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Math Help - Life Contingencies

  1. #1
    Junior Member plm2e's Avatar
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    Life Contingencies

    please see attached word doc for problem, I could not get all the symbols to properly display on the thread.

    there are 2 problems, 1st uses the survival function and the second is using de moivre's law. any help would be appreciated
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  2. #2
    Flow Master
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    Quote Originally Posted by plm2e View Post
    please see attached word doc for problem, I could not get all the symbols to properly display on the thread.



    there are 2 problems, 1st uses the survival function and the second is using de moivre's law. any help would be appreciated
    Many members do not open attached files of any sort - and with good reason.

    It would be easier for everyone if you just typed your questions rather than attaching them as a word file. That way they can be easily seen and quoted.

    If you want to help, it's in your own self-interest to make things as easy as possible for people.
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  3. #3
    Junior Member plm2e's Avatar
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    ok, sorry about that. I was trying to make it look neat.


    You are given that
    s(x) = 1 - .01x^2 for 0 ≤ x ≤ 10 and l0 = 1000.
    Find a.) μ(4) b.) e8 (the is supposed to be above the e)




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  4. #4
    Junior Member plm2e's Avatar
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    Quote Originally Posted by plm2e View Post
    ok, sorry about that. I was trying to make it look neat.


    You are given that
    s(x) = 1 - .01x^2 for 0 ≤ x ≤ 10 and l0 = 1000.
    Find a.) μ(4) b.) e8 (the is supposed to be above the e)





    i think i figured it out if anyone can check my work.

    a.
    μ(4) = -s'(x)/s(x) = .08/.84 = .0952

    b.
    e8 = E[T(8)] = (from 0 to infin) tp8 dt
    = [1-.01(8+t)^2] / [1-.01(8^2)]
    = .36 1-.64-.16t-.01t^2 dt =
    = [.36t-.08t^2-(.01/3)t^3]l (from 0 to 2) *[(.36)]
    = (.72-.32-.27)*.36 = .1344
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  5. #5
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    a. I'd prefer the prettier 2/21.

    b. Does that make sense? Curtate survival to age 9 is s(9)/s(8) = 0.528
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