Q.7. 5 - A First Course in Probability

Question

The county hospital is located at the center of a square whose sides are $3$ miles wide. If an accident occurs within this square, then the hospital sends out an ambulance. The road network is rectangular, so the travel distance from the hospital, whose coordinates are $(0,0)$, to the point $(x, y)$ is $\left|x\right|+\left|y\right|$. If an accident occurs at a point that is uniformly distributed in the square, find the expected travel distance of the ambulance. 

Step-by-Step Solution

Answer

The expected travel distance of the ambulance value found to be $1.5$

Step 1: Given Information

Hospital is located at the center of a square whose sides are $3$ miles wide.

The coordinates of a hospital are $(0,0)$

The coordinates of an accident are $(x, y)$.

Step 2: Explanation

The road network is rectangular, hence the travel distance from the hospital to the accident point is$\left|x\right|+\left|y\right|$.

Since sides of the square are $3$ miles, Random variable $X$is uniformly distributed with parameters $-1.5$ and $1.5$.

Random variable $Y$ is also uniformly distributed with parameters $-1.5$ and $1.5$.

Step 3: Explanation

Let us find the expected travel distance of the ambulance,

$E=E\left(\right|x|+|y\left|\right)$

$=E\left(\right|x\left|\right)+E\left(\right|y\left|\right)$

$={\int }_{-1.5}^{1.5}\left|x\right|f\left(x\right)dx+{\int }_{-1.5}^{1.5}\left|y\right|f\left(y\right)dy$

Add the values.

$={\int }_{-1.5}^{1.5}\frac{\left|x\right|}{1.5-(-1.5)}dx+{\int }_{-1.5}^{1.5}\frac{\left|y\right|}{1.5-(-1.5)}dy$

$=\frac{1}{3}{\int }_{-1.5}^{1.5}\left|x\right|dx+\frac{1}{3}{\int }_{-1.5}^{1.5}\left|y\right|dy$.

Step 4: Substitute the value

Substitute the value

$=\frac{1}{3}\left\{\left[{\int }_{-1.5}^0-xdx+{\int }_0^{1.5}xdx\right]+\left[{\int }_{-1.5}^0-ydy+{\int }_0^{1.5}ydy\right]\right\}$

$=\frac{1}{3}\left\{{\left[\frac{-x^2}{2}\right]}_{-1.5}^0+{\left[\frac{x^2}{2}\right]}_0^{1.5}+{\left[-\frac{y^2}{2}\right]}_{-1.5}^0+{\left[\frac{y^2}{2}\right]}_0^{15}\right\}$

$=\frac{1}{3}\left\{-\left[\frac{-(-1.5{)}^2}{2}\right]+\left[\frac{(1.5{)}^2}{2}\right]+\left[\frac{(-1.5{)}^2}{2}\right]+\left[\frac{(1.5{)}^2}{2}\right]\right\}$

Multiply the value

$=\frac{1}{3}\times 4\times \frac{(1.5{)}^2}{2}$

$=1.5$.

Step 5: Final answer

The expected travel distance of the ambulance value found to be $1.5$ .