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Poisson Process

  1. Mar 4, 2005 #1
    Suppose we have a highway with vehicles comming from left(Point A) an from the right(Point B). And in some point we have a Restaurant.

    The number of vehicles passing the point A in an hour follows the Poisson distribution with mean 60; 20% of these vehicles are trucks. The number of vehicles passing B in an hour is also Poisson with mean 80; 30% of these are trucks. In general, 10% of all vehicles stop at the restaurant. The number of persons in a truck is one; the number of passengers in a car is equal to 1, 2, 3, 4, or 5 with respective probabilities 0.30, 0.30, 0.20, 0.10 and 0.10.

    Find the expected value E[Z] of the number of persons Z arriving at the restaurant within that on hour.

    Any suggestions?
  2. jcsd
  3. Mar 4, 2005 #2
    I think you can ignore the poisson distribution here and just work with its mean. Otherwise (or if you must work it formally) it is somewhat more complicated, but intuitively I think you can.

    So you find the number of people arriving at the restaurant passing point A, then the number of people arriving at the restaurant passing point B, then add them.

    For point A you have 60 vehicles: 12 trucks and 48 cars. 10% of the trucks=1.2 trucks for 1.2 people. 10% of the cars=4.8 cars, so the # of people coming in cars is 4.8 * E[C] where C is the # of people in a car. Point B would be done similarly.
  4. Mar 4, 2005 #3
    I think you are right. It may work since:
    N(t) = n, the number of events up to time t is n.
    P{N(t)=n} = exp(-lamba*t)(lamba*t)^n/n!
    E[N(t)] = lamba*t
    Then, E[Z]=lamba*t
    Since t = 1 hout, then we can find E[Z] by computing the total number of people that arrive in 1 hour to points A and B, for trucks and cars respectively and then its 10% goes to the Restaurant.

  5. Mar 4, 2005 #4
    You can't just assume that Z will also have the poisson distribution. I'm pretty sure it will, but you can't just assume that.
  6. Mar 5, 2005 #5
    Well, if Z is the number of people arriving at each point A and B, and the arrivals have Poisson distribution, we can not think about Z as a Poisson r.v.?
  7. Mar 6, 2005 #6
    You don't know that the arrivals have the Poisson distribution right off the bat. You only know that the vehicles have the Poisson distribution.
  8. Mar 7, 2005 #7
    Oh, what wrong I am... you are right, finally understood the point. Thank you.
    We only know arrivals of vehicles have a Poisson distribution with rate lamba vehicles/hour.
    So, for this reason we only compute the arrival of vehicles, and considering the number of passenger for each type of vehicle we can compute the expected value of, say Z, the number of people who go to the restaurant. That's all. We do not know what distribution has the arrival of people to the restaurant, only the vehicles at points A and B.

    Ok, now everything is clear. Thank you for your help. Sorry if I bothered you.
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