{25}{\displaystyle \underset{0}{\overset{2}{\int }}ydy}+{\displaystyle \underset{6}{\overset{10}{\int }}\left[\frac{2}{5}-\frac{1}{25}y\right]dy}\\ =\frac{1}{25}{\frac{y^2}{2}]}_0^2+{\frac{2}{5}y-\frac{1}{25}\times \frac{y^2}{2}]}_6^{10}\end{array}$

                            $\begin{array}{c}={\frac{y^2}{50}]}_0^2+{\frac{2}{5}y-\frac{y^2}{50}]}_6^{10}\\ =\left(\frac{2^2}{50}-\frac{0^2}{50}\right)+\left\{\left(\frac{2\left(10\right)}{5}-\frac{{10}^2}{50}\right)-\left(\frac{2\left(6\right)}{5}-\frac{6^2}{50}\right)\right\}\\ =\frac{4}{50}+\left\{\left(\frac{20}{5}-\frac{100}{50}\right)-\left(\frac{12}{5}-\frac{36}{50}\right)\right\}\\ =0.08+\left\{\left(4-2\right)-\left(2.4-0.72\right)\right\}\end{array}$

$\begin{array}{c}=0.08+\left\{2-1.68\right\}\\ =0.08+0.32\\ =0.4\end{array}$

Thus, the probability that the total waiting time is either less than 2 min or more than 6 minis 0.4.