The number of hours between successive train arrivals at thestation is uniformly distributed on (0, 1). passengers arriveaccording to a poisson process with rate 7 per hour. suppose atrain has just left the station. let x denote the number of peoplewho get on the next train. find

(a) e(x)

(b) var(x)

a)

b)

Step-by-step explanation:

A uniform distribution, sometimes also known as a rectangular distribution, is a distribution that has constant probability and is defined on a interval [a,b].

The Poisson distribution is the discrete probability distribution in order to describe the number of events occurring in a given time period.   And is defined with a parameter calles usually .

Let T the random variable that represent the number of hours between successive train arrivals at the station, and we know that the distribution for T is given by:

The expected value and the variance for an uniform random variable is given by:

ANd let X the random variable that represent the  number of people who get on the next train, so the conditional distribution X|T is given by:

Since we have a Poisson distribution we can find the expected value and the variance for the random variable X|T

We can use this result in order to answer the questions like this:

Part a

Using the properties for expected value we have this:

Part b

We are interested in Var (X) and in order to find this we can use the following property from conditional probability:

And since we already found Var[X|T] and E[X|T] we have this:

1. credits

2. dual - aspect

hope i : )

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )