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Math Help - Chi-square distribution

  1. #1
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    Chi-square distribution

    If 15 observations are taken independently from a chi-square distribution with 15 degrees of freedom, find the probability that at least two of the sample items exceed 8.547.

    Can anyone help with this?
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  2. #2
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    Let X be a random variable with distribution \chi^2_{15}. A success is when X>8.547. Now repeat 15 times (each trial is independent), denoting the i^{th} trial as X_i.

    \mathrm{P}(X_i>8.547) (i.e. the probabilities of success) are constant with respect to i. In other words, you have 15 independent trials, success is constant ( p=\mathrm{P}(X>8.547)), and you are asked what is the probability of at least 2 successes. This smells like a binomial distribution if you let N be the number of successes.

    You use the \chi^2_{15} distribution only to calculate p.

    p=\mathrm{P}(X>8.547)=0.10 (from \chi^2 table)

    \mathrm{P}(N>=2)=1-\mathrm{P}(N\leq 1)=1-\mathrm{P}(N= 0)-\mathrm{P}(N= 1)=\ldots

    Make sense?
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  3. #3
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    Yea a little bit, I see if I can figure it out and if not, I can show you what I've done. Then we can figure it out.
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