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Math Help - PDF to CDF question

  1. #1
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    PDF to CDF question

    Hi I'm having problems with part c of this question



    and I want to define fully the CDF. I know I have to intergrate the functions in turn

    so the first bit between 0 and 2 is \frac{1}{2}t^{2}

    and I begin doing the next part between 2 and 6 \frac{1}{12}6t \frac{1}{2}t^{2}-10 but in the mark scheme it's

    \frac{1}{24}6t -\frac{1}{2}t^{2}-12

    do you have to do somthing with the other limits first? I've missed a small step somwhere I think :\

    Thanks
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  2. #2
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    Quote Originally Posted by thomas49th View Post
    Hi I'm having problems with part c of this question



    and I want to define fully the CDF. I know I have to intergrate the functions in turn

    so the first bit between 0 and 2 is \frac{1}{2}t^{2}

    and I begin doing the next part between 2 and 6 \frac{1}{12}6t \frac{1}{2}t^{2}-10 but in the mark scheme it's

    \frac{1}{24}6t -\frac{1}{2}t^{2}-12

    do you have to do somthing with the other limits first? I've missed a small step somwhere I think :\

    Thanks
    Sorry but nothing you've posted is correct. Not even the answer from the marking scheme.

    For x < 0: F(x) = 0.

    For 0 \leq x \leq 2: F(x) = \int_0^x \frac{t}{6} \, dt = \frac{x^2}{12}.

    For 2 \leq x \leq 6: F(x) = \int_0^2 \frac{t}{6} \, dt + \int_2^x \frac{6 - t}{12} \, dt = \frac{1}{3} + \frac{12x - x^2 - 20}{24} = \frac{12x - x^2 - 12}{24}.

    Note that F(6) = 1.

    For x > 6: F(x) = 1.
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  3. #3
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    Of course! Thanks
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