Suppose passengers arrive at the MTA train station following a Poisson distribution with parameter 9 and...

Suppose passengers arrive at the MTA train station following a Poisson distribution with parameter 9 and the unit of time 1 hour.

Next train will arrive either 1 hour from now or 2 hours from now, with a 50/50 probability.

i. E(train arrival time)

ii. E(number of people who will board the train)

iii. var(number of people who will board the train)

Expert Solution

(i) The expectation of train arrival time is = 1 x 0.5 2 x 0.5 = 1.5 (ii Given that the train arrives at time t, the number of passengers arrived will follow Poisson distribution with parameter 9t. Hence E[Number of people arrived to board the train|The train has arrived at time t =9t E[Number of people arrived to board the train] -9 x E[The train has arrived at time t 9 x 1.5 = 13.5 (iii) Similarly Variance of number of people who have arrived given that the train arrives at time t is 9t. Hence Variance of number of people who have arrived is =9 x E[The train has arrived at time t]9 x Var[The train has arrived at time t =9 x 1.59 x [(1 - 1.5)2 x 0.5+ (2 1.5)2 x 0.5] = 13.5 9 x 0.25 = 15.75

(i) The expectation of train arrival time is =1 x 0.52 x 0.5 = 1.5 (ii Given that the train arrives at time t, the number of passengers arrived will follow Poisson distribution with parameter 9t. Hence E[Number of people arrived to board the train|The train has arrived at time t =9t .E[Number of people arrived to board the train] 9 x E[The train has arrived at time t] 9 x 1.5 = 13.5 (iii) Similarly Variance of number of people who have arrived given that the train arrives at time t is 9t. Hence Variance of number of people who have arrived is 9 x E[The train has arrived at time t] + 81 x Var[The train has arrived at time t =9 x 1.5 + 81 x [(1 - 1.5)2 x 0.5 (2 1.5)2 x 0.5 13.5 9 x 0.25 = 33.75

orchestra answered 1 year ago

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