1. ## probability density function updated!

find the p(y<3.2) of the probability density function below

$\displaystyle f(x)= \frac{1}{8}(y+1)$ for 2<y<4

0 elsewhere

tried to solve the problem above by $\displaystyle \int\frac{1}{8}(y+1)$ at 0<y<3.2 as boundary, but got a value greater then 1 which i think is not possible. how should i

2. ## Re: probability density function!

Looks like you have not posted the complete question. It says 0 elsewhere, implying there was more information given somewhere.

3. ## Re: probability density function!

Originally Posted by lawochekel
find the p(y<3.2) of the probability density function below

$\displaystyle f(x)= \frac{1}{8}(y+1)$

0 elsewhere

tried to solve the problem above by $\displaystyle \int\frac{1}{8}(y+1)$ at 0<y<3.2 as boundary, but got a value greater then 1 which i think is not possible. how should i
That is not a probability function as written. You forgot to give the domain.
Elsewhere, WHAT?

4. ## Re: probability density function updated!

Originally Posted by lawochekel
find the p(y<3.2) of the probability density function below

$\displaystyle f(x)= \frac{1}{8}(y+1)$ for 2<y<4

0 elsewhere
You edited the OP.
$\displaystyle \int_2^{3.2} {f(y)dy} = ?$