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    Hi, Im a bit stuck, I have to find the CDF for this:

    f(x) 1/3 0<= x <1
    8x^3 /45 1<= x <= 2
    0 otherwise

    I integrated these but then didnt get the right answer at all (I only have part of the answer written down which is a slight problem!)

    thanks for your help

    franks xxx
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    F(x)=
    0 for x<0
    x/3 for 0<x<1
    1/3 + 2x^4/45 for 1<x<2
    1 for x>2
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    Is this question from the S2 book?
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    hmmm the answer I have says for 1<x<2, 13/45 + 2x^4/45
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    I think your answers may be wrong as INT(0 and 1){1/3dx} = x/3 = 1/3
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    I think your answers may be wrong as INT(0 and 1){1/3dx} = x/3 = 1/3
    I thought you're meant to use the imits with x0 ...?
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    Ah, I think I've sorted it out, could someone check this please:

    Integral of 1/3 limits x0 and 0 = 1/3x
    Integral of 8/45 x^3 limits x0 and 1 = 2/45x^4 - 2/45 and add 1/3 as it is cumulative which = 13/45 + 2x^4/45

    so CDF is:
    F(x)=
    0 for x<0
    x/3 for 0<x<1
    13/45 + 2x^4/45 for 1<x<2
    1 for x>2
 
 
 
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Updated: April 15, 2006
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