Find Probability of 3 Samples from Chi-Square Distribution Exceeding 7.779

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SUMMARY

The discussion focuses on calculating the probability of at most 3 samples exceeding 7.779 from a chi-square distribution with 4 degrees of freedom. The participant identifies that the cumulative distribution function (CDF) value for 7.779 is approximately 0.9, but the expected answer is 0.9444. The discrepancy arises from the need to apply the binomial distribution to account for the "at most 3" condition, which requires further analysis of the probabilities of 0, 1, 2, and 3 samples exceeding the threshold.

PREREQUISITES
  • Understanding of chi-square distribution and its properties
  • Familiarity with cumulative distribution functions (CDF)
  • Knowledge of binomial probability calculations
  • Basic statistical concepts related to hypothesis testing
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  • Learn how to calculate cumulative distribution functions for chi-square distributions
  • Explore binomial probability formulas and their applications in statistical problems
  • Review examples of probability calculations involving multiple independent samples
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Students in statistics, data analysts, and anyone involved in statistical modeling or hypothesis testing who needs to understand probability calculations related to the chi-square distribution.

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Homework Statement



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

Homework Equations


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

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