How to solve for probability given density function

1. Feb 4, 2012

eric.mercer92

1. The problem statement, all variables and given/known data
The probability density function of a random variable y is:
f(y) = 100ye$^{-10y}$, if y>0
f(y) = 0 otherwise

What is the probability that 45y <= 10?

2. Relevant equations
E(y) = ∫yf(y)dy
Var(y) = ∫(y-E(y))f(y)dy

3. The attempt at a solution
I solved for the expected value of y and got E(y) = 0.2
I then got Var(y) = 0.02
I don't think this is a normal distribution because the PDF is not a normal
PDF. I don't know how to solve for the probability of 45y <= 10

Any help will be greatly appreciated!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 4, 2012

4570562

Not quite sure, but maybe you do ∫f(y) dy from 0 to 10/45?

3. Feb 4, 2012

eric.mercer92

I thought about that, but it just didn't seem right, after all, if one of the y's is over 10/45, the other y's can be less than that to make up for it.

4. Feb 4, 2012

Curious3141

You shouldn't be worrying about what sort of distribution this represents. You're already given a pdf. Compute the cdf (cumulative distribution function). If the pdf is $f(y)$, the cdf is $\int_{-\infty}^y f(y)dy$. Remember that from -∞ to 0, f(y) = 0, so break it up into two integrals, one from -∞ to 0 (which vanishes), and the other from 0 to y.

Do the integration by parts and work out the definite integral in terms of y. Then it's as simple as substituting y = 10/45 into that.

The answer I get lies between 0.6 and 0.7, if you wish to check yours.

5. Feb 4, 2012

eric.mercer92

Ahh yes, I see, thank you very much. I got the same answer.
Thanks again!

6. Feb 4, 2012

No problemo.