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Math Help - Central Limit Theorem Approximation

  1. #1
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    Central Limit Theorem Approximation

    The Internal Revenue Services uses a letter-opening machine to remove checks from envelopes. The machine has a probability p=0.03 of failure for any envelope. On a certain day, n=500 envelopes are independently handled by the machine. Let Y be the number of envelopes for which the machine fails to remove a check. Using the Central Limit Theorem (in this case DeMoivre's Theorem) to approximate P{Y \leq10}
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Janu42 View Post
    The Internal Revenue Services uses a letter-opening machine to remove checks from envelopes. The machine has a probability p=0.03 of failure for any envelope. On a certain day, n=500 envelopes are independently handled by the machine. Let Y be the number of envelopes for which the machine fails to remove a check. Using the Central Limit Theorem (in this case DeMoivre's Theorem) to approximate P{Y \leq10}
    In a case like this Bin(n,p) \approx Normal(np, np(1-p)).

    P(Y \leq 10) = P\Big(\frac{Y - np}{\sqrt{np(1-p)}} \leq \frac{10 - np}{\sqrt{np(1-p)}}\Big) \approx 1- \Phi\Big(\frac{\color{red}{10.5}\color{black}{} - np}{\sqrt{np(1-p)}}\Big).

    You have n and p, now plug in and look up in the tables.
    Last edited by Anonymous1; May 12th 2010 at 09:23 PM. Reason: 10.5
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  3. #3
    MHF Contributor matheagle's Avatar
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    1 you should use 10.5 instead of 10.
    2 I might use a Poisson approximation since p is near 0.
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  4. #4
    MHF Contributor matheagle's Avatar
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    P(Bin \le 10) \approx P(Normal\le 10.5)=P\left(Z\le {10.5-np\over<br />
{\sqrt{np(1-p)}}}\right)

    The 10.5 is the continuity correction, where we are overlaying a curve (normal) over the rectangles (binomial).
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