Respuesta :
Answer:
The value of [tex]E(X) =\frac{7}{2}[/tex].
The value of [tex]Var (X)= \frac{91}{12}[/tex].
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
[tex]E(X) = E(N(T))[/tex]
[tex]E(X) = E(E(N(T)|T)) [/tex]
[tex]E(X) = E(\lambda T) = \frac{7}{2}[/tex]
Hence, the value of [tex]E(X) =\frac{7}{2}[/tex].
Part (B)
Now we need to find VAR[X]
[tex]Var (X) = Var (E(N(T)|T)) +E(Var (X|T))[/tex]
[tex]Var (X)= E(\lambda T)+Var (\lambda T)[/tex]
[tex]Var (X)= \frac{49}{12}+\frac{7}{2}[/tex]
[tex]Var (X)= \frac{91}{12}[/tex]
Hence, the value of [tex]Var (X)= \frac{91}{12}[/tex]
