Chapter 4.4, Problem 48E

If a point is randomly located in an interval (a, b) and if Y denotes the location of the point, then Y is assumed to have a uniform distribution over (a, b). A plant efficiency expert randomly selects a location along a 500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is

1. a within 25 feet of the end of the line?
2. b within 25 feet of the beginning of the line?
3. c closer to the beginning of the line than to the end of the line?

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval $\left({\theta }_1,{\theta }_2\right)$ is as follows:

$f\left(y\right)=\frac{1}{{\theta }_2-{\theta }_1}{\theta }_1\le y\le {\theta }_2$

Here, the interval will be (0, 500) and the length of the interval is 500.

$f\left(y\right)=\{\begin{array}{l}\frac{1}{500-0}0\le y\le 500\\ 0\text{elsewhere}\end{array}$

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

$P\left(500-25<Y<500\right)=P\left(475<Y<500\right)$

$\begin{array}{c}P\left(a<Y<b\right)={\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}f\left(y\right)dy}\\ P\left(475<Y<500\right)={\displaystyle \underset{475}{\overset{500}{\int }}\frac{1}{500-0}dy}\\ ={\displaystyle \underset{475}{\overset{500}{\int }}\frac{1}{500}dy}\\ =\frac{1}{500}{\left[y\right]}_{475}^{500}\end{array}$

                             $\begin{array}{c}=\frac{1}{500}\left[500-475\right]\\ =\frac{25}{500}\\ =0.05\end{array}$

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval $\left({\theta }_1,{\theta }_2\right)$ is as follows:

$f\left(y\right)=\frac{1}{{\theta }_2-{\theta }_1}{\theta }_1\le y\le {\theta }_2$

Here, the interval will be (0, 500) and the length of the interval is 500.

$f\left(y\right)=\{\begin{array}{l}\frac{1}{500-0}0\le y\le 500\\ 0\text{elsewhere}\end{array}$

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

$\begin{array}{c}P\left(a<Y<b\right)={\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}f\left(y\right)dy}\\ P\left(0<Y<25\right)={\displaystyle \underset{0}{\overset{25}{\int }}\frac{1}{500-0}dy}\\ ={\displaystyle \underset{0}{\overset{25}{\int }}\frac{1}{500}dy}\\ =\frac{1}{500}{\left[y\right]}_0^{25}\end{array}$

                      $\begin{array}{c}=\frac{1}{500}\left[25-0\right]\\ =\frac{25}{500}\\ =0.05\end{array}$

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval $\left({\theta }_1,{\theta }_2\right)$ is as follows:

$f\left(y\right)=\frac{1}{{\theta }_2-{\theta }_1}{\theta }_1\le y\le {\theta }_2$

Here, the interval will be (0, 500) and the length of the interval is 500.

$f\left(y\right)=\{\begin{array}{l}\frac{1}{500-0}0\le y\le 500\\ 0\text{elsewhere}\end{array}$

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

$\begin{array}{c}P\left(a<Y<b\right)={\displaystyle \underset{{\theta }_1}{\overset{{\theta }_2}{\int }}f\left(y\right)dy}\\ P\left(0<Y<250\right)={\displaystyle \underset{0}{\overset{250}{\int }}\frac{1}{500-0}dy}\\ ={\displaystyle \underset{0}{\overset{250}{\int }}\frac{1}{500}dy}\\ =\frac{1}{500}{\left[y\right]}_0^{250}\end{array}$

                      $\begin{array}{c}=\frac{1}{500}\left[250-0\right]\\ =\frac{250}{500}\\ =0.50\end{array}$

Thus, the probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.