I've gotten a weird answer after doing the problem but I'm stuck as to where I messed up.(adsbygoogle = window.adsbygoogle || []).push({});

The density function is this:

[tex]f_{X} (x) = \frac{1}{6}x[/tex] for [tex]0<x\leq2[/tex]

[tex] = \frac{1}{3}(2x-3)[/tex] for 2<x<3

and 0 otherwise

And the question is to find the distribution function.

So integrating for the first part from 0 to x:

[tex]\int \frac{1}{6}u du = \frac {1}{6} \frac{x^2}{2} = \frac{x^2}{16}[/tex]

for 0<x<=2

I have a big problem with the second, part. This is what I did (integrating from 2 to x):

[tex]\int \frac{1}{3} (2u-3) du = \frac{1}{3} [u^2-3u] = \frac{1}{3} (x^2-3x - (2^2-6)) = \frac{x^2}{3} - x + \frac{2}{3}[/tex]

for 2<x<3

I know my answer for the second part is wrong as when x=2, the distribution function = 0 and when x=3, the distribution function = 2/3. But the distribution shouldn't be broken up like that at x=2 and supposedly at x=3, it should = 1. So did I forget to do something to the end points or did I not integrate properly?

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# Homework Help: Determining the distribution function

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