Twiglet
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#1
Report Thread starter 13 years ago
#1
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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ssmoose
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#2
Report 13 years ago
#2
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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Kolya
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#3
Report 13 years ago
#3
Is this question from the S2 book?
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Twiglet
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#4
Report Thread starter 13 years ago
#4
hmmm the answer I have says for 1<x<2, 13/45 + 2x^4/45
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ssmoose
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#5
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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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Twiglet
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#6
Report Thread starter 13 years ago
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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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Twiglet
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#7
Report Thread starter 13 years ago
#7
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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