Question

In a certain city, the daily consumption of water (in millions of liters) follows approximately a gamma distribution with α = 2 and β = 3. If the daily capacity of that city is 9 million liters of water, what is the probability that on any given day the water supply is inadequate?

Answer 1

Answer:

The probability that on any given day the water supply is inadequate $0.1991$

Step-by-step explanation:

Given

α = 2 and β = 3

As per Gamma distribution Function

$P(X>9)= 1-P(X\leq 9)$

Expanding the function and putting the given values, we get -

$[tex]1 - \int\limits^9_0 {f(x;2,3)} \, dx \\1- \int\limits^0_0 {\frac{1}{9}xe^{\frac{-x}{3} } \, dx\\\\= 1- \frac{1}{9} [x(-3e^{\frac{-x}{3}}) -\int\limits {(-3e^{\frac{-x}{3}})} \, dx]^9_0= 1- 1/9 [x(-3e^{\frac{-x}{3}}) -9e^{\frac{-x}{3}})} \, dx]^9_0\\1-((\frac{-1}{3} *9*e^({\frac{-9}{3}})- e^(\frac{-9}{3}))- ((\frac{-1}{3} *0*e^(\frac{-9}{3})-e^{\frac{-0}{3}})\\1-((-3e^{\frac{-9}{3} }-e^{\frac{-9}{3}}-(0-1))\\1-(1-4e^{-3})\\1-(1-0.1991)\\1-0.8009\\0.1991$

The probability that on any given day the water supply is inadequate $0.1991$