1. ## cumulative distribution

Hi ^^

This question asks me to compute the cdf of Y from this pdf :
f(y) { y/25 for 0 =< y < 5 ; (2/5) - (y/25) for 5 =< y =< 10 ; 0 for otherwise }

After I integrate the function (because F'(x) = f(x)), I got:

F(y) { 0 for y<0 ; y^2/50 for 0 =< y < 5 ; (2y/5) - y^2/50 for 5=<y=<10 ; 1 for y => 10}
this cdf is weird because if i plug 10 in, like (2(10)/5) - (10)^2/50) i got 2 instead of 1 . I don't understand, would someone like to explain to me thank you very much ^^

2. You need to go back to the definition of the cdf, which is $P(Y\le y)$

If y<0, then F(y)=0.

If 0<y<5, then you are correct, $P(Y\le y)=\int_{-\infty}^y f(t)dt=\int_0^y tdt/25=y^2/50$

HOWEVER, and it's good you check your work...

If 5<y<10, then $P(Y\le y)=\int_{-\infty}^y f(t)dt=\int_0^5 tdt/25 +\int_5^y \biggl({2\over 5}-{t\over 25}\biggr) dt$ =.5 + whatever.

And, for y>10, you better get 1, which I did, since this is a valid denisty.
That was the first thing I did, to see if it integrated to one.
Too often students write down incorrect 'densities' and I spend forever on the problem.

3. From what I can make out from the question...

$\int_0^5 \frac{y}{25} dy = \frac{5^2}{50} - \frac{0^2}{50}$

and

$\int_5^{10} \frac{2}{5}-\frac{y}{25} dy = [\frac{2(10)}{5}-\frac{10^2}{50}] - [\frac{2(5)}{5}-\frac{5^2}{50}]$

should sum to 1.

4. Mr Pickles, it is a valid density, but he's misunderstanding what a CDF is.

5. i see my mistake thank you very much ^^