Chapter 4, Problem 101SE

The completion time X for a certain task has cdf F(x) given by

     $\{\begin{array}{ll}0\hfill & x<0\hfill \\ \frac{x^3}{3}\hfill & 0\le x\le 1\hfill \\ 1-\frac{1}{2}\left(\frac{7}{3}-x\right)\left(\frac{7}{4}-\frac{3}{4}x\right)\hfill & 1\le x\le \frac{7}{3}\hfill \\ 1\hfill & x>\frac{7}{3}\hfill \end{array}$

a. Obtain the pdf f (x) and sketch its graph.

b. Compute P(.5 ≤ X ≤ 2).

c. Compute E(X).

To determine

Obtain the probability density function $f\left(x\right)$ and its graph.

Answer to Problem 101SE

The probability density function is X is given below:

$f\left(x\right)=\{\begin{array}{l}{x}^{2}0\le x<1\\ \frac{7}{4}-\frac{3}{4}x1\le x<\frac{7}{3}\\ 0\text{Otherwise}\end{array}$

The graph of $f\left(x\right)$ is given below:

Explanation of Solution

Given info:

Thecumulative distribution function of completion time X for the certain task is given by:

$F\left(x\right)=\{\begin{array}{l}0x<0\\ \frac{x^3}{3}0\le x<1\\ 1-\frac{1}{2}\left(\frac{7}{3}-x\right)\left(\frac{7}{4}-\frac{3}{4}x\right)1\le x\le \frac{7}{3}\\ 1x>\frac{7}{3}\end{array}$

Calculation:

The probability density function $f\left(x\right)$ is obtained as shown below:

For the interval $x<0$

$\begin{array}{c}\frac{dF\left(x\right)}{dx}=F^{\prime }\left(x\right)\\ =\frac{d}{dx}\left(0\right)\\ =0\end{array}$

For the interval $0\le x<1$,

$\begin{array}{c}\frac{dF\left(x\right)}{dx}=F^{\prime }\left(x\right)\\ =\frac{d}{dx}\left(\frac{x^3}{3}\right)\\ =\frac{3x^2}{3}\\ =x^2\end{array}$

For the interval $1\le x<\frac{7}{3}$,

$\begin{array}{c}\frac{dF\left(x\right)}{dx}=F^{\prime }\left(x\right)\\ =\frac{d}{dx}\left(1-\frac{1}{2}\left(\frac{7}{4}-x\right)\left(\frac{7}{4}-\frac{3}{4}x\right)\right)\\ =\frac{d}{dx}\left(1-\frac{1}{2}\left(\frac{49}{12}-\frac{7x}{4}-\frac{21x}{12}-\frac{3x^2}{4}\right)\right)\\ =\frac{d}{dx}\left(1-\left(\frac{49}{24}-\frac{7x}{8}-\frac{21x}{24}-\frac{3x^2}{8}\right)\right)\end{array}$

On simplifying,

$\begin{array}{c}\frac{dF\left(x\right)}{dx}=\frac{d}{dx}\left(1-\frac{49}{24}+\frac{7x}{8}+\frac{21x}{24}+\frac{3x^2}{8}\right)\\ =\frac{7}{8}+\frac{21}{24}-\frac{3\times 2x}{8}\\ =\frac{21+21}{24}-\frac{3x}{4}\\ =\frac{42}{24}-\frac{3x}{4}\\ =\frac{7}{4}-\frac{3}{4}x\end{array}$

For the interval $2\le x$

$\begin{array}{c}\frac{dF\left(x\right)}{dx}=F^{\prime }\left(x\right)\\ =\frac{d}{dx}\left(1\right)\\ =0\end{array}$

Thus, the probability density function is X is,

$f\left(x\right)=\{\begin{array}{l}{x}^{2}0\le x<1\\ \frac{7}{4}-\frac{3}{4}x1\le x<\frac{7}{3}\\ 0\text{Otherwise}\end{array}$

From the equation, it can be observed that the values of x lie between 0 and 2.33.

For $0\le x<1$,

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

For x at 0:

Substitute 0 for x in the equation $f\left(x\right)=x^2$

$\begin{array}{c}f\left(x\right)=0^2\\ =0\end{array}$

For x at 0.5:

Substitute 0.5 for x in the equation $f\left(x\right)=x^2$

$\begin{array}{c}f\left(x\right)={0.5}^2\\ =0.25\end{array}$

Similarly the remaining points are obtained as follows:

x	$f\left(x\right)$
0	0
0.3	0.09
0.5	0.25
0.7	0.49

Table 1

Since, $f\left(x\right)=x^2$ it is a parabola with vertex at the origin.

