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? 


1
Expert's answer
2022-02-22T11:52:55-0500

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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