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Math Help - chebyshev inequality

  1. #1
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    chebyshev inequality

    Hi,

    A product is sold by box of 5000 products. The mean and variance for the number of defective products in one box is respectivly 10 and 5.76. Use chebyshev inequality to find a boundary for the probability that a box contains 8 to 12 defective products.

    What i did:

    P(8 <= X <= 12) = P(|X - 10| <= 2) = 1 - P(|X -10| > 2), and with chebyshev inequality i find P(|X - 10| > 2) <= 5.76/2^2 = 1.44?? What is my mistake? Thank you so much
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  2. #2
    MHF Contributor harish21's Avatar
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    If X is a real valued random variable with mean \mu
    and variance \sigma^{2}, then your chebyshef's inequality is given by:

    P\{|X - \mu| < k \sigma\} \geq 1 - k^{-2}



    use this to find what your question is asking
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by harish21 View Post
    If X is a real valued random variable with mean \mu
    and variance \sigma^{2}, then your chebyshef's inequality is given by:

    P\{|X - \mu| < k \sigma\} \geq 1 - k^{-2}



    use this to find what your question is asking
    Not a real criticism, but for other reasons I would prefer:

    P\left(\left|\dfrac{X - \mu}{\sigma}\right| < k \right) \geq 1 - k^{-2}

    CB
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  4. #4
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    yes but the problem is that k < 1...

    P(|X - u| < ka) = P(|X - 10| < 2) where ka=2 and a = 2.4 so k=5/6, then 1 - (5/6)^-2 = - 0.44 ...
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by mathworld View Post
    yes but the problem is that k < 1...

    P(|X - u| < ka) = P(|X - 10| < 2) where ka=2 and a = 2.4 so k=5/6, then 1 - (5/6)^-2 = - 0.44 ...
    And the interpretation is that the bound is zero.

    CB
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