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Binomial -> Poisson distribution question

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data

    A teacher has an infinite flow of papers to mark. They appear in his office at random times, at an average rate of 10 a day. On average 10% of the manuscripts are free from errors. What is the probability that the teacher will see exactly one error-free manuscript (a) after he has marked 10 of them? (b) after a day?
    2. Relevant equations
    Binomial dist.:
    [itex]P(k) = \frac{N!}{k!(N-k)!}p^{k}(1-p)^{N-k}[/itex]

    Poisson dist.:
    [itex]P(k) = \frac{\mu^{k}}{k!}e^{-\mu}[/itex]

    3. The attempt at a solution
    a)
    The first part seemed straight forward, just plug on following values
    No. of trials: N = 10
    No. of successes: k = 1
    Prob. of success: p = 0.1

    This gives an answer ~0.4 (1 s.f.)

    b) The second part I don't really understand. I know it goes to a Poisson distribution because I asked my teacher if that is so. I thought P-distributions are for very large N? Can we assume that here? Mean I know the no. of papers i infinite, but the per day average is only 10?

    Secondly is the mean value per day μ, the same as the expectation value per 10 papers? I don't think so, but here goes:
    let μ = 0.1 and using second equation -> P(1) = ~0.2 (1 s.f)

    I thought given the same parameters, the Poisson and Binomial distributions should give very similar results.
     
  2. jcsd
  3. Oct 19, 2015 #2

    DrDu

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    No, P-distributions are not restricted in any sense to very large N.
     
  4. Oct 19, 2015 #3
    Sorry, I meant that the Poisson is the limit of the binomial distribution for large N, but here N = 10?
     
  5. Oct 19, 2015 #4

    Ray Vickson

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    No, N is not 10, it is ∞. The question said the teacher marks an infinite flow of papers.

    However, I think the problem leaves out some important and significant information: it says the papers appear at random times and at a given average rate. We are likely being asked to assume that "random times" mean "independent, identically distributed inter-arrival times", but even then we do not really have a rigorous justification of the Poisson arrival model.

    Nevertheless, I would bet the person setting the problem wants us to assume Poisson arrivals, despite not having enough information to justify it. I think that is acceptable in an introductory course.
     
    Last edited: Oct 19, 2015
  6. Oct 19, 2015 #5

    DrDu

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    That you can derive the Poisson distribution as a limit of some other distribution doesn't say nothing about it's range of validity.
     
  7. Oct 19, 2015 #5

    DrDu

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    That you can derive the Poisson distribution as a limit of some other distribution doesn't say nothing about it's range of validity.
     
  8. Oct 19, 2015 #6

    Yes, that's correct. I'm still none the wiser about part 2. If we assume Poisson distributed data, it still needs a mean μ over one day and this what I don't get. It seems like μ = Np = 0.1, but over the day the number of papers N that arrive may be less or greater than 10??
     
  9. Oct 19, 2015 #7

    RUber

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    I agree with your part (a).
    For part (b) It looks like you will have an infinite sum. k = 1, and P(k) will still follow the binomial distribution.
    However, you need to handle N.
    N is Poisson distributed.
    Total probability ##P = \sum_{N=0}^\infty P_{poisson}(N) *P_{binom}(k|N)##
    This sum will necessarily converge. It may take you some additional work to find it. Otherwise, just plug it into Excel and see what happens when you start getting N up near 20 or so.
     
  10. Oct 19, 2015 #8

    Ray Vickson

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    The average arrival rate is 10 per day = 10/24 per hour = 10/(24*60) per minute. So, in a day of 24*60 = 1440 minutes, we have an arrival probability of p = 10/1440 = 0.006944 in each minute. This is a case of a binomial with a large N (N = 1440) and a small p (p = 10/1440), so leads right away to the Poisson limit.

    And, of course, the number of arrivals in any particular day can be any integer from 0 to ∞, but will very likely not differ too much from 10.
     
  11. Oct 19, 2015 #9
    thank you that makes sense. so if we're work out the probability of an error free paper per minute, then each trial can be a time interval of 1 min and still with two possible outcomes - success or failure.
     
  12. Oct 19, 2015 #10

    Ray Vickson

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    Warning: the analysis I gave you was for paper arrivals of all types---whether or not they have any errors.

    If you want to look at only papers with no errors, you need to use a different rate and a different Poisson process. As before, assuming the error/non-error probabilities are independent of everything else, the average number of error-free papers is (1/10)*(1) = 1 per day = 1/24 per hour = 1/(24*60) = 1/1440 per minute = 1/(24*60*60) = 1/86,400 per second, etc. So, in a day of N = 86,400 seconds, with an error-free arrival probability of 1/86,400 each second (and assuming independence) we are in the very large-N, very small-p limit, giving a Poisson random variable ##X## with mean ##\lambda = 86,400/86,400 = 1## (per day). So, the probability of exactly 1 error-free paper in a day is ##P(X = 1)##, where ##X\sim \: \text{Poisson}(\lambda = 1)##.
     
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