SOLUTION: At a construction site, the daily requirement of gneiss (in metric tons) is a random variable having a gamma distribution with α = 2 and β = 5. If their supplier’s daily supp

Question 1177278: At a construction site, the daily requirement of gneiss
(in metric tons) is a random variable having a gamma distribution with α = 2 and β = 5. If their supplier’s
daily supply capacity is 25 metric tons, what is the
probability that this capacity will be inadequate on
any given day?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**Understanding Gamma Distribution**
* A random variable X has a gamma distribution with parameters α and β if its probability density function (pdf) is:
* f(x) = (1 / (β^α * Γ(α))) * x^(α-1) * e^(-x/β) for x > 0
* In this case, α = 2 and β = 5.
* The expected value is E(X) = αβ = 2 * 5 = 10.
* The variance is Var(X) = αβ² = 2 * 5² = 50.
**The Problem**
We need to find the probability that the daily requirement (X) exceeds the supplier's capacity (25 metric tons). In other words, we need to find P(X > 25).
**Calculating P(X > 25)**
* P(X > 25) = ∫[25, ∞] f(x) dx
* P(X > 25) = ∫[25, ∞] (1 / (5² * Γ(2))) * x^(2-1) * e^(-x/5) dx
* P(X > 25) = ∫[25, ∞] (1 / 25) * x * e^(-x/5) dx
Since Γ(2) = 1! = 1.
Now, we need to evaluate the integral:
* P(X > 25) = (1/25) * ∫[25, ∞] x * e^(-x/5) dx
We'll use integration by parts:
* Let u = x, dv = e^(-x/5) dx
* Then du = dx, v = -5e^(-x/5)
* ∫ x * e^(-x/5) dx = -5xe^(-x/5) - ∫ -5e^(-x/5) dx
* ∫ x * e^(-x/5) dx = -5xe^(-x/5) + 5 ∫ e^(-x/5) dx
* ∫ x * e^(-x/5) dx = -5xe^(-x/5) - 25e^(-x/5)
Now, we evaluate the definite integral:
* ∫[25, ∞] x * e^(-x/5) dx = [-5xe^(-x/5) - 25e^(-x/5)] from 25 to ∞
As x approaches infinity, e^(-x/5) approaches 0, so the upper limit is 0.
* = 0 - [-5(25)e^(-25/5) - 25e^(-25/5)]
* = 125e^(-5) + 25e^(-5)
* = 150e^(-5)
Now, plug this back into the probability equation:
* P(X > 25) = (1/25) * 150e^(-5)
* P(X > 25) = 6e^(-5)
* P(X > 25) ≈ 6 * 0.006737947
* P(X > 25) ≈ 0.040427682
**Therefore, the probability that the capacity will be inadequate on any given day is approximately 0.0404.**

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