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Math Help - Integration to find pdf/cdf

  1. #1
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    Integration to find pdf/cdf

    I have a pdf:

    0 x<=4
    kx 4<x<=9
    0 x>9

    I found k=2/65

    So I thought the cdf would be (x^2)/32 but apparently not

    Can someone please help me with the cdf, thanks
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  2. #2
    Moo
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    P(I'm here)=1/3, P(I'm there)=t+1/3
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    Hello,

    For t<4, the cdf is 0.

    For t>9, the cdf is 1.

    For t\in[4,9], the cdf is \int_4^t kx ~dx

    Compute that again
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  3. #3
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    ok thanks i've got that now. So for a cdf where

    0<x<4 & 4<x<8 have different functions, what limits would i use?
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  4. #4
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    would the limits just be 0 - t and 4 - t?
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  5. #5
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    can someone please help me as I am almost sure that the limits for 0 to 4 and 4 to 8 would be either 0 to t and 4 to t or 0 to t and t to 8 but I dont know which 1, thanks
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  6. #6
    MHF Contributor matheagle's Avatar
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    from 0 to 4, there is no probability
    I don't know why you keep writing 8, but the probabilities exist between 4 and 9.

    Hence F(x)=0 for any x less than 4, since f(t)=0 in that region..........

    F(x)=\int_{-\infty}^xf(t)dt=\int_{-\infty}^x0dt=0

    If 4<x<9, then F(x)=\int_{-\infty}^xf(t)dt=\int_{-\infty}^40dt+\int_4^x ktdt

    and after 9, we have

    F(x)=\int_{-\infty}^xf(t)dt=\int_{-\infty}^40dt+\int_4^9 ktdt+\int_9^x 0dt=0+1+0=1
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