Answer to Question #299753 in Statistics and Probability for nih

Question

In a certain city, the daily consumption of electric power, X (in millions of kilowatt hours) is a random variable where X ~ Gamma (α, β) with a mean of 6 and a variance of 12. i) Find α and β. ii) Given a month has 30 days, how many kilowatt hours should we expect the city to consume during the month? 

Answer 1

$i)$

The formula for the mean and variance of the Gamma distribution are given as,

$E(x)=\alpha\beta$ and $var(x)=\alpha\beta^2$.

Now,

$\alpha\beta=6$ and $\alpha\beta^2=12\implies (\alpha\beta)\beta=12$

Since $\alpha\beta=6\implies 6\beta=12\implies \beta=2$

$\alpha\beta =6$ but $\beta=2\implies 2\alpha=6\implies \alpha=3$

Therefore, the values of $\alpha$ and $\beta$ are 3 and 2 respectively.

$ii)$

For this part, the distribution for the $n=30$ independent days will be a Gamma distribution with parameters $(\displaystyle\sum^{30}_{i=1}\alpha_i, \beta )$. That is, $Gamma (\displaystyle\sum^{30}_{i=1}\alpha_i, \beta )=Gamma((60),2)$ . Its mean is $60\times 2=120$.

Therefore, we expect that 120 Kilowatt hours will be consumed by the city during this month .