- 2073 Chaitra · 5 marks
Gamma: water supply inadequate
Short answer
Daily consumption X ~ Gamma(α=2, β=3) million gallons with pdf f(x)=(x/9)e^(−x/3). Supply is inadequate if X>9. P(X>9)=∫_9^∞ f(x) dx = 4e^(−3) ≈ 0.1991 (19.91%).
The daily consumption of water in a certain place follow a gamma distribution with parameters α = 2 and β = 3. If the daily capacity of this city is 9 million gallon of water, what is the probability that on any given day the water supply is inadequate?
The daily consumption X (million gallons) follows a gamma distribution with shape α = 2 and scale β = 3, whose pdf is
f(x) = x^(α−1) e^(−x/β) / (β^α Γ(α)) = x e^(−x/3) / 9, x > 0
(Mean = αβ = 6 million gallons.) The supply is inadequate when consumption exceeds the capacity of 9 million gallons.
P(X > 9) = ∫₉^∞ (x/9) e^(−x/3) dx
Integrate by parts: ∫ x e^(−x/3) dx = −3x e^(−x/3) − 9 e^(−x/3)
∫₉^∞ x e^(−x/3) dx = [−3x e^(−x/3) − 9 e^(−x/3)] from 9 to ∞
= 0 − (−27 e^(−3) − 9 e^(−3))
= 36 e^(−3)
P(X > 9) = 36 e^(−3) / 9 = 4 e^(−3)
= 4 × 0.04979 = 0.1991
Answer: the probability that the water supply is inadequate on a given day is about 0.1991 (19.91%).
✓ Answer checked by Claude Sonnet 5.5
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