Gamma distribution and probability less then expected value? 
Let $X\sim \operatorname{Gamma}(\alpha = 7, \beta)$, then $P(X > E(X))$ is:

A) 0.35
B) 0.45 
C) 0.55 
D) 0.65
The answer is 0.45.

This is what I have so far: 

$E(X)=\alpha\beta$ 
so I want $P(X > \alpha \beta)$, $\alpha=7$ so I can use the table for this if I divide by beta, but I don't have beta's original value. From the table I'm deducting that $X=7$..

My main question is, when I divide by $\beta$, am I also dividing the mean and variance by Beta? if that's the actual case I'm guessing $$P(X > E(X))= P(X > 7)$$ which then would make sense since $\alpha$ and $x$ are both 7. So is this the general rule when dividing by beta to make it equal to 1 to use the tables?

Thank you!!! sorry for the rather lengthy question
 A: The value of $\beta$ doesn't matter at all since $\beta$ is a scale parameter, so altering $\beta$ merely multiplies both sides of the inequality $X> \operatorname{E}(X)$ by the same number. Thus for simplicity we can assume $\beta=1.$
The probability distribution is then
$$
\frac 1 {\Gamma(\alpha)} x^{\alpha-1} e^{-x} \, dx \text{ for } x\ge 0.
$$
The expected value is $\alpha.$ So
\begin{align}
\Pr(X>\operatorname{E}(X)) = {} & \Pr(X>\alpha) = \int_\alpha^\infty \frac 1 {\Gamma(\alpha)} x^{\alpha-1} e^{-x}\,dx = \frac 1 {720} \int_7^\infty x^6 \Big(e^{-x} \, dx\Big) \\[10pt]
= {} & \frac 1 {720} \int u\,dv = \frac 1 {720} \left( uv - \int v\,du \right) \\[10pt]
& \text{where } u = x^6 \text{ and } dv = e^{-x} \, dx.
\end{align}
When you do this integration by parts, you will get an expression involving
$$
\int_7^\infty 6x^5 e^{-x} \, dx
$$
and you will have to integrate by parts again, and iterate that until the power of $x$ is $0$.
Another way to look at it is this: Let $X$ be the waiting time until the $7$th arrival in a Poisson process. Then $X$ has this distribution. But the following events are the same:
$$
\Big[ X > 7 \Big] \quad \text{and} \quad \Big[ \text{fewer than 7 arrivals before time 7} \Big]
$$
The number of arrivals before time $7$ has a Poisson distribution with expected value $7$, so the probability is
$$
\Pr(\text{fewer than 7 arrivals}) = \sum_{n=0}^6 \frac{7^n e^{-7}}{n!}.
$$
