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Math Help - [SOLVED] Probability of General Intervals, Life expectancy

  1. #1
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    [SOLVED] Probability of General Intervals, Life expectancy

    Example: Life Expectancy
    Suppose we believe the probability of dying before the age of
    t0 is defined by:

    P(0 \leq t < t_{0}) = \int_0^{t_{0}} \alpha(t)dt

    with,


    \alpha(t)  = \left\{ \begin{array}{cc}<br />
        At^2 (100 - t)^2, &\mbox{if}\ 0 \leq t < 100\\<br />
        0 & \mbox{if}\ t \geq 100\end{array} \right.


    The condition,

    \int_0^\infty \alpha(t) dt = 1\  requires\  A = 3 * 10^{-9}

    Exercise: Verify this!

    What is the probability of someone dying between 60 and 70
    Answer
    P(60 \leq t < 70) = \int_{60}^{70} (3 * 10^{-9})t^2(100-t)^2 dt = 0.154

    Exercise: Verify this.

    I cannot seem to get these answers correctly. Any help on how to do it?
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  2. #2
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    Ok, I have figured these out. Thanks anyway.
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