Question

In a city, the daily consumption of Electric power in millions of ‘kwh’ can be considered as a random variable following Gamma Distribution with α= 3, λ= ½. If the power plant in that city has a daily consumption of 12 million KWs, what is the probability that this supply of power is ‘insufficient’ in any given day?[ Hint: Find P(X>12)]

Answer 1

We can find P(X>12) in a way

$P(X>12) = 1 - P(X \leq 12)$ .

By definition, $P(X\leq x)$ $=F(x)$ - cumulative distribution function(CDF). For gamma distribution CDF is

$P(X\leq x) = F(x; \alpha, \lambda) = \frac{\gamma(\alpha, \lambda x)}{\Gamma(\alpha)}$

where where ${\displaystyle \gamma (\alpha ,\lambda x)}$ is the lower incomplete gamma function and $\Gamma(\alpha)$ is ordinary gamma function.

$P(X\leq12) = \frac{\gamma(3, 6)}{\Gamma(3)} = 0.938.$

So, $P(X>12) = 1 -0.938 = 0.062$