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Central Limit PROBLEM

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  • Central Limit PROBLEM

    ok, my question pretains mainly to 123 problemsb 80 and 81.

    Problem 80: Charity receives 2025 contributions. Contributions are ind. and dist. with mean 3125 and std dev 250

    Problem 81: Claims filed under auto insurance are normal with mean 19400 and std dev 5000.

    In the solution for problem 80, the mean and variance were both multiplied by n and then plugged into the P(Z<X)=.9

    In the solution to problem 81, just the variance is divided by n, buyt divided, not multiplied.

    what was the difference in the problems. (I.E. why in problem 80 did we not divide by n for teh variance)

    Thanks,

    Chris

  • #2
    Originally posted by EXAMER
    ok, my question pretains mainly to 123 problemsb 80 and 81.

    Problem 80: Charity receives 2025 contributions. Contributions are ind. and dist. with mean 3125 and std dev 250

    Problem 81: Claims filed under auto insurance are normal with mean 19400 and std dev 5000.

    In the solution for problem 80, the mean and variance were both multiplied by n and then plugged into the P(Z<X)=.9

    In the solution to problem 81, just the variance is divided by n, buyt divided, not multiplied.

    what was the difference in the problems. (I.E. why in problem 80 did we not divide by n for teh variance)

    Thanks,

    Chris
    An alternate way to look at 81 is to say:-

    Pr(claim > 20000) = 1 - Pr( Z< ((20000.25 - 19400.25) / sqrt(25).5000))

    If you run that through your calculator you will get the correct answer, using the same method as in 80.

    Hope this helps.
    Proper prior planning prevents poor performance!

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    • #3
      perhaps i can answer my own question...

      is it that in problem 80, it says that the contributions are "identically distributed" meaning that the mean and standard deviation are for each one. therefore, if you wanted to get the total mean and total std deviation for the sum of all independant contributions, then you have to add them all up? Since there is independance, there is no covariance, so the variances just add up like the expected values.

      in problem 81, all claims contribute to the distribution with mean and standard deviation. therefore, if we want a sample of 25, then we need to divide the variance by n as outlined in the central limit theorem when finding the probability of a sample?

      Chris

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      • #4
        #80 uses the C.L.T. to approximate the SUM of a sample of iid rv's, #81 uses the C.L.T. to approximate the MEAN of a sample of iid rv's.
        Last edited by bmathew22; February 19 2006, 11:56 PM.

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