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The Poisson Distribution

  1. Dec 11, 2016 #1
    1. The problem statement, all variables and given/known data
    On average, 2 students per hour come into the class. What is the probability that the time between two consecutive arrivals is in the interval <10 minutes; 50 minutes>.

    2. Relevant equations
    p(k)=P(Y=k)=((lambda*t)k*(e-lambda*t)/k!

    3. The attempt at a solution
    I've tried using the Poisson distribution with k=2 t=40/60=2/3 and lambda=4/3 but the result is wrong. Thanks for helping!
     
  2. jcsd
  3. Dec 11, 2016 #2

    haruspex

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    Try to justify those parameters.
    In the equation you quote, how would you describe the meanings of k and λ?
     
  4. Dec 11, 2016 #3
    I think k is number of events that can happen in selected time unit...two student arrivals in this case and λ is average number of events that can happen in an hour...is it 2 as well? I'm not sure.
     
  5. Dec 11, 2016 #4

    haruspex

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    Yes, λ here is 2 per hour.
    But the equation you quote is not suited to the task, directly. It gives you the probability of a specific number of arrivals in a specific period of time. You want the probability that the time between a particular pair of consecutive arrivals is in some range.

    You might as well suppose that the first of the two arrivals has just occurred. How can you rephrase the question a bit more simply?
     
  6. Dec 11, 2016 #5
    I could use the Exponential distribution and determine the time of waiting for another arrival?
     
  7. Dec 11, 2016 #6
    2*e(-2*t) where t=2/3?
     
  8. Dec 11, 2016 #7

    haruspex

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    No, you need to do some more thinking before trying to plug numbers into an equation.
    Suppose that one student has just arrived. Rephrase the question in terms of a single arrival.
     
  9. Dec 11, 2016 #8
    Well I need to find out if the time of waiting for arrival of the second student is in the interval <= 50 minutes. And the arrival of the first student has to be in the interval >= 10 minutes.
     
  10. Dec 11, 2016 #9

    haruspex

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    No, we can start the clock when the first student arrives. What are the constraints on the arrival time of the next student?
     
  11. Dec 11, 2016 #10
    The second student has to come 40 minutes later at most after the first student.
     
  12. Dec 11, 2016 #11

    haruspex

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    No, that is not what the question says. Read it again:
    Edit: maybe the word interval is confusing you. Here it just means a range of values: the time between two consecutive arrivals is in the range from 10 minutes to 50 minutes.
     
    Last edited: Dec 11, 2016
  13. Dec 11, 2016 #12
    The second student has to come in the fiftieth minute at the latest?
     
  14. Dec 11, 2016 #13

    haruspex

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    And the earliest?
     
  15. Dec 11, 2016 #14
    The fiftieth minute minus the time when the first student came?
     
  16. Dec 11, 2016 #15

    haruspex

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    No, what is the earliest, after the first student, that the second student can arrive (to fit in the given window)?

    Let's make it more concrete. Say the first student arrives at 9am. Between what two times is the second student to arrive for the gap between them to be from 10 to 50 minutes?

    It's late here.... off to bed.
     
    Last edited: Dec 11, 2016
  17. Dec 11, 2016 #16
    9:10-9:50?
     
  18. Dec 11, 2016 #17

    haruspex

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    Right.
    So how many students arrive between 9 and 9:10?
     
    Last edited: Dec 11, 2016
  19. Dec 11, 2016 #18
    0 (and between 9:50-10 too)...but I still don't know how to solve it.
     
  20. Dec 11, 2016 #19

    haruspex

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    Right. Using your equation, what is the probability of that?
    No. We might come back to that later.
    How many students arrive between 9:10 and 9:50?
     
  21. Dec 11, 2016 #20
    9:00-9:10 P(Y=0)=((2*1/6)0*(e-1/3))/0!=0.716
    9:10-9:50 1 student P(Y=1)=((2*2/3)1*(e-4/3)/1!=0.351
     
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