Question

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]?

Answer 1

Answer:

The value of $E(X) =\frac{7}{2}$.

The value of $Var (X)= \frac{91}{12}$.

Step-by-step 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

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

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

$E(X) = E(\lambda T) = \frac{7}{2}$

Hence, the value of $E(X) =\frac{7}{2}$.

Part (B)

Now we need to find VAR[X]

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

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

$Var (X)= \frac{49}{12}+\frac{7}{2}$

$Var (X)= \frac{91}{12}$

Hence, the value of $Var (X)= \frac{91}{12}$