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PMF Probability

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Of the people passing through an airport metal detector, 0.5% activate it; Let X denote the number among a randomly selected group of 500 who activate it.

    1) What is the PMF of X
    i) Using th CLT (approximate PMF)
    ii) Using the exact distribution of X

    2) P(X = 5) using i and ii

    3) P(X<=5) using i and ii


    2. Relevant equations



    3. The attempt at a solution

    So its been a while since Ive taking a probability class but I thought this was a Binomial Distribution problem where n = X, p = .005 and n = 500

    P(X = x) = (500 Permutation X) * (.005^X) * (1-.005)^(500-X)

    but the output im getting for X = 1,2,3... = .20, .51, 1.29 which are clearly wrong.

    My thought for X = 1 would be P(X=1) = (1/500)*.005 = .00001 which seams reasonable to me but Im confusing myself when I go on to P(X = 2)

    Any hints would be appreciated.
     
  2. jcsd
  3. Sep 11, 2014 #2
    500 Permutation X is not part of the Binomial Distribution.

    Your first term is correct (The .20 one), but the second and third are wrong.
     
    Last edited: Sep 11, 2014
  4. Sep 11, 2014 #3

    Ray Vickson

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    You should also have a value for x = 0.

    Anyway, why do you think that P(X=1) = (1/500)*.005 ? You already gave a formula for P(X=x); what does it give you when x = 1? (It WON'T be what you wrote!)
     
  5. Sep 11, 2014 #4
    Oops, should be Combination im guessing???

    Using that I get P(X = 1,2,3,...) = .20, .26, .21, .13, ...

    These just don't seem correct to me. Seem way to high??
     
  6. Sep 11, 2014 #5

    Ray Vickson

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    They are OK, but you also need to look at P(X=0). After all, it is possible that all 500 fail to activate the device. And, of course, P(X = 0) is a definite part of the binomial distribution.
     
  7. Sep 11, 2014 #6
    Ok thank you. So I have the PMF function. The problem asks to compute probabilities using...

    1) the CLT (approximation)
    2) exact distribution

    What is the difference. I thought the CLT allowed the use of binomial dist because of the large n. How do I "approximate PMF" using CLT.
     
  8. Sep 11, 2014 #7

    Ray Vickson

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    Whoever is asking you to use the CLT result is asking you to perform a disastrously bad approximation in this case. Your problem does not at all fit the criteria for getting a reasonable approximation via the CLT. The CLT is a LIMIT theorem; the question is whether you can use the limit as a good approximation when you are not taking something to ∞, but merely to a large value (500 in this case). Under certain circumstances the answer is YES, but not in this case.

    However, for this problem there is another type of limit result that can be used instead. Google 'limits of binomial distribution'.

    Note added in edit: well, maybe 'disastrously bad' is too strong; just plain 'not very good' is a better description.
     
    Last edited: Sep 11, 2014
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