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Gamma: power supply inadequate
Short answer
For Gamma(2,3), inadequate supply is P(X>12). Integrate (x/9)e^(-x/3) from 12 to infinity; result is 5e^-4, about 0.0916.
The daily consumption of electric power in a certain city follow a gamma distribution with α = 2 and β = 3. If the power plant of this city has daily capacity of 12 million kilowatt hours, what is the probability that this power supply will be inadequate on any given day?
Let X = daily power consumption (million kWh), X ~ Gamma with α = 2 and β = 3, with pdf
f(x) = x^(α−1) e^(−x/β) / (βᵅ Γ(α)) = x e^(−x/3) / (9 · 1) = (x/9) e^(−x/3), x > 0 (since Γ(2) = 1, β² = 9).
The supply is inadequate when consumption exceeds the capacity of 12:
P(X > 12) = ∫₁₂^∞ (x/9) e^(−x/3) dx
Put u = x/3, so x = 3u and dx = 3 du. The limits become u = 4 to ∞:
P = ∫₄^∞ (3u/9) e^(−u) · 3 du = ∫₄^∞ u e^(−u) du
Integrating by parts, ∫ u e^(−u) du = −(u + 1) e^(−u):
P = [−(u + 1) e^(−u)]₄^∞ = (4 + 1) e⁻⁴ = 5 × 0.018316 = 0.0916
Answer: the probability that the supply is inadequate is 0.0916 (about 9.2 %).
✓ Answer checked by Claude Sonnet 5.5
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