In a store, a cashier only helps people when 5 people are in the queue. The people arrive at the cash desk according to a poisson process with a rate of 1 people per 2 minutes. What is the expected time that the cashier helps a customer?
I thought, you just do 5 x 2 = 10 minutes as expected value to have 5 people at the cash desk. But you could also see it als independent variables which are exponentially distributed. Can someone explain me which method I need to use?
People arrive at an average rate of one per two minutes, so the expected time to arrival of the fifth customer is indeed ten minutes. This is because the interarrival times are iid exponential random variables.
The Law of Total Expectation says the expected time to the fifth arrival equals the sum of the expected inter arrival times. $$\mathsf E(T_5)=\mathsf E(T_1)+\mathsf E(T_2{-}T_1)+\mathsf E(T_3{-}T_2)+\mathsf E(T_4{-}T_3)+\mathsf E(T_5{-}T_4)$$
Now, the memoryless property of the Poisson Process entails that each of those interarrival times is an independent and identically exponentially distributed random variable. $$\mathsf E(T_5)=5\mathsf E(T_1)$$
And we know that the expected time to the first arrival is two minutes.$$\mathsf E(T_5)=10$$