Question #299753

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? 


Expert's answer

i)i)

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

E(x)=αβE(x)=\alpha\beta and var(x)=αβ2var(x)=\alpha\beta^2.

Now,

αβ=6\alpha\beta=6 and αβ2=12    (αβ)β=12\alpha\beta^2=12\implies (\alpha\beta)\beta=12

Since αβ=6    6β=12    β=2\alpha\beta=6\implies 6\beta=12\implies \beta=2

αβ=6\alpha\beta =6 but β=2    2α=6    α=3\beta=2\implies 2\alpha=6\implies \alpha=3

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

ii)ii)

For this part, the distribution for the n=30n=30 independent days will be a Gamma distribution with parameters (i=130αi,β)(\displaystyle\sum^{30}_{i=1}\alpha_i, \beta ). That is, Gamma(i=130αi,β)=Gamma((60),2)Gamma (\displaystyle\sum^{30}_{i=1}\alpha_i, \beta )=Gamma((60),2) . Its mean is 60×2=12060\times 2=120.

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


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