The number of hours between successive train arrivals at the station is uniformlydistributed on [0;1]. Passengers arrive according to a Poisson process with rate 7 perhour. Suppose a train has just left the station. LetXdenote the number of peoplewho get on the next train. Denote byTthe arrival time of the next train.(a)GivenT= 0:4, what is the conditional expectationE(XjT= 0:4).(b)FindE(X). Hint: useE(X) =E[E(XjT)].

E ( X ) = 3.5

Var ( X ) = 91 / 12

Step-by-step explanation:

Solution:-

- Let X = M (T), where M (T) is the Random variable that denotes the number of arrivals in time T for the Poisson process.  The parameter λ = rate of 7 per hour

- To find E ( X ), condition X on the random arrival time T of the next train.

- Note that if Y ~ Poisson ( μ ), then the T is defined by uniform distribution over the interval [ 0 , 1 ] :

                           E ( Y ) = Var ( Y ) = μ

                           E ( T ) = 1 / 2

                           Var ( T ) = 1 / 12

- We have, N ( T ) ~ Poisson ( λt ), where t ≥ 0 and λ = 7. Thus,

                           E ( N ( T ) / T ) =  λ*T

                           Var ( N ( T ) / T ) =  λ*T.

Therefore,

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

                            = E ( λ*T ) = λ* E ( T ) = 7/ 2 = 3.5

- For two Random Variables U and V,

                           Var ( U ) = E ( Var ( U / V ) ) + Var ( E ( U / V ) )

Therefore,

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

                           =  E ( λ*T) + Var ( λ*T )

                           = λ* E ( T ) + λ^2* Var ( T )

                           = 3.5 + 7^2 / 12

                           = 91 / 12

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation: