Finding the probability of <25 of a Gamma Distribution

Question

I'm given a random variable X that follows a Gamma Distribution with mean 20 and standard deviation 10. I already figured out that $\therefore \theta = 5, \alpha = 4$.

I am trying to find the $P(X \leq 25)$.

So far I have done $P(X \leq 25) = 1 - P(X > 25)= $\begin{equation*} 1 - \int_{25}^{\infty} \frac{1}{\Gamma(\alpha)\theta^\alpha}x^{\alpha-1}e^{\frac{-x}{\theta}} \end{equation*}

However after this I am stuck on what steps to solve the integral.

Answer 1

Well, recalling that $\Gamma(\alpha) = (\alpha-1)!$ when $\alpha = 1,2,3,\dots$ we have \begin{align} P(X>25) &= \frac{1}{(3!)5^4} \int_{25}^\infty x^3 e^{-x/5} dx \\ &= \frac{1}{(3!)5^4} \int_0^\infty 5^3(t+5)^3e^{-(t+5)} (5dt) \qquad (t+5=x/5) \\ &= \frac{1}{e^5(3!)} \Big( \int_0^\infty t^3e^{-t}dt + 15 \int_0^\infty t^2e^{-t}dt + 75 \int_0^\infty te^{-t}dt + 125 \int_0^\infty e^{-t}dt \Big) \\ &= \frac{1}{6e^5} \big( 3! + 15(2!) + 75(1!) + 125(0!) \big) \\ &= \frac{118}{3e^5} \end{align}

Answer 2

hint

The integral you need to solve can be reduced to $\frac{1}{6}\int_5^\infty u^3 e^{-u}du$ which can be approached by integration by parts.

Alternatively, a Gamma with $\alpha=4$ and $\theta=5$ the waiting time distribution for the fourth event for a Poisson process with rate $\lambda = 1/5,$ so it's the same as the probability this process has four or more events within time $t=25.$ And the number of events within time $t=25$ is Poisson with mean $25\cdot \frac{1}{5} = 5.$