In a certain city, the daily consumption of electrical energy, in millions of kilowatts per hour, can be considered as a random variable with gamma distribution of parameters  = 3 and  = 0.5 The power plant of this city has a daily capacity of 10 million KW/hour

What is the probability that this supply is:
to. …insufficient on any given day?
b. …between 3 and 8 million KW/Hour is consumed? C. find E(x) and V(x)​

1 answer

To solve this problem, we need to use the gamma distribution formulas and apply them to the given parameters.

a. The probability of the supply being insufficient on any given day can be found by calculating the cumulative distribution function (CDF) of the gamma distribution. The CDF is given by the formula:
CDF(x) = 1 - Γ(α, λ * x)

where Γ(α, λ * x) is the lower incomplete gamma function.

In this case, the daily capacity is 10 million KW/hour, so we are interested in finding the probability of the supply being less than or equal to 10.

Using the given parameters (α = 3, λ = 0.5), we can calculate the CDF as follows:
CDF(10) = 1 - Γ(3, 0.5 * 10)

Using a calculator or statistical software, we can find that Γ(3, 5) ≈ 0.546

Therefore, CDF(10) = 1 - 0.546 ≈ 0.454

So the probability of the supply being insufficient on any given day is approximately 0.454.

b. To find the probability that between 3 and 8 million KW/hour is consumed, we need to calculate the difference in CDF values at these two points. Using the same formula as above, we can find:

CDF(8) = 1 - Γ(3, 0.5 * 8)
CDF(3) = 1 - Γ(3, 0.5 * 3)

Calculating these values using a calculator or statistical software, we find:
CDF(8) ≈ 0.958
CDF(3) ≈ 0.317

Therefore, the probability of the daily consumption being between 3 and 8 million KW/hour is approximately 0.958 - 0.317 ≈ 0.641.

c. The expected value (mean) of a gamma distribution is given by the formula:
E(X) = α / λ

In this case, α = 3 and λ = 0.5, so we can calculate:
E(X) = 3 / 0.5 = 6

Therefore, the expected daily consumption of electrical energy is 6 million KW/hour.

The variance of a gamma distribution is given by the formula:
V(X) = α / (λ^2)

In this case, α = 3 and λ = 0.5, so we can calculate:
V(X) = 3 / (0.5^2) = 12

Therefore, the variance of the daily consumption of electrical energy is 12 million KW/hour.