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Math Help - [SOLVED] p.d.f; how do I find the c.d.f?

  1. #1
    Super Member Quacky's Avatar
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    [SOLVED] p.d.f; how do I find the c.d.f?

    I have a p.d.f given by

    f(x)=\frac{3}{32}(4-x^2) for -2\leq x\leq 2
    f(x)=0 otherwise.

    The question asks me to find the c.d.f. In the textbook, it says I do this by integrating the function, which gives me:

    \frac{3}{8}x-\frac{1}{32x^3}
    The answer is \frac{3}{8}x-\frac{1}{32x^3}+\frac{1}{2}

    but I can't spot my mistake - so where has the half come from?
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Quacky View Post
    I have a p.d.f given by

    f(x)=\frac{3}{32}(4-x^2) for -2\leq x\leq 2
    f(x)=0 otherwise.

    The question asks me to find the c.d.f. In the textbook, it says I do this by integrating the function, which gives me:

    \frac{3}{8}x-\frac{1}{32x^3}
    The answer is \frac{3}{8}x-\frac{1}{32x^3}+\frac{1}{2}

    but I can't spot my mistake - so where has the half come from?
    For x < -2, F(x) = 0.

    For -2 \leq x \leq 2, F(x) = \left[ \frac{3}{8} u - \frac{1}{32} u^3 \right]_{u=-2}^{u=x} = ....

    For x > 2, F(x) = 1.
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  3. #3
    Super Member Quacky's Avatar
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    Thankyou for the explanation
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