Chapter 4, Problem 99SE

a.

To determine

Obtain the cumulative distribution function $F\left(y\right)$.

Construct the graph of cumulative distribution function $F\left(y\right)$.

Answer to Problem 99SE

The cumulative distribution function $F\left(y\right)$ is given by:

$\underset{\_}{F\left(y\right)=\{\begin{array}{l}0y<0\\ \frac{y^2}{48}-\frac{y^3}{864}0\le y\le 12\\ 1y>12\end{array}}$

The graph of $F\left(y\right)$ is given by:

Explanation of Solution

Given info:

It is noted that Y denotes the distance from the left of the occurrence of break.

The pdf of the X is given by:

$f\left(y\right)=\{\begin{array}{l}\left(\frac{1}{24}\right)y\left(1-\frac{y}{12}\right)0\le y\le 12\\ 0\text{otherwise}\end{array}$

Calculation:

The cumulative distribution function $F\left(y\right)$ is obtained as given below:

For $y<0$,

Since the interval of the $f\left(y\right)$ lies between $0\le y\le 12$, so, $F\left(y\right)=0$

For $0\le y\le 12$,

$\begin{array}{c}F\left(y\right)={\displaystyle {\int }_0^y\left(\frac{1}{24}\right)u\left(1-\frac{u}{12}\right)du}\\ =\frac{1}{24}{\displaystyle {\int }_0^{y}u\left(1-\frac{u}{12}\right)du}\\ =\frac{1}{24}{\displaystyle {\int }_0^y\left(u-\frac{u^2}{12}\right)du}\\ =\frac{1}{24}{\left[\frac{u^2}{2}-\frac{u^3}{36}\right]}_0^y\\ =\frac{1}{24}\left\{\left(\frac{y^2}{2}-\frac{y^3}{36}\right)-0\right\}\\ =\frac{y^2}{48}-\frac{y^3}{864}\end{array}$

For $y>12$, $F\left(y\right)=1$

Thus, the cumulative distribution function $F\left(y\right)$ is given by:

$\underset{\_}{F\left(x\right)=\{\begin{array}{l}0y<0\\ \frac{y^2}{48}-\frac{y^3}{864}0\le y\le 12\\ 1y>12\end{array}}$

The graph of the $F\left(y\right)$ obtained as follows:

For y at –5:

Substitute 0 for x in the equation $F\left(y\right)=\frac{y^2}{48}-\frac{y^3}{864}$

$\begin{array}{c}F\left(y\right)=\frac{-5^2}{48}-\frac{-5^3}{864}\\ =\frac{25}{48}+\frac{125}{864}\\ =0.52+0.1447\\ =0.6647\end{array}$

Similarly the remaining points are obtained as follows:

x	$f\left(x\right)$
–5	0.6647
0	0
5	0.6647
10	0.925
12	1

Table 1

Procedure for drawing the density curve of the variable $F\left(y\right)=\frac{y^2}{48}-\frac{y^3}{864}$ for $0<x<12$ otherwise $x=0$ 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 $F\left(y\right)$ is given below:

Figure 1

In the above graph, y takes the values from the interval of $\left[0,12\right]$ and $F\left(y\right)$ represents the probability density function.

Thus, the graph of $F\left(y\right)$ has been obtained.

b.

To determine

Find the $P\left(Y\le 4\right)$,$P\left(Y>6\right)$ and $P\left(4\le Y\le 6\right)$.

Answer to Problem 99SE

The value of $P\left(Y\le 4\right)$ is 0.259.

The value of $P\left(Y>6\right)$ is 0.5.

The value of $P\left(4\le Y\le 6\right)$ is 0.2441.

Explanation of Solution

Calculation:

The cdf of a continuous distribution is $F\left(x\right)=P\left(X\le x\right)=P\left(X<x\right)$.

The value of $P\left(Y\le 4\right)$ is:

$\begin{array}{c}P\left(Y\le 4\right)=F\left(4\right)\\ =\frac{4^2}{48}-\frac{4^3}{864}\\ =\frac{16}{48}-\frac{64}{864}\\ =0.33-0.0741\\ =0.2559\end{array}$

Hence, the $P\left(Y\le 4\right)$ is­0.2559.

The value of $P\left(Y>6\right)$ is:

$\begin{array}{c}P\left(Y>6\right)=1-F\left(6\right)\\ =1-\left\{\frac{6^2}{48}-\frac{6^3}{864}\right\}\\ =1-\left\{\frac{36}{48}-\frac{216}{864}\right\}\\ =1-\left\{0.75-0.25\right\}\\ =1-0.5\\ =0.5\end{array}$

Hence, the $P\left(X\le 2\right)$ is­0.665.

