# Gamma Distribution Help!!

• Oct 30th 2009, 09:20 AM
statman101
Gamma Distribution Help!!
Suppose the reaction time X of a randomly selected individual to a certain stimulus is a standard gamma distribution with (alpha=2)

What is F(5;2) using gamma

show steps i dont understand!
• Oct 30th 2009, 09:39 AM
theodds
I'm guessing by standard gamma they mean the scale parameter (i.e. beta) is equal to 1. Since alpha = 2, we can just write down the density $\displaystyle f_X (x) = \frac{1}{\Gamma(2)} x^{2 - 1}e^{-x} = xe^{-x}$. You can solve for F(x) in the usual way using integration by parts. If you haven't had calculus, or are just being blindsided by this (since this is the pre-university forum), just post so and I'll give a different explanation (at some point, I have to go to class).
• Oct 30th 2009, 10:36 AM
statman101
you put 2 in for x, but didnt put 2 in for e^-x
• Oct 30th 2009, 11:17 AM
theodds
Nah, I put in 2 for alpha everywhere. Or are you talking about the exponential not having a parameter? That is where beta would normally be, but since it's 1, it's just $\displaystyle e^{-x}$. Just to be sure we're clear, the density for a Gamma(alpha, beta) is $\displaystyle f(x) = \frac{1}{\Gamma(\alpha) \beta^{\alpha}} x^{\alpha - 1} e^{\frac{-x}{\beta}}$. Plug 1 for beta and 2 for alpha and you get the density I posted.
• Oct 30th 2009, 11:24 AM
statman101
can u show in steps the function evaluated at F(5;2)

its supposed to equal
.960
• Oct 30th 2009, 11:43 AM
theodds
To get F(5), use the fact that $\displaystyle F(x) = \int_0 ^{x} f(t)dt$. I did this and got .960. After integration by parts you get $\displaystyle F(x) = 1 - e^{-x}(x + 1)$