For $1\le x<2.33$,

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

For x at 1:

Substitute 0 for x in the equation $f\left(x\right)=\frac{7}{4}-\frac{3}{4}x$

$\begin{array}{c}f\left(x\right)=\frac{7}{4}-\frac{3}{4}\left(1\right)\\ =1.75-0.75\\ =1\end{array}$

For x at 2:

Substitute 2 for x in the equation $f\left(x\right)=\frac{7}{4}-\frac{3}{4}x$

$\begin{array}{c}f\left(x\right)=\frac{7}{4}-\frac{3}{4}\left(2\right)\\ =1.75-1.5\\ =0.25\end{array}$

Similarly the remaining points are obtained as follows:

x	$f\left(x\right)$
1	1
2	0.25

Table 2

Procedure for drawing the density curve of the variable $f\left(x\right)$ is as follows:

• Let horizontal axis take x values and vertical axis take $f\left(x\right)$ values
• Plot the points obtained from Table 1 and Table 2.

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

Figure 1

Thus, the graph for the probability density function of $f\left(x\right)$ is obtained.

To determine

Calculate $P\left(0.5\le X\le 2\right)$.

Answer to Problem 101SE

The value of $P\left(0.5\le X\le 2\right)$ is 0.917.

Explanation of Solution

Calculation:

Let X be continuous random variable with pdf $f\left(x\right)$ and cdf $F\left(x\right)$

For any number a and b,

$\begin{array}{c}P\left(X<a\right)=F\left(a\right)\\ P\left(X>a\right)=1-F\left(a\right)\\ P\left(a\le X\le b\right)=F\left(b\right)-F\left(a\right)\end{array}$

The value of $P\left(0.5\le X\le 2\right)$ is obtained as shown below:

$\begin{array}{c}P\left(0.5\le X\le 2\right)=P\left(X\le 2\right)-P\left(X\le 0.5\right)\\ =F\left(2\right)-F\left(0.5\right)\\ =\left\{1-\frac{1}{2}\left(\frac{7}{3}-\left(2\right)\right)\left(\frac{7}{4}-\frac{3}{4}\left(2\right)\right)\right\}-\left\{\frac{{0.5}^3}{3}\right\}\\ =\left\{1-\left(\frac{7}{6}-1\right)\left(\frac{7}{8}-\frac{3}{4}\right)\right\}-\left\{\frac{{0.5}^3}{3}\right\}\end{array}$

$\begin{array}{c}=\left\{1-\left(\frac{1}{3}\right)\left(\frac{7}{8}-\frac{3}{4}\right)\right\}-\left\{\frac{{0.5}^3}{3}\right\}\\ =\left\{1-\left(\frac{1}{3}\right)\left(\frac{1}{8}\right)\right\}-\left\{\frac{0.125}{3}\right\}\\ =\left\{1-0.04167\right\}-\left\{0.04167\right\}\\ =0.917\end{array}$

Thus, the $P\left(0.5\le X\le 2\right)$ is0.917.

To determine

Calculate $E\left(X\right)$.

Answer to Problem 101SE

The value of $E\left(X\right)$ is 1.21.

Explanation of Solution

Calculation:

The probability density function is X is,

$f\left(x\right)=\{\begin{array}{l}{x}^{2}0\le x<1\\ \frac{7}{4}-\frac{3}{4}x1\le x<\frac{7}{3}\\ 0\text{Otherwise}\end{array}$

The expected value of X is obtained by,

$\begin{array}{c}{\mu }_X=E\left(X\right)\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}xf\left(x\right)dx}\end{array}$

The expected value is obtained as given below:

$\begin{array}{c}{\mu }_X={\displaystyle \underset{0}{\overset{1}{\int }}xf\left(x\right)dx}+{\displaystyle \underset{1}{\overset{\frac{7}{3}}{\int }}xf\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}x.x^{2}dx}+{\displaystyle \underset{1}{\overset{\frac{7}{3}}{\int }}x\left(\frac{7}{4}-\frac{3}{4}x\right)dx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}+{\displaystyle \underset{1}{\overset{\frac{7}{3}}{\int }}\left(\frac{7}{4}x-\frac{3}{4}{x}^2\right)dx}\\ ={\frac{x^4}{4}]}_0^1+{\left(\frac{7}{4}\left(\frac{x^2}{2}\right)\right)-\left(\frac{3}{4}\left(\frac{x^3}{3}\right)\right)]}_1^{\frac{7}{3}}\\ =\left(\frac{1^4}{4}-0\right)+\left\{\left(\frac{7}{4}\left(\frac{{\left(\frac{7}{3}\right)}^2}{2}\right)\right)-\left(\frac{3}{4}\left(\frac{{\left(\frac{7}{3}\right)}^3}{3}\right)\right)-\left(\left(\frac{7}{4}\left(\frac{1^2}{2}\right)\right)-\left(\frac{3}{4}\left(\frac{1^3}{3}\right)\right)\right)\right\}\end{array}$

$\begin{array}{c}=0.25+\left\{\left(\left(0.875\right)\left(5.44\right)\right)-\left(0.75\left(4.24\right)\right)-\left[\left(1.75\left(0.5\right)\right)-\left(0.75\left(0.33\right)\right)\right]\right\}\\ =0.25+\left\{\left[4.76-3.18\right]-\left[0.875-0.25\right]\right\}\\ =0.25+\left\{1.585-0.625\right\}\\ =0.25+0.96\\ =1.21\end{array}$

Thus, the $E\left(X\right)$ is1.21.