The value of $P\left(4\le Y\le 6\right)$ is:

$\begin{array}{c}P\left(4\le Y\le 6\right)=P\left(Y\le 6\right)-P\left(Y\le 4\right)\\ =F\left(6\right)-F\left(4\right)\\ =0.5-0.2559\\ =0.2441\end{array}$

Hence, the $P\left(-1\le X\le 2\right)$ is0.2441.

c.

To determine

Find the $E\left(Y\right)$,$E\left(Y^2\right)$ and $V\left(Y\right)$.

Answer to Problem 99SE

The value of $E\left(Y\right)$ is 6 inches.

The value of $E\left(Y^2\right)$ is 43.2.

The value of $V\left(Y\right)$ is 7.2 inches.

Explanation of Solution

Calculation:

Expected mean:

$E\left(x\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}xf\left(x\right)dx}$

The value of $E\left(Y\right)$ is obtained as shown below:

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}yf\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{12}{\int }}y\left(\frac{1}{24}\right)y\left(1-\frac{y}{12}\right)dy}\\ =\frac{1}{24}{\displaystyle \underset{0}{\overset{12}{\int }}{y}^2\left(1-\frac{y}{12}\right)dy}\\ =\frac{1}{24}{\displaystyle \underset{0}{\overset{12}{\int }}\left(y^2-\frac{y^3}{12}\right)dy}\end{array}$

$\begin{array}{c}=\frac{1}{24}{\left[\frac{y^3}{3}-\frac{y^4}{48}\right]}_0^{12}\\ =\frac{1}{24}\left\{\left(\frac{{12}^3}{3}-\frac{{12}^4}{48}\right)-0\right\}\\ =\frac{1}{24}\left(576-432\right)\\ =6\end{array}$

Thus, the value of $E\left(Y\right)$ is 6 inches.

The value of $E\left(Y^2\right)$ is obtained as shown below:

$\begin{array}{c}E\left(Y^2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{y}^{2}f\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{12}{\int }}{y}^2\left(\frac{1}{24}\right)y\left(1-\frac{y}{12}\right)dy}\\ =\frac{1}{24}{\displaystyle \underset{0}{\overset{12}{\int }}{y}^3\left(1-\frac{y}{12}\right)dy}\\ =\frac{1}{24}{\displaystyle \underset{0}{\overset{12}{\int }}\left(y^3-\frac{y^4}{12}\right)dy}\end{array}$

$\begin{array}{c}=\frac{1}{24}{\frac{y^4}{4}-\frac{y^5}{60}]}_0^{12}\\ =\frac{1}{24}\left\{\left(\frac{{12}^4}{4}-\frac{{12}^5}{60}\right)-0\right\}\\ =\frac{1}{24}\left(5,184-4,147.2\right)\\ =43.2\end{array}$

Thus, the value of $E\left(Y^2\right)$ is 43.2.

The value of $V\left(Y\right)$ is:

$\begin{array}{c}V\left(Y\right)=E\left(Y^2\right)-{\left[E\left(Y\right)\right]}^2\\ =43.2-{\left(6\right)}^2\\ =43.2-36\\ =7.2\end{array}$

Thus, the variance of Y is 7.2 inches.

d.

To determine

Find the probability that the break point occurs more than 2 in. from the expected break point.

Answer to Problem 99SE

The probability that the break point occurs more than 2 in. from the expected break pointis 0.518.

Explanation of Solution

Calculation:

It is known that the break point occurs more than 2 in. from the expected break pointis $\left(Y<4\text{or }Y>8\right)$.

The probability that the break point occurs more than 2 in. from the expected break pointis obtained as shown below:

$\begin{array}{c}P\left(Y<4\text{or }Y>8\right)=1-P\left(4\le Y\le 8\right)\\ =1-\left\{P\left(Y\le 8\right)-P\left(Y\le 4\right)\right\}\\ =1-\left\{F\left(8\right)-F\left(4\right)\right\}\\ =1-\left\{\frac{8^2}{48}-\frac{8^3}{864}-0.2559\right\}\\ =1-\left[\left\{\frac{64}{48}-\frac{512}{864}\right\}-0.2559\right]\end{array}$

$\begin{array}{c}=1-\left[\left\{1.33-0.5926\right\}-0.2559\right]\\ =1-\left[0.7407-0.2559\right]\\ =1-0.4817\\ =0.518\end{array}$

Thus, the probability that the break point occurs more than 2 in. from the expected break pointis 0.518.

e.

