Chapter 4.2, Problem 18E

a.

To determine

Estimate the value of c.

Answer to Problem 18E

The value of c is 1.2.

Explanation of Solution

The probability density function is obtained below:

$f\left(y\right)=\{\begin{array}{l}0.2-1\le y\le 0,\\ 0.2+cy0<y\le 1,\\ 0,\text{elsewhere}\end{array}$

The value of k is calculated below:

$\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}=1\\ {\displaystyle \underset{-1}{\overset{0}{\int }}0.2dy}+{\displaystyle \underset{0}{\overset{1}{\int }}\left(0.2+cy\right)dy}=1\\ {\left(0.2y\right)}_{-1}^0+{\left(0.2y+c\frac{y^2}{2}\right)}_0^1=1\\ \left(0-0.2\left(-1\right)\right)+\left(0.2\left(1\right)+c\frac{1^2}{2}-0\right)=1\end{array}$

         $\begin{array}{c}0.2+0.2+\frac{c}{2}=1\\ \frac{c}{2}=1-0.4\\ \frac{c}{2}=0.6\\ c=1.2\end{array}$

Thus, the value of c is 1.2.

b.

To determine

Find the value of $F\left(y\right)$.

Answer to Problem 18E

The value of $F\left(y\right)$ is $F\left(y\right)=\{\begin{array}{l}0y\le -1\\ 0.2\left(1+y\right)-1<y\le 0\\ 0.2\left(1+y+3y^2\right)0<y\le 1\\ 1y>1\end{array}$

Explanation of Solution

The value of $F\left(y\right)$ for $y\le -1$ is obtained below:

$\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}={\displaystyle \underset{-\infty }{\overset{-1}{\int }}0dy}\\ =0\end{array}$

The value of $F\left(y\right)$ for $-1<y\le 0$ is obtained below:

$\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}={\displaystyle {\int }_{-\infty }^{-1}f\left(y\right)dy}+{\displaystyle {\int }_{-1}^{y}f\left(y\right)dy}\\ ={\displaystyle {\int }_{-\infty }^{-1}0dy}+{\displaystyle {\int }_{-1}^{y}0.2dy}\\ =0+{\left[0.2y\right]}_{-1}^y\\ =0.2\left(y+1\right)\end{array}$

The value of $F\left(y\right)$ for $0<y\le 1$ is obtained below:

$\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}={\displaystyle {\int }_{-\infty }^{-1}f\left(y\right)dy}+{\displaystyle {\int }_{-1}^{0}f\left(y\right)dy}+{\displaystyle {\int }_0^{y}f\left(y\right)dy}\\ ={\displaystyle {\int }_{-\infty }^{-1}0dy}+{\displaystyle {\int }_{-1}^{0}0.2dy}+{\displaystyle {\int }_0^y\left(0.2+1.2y\right)dy}\\ =0+{\left[0.2y\right]}_{-1}^0+{\left(0.2y+1.2\frac{y^2}{2}\right)}_0^y\\ =0.2\left(0+1\right)+\left(0.2y+1.2\frac{y^2}{2}-0\right)\\ =0.2+0.2y+0.6y^2\\ =0.2\left(1+y+3y^2\right)\end{array}$

The value of $F\left(y\right)$ for $y>1$ is obtained below:

$\begin{array}{c}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(y\right)dy}={\displaystyle {\int }_{-\infty }^{-1}f\left(y\right)dy}+{\displaystyle {\int }_{-1}^{0}f\left(y\right)dy}+{\displaystyle {\int }_0^{1}f\left(y\right)dy}+{\displaystyle {\int }_1^{y}f\left(y\right)dy}\\ ={\displaystyle {\int }_{-\infty }^{-1}0dy}+{\displaystyle {\int }_{-1}^{0}0.2dy}+{\displaystyle {\int }_0^1\left(0.2+1.2y\right)dy}+{\displaystyle {\int }_1^{y}0dy}\\ =0+{\left[0.2y\right]}_{-1}^0+{\left(0.2y+1.2\frac{y^2}{2}\right)}_0^1+0\\ =0.2\left(0+1\right)+\left(0.2\left(1\right)+1.2\frac{{\left(1\right)}^2}{2}-0\right)\\ =0.2+0.2+0.6\\ =1\end{array}$

Thus, the value of F(y) is $F\left(y\right)=\{\begin{array}{l}0y\le -1\\ 0.2\left(1+y\right)-1<y\le 0\\ 0.2\left(1+y+3y^2\right)0<y\le 1\\ 1y>1\end{array}$

c.

