# Thread: PDF to CDF question

1. ## PDF to CDF question

Hi I'm having problems with part c of this question

and I want to define fully the CDF. I know I have to intergrate the functions in turn

so the first bit between 0 and 2 is $\frac{1}{2}t^{2}$

and I begin doing the next part between 2 and 6 $\frac{1}{12}6t \frac{1}{2}t^{2}-10$ but in the mark scheme it's

$\frac{1}{24}6t -\frac{1}{2}t^{2}-12$

do you have to do somthing with the other limits first? I've missed a small step somwhere I think :\

Thanks

2. Originally Posted by thomas49th
Hi I'm having problems with part c of this question

and I want to define fully the CDF. I know I have to intergrate the functions in turn

so the first bit between 0 and 2 is $\frac{1}{2}t^{2}$

and I begin doing the next part between 2 and 6 $\frac{1}{12}6t \frac{1}{2}t^{2}-10$ but in the mark scheme it's

$\frac{1}{24}6t -\frac{1}{2}t^{2}-12$

do you have to do somthing with the other limits first? I've missed a small step somwhere I think :\

Thanks
Sorry but nothing you've posted is correct. Not even the answer from the marking scheme.

For x < 0: F(x) = 0.

For $0 \leq x \leq 2$: $F(x) = \int_0^x \frac{t}{6} \, dt = \frac{x^2}{12}$.

For $2 \leq x \leq 6$: $F(x) = \int_0^2 \frac{t}{6} \, dt + \int_2^x \frac{6 - t}{12} \, dt = \frac{1}{3} + \frac{12x - x^2 - 20}{24} = \frac{12x - x^2 - 12}{24}$.

Note that $F(6) = 1$.

For x > 6: F(x) = 1.

3. Of course! Thanks