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The number of hours between successive train arrivals at the station is uniformly distributed on (0, 1). Passengers arrive according to a Poisson process with rate 7 per hour. Suppose a train has just left the station. Let X denote the number of people who get on the next train. What are the E[X] and VAR[X]?

User Jgb
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Answer:

The value of
E(X) =(7)/(2).

The value of
Var (X)= (91)/(12).

Explanation:

Consider the provided information.

The number of hours between successive train arrivals at the station is uniformly distributed on (0, 1). Passengers arrive according to a Poisson process with rate 7 per hour.

Let X denote the number of people who get on the next train.

Part (A)

Here, X = N(T), N(t) ∼ Poisson(λt), and λ = 7

Therefore, E(N(T)|T) = λT

Now find E(X) as shown below


E(X) = E(N(T))


E(X) = E(E(N(T)|T))


E(X) = E(\lambda T) = (7)/(2)

Hence, the value of
E(X) =(7)/(2).

Part (B)

Now we need to find VAR[X]


Var (X) = Var (E(N(T)|T)) +E(Var (X|T))


Var (X)= E(\lambda T)+Var (\lambda T)


Var (X)= (49)/(12)+(7)/(2)


Var (X)= (91)/(12)

Hence, the value of
Var (X)= (91)/(12)

User Cajuuh
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