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Math Help - Normal approximation: Method

  1. #1
    Yan
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    Normal approximation: Method

    a fair coin is tossed 10,000 times. Find a number m such that the chance of the number of heads being between 5000-m and 5000+m is approximately 2/3.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Yan View Post
    a fair coin is tossed 10,000 times. Find a number m such that the chance of the number of heads being between 5000-m and 5000+m is approximately 2/3.
    In 10000 tosses of a fair coin the mean number of heads is \mu=5000, and the standard deviation of the number of heads is \sigma=\sqrt{10000\times 0.25}

    Also the number of heads is approximatly normally distributed \sim N(\mu,\sigma^2).

    Now a symetrical interval that contains 2/3 of a normal distribution is \mu \pm 0.965 \sigma (you get this from looking up a tail probability of 0.16667 backwards so to speak in a table of the standard normal distribution).

    Now finish this yourself.

    CB
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