To determine

Draw a graph of f(y)

Draw a graph of F(y)

Answer to Problem 18E

The graph of f (y) and F (y) is given below:

In the above graph the solid line indicates the f (y) and the dotted line indicates the F(y).

Explanation of Solution

Calculate the values of the density function as shown in table below:

y	$f\left(y\right)$
–1	0.2
–0.75	0.2
–0.5	0.2
–0.25	0.2
0	0.2
0.25	0.5
0.5	0.8
0.75	1.1
1	1.4

Calculate the values of the distribution function as shown in table below:

y	$F\left(y\right)$
–1	0
–0.75	0.05
–0.5	0.1
–0.25	0.15
0	0.2
0.25	0.2875
0.5	0.45
0.75	0.6875
1	1

To sketch the graph, provide the values of y in vertical axis and the values of f (y) and F (y) in horizontal axis, and join the points to obtain the respective graphs of f (y) and F (y).

d.

To determine

Find the value $F\left(-1\right)$ by using F(y).

Find the value $F\left(0\right)$ by using F(y).

Find the value $F\left(1\right)$ by using F(y).

Answer to Problem 18E

The value $F\left(-1\right)$ by using F(y) is 0.

The value $F\left(0\right)$ by using F(y) is 0.2.

The value $F\left(1\right)$ by using F(y) is 1.

Explanation of Solution

From Part b, it can be observed that $F\left(y\right)=0$ for $y\le -1$.

The value $F\left(-1\right)$ by using F(y) is obtained below:

$F\left(-1\right)=0$

The value $F\left(0\right)$ by using F(y) is obtained below:

$\begin{array}{c}F\left(0\right)=0.2\left(1+\left(0\right)\right)\\ =0.2\left(1\right)\\ =0.2\end{array}$

The value $F\left(1\right)$ by using F(y) is obtained below:

$\begin{array}{c}F\left(y\right)=0.2\left(1+y+3y^2\right)\\ F\left(1\right)=0.2\left(1+\left(1\right)+3{\left(1\right)}^2\right)\\ =0.2\left(1+1+3\right)\\ =0.2\left(5\right)\\ =1\end{array}$

Thus, the value $F\left(-1\right)$ by using F(y) is 0.

Thus, the value $F\left(0\right)$ by using F(y) is 0.2.

Thus, the value $F\left(1\right)$ by using F(y) is 1.

e.

To determine

Find the value of $P\left(0\le Y\le 0.5\right)$.

Answer to Problem 18E

The value of $P\left(0\le Y\le 0.5\right)$ is 0.25

Explanation of Solution

The value of $P\left(0\le Y\le 0.5\right)$.is obtained below:

$\begin{array}{c}F\left(y\right)=0.2\left(1+y+3y^2\right)\\ P\left(a\le Y\le b\right)=F\left(b\right)-F\left(a\right)\\ P\left(0\le Y\le 0.5\right)=F\left(0.5\right)-F\left(0\right)\\ =0.2\left(1+0.5+3{\left(0.5\right)}^2\right)-0.2\\ =0.2+0.1+0.15-0.2\\ =0.25\end{array}$

Thus, the value of $P\left(0\le Y\le 0.5\right)$ is 0.25.

f.

To determine

Find the value of $P\left(Y>0.5|Y>0.1\right)$.

Answer to Problem 18E

The value of $P\left(Y>0.5|Y>0.1\right)$ is 0.71.

Explanation of Solution

The value of $P\left(Y>0.5|Y>0.1\right)$ is obtained below:

$\begin{array}{c}P\left(Y\ge 0.5|Y\ge 0.1\right)=\frac{P\left(Y\ge 0.5\cap Y\ge 0.1\right)}{P\left(Y\ge 0.1\right)}\\ =\frac{P\left(Y\ge 0.5\right)}{P\left(Y\ge 0.1\right)}\\ =\frac{1-F\left(0.5\right)}{1-F\left(0.1\right)}\end{array}$

                                     $\begin{array}{c}=\frac{1-0.2\left(1+0.5+3{\left(0.5\right)}^2\right)}{1-0.2\left(1+0.1+3{\left(0.1\right)}^2\right)}\\ =\frac{1-0.45}{1-0.226}\\ =\frac{0.55}{0.774}\\ =0.71\end{array}$

Thus, the value of $P\left(Y>0.5|Y>0.1\right)$ is 0.71.