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Poisson Probability Distribution Problem

  1. Oct 12, 2015 #1
    1. The problem statement, all variables and given/known data
    An article suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is 0.4 year.

    a). How many loads can be expected to occur during a 4-year period?

    b). What is the probability that more than 13 loads occur during a 4-year period?

    c). How long must a time period be so that the probability of no loads occurring during that period is at most 0.3?

    2. Relevant equations


    3. The attempt at a solution
    I tried setting up the equation as a Poisson probability distribution for a). as (e^(-0.4)*0.4^(4)) / (4!) but I wasn't sure if this was correct. I couldn't move on to b or c without knowing a. If anyone could help give me some direction with good details that would be appreciated! I want to be able to learn the material and reasoning, not simply obtain the answer.

    Also, I apologize if this is not in the right category. I didn't see any homework-related sub-forums for probability and stats.

    Thanks in advance!
     
  2. jcsd
  3. Oct 12, 2015 #2
    Is a). actually just 4/0.4?
     
  4. Oct 12, 2015 #3

    Ray Vickson

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    Your ##\tau = 0.4## yr. is a time, not a rate. The parameter in the Poisson distribution is dimensionless:
    [tex] \text{parameter} = \text{mean number} = \text{rate} \times \text{time}. [/tex]
    In your problem, the rate is ##\lambda = 1/\tau = 1/0.4 = 2.5## events per year.

    Problems in probability and/or statistics are usually posted here or in the "Calculus and Beyond" forum, depending on the level of the question and the mathematical tools needed to deal with it. Occasionally they appear in the Elementary or Advanced Physics forums, especially if they have something to do with experimental error analysis or statistical mechanics and the like.
     
  5. Oct 12, 2015 #4
    Ah, that makes sense, so in this case what is x? Or is x the value that I am solving for?
     
  6. Oct 12, 2015 #5
    So far my equation is [itex]\frac{e^{-10}10^x}{x!}[/itex]
     
    Last edited: Oct 12, 2015
  7. Oct 12, 2015 #6
    Oh, so ##\mu## (aka ##2.5 \times 4##) is the expected value, so that should be the value that I'm looking for for a)?
     
    Last edited: Oct 12, 2015
  8. Oct 12, 2015 #7

    Ray Vickson

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    There was no letter "x" in the above; there was the multiplication sign ##\times##. Sorry if that confused you.
     
  9. Oct 12, 2015 #8
    I was referring to the equation that I made in the following post. But I realized that the value that I'm looking for is ##\mu## anyway, which if I'm understanding correctly, should be 10.

    I'm confused about how to go about b). It gave me a table of CDF values listed for ##\mu## and x, so and it wants P(X>13). My logic is that that would simply sum to infinity with infinite terms, so I did 1-P(X≤13). Since the table is CDF, that would mean I simple take 1-P(13) wouldn't it?
     
  10. Oct 13, 2015 #9

    Ray Vickson

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    Yes, that would be the way to do it.
     
  11. Oct 13, 2015 #10
    Alright. And for C, do I just solve for ##\mu## from the Poisson equation? And then divide by the rate?
     
  12. Oct 13, 2015 #11
    Hmm, or could I solve C using the CDF tables?
     
  13. Oct 13, 2015 #12

    Ray Vickson

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    Solve for ##\mu## how? What would be the equation you want to solve?
     
  14. Oct 13, 2015 #13
    Well it tells me that the probability would be 0.3 so I was thinking I could set it up as
    [itex]\frac{e^{-\mu}\mu^x}{x!}=0.3[/itex]
    But then again I don't have a specified x...
     
  15. Oct 13, 2015 #14
    Oh wait x would just be 0 because it's referring to no loads
     
  16. Oct 13, 2015 #15

    Ray Vickson

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    Yes, exactly.
     
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