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Math Help - Chebychev inequality/Probability question

  1. #1
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    Chebychev inequality/Probability question

    A box contains 100 tickets labeled with numbers. The average of the labels is -59.7 and the SD of the labels is 1.9. Fourteen tickets will be drawn independently at random with replacement from the box. The chance that the absolute value of the difference between the sample sum of the labels and -835.8 exceeds 44.79 is at most _______?

    Chebychev's inequality states that the "fraction of elements in the list that are k or more SDs away from the mean of the list is less than or equal to 1/k^2." It also says that the probability that "X is k or more SEs (standard error) away from E(X) (expected value) is less than or equal to 1/k^2."
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    Quote Originally Posted by steve123 View Post
    A box contains 100 tickets labeled with numbers. The average of the labels is -59.7 and the SD of the labels is 1.9. Fourteen tickets will be drawn independently at random with replacement from the box. The chance that the absolute value of the difference between the sample sum of the labels and -835.8 exceeds 44.79 is at most _______?

    Chebychev's inequality states that the "fraction of elements in the list that are k or more SDs away from the mean of the list is less than or equal to 1/k^2." It also says that the probability that "X is k or more SEs (standard error) away from E(X) (expected value) is less than or equal to 1/k^2."
    The mean of the sum of fourteen labels is 14 \times (-59.7)=835.8 and the standard deviation of the sum is \sqrt{14} \times 1.9\approx 7.11.

    So 44.79 is 6.3 SD's, so the required bound is 1/6.3^2 \approx 2.5 \%

    CB
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