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Poisson distribution (would you please verify?)

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data

    1. Suppose that the number of telephone calls an operator receives from 9:00 to 9:05 A.M. follows a Poisson distribution with mean 3. Find the probability that the operator will receive:
    a. no calls in that interval tomorrow.
    b. three or more calls in that interval the day after tomorrow.


    2. Find the number of chocolate chips a cookie should contain
    on the average if it is desired that the probability of a cookie
    containing at least one chocolate chip be .99.




    Thank you very much for any help!


    2. Relevant equations



    3. The attempt at a solution
    Problem (1):
    Noting probability of Poisson distribution with p, the number of calls with x, and the mean with m, we have
    a) p=[(m^x)*(e^-m)]/x! = e^-m = e^-3
    b) x=3 => p=1-p(x=2) = 1- [(m^2)*(e^-m)]/2! = 1- [(3^2)*(e^-3)]/2

    Problem (2):
    Preserving the notation above, p=.99 , x=1 =>
    .99 = [(m^x)*(e^-m)]/x! = m*(e^-m)
    If so, do I take ln of both sides to find m?


    Thank you very much for any help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 18, 2009 #2

    Mark44

    Staff: Mentor

    1a looks fine, but your notation could use some work. Using m = 3, P(X = 0) = m0e-m/0! = e-3
    For 1b, it asks for the probability that X is >= 3, not just P(X = 3). P(X >= 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]
     
    Last edited: Nov 18, 2009
  4. Nov 18, 2009 #3
    Thank you very much, Mark!
    Would you please extend your help?
    1. I don't understand how I would arrive at e^3 instead of e^-3
    2. Would you have any suggestions for Problem 2 (the cookie)?

    Thank you to anyone who would help!
     
  5. Nov 18, 2009 #4

    Mark44

    Staff: Mentor

    It is e^(-3). I didn't carry the minus sign through from the previous step. (It's fixed, now.)
    P(X >= 1) = .99, so P(X = 0) = 1 - P(X >= 1) = 1 - .99 = .01
    But P(X = 0) = .01
    ==> m0e-m/0! = .01
    Can you solve this for m?
     
  6. Nov 18, 2009 #5
    Thanks Mark!
    Before I saw your reply, this is what I did:

    p = .99 + p(x=0) = .99 + [(m^0)*(e^-m)]/0! = .99 + e^-m = 1 (1 is total probability = certainty)

    If so, then m = -ln(0.01)

    How wrong am I?
     
  7. Nov 18, 2009 #6

    Mark44

    Staff: Mentor

    You're not wrong - that's correct. According to this problem, a cookie should have between 4 and 5 chocolate chips.
     
  8. Nov 18, 2009 #7
    Thank you very much, Mark, for all your help!
     
  9. Nov 18, 2009 #8
    Problems resolved.
     
  10. Nov 18, 2009 #9

    Mark44

    Staff: Mentor

    I hope you take heed of the guidance I was providing on writing notation so that it makes sense.
     
  11. Nov 18, 2009 #10
    Thank you again, Mark!
    Good thing you refreshed the point; I had almost forgotten.
    Of course, I see that it wouldn't make sense without specifying the P(x=0) how I arrive at the end form.

    I hope you are around if I need help with other things in the future.

    Thanks again, and best to you.
     
  12. Nov 18, 2009 #11

    Mark44

    Staff: Mentor

    You're welcome. I'm around here quite a bit, and when I'm not, there are lots of other very capable people who enjoy helping out.
    Mark
     
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