Answer to Question #132461 in Statistics and Probability for MANDALAPU SANJANA

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

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

To describe $P(X \leq k)$ we use 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.

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

Although $\Gamma(x)$ for integer numbers can be easily calculated, ${\displaystyle \gamma (3 ,6)}$ has to be found from tables or calculated numerically (e.g. here: https://www.wolframalpha.com/input/?i=Gamma%5B3%2C0%2C6%5D%2FGamma%5B3%5D).

 $P(X>12) = 1 - P(X \leq 12) = 1 - 0.938 =0.062$

Answer: $P(X>12) = 0.062$