91Ó°ÊÓ

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$(x,y)$.

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$1.5$.

Let us find the expected travel distance of the ambulance,

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

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

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

Add the values.

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

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

Substitute the value

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

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

Multiply the value

role="math" localid="1647250321900" $=\frac{1}{3}×4×\frac{(1.5{)}^2}{2}$

$=1.5$.

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