Using pdf to find the Probability of a range

rhyno89

1. The problem statement, all variables and given/known data

For a given cdf: Fy(y) = 0 for y less than 0
= y^2 for the range 0 to 1
= 1 for all values greater than or equal to 1

the problem asks to find P(1/2 < Y < 3/4) first using the cdf and then using the pdf



2. Relevant equations



3. The attempt at a solution

Okay so the first part of this i was able to do easily. Since its giving the cdf you just have to plug in the two values of the range and the end result is 5/16

Using the pdf to find the answer is where i ran into trouble

I kno that the relationship b/w the cdf and the pdf is that the cdf is the anti derivative of the pdf so i began by differentiating the values of Y over the range.

This of course gave me 0, 2y and 0 for the different ranges.

Because of this however, the values for either of the two ranges give me a probability greater than 1 and that alone violates the laws around it so that doesnt work even if the answer would have matched up, which it doesnt anyway. So im guessing the problem is how I calculated the pdf...any tips?

Thanks

rock.freak667

Wouldn't it just be

[tex]\int_\frac{1}{2} ^\frac{3}{4} 2y dy [/tex]

I don't think this gives an answer greater than one.

rhyno89

i thought i had to do something like that since it was an interval and not merely a finite set of numbers. This kinda seems like a pointless excercise since this would have me differentiating the cdf to get the pdf just to re integrate it and return it to the cdf plugging in the same numbers that I had to do for part a.
I was also looking through the book and I dont see any other way of getting it so i think ill run with this...thanks for the quick response