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Stochastic Processes - Poisson Process question

  1. Apr 27, 2009 #1
    I had this problem on my last midterm and received no credit for these parts.

    1. Express trains arrive at Hiawatha station according to a Poisson process at rate 4 per hour, and independent of this, Downtown local buses arrive according to a Poisson process at rate 8 per hour.

    a. Given that 10 Express trains arrive during the morning hours of 9:00-11:00 am, what is the expected number of Express trains that will arrive during the afternoon hours of 1:00-3:00pm

    The attempt at a solution
    We want E[# of trains]

    Given 9:00-11:00 => 10 trains
    <=>\int_{9}^{11}4c dr = 10 => c=5/4
    \int_{13}^{15}4c dr = ? = 10 <=> E[# of trains] = 10

    Sorry, latex typesetting was not working.

    b. Suppose your friend arrives at the station and decides to take the first transportation that arrives. Given she takes an Express train (it arrives first), what is the expected amount of time she waited for it to arrive?

    The attempt at a solution

    c. Suppose each Downtown bus carries passengers, the number of which has a probability distribution with mean 24. Find the expected value of the number of passengers that ride the Downtown bus line during 12 hours.

    The attempt at a solution
    E[# of passengers]=mu(P(T \leq 12))=24(1-exp(-12lambdabus)=24

    d. What is the probability that a total of exactly 12 trains and buses arrive in a given 1-hour period?

    The attempt at a solution
    P(#trains=12 & #buses=12 | time = 1 hr)
    average 4 trains per hour
    average 8 buses per hour

    Now, I am stuck.
    Last edited: Apr 27, 2009
  2. jcsd
  3. Apr 27, 2009 #2
    Well, I got some help from elsewhere, but thought I might post the solutions anyway.

    part a. solution
    Expected number of trains from 1 to 3
    =(2hrs) x (4 per hour) = 8 trains.

    part b. solution
    E[T]=1/(lambdatrains + lambdabuses)=1/12=5minutes.

    part c. solution
    (24 passengers/bus) x (8 buses/hour) x (12 hours)= 2,304 passengers.

    part d. solution
    P(N(t)=n) = [(lambda*t)^n][exp(-lambda*t)]/n!
    P(N(1)=12) = [(12*1)^12][exp(-12*1)]/12! = 0.1144
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