1. Density function

Given: Suppose that a R.V Y has a probability density function by:

f(y) = k * (y^3) * e^(-y/2) when y > 0

f(y) = 0 elsewhere

a) Find the value of k that makes f(y) a density function:

b) What are the mean and standard deviation of Y?

Sol.

For part a) I started out the same way as in previous similar problem.
I integrate f(y)=k*(y^3) * e^(-y/2) from 0 to infinity and make it equal to 1. I get this:

-2 k (48 + x (24 + x (6 + x)))
------------------------------
e^x/2

If we solve it from 0 to infinity it comes out to be 0... I checked the integral with a computer program, it looks right. I wonder maybe it should be from 1 to infinity? Then it is possible to find k.

2. I think I found the problem, when we integrate from 0 to infinity we get
2*k*48 = 1
so k= 1/96,

3. Originally Posted by somestudent2
Given: Suppose that a R.V Y has a probability density function by:

f(y) = k * (y^3) * e^(-y/2) when y > 0

f(y) = 0 elsewhere

a) Find the value of k that makes f(y) a density function:

b) What are the mean and standard deviation of Y?

Sol.

For part a) I started out the same way as in previous similar problem.
I integrate f(y)=k*(y^3) * e^(-y/2) from 0 to infinity and make it equal to 1. I get this:

-2 k (48 + x (24 + x (6 + x)))
------------------------------
e^x/2

If we solve it from 0 to infinity it comes out to be 0... I checked the integral with a computer program, it looks right. I wonder maybe it should be from 1 to infinity? Then it is possible to find k.
The expression evaluates to -96k at x=0, and 0 at infinity, so your integral is 96k.

RonL

4. cool, thank you sir. I am happy that at least some of my answers match up.