How to solve for probability given density function

eric.mercer92
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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!
 
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Not quite sure, but maybe you do ∫f(y) dy from 0 to 10/45?
 
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.
 
eric.mercer92 said:

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.
 
Ahh yes, I see, thank you very much. I got the same answer.
Thanks again!
 
eric.mercer92 said:
Ahh yes, I see, thank you very much. I got the same answer.
Thanks again!

No problemo. :smile:
 
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