To determine

Find the expected length of the shorter segment at the occurrence of break.

Answer to Problem 99SE

The expected length of the shorter segment at the occurrence of break is 3.75 inches.

Explanation of Solution

Calculation:

The shorter segment at the occurrence of break is $\min \left(Y,12-Y\right)$.

The expected length of shorter segment at the occurrence of break is obtained by:

$\begin{array}{c}E\left(\min \left(Y,12-Y\right)\right)={\displaystyle \underset{0}{\overset{12}{\int }}\min \left(Y,12-Y\right)f\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{6}{\int }}\min \left(Y,12-Y\right)f\left(y\right)dy}+{\displaystyle \underset{6}{\overset{12}{\int }}\min \left(Y,12-Y\right)f\left(y\right)dy}\end{array}$

Now, when $0\le x<6$, it is evident that $y>0$, which implies that $\min \left(Y,12-Y\right)=Y$. When $6\le x<12$, it is evident that $12-Y>0$, which implies that $\min \left(Y,12-Y\right)=12-Y$

$\begin{array}{c}E\left(\min \left(Y,12-Y\right)\right)={\displaystyle \underset{0}{\overset{6}{\int }}yf\left(y\right)dy}+{\displaystyle \underset{6}{\overset{12}{\int }}\left(12-y\right)f\left(y\right)dy}\\ ={\displaystyle \underset{0}{\overset{6}{\int }}y\left(\frac{1}{24}\right)y\left(1-\frac{y}{12}\right)dy}+{\displaystyle \underset{6}{\overset{12}{\int }}\left(12-y\right)\left(\frac{1}{24}\right)y\left(1-\frac{y}{12}\right)dy}\end{array}$

$\begin{array}{c}=\left(\frac{1}{24}\right){\displaystyle \underset{0}{\overset{6}{\int }}{y}^2\left(1-\frac{y}{12}\right)dy}+\left(\frac{1}{24}\right){\displaystyle \underset{6}{\overset{12}{\int }}\left(12y-y^2\right)\left(1-\frac{y}{12}\right)dy}\\ =\left(\frac{1}{24}\right){\displaystyle \underset{0}{\overset{6}{\int }}\left(y^2-\frac{y^3}{12}\right)dy}+\left(\frac{1}{24}\right){\displaystyle \underset{6}{\overset{12}{\int }}\left(12y-y^2\right)\left(1-\frac{y}{12}\right)dy}\\ =\left(\frac{1}{24}\right){\displaystyle \underset{0}{\overset{6}{\int }}\left(y^2-\frac{y^3}{12}\right)dy}+\left(\frac{1}{24}\right){\displaystyle \underset{6}{\overset{12}{\int }}\left(12y-\frac{12y^2}{12}-y^2+\frac{y^3}{12}\right)dy}\end{array}$

$\begin{array}{c}=\frac{1}{24}{\left[\frac{y^3}{3}-\frac{y^4}{48}\right]}_0^6+\left(\frac{1}{24}\right){\displaystyle \underset{6}{\overset{12}{\int }}\left(12y-2y^2+\frac{y^3}{12}\right)dy}\\ =\frac{1}{24}{\left[\frac{y^3}{3}-\frac{y^4}{48}\right]}_0^6+\left(\frac{1}{24}\right){\left(6y^2-\frac{2}{3}{y}^3+\frac{y^4}{48}\right)]}_6^{12}\\ =\frac{1}{24}\left\{\left[\left(\frac{6^3}{3}-\frac{6^4}{48}\right)-0\right]+\left[\begin{array}{l}\left(6{\left(12\right)}^2-\frac{2}{3}{\left(12\right)}^3+\frac{{12}^4}{48}\right)-\\ \left(6{\left(6\right)}^2-\frac{2}{3}{\left(6\right)}^3+\frac{6^4}{48}\right)\end{array}\right]\right\}\\ =\frac{1}{24}\left\{\left[45\right]+\left[6\left({12}^2-6^2\right)-\frac{2}{3}\left({12}^3-6^3\right)+\frac{1}{48}\left({12}^4-6^4\right)\right]\right\}\end{array}$

$\begin{array}{c}=\frac{1}{24}\left\{\left[194.4\right]+648-1008+405\right\}\\ =\frac{1}{24}\left\{45+45\right\}\\ =\frac{90}{24}\\ =3.75\end{array}$

Thus, the expected length of the shorter segment at the occurrence of break is 3.75 inches.