A bus travels between the two cities A and B, which are $100$miles apart. If the bus has a breakdown, the distance from the breakdown to city A has a uniform distribution over $(0,100)$. There is a bus service station in city A, in B, and in the center of the route between A and B. It is suggested that it would be more efficient to have the three stations located $25,50,\mathrm{and}75$miles, respectively, from A. Do you agree? Why?

Here, it is given that:

The distance between City A and B $=100miles$

In case of breakdown of the bus, the distance from the breakdown to city A has a uniform distribution over$\left(0,100\right)$.

Let, the distance between the place where bus breaks down to city A be $X$and the distance to the nearest service station be Y.

The random variable X follows uniform distribution with $\left(0,100\right)$.

The probability density function of uniform distribution can be defined as:

$f\left(x\right)=\frac{1}{B-A}=\frac{1}{100}$

If three stations are located $0,50\mathrm{and}100$miles, then the towing distance will be given by:

role="math" localid="1646658782747" $Y=g_1\left(x\right)=\left\{\begin{array}{l}X0<X<25\\ \left|X-50\right|25\le X\le 75\\ 100-X75<X<100\end{array}\right.$

The expected time under the current scheme is:

role="math" localid="1646659067639" $$\begin{gathered} E\left(Y\right)={\int }_0^{100}{g}_1\left(x\right)f\left(x\right)dx \\ ={\int }_0^{25}x\left(\frac{1}{100}\right)dx+{\int }_{25}^{75}\left|x-50\right|\left(\frac{1}{100}\right)dx+{\int }_{75}^{100}\left(100-x\right)\left(\frac{1}{100}\right)dx \\ =\frac{1}{100}\left[{\left(\frac{x^2}{2}\right)}_0^{25}+{\left(50x-\frac{x^2}{2}\right)}_{25}^{50}+{\left(\frac{x^2}{2}-50x\right)}_{50}^{75}+{\left(100x-\frac{x^2}{2}\right)}_{75}^{100}\right] \\ =\frac{1}{100}\left[\frac{625}{2}+\frac{625}{2}+\frac{625}{2}+\frac{625}{2}\right] \\ =12.5 \end{gathered}$$

If the service stations are located at $25,50,\text{and}100$miles then the towing distance will be given by:

$Y=g_1\left(x\right)=\left\{\begin{array}{l}\left|X-25\right|0<X<37.5\\ \left|X-50\right|37.5\le X\le 62.5\\ \left|X-75\right|62.5<X<100\end{array}\right.$

The expected time under the new scheme is:

$$\begin{gathered} E\left(Y\right)={\int }_0^{100}{g}_2\left(x\right)f\left(x\right)dx \\ ={\int }_0^{25}\left|25-x\right|\left(\frac{1}{100}\right)dx+{\int }_{25}^{75}\left|x-50\right|\left(\frac{1}{100}\right)dx+{\int }_{75}^{100}\left|x-75\right|\left(\frac{1}{100}\right)dx \\ ={\int }_0^{25}\left(25-x\right)dx+{\int }_{25}^{37.5}\left(x-25\right)dx+{\int }_{37.5}^{50}\left(50-x\right)dx+{\int }_{50}^{62.5}\left(x-50\right)dx+{\int }_{62.5}^{75}\left(75-x\right)dx+{\int }_{75}^{100}\left(x-75\right)dx \\ =\frac{1}{100}\left[{\left(25x-\frac{x^2}{2}\right)}_0^{25}+{\left(\frac{x^2}{2}-25x\right)}_{25}^{37.5}+{\left(50x-\frac{x^2}{2}\right)}_{37.5}^{50}+{\left(\frac{x^2}{2}-50x\right)}_{50}^{62.5}+{\left(75x-\frac{x^2}{2}\right)}_{62.5}^{75}+{\left(\frac{x^2}{2}-75x\right)}_{75}^{100}\right] \\ =\frac{1}{100}\left[\frac{625}{2}+78.125+78.125+78.125+78.125+\frac{625}{2}\right] \\ =9.375 \end{gathered}$$

The expected time of new scheme is lesser than the expected time of the current scheme. Therefore, the new scheme is better.