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Thread: Chebyshev Inequality

  1. #1
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    Chebyshev Inequality

    Chebyshev Inequality-topic10-5.jpg
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  2. #2
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    Re: Chebyshev Inequality

    let $\bar{X}$ be the sample mean. It has variance $\sigma^2 = \dfrac{\sigma_p^2}{n}$
    and thus $\sigma = \dfrac{\sigma_p}{\sqrt{n}}$

    Assume worst case and take $\sigma_p^2 = 5$

    $\sigma=\sqrt{\dfrac{5}{n}}$

    by Chebyshev

    $P[|\bar{X}-\mu| \geq k \sigma] \leq \dfrac{1}{k^2}$

    we want to ensure with 95% confidence that $|\bar{X}-\mu| \leq 1$ so

    $1 = k \sigma$

    $k = \dfrac{1}{\sigma} = \sqrt{\dfrac{n}{5}}$

    and we need $\dfrac{1}{k^2} \leq 0.05 \Rightarrow k \approx 4.472$

    $\sqrt{\dfrac{n}{5}} \approx 4.472$

    $n = 100$
    Thanks from Amanda2018
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