# Gamma Distribution (Statistics)

A random variable X is gamma distributed with a=3, b=2 (alpha/beta). Determine the probability that 1$$\leq$$X$$\leq$$2

The first thing I did was plug a & b into the gamma distribution formula 1/(b^a $$\Gamma$$a * x^(a-1) * e^(-x/b)

Which I ended up with

1/16(x^2 * e^(-x/2)

Which I then though I'd take the integral from 1 to 2

$$\int$$ 1/16(x^2 * e^(-x/2) dx = 1/16(-2x^2 * e^-.5 - 8x * e^-.5 - 16e^-.5) & then I evaluated that from 1 to 2. Which gave .06...

The answer in the book gives (b-a)^4/80

So where am I going wrong & am I just approaching this problem in totally the wrong way?

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Try this for Gamma density.

$$f(x)= \frac {\beta^\alpha}{\Gamma(\alpha)}x^{\alpha -1} e^{-\beta x}$$

The books answer would be negative ???

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The book answer wouldn't be negative, it'd be ((2-3)^4)/80.

Also tried the formula, it does not work, it gives a probability over 11, which clearly can't be. The only difference between the formula I posted & yours is you have e^-bx & I had e^(-x/b).

exk
I am assuming montoyas7940 looked up the gamma pdf on wikipedia (as I did) and I think that parametrization is correct for your application.

I get:

$\int_{1}^{2} 4x^{2}e^{-2x}dx$

That doesn't evaluate to 1/80.

Try this maybe: P(1<X<2) = P(X<2)+P(X<1)-1

Oh, I see. I assumed an extra set of parentheses in the books answer. I used the gamma density from "Probability for Risk Management".

Unless I fat fingered a key on my calculator,

$\int_{1}^{2} 4x^{2}e^{-2x}dx$

is not .6 either.