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Math Help - Chebyshev's Theorem

  1. #1
    Yan
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    Use the Chebyshev's theorem and 1-(p(1-p)/(nc^2)) to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.4 and 0.6

    Mr F edit: Merged post from a new thread:

    Use the Chebyshev's theorem and Binomial distribution to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.4 and 0.6.
    Last edited by mr fantastic; February 15th 2009 at 12:02 PM.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by Yan View Post
    Use the Chebyshev's theorem and Binomial distribution to verify that the probability is at least 35/36 that in 900 flips of a balanced coin the proportion of heads will be between 0.4 and 0.6.
    Let X be the number of heads that you get. This follows a binomial distribution of parameters (n,p), where p=0.5, since it's a balanced coin.

    The proportion H of heads is given by H=\frac{X}{900}


    You want to prove that P\{0.4<H<0.6\} \geq \frac{35}{36}

    The mean of a binomial distribution is np.
    So \mu(X)=900 \cdot 0.5=450

    \boxed{0.4<H<0.6} \leftrightarrow [900 \cdot 0.4<X<900 \cdot 0.6] \leftrightarrow [360<X<540] \leftrightarrow \boxed{-90<X-\mu(X)<90}

    So P\{0.4<H<0.6\}=P\{-90<X-\mu(X)<90\}=P\{|X-\mu(X)|<90\} =1-P\{|X-\mu(X)|>90\}

    But we know by Chebyshev's inequality :
    P\{|X-\mu(X)|>90\} \leq \frac{V(X)}{90^2}

    The variance of a binomial distribution is np(1-p).
    So V(X)=900 \cdot 0.5^2=225

    Hence P\{|X-\mu(X)|>90\} \leq \frac{225}{90^2}=\frac{1}{36}



    ------> P\{0.4<H<0.6\}=1-P\{|X-\mu(X)|>90\} \geq 1-\frac{1}{36}=\frac{35}{36}
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