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Approximating distributions with other distributions

  1. May 18, 2013 #1
    In shipment A, there are 990 correct and 10 faulty units. In shipment B, there are 1940 correct and 60 faulty units.

    100 units out of each shipment is inspected. Calculate with an APPROPRIATE approximation the probability of finding five or more faulty units.

    Emphasis from book, not me.

    My solution was

    A = Bin(100,0.01) ≈ N(1,0.1)
    B = Bin(100,0.03) ≈ N(3,1.71)

    A+B ≈ N(4,1.71)

    P(A+B > 4.5) = 1-P(A+B < 4.5) = /normal table/ = 0.386

    Which produced a fairly good answer, but the book also made it clear that they were looking for a poisson approximation. I am not entirely clear on why that is. Anyone?
     
  2. jcsd
  3. May 18, 2013 #2

    I like Serena

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    Perhaps what they wanted, is for you to use a Poisson approximation with mean (1+3), in which case you get:
    A = Bin(100,0.01) ≈ Poisson(1)
    B = Bin(100,0.03) ≈ Poisson(3)
    A+B ≈ Poisson(1+3)

    P(A+B >_ 5) = 1 - P(A+B <_ 4) = 1 - POISSON.DISTR(4, 1+3, cumulative) = 0.389.

    Either way, they may have wanted you to associate the probability on "faulty" units with Poisson, which would be appropriate.

    Btw, I get slightly different results.
    My result for your normal approximation based on the binomial distribution is 0.419.
    The same normal approximation based on the Poisson distribution is 0.420.
     
    Last edited: May 18, 2013
  4. May 18, 2013 #3
    That is almost certainly what they wanted. But there doesn't seem to be any great motive for why a Poisson approximation would be BETTER than a normal approximation, right?

    Btw my book says Bin(n,p) ~ N[np,sqrt(np(1-p)]. Wikipedia says Bin(n,p) ~ N[np,np(1-p)]. Is this a notation difference? My book uses N(a,b) where a is expected value, b is standard deviation.
     
  5. May 18, 2013 #4

    I like Serena

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    Actually, the normal distribution is not a very good approximation where the tails of the distribution are concerned.
    The Poisson distribution has a more natural resemblance to the binomial distribution, on top of being appropriate for fault behavior.

    At any rate, I wouldn't call your normal approximation wrong. It's still a good approximation in this case.


    Yes, these are indeed notational differences.
    The parameters of the normal distribution are expected value (##\mu##) and standard deviation (##\sigma##).
    Whenever you specify the normal distribution, you need to make sure to eliminate any ambiguity where the standard deviation is concerned.
    As you can see in the wiki article, they've made sure to specify ##\mathcal N(\mu, \sigma^2)## first, before specifying anything with it.
    Likely your book will have specified N[μ,σ] first, before making any statements using it.
     
  6. May 18, 2013 #5

    Ray Vickson

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    The normal approx. is not advised for a large-n, small-p case like yours. The Poisson is much better.

    Note: you should have A ~ N(1,0.995), giving A+B~N(4,1.975) and P(>=4.5)=0.40006387.

    The results are for P(>=5) are:
    exact binomial = 0.37115~0.37
    Poisson = 0.37116~0.37
    Normal = 0.40006~0.40

    Note: these are the Poisson approx. to the binomial, and the normal approx. to the same binomial.

    My home internet connection is down, so I am using my I-phone with all its difficulties and limitations. (Computations done off-line.)
     
    Last edited: May 18, 2013
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