Question

Assume that in a traffic junction, the cycle of the traffic signal lights is 2 minutes of green (vehicle does not stop) and 3minutes of red (vehicle stops). Consider that the arrival tme of vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time (in minutes) for the vehicle at the junction is

• A. 0.9

Answer 1

The correct option is A 0.9

Let time of arrival be a random variable x.

then, its p.d.f. is $f\left(x\right)=(\begin{array}{cc}\frac{1}{5},& 0\le x\le 5\\ 0,& \mathrm{else}\mathrm{where}\end{array}$
Assume waiting time be g(x), a function of arrival time.

$g\left(x\right)=(\begin{array}{ccc}0,& 0\le x\le 2& \left(\mathrm{Green}\mathrm{light}\right)\\ 5-x,& 2\le x\le & \left(\mathrm{Red}\mathrm{light}\right)\end{array}$
Average waiting time

$=E\left(g\right(x\left)\right)={\int }_{-\infty }^{\infty }g\left(x\right)f\left(x\right)dx$

$={\int }_0^{5}g\left(x\right).\frac{1}{5}dx+{\int }_{-\infty }^{\infty }g\left(x\right).0dx$

$={\int }_0^{2}0\times \frac{1}{5}dx+{\int }_2^5(5-x)\times \frac{1}{5}dx$

$={[x-\frac{x^2}{10}]}_2^5$

$=\frac{9}{10}=0.9$