Math Help Forum: Weibull distribution

Having trouble on this question:
The random variable Y with a density function given by
f(y)=[[(m)y^(m-1)]/(a)]*e^[(-y^(m))/(a)] where y is between 0 and infinity and a,m are greater than zero
is said to have a Weibull distribution. Find the mean and variance for a Weibull distributed random variable with m=2.

I can find the answers on the Internet somewhere, but how do you actually derive them?

Thanks for any help.
Jack

OK. I found the mean, but could someone please help me with the variance? I'm stuck.

Thanks in anticipation.

Originally Posted by JackStolerman

Having trouble on this question:
The random variable Y with a density function given by
f(y)=[[(m)y^(m-1)]/(a)]*e^[(-y^(m))/(a)] where y is between 0 and infinity and a,m are greater than zero
is said to have a Weibull distribution. Find the mean and variance for a Weibull distributed random variable with m=2.

I can find the answers on the Internet somewhere, but how do you actually derive them?

Thanks for any help.
Jack

You have density:



where and

Then the mean is:



which you say you can do.

The variance is given by:



Now the integral on the right in this last equation should be do-able
using the same technique you used to evaluate the mean.

RonL