Mathematics: Gate EE 17

Question

Assume that in a traffic junction, the cycle of traffic signal lights is 2 minutes of green(vehicle does not stop) and 3 minutes of red (vehicle stops). Consider the arrival time 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 _________

Comments

Then what will be our a and b in that case.I was actually trying but not getting it.If i directly take 3,0 then it will come as (3+0)/2= 1.5 whihc is incorrect. Whats the correct way if i want to use this formula?

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sorry, here it cant be used because (a+b)/2 can be used only when probability is uniformly distributed over [a,b] and you have to find mean from [a,b]...here in this qsn probability is uniformly ditributed over [0,5] and you have to find mean between [0,3]...

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The mean between [0,3] = 3/5=0.6 and answer is 0.9 . How it will be solve correctly ?

Answer 1

Here in the span of $5$ minutes there is $2$ minute of $"no"$ waiting time and $3$ minute of waiting time that depends on the arrival,  it could take any value from $0-3$ minutes.

So I am assuming that starting $3$ minutes its RED and last $2$ min its GREEN.

$\therefore $ the function of waiting time is $\color{Blue }{3-x}$ $\{$ if a person arrive at time 0 it will wait for 3 minutes and soon and so forth.$\}$

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Since its uniformly distributed expected waiting time will be :
$\Rightarrow $  $\Large \frac{1}{5-0}\int_{0}^{3}(3-x) \ dx$

$\Rightarrow $$\Large \frac{1}{5}[9-\frac{9}{2}]$

$\Rightarrow $$\Large \frac{9}{10}=\color{Red}{0.9}$