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Math Help - Normal approximation question?

  1. #1
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    Normal approximation question?

    A number is randomly selected from a box with numbers -2, -2, -2, 0, 2, 2 and the number obtained is the amoun of dollar you win or lose

    E(x) = -1/3 and Stdv = 1.795

    Then after you play 30 games, what is the chance that you will be in will be losing money?
    answer is0.8454
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  2. #2
    Senior Member Sambit's Avatar
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    Thumbs up

    let,  X_i = the amount of money I win (or lose) in the  ith turn. so, after 30 turns, I lose money only if  X_1 + X_2 + X_3 + ...... + X_{30} < 0 ie,  \sum_{i=1}^{30} X_i < 0 , the probability of which is given by  P(\sum_{i=1}^{30} X_i < 0) = P(\frac{(\sum_{i=1}^{30} X_i) - (30 * -1/3)}{1.795*\sqrt{30}} < \frac{0 - (30 * -1/3)}{1.795*\sqrt{30}}) = \Phi(\frac{0 - (30 * -1/3)}{1.795*\sqrt{30}}) = \Phi(1.01713) = 0.845454

    hence the answer.

    note that, if X_i \sim N(\mu , \sigma^2) , then \sum_{i=1}^{30} X_i \sim N(30\mu , 30\sigma^2) , where X_i's are i.i.d. variables.
    then, by standardization, we get the result.
    Last edited by Sambit; October 22nd 2010 at 12:45 AM.
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