PDF to CDF question

**thomas49th** · May 25th 2009, 11:56 AM

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

**mr fantastic** · May 25th 2009, 07:52 PM

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.

**thomas49th** · May 26th 2009, 02:42 AM

Of course! Thanks

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