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Chi Squared

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data

    If 15 Observations are taken independently from a chi-square distribution with 4 degrees of freedom, find the probability that at most 3 of the 15 sample items exceed 7.779



    2. Relevant equations



    3. The attempt at a solution

    This problem should be quite simple. I find that in the back of my text that the chi square distribution of 7.779 is .9... but the answer is .9444... I am at a loss at why it is not exactly .9 and where to go from here. I would appreciate any help. Thanks.
     
    Last edited: Nov 14, 2007
  2. jcsd
  3. Nov 16, 2007 #2

    EnumaElish

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    Have you taken into consideration that the problem asks for "at most 3"?
     
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