People Lining up at Bus Station

People line up at a bus station according to a Poisson distribution with 5 people / minute. The next bus will arrive some time in the next 2 minutes, with "some time" being a uniform distribution. How many people will get onto the next bus? What is the standard deviation of the number of people?


Solution:

The expected number of people waiting at time t is: E[Poisson(n)|t] = 5t.
Taking the expectation of this value over t = [0,2] we have: E[ E[Poisson(n)|t] ] = E [ 5t ] = 5.

So expect 5 people to get onto the next bus.

The standard deviation: use the variance formula, Var(Y) = Var( E[Y|X] ) + E[ Var(Y|X) ] with Y = number of people and X = time.

We have: Var(people) = Var ( E[People | t ] ) + E[ Var( People | t ) ]
The first term = Var ( E[Poisson(n, t)] ) = Var ( 5t ) = Int(0,2) { 1/2 * (5t - 5)² }dt = 0.5 * 25 * 1/3 * (t-1)³ | (2, 0) = 25/3.
Second term = E[ Var( Poisson (n,t) | t ] = E[ 5t ] = 5. Variance of Poisson process is just 5 for this case.


Total Variance: 5+ 25/3 = 13 1/3.

So standard deviation is ~ 3.65.