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Math Help - Normal approximation of a die.

  1. #1
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    Normal approximation of a die.

    Final the exact probability of rolling 2 or 3 sixes when rolling a normal six-sided die 12 times. What is the normal approximation for this situation?
    Find the exact probability of rolling 14 - 17 sixes when rolling a normal six-sided die 100 times. What is the normal approximation for this situation? Compare the two cases.
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  2. #2
    MHF Contributor matheagle's Avatar
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    These are binomial probabilities.
    The p=1/6 and in the first case n=12 and has a lousy approximation via the central limit theorem since n is small.

    The answer to the first one is

     {12\choose 2} \biggl({1\over 6}\biggr)^2 \biggl({5\over 6}\biggr)^{10} +{12\choose 3} \biggl({1\over 6}\biggr)^3 \biggl({5\over 6}\biggr)^9.

    The second is

     \sum_{k=14}^{17}{100\choose k} \biggl({1\over 6}\biggr)^k \biggl({5\over 6}\biggr)^{100-k}

    where I am assuming that you're including the 14 and 17.
    And instead of a normal die you mean a fair die.
    You can approximate this via the CLT, the mean is 100/6 and the variance is 100(1/6)(5/6).
    The normal superimposed over this binomial is  P(13.5<X< 17.5).
    Now subtract the mean and divide by the st deviation and use your st normal table.
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