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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- Mar 11th 2010, 01:07 PMjpquinn91Integration 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 - Mar 11th 2010, 01:11 PMMoo
Hello,

For $\displaystyle t<4$, the cdf is 0.

For $\displaystyle t>9$, the cdf is 1.

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

Compute that again :) - Mar 11th 2010, 01:21 PMjpquinn91
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? - Mar 11th 2010, 01:48 PMjpquinn91
would the limits just be 0 - t and 4 - t?

- Mar 12th 2010, 02:49 PMjpquinn91
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

- Mar 12th 2010, 03:50 PMmatheagle
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..........

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

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

and after 9, we have

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