Parameterization (scale form): \(X \sim \text{Gamma}(\alpha,\theta)\) with PDF \[ f(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\,\theta^{\alpha}},\quad x>0 \] Mean \(=\alpha\theta,\ \text{Var}= \alpha\theta^2\).
Excel uses beta = θ as the scale parameter.
Scenario: Calls follow a Poisson process with rate \(\lambda = 12\) per hour \(=\;0.2\) per minute. Let \(T_3\) = waiting time (minutes) for the third call. Then \(T_3 \sim \text{Gamma}(\alpha=3,\ \lambda=0.2)\) (Erlang) or equivalently \(T_3 \sim \text{Gamma}(\alpha=3,\ \theta=5)\).
Formula:
\[ F_{T_3}(t)=1-e^{-\lambda t}\left(1+\lambda t+\frac{(\lambda t)^2}{2!}\right) \]
Substitute: \(t=10,\ \lambda t=2\):
\[ F_{T_3}(10)=1-e^{-2}(1+2+2^2/2)=1-e^{-2}\cdot 5 \approx 0.3233 \] Answer: \(P(T_3\le 10)=0.3233\)
Excel: =GAMMA.DIST(10,3,5,TRUE) → 0.3233
\[ F_{T_3}(20)=1-e^{-4}(1+4+4^2/2)=1-e^{-4}\cdot 13\approx 0.7619 \] \[ P(T_3>20)=1-0.7619=0.2381 \] Answer: \(P(T_3>20)=0.2381\)
Excel: =1 - GAMMA.DIST(20,3,5,TRUE) → 0.2381
Moments: \(E[T_3]=15\) min, \(\text{SD}=8.66\) min.
If arrivals are Poisson(\(\lambda\)):
P(Tk ≤ t) = P(N(t) ≥ k)
\[ P(T_k \le t)=1-\sum_{j=0}^{k-1}\frac{(\lambda t)^j e^{-\lambda t}}{j!} \]
Excel: =1 - POISSON.DIST(k-1, λ*t, TRUE)
Scenario: A pump’s lifetime \(X\) (hours) follows \(X \sim \text{Gamma}(\alpha=2.5,\ \theta=1000)\).
\[ P(X>2000)=1-F_X(2000) \] Answer: \(P(X>2000)=0.5494\)
Excel: =1 - GAMMA.DIST(2000, 2.5, 1000, TRUE) → 0.5494
Excel uses beta = θ (scale). If given rate λ, use θ=1/λ.