How to solve for probability given density function

[eric.mercer92]

Homework Statement

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?

Homework Equations

E(y) = ∫yf(y)dy
Var(y) = ∫(y-E(y))f(y)dy

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!

 

[4570562]

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

 

[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.

 

[Curious3141]

Homework Statement

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?

Homework Equations

E(y) = ∫yf(y)dy
Var(y) = ∫(y-E(y))f(y)dy

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!

Homework Statement

Homework Equations

The Attempt at a Solution

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.

 

[eric.mercer92]

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

 

[Curious3141]

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

No problemo.