# Thread: the probability that this power supply will be inadequate on any given day

1. ## the probability that this power supply will be inadequate on any given day

In a certain city, the daily consumption of electric power in millions of kilowatt-hours can be treated as a random variable having a gamma distribution with α=3 and β=b. If the power plant of this city has a daily capacity of 12 million kilowatt-hours, what is the probability that this power supply will be inadequate on any given day?

2. Hello,
Originally Posted by Yan
In a certain city, the daily consumption of electric power in millions of kilowatt-hours can be treated as a random variable having a gamma distribution with α=3 and β=b. If the power plant of this city has a daily capacity of 12 million kilowatt-hours, what is the probability that this power supply will be inadequate on any given day?
Let X be the rv representing the daily consumption (in millions of kw-h).

So you are looking for $\mathbb{P}(X>12)=1-\mathbb{P}(X\leqslant 12)$

The pdf of a gamma distribution with parameters $(\alpha,\beta)$ is:
$g(x,\alpha,\beta)=x^{\alpha-1} \cdot \frac{\beta^\alpha e^{-\beta x}}{\Gamma(\alpha)}$

So here, it's $g(x)=\frac{b^3}{\Gamma(3)} \cdot x^2 e^{-bx}$

$\mathbb{P}(X\leqslant 12)=\int_0^{12} g(x) ~dx=\frac{b^3}{\Gamma(3)} \int_0^{12}x^2 e^{-bx} ~dx$

make twice an integration by parts and you'll be done.

3. sorry! i made a mistake on the question. the β is equal to 2 not b. β=2.
and can you explain why i have to make twice an integration by parts? thanks!!!

4. is the equation should be

The pdf of a gamma distribution with parameters $(\alpha,\beta)$ is:
$g(x,\alpha,\beta)=x^{\alpha-1} \cdot \frac{\beta^{-\alpha} e^{-x/\beta }}{\Gamma(\alpha)}$?

5. Originally Posted by Yan
is the equation should be

The pdf of a gamma distribution with parameters $(\alpha,\beta)$ is:
$g(x,\alpha,\beta)=x^{\alpha-1} \cdot \frac{\beta^{-\alpha} e^{-x/\beta }}{\Gamma(\alpha)}$?
I don't know... the notations are made this way in the wikipedia.

The pdf I gave you is if we're working on $(\alpha,\beta)$. The pdf you gave is if we're working on $(k,\theta)$

You can see that $\alpha=k$ and $\beta=\frac 1 \theta$

So here, it would depend on how you have been taught

As for your mistake of b=2, it doesn't change anything to the result, since b was a constant, and that you can substitute it by any positive value you want

And for the double integration by parts, it's because we have $x^2 e^{-2x}$ in the integrand. We can't calculate an antiderivative of it. So make an integration by parts : the power x^2 will be transformed into x and the exponential remains. But we still cannot find an antiderivative. So integrate again by parts : the power x will be transformed to a constant and the exponential remains. But then you can compute an antiderivative, because the integrand will be an exponential.
Just do it and you'll see (each time, take the u part as x^2 or x and the dv part as the exponential)

6. LOL
I was about to warn you.
IN SOME books $\beta$ is in the numerator
and in other books it's in the denominator.
I was going ask the person who submitted this thread to clarify.
By just saying $\beta$ is such and such it's not clear.