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Suppose a random variable has a gamma distribution with $\alpha = 0.8$ and $\beta = 2.4$

How can we calculate $P(Y > 3)$?

My book says when $d$ and $c$ are such that $0 < c < d < \infty$ the integral of the gamma pdf cannot be directly integrated:

$$\int_{c}^{d}\frac{{y^{\alpha-1}e^{-y/\beta}}}{\beta^\alpha\Gamma(\alpha)}dy $$

But for $P(Y > 3)$ would I want to calculate the following? $$1-\int_{0}^{3}\frac{{y^{-.2}e^{-y/2.4}}}{2.4^.8\Gamma(.8)}dy$$ Since in this case $0=c<d<\infty$. Does that make things easier?

Also, for $\Gamma(.8)$, wolfram gives me $1.16$ but my TI-84 gives me $0$? Does anyone know how to do it correctly on the calculator?

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    $\begingroup$ Correct, the survival function is given by the second formula ($1 - \int_0^3$). It's the incomplete gamma function. I suppose you'll have to implement your own numerical routine on the calculator for $\Gamma(x), \, 2 x \not \in \mathbb N$. $\endgroup$ – Maxim Apr 2 at 19:15

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