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