I'm trying to solve this question:
Let T be a random variable with Gamma(r=7, LAMBDA) distribution, where r is the shape parameter and LAMBDA the rate parameter. What is P(T > E[T])?
I'm trying to plug this into a gamma distribution graph calculator online, but I'm not sure what exactly to put in the place of lambda since it's not given? Also, what exactly is E(T) in this case?
Refer to the Wikipedia article for the gamma distribution. Specifically, look at the second column of the table on the right-hand side, which shows that the probability density function for shape $\alpha$ and rate $\beta$ is given by
$$f_X(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}, \quad x > 0.$$ Under this parametrization, the mean or expected value is $$\operatorname{E}[X] = \alpha/\beta.$$
Rewriting this in terms of your variables, $X = T$, $\alpha = r = 7$, $\beta = \lambda$,
$$F_T(t) = \frac{\lambda^7}{6!} t^6 e^{-\lambda t}, \quad t > 0, \\ \operatorname{E}[T] = 7/\lambda.$$ You want to compute $$\Pr[T > 7/\lambda].$$ You can do this either by writing the integral directly: $$\Pr[T > 7/\lambda] = \int_{t = 7/\lambda}^\infty \frac{\lambda^7}{6!} t^6 e^{-\lambda t} \, dt, \tag{1}$$ or you can define a new, transformed random variable, say $W = \lambda T$, whose density is given by $$f_W(w) = \frac{1}{\lambda} f_T(w/\lambda) = \frac{1}{\lambda} \cdot \frac{\lambda^7}{6!} (w/\lambda)^6 e^{- \lambda (w/\lambda)} = e^{-w} \frac{w^6}{6!}.$$ And now you can see that $$\Pr[T > 7/\lambda] = \Pr[W > 7] = \int_{w=7}^\infty e^{-w} \frac{w^6}{6!} \, dw, \tag{2}$$ and the $\lambda$ drops out entirely. You could also have seen this by performing the variable substitution $w = \lambda t$, $dw = \lambda \, dt$ in Equation $(1)$. This is why $\lambda$ is not given, because the answer doesn't depend on it.
I leave the computation of the integral as an exercise.