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Thread: AP STATISTICS Random Variables

  1. #1
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    AP STATISTICS Random Variables

    While he was a prisonerof war during World War II, John Kerrich tossed a coin10,000 times. He got 5067 heads. If the coin is perfectlybalanced, the probability of a head is 0.5.







    a.) Is there reason to think that Kerrich's coin was notbalanced? To answer this question, use a Normaldistribution to estimate the probability that tossinga balanced coin 10,000 times would give a countof heads at least this far from 5000 (that is, at least5067 heads or no more than 4933 heads).


    can anyone enlighten me and tell me what the question is asking???? would really appreciate it.. thank you
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    Re: AP STATISTICS Random Variables

     P\left(X\leq4933\right) + P\left(X \geq 5067\right)

    Note: By symmetry, P\left(X\leq4933\right) = P\left(X \geq 5067\right). However, I'll proceed with the individual calculations.


     {\tiny P\left(X\leq 4933\right) = P\left(Z\leq\frac{4933-\mu}{\sigma}\right) = P\left(Z\leq\frac{4933-np}{\sqrt{npq}}\right) = P\left(Z\leq\frac{4933-10000\cdot0.5}{\sqrt{10000\cdot0.5^2}}\right) =  P(Z\leq-1.34)=1-P(Z\leq1.34) \approx1-0.9099 = 0.0901}


     {\tiny P\left(X\geq 5067\right) = 1 - P\left(Z\leq\frac{5067-\mu}{\sigma}\right) = 1-P\left(Z\leq\frac{5067-np}{\sqrt{npq}}\right) = 1-P\left(Z\leq\frac{5067-10000\cdot0.5}{\sqrt{10000\cdot0.5^2}}\right) =  1-P(Z\leq1.34) \approx 1-0.9099 = 0.0901}

    Therefore,  P\left(X\leq4933\right) + P\left(X \geq 5067\right) \approx 2(0.0901)=0.1802 .
    Last edited by abender; Dec 14th 2012 at 10:45 PM.
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