Finding the probability of $X>3$ of a Gamma

Question

I'm given that X follows a gamma distribution and$E(X) = 8 \text{ and Var}(X)=32$

So what I tried to do was $$k\theta=8 \text{ and }k\theta^2=32 \text{ so }\theta=4,k=2$$So now I did $$\int_3^{\infty}{1\over16\Gamma(2)}xe^{-x\over4}={1\over 4}\int_3^{\infty}{x\over4}e^{-x\over 4}={1\over4}e^{-3\over 4}$$ but this wasn't correct. What am I missing?

Answer 1

Show your work for the integral -- You performed it wrong (it should come out to be $\frac{7}{4} e^{-3/4}$).

Answer 2

Given the Gamma distribution:

$$f(x) = \frac{x^{a-1} e^{-\frac{x}{b}}}{b^a \Gamma (a)}$$

where $E[X]=8$ and $V[X]=32$, then clearly $a=2$ and $b=4$.

To find $P(X > 3)$, we do the following:

$$\int_3^{\infty } \frac{x^{a-1} e^{-\frac{x}{b}}}{b^a \Gamma (a)} \, dx=\frac{\left(\frac{1}{b}\right)^{-a} b^{-a} \Gamma \left(a,\frac{3}{b}\right)}{\Gamma (a)}$$.

For $a=2$ and $b=4$, we find

$$P(X > 3)=\Gamma \left(2,\frac{3}{4}\right)=0.826641$$