📘 Weibull Distribution

🧭 When to Use the Weibull

• Modeling failure times / lifetimes (reliability). Always positive.
• Failure rate (hazard) may be decreasing, constant, or increasing:
  • \(\beta < 1\): hazard decreases with time (infant mortality / early failures).
  • \(\beta = 1\): exponential (constant hazard).
  • \(\beta > 1\): hazard increases with time (wear-out).
• Flexible alternative to exponential/gamma when shape is uncertain but monotone hazard is reasonable.

📐 Definitions & Key Formulas

We write \(X \sim \text{Weibull}(\beta,\delta)\) with shape \(\beta\) and scale \(\delta\) (both > 0).

• PDF: \[ f(x) = \frac{\beta}{\delta}\left(\frac{x}{\delta}\right)^{\beta - 1} \exp\!\left( -\left(\frac{x}{\delta}\right)^\beta \right), \quad x>0 \]
• CDF: \[ F(x) = 1 - \exp\!\left( -\left(\frac{x}{\delta}\right)^\beta \right), \quad x>0 \]
• Survival: \(S(x)=P(X>x)=\exp\!\left(-\left(\frac{x}{\delta}\right)^\beta\right)\)
• Mean: \(\mathbb{E}[X] = \delta\;\Gamma\!\left(1+\frac{1}{\beta}\right)\)
• Variance: \(\mathrm{Var}(X) = \delta^2\!\left[\Gamma\!\left(1+\frac{2}{\beta}\right) - \Gamma\!\left(1+\frac{1}{\beta}\right)^2\right]\)
• Quantile (inverse CDF): For \(0<p<1\), \[ x_p = \delta\,\big[-\ln(1-p)\big]^{1/\beta} \]

Excel parameter mapping: WEIBULL.DIST(x, alpha, beta, cumulative) uses alpha = shape = \(\beta\), beta = scale = \(\delta\). Quantiles: WEIBULL.INV(p, alpha, beta).

🛠️ Worked Example 1 — Tail Probability

Scenario. A bearing lifetime \(X\) (hours) follows \(\text{Weibull}(\beta=1.5,\ \delta=5000)\). Find \(P(X>6000)\) and interpret.

Step 1 — Use survival function

\[ P(X>x)=\exp\!\left(-\left(\frac{x}{\delta}\right)^\beta\right) \]

Step 2 — Substitute numbers

\[ P(X>6000)=\exp\!\left(-\left(\frac{6000}{5000}\right)^{1.5}\right) = \exp\!\big(-1.2^{1.5}\big) \]

\(1.2^{1.5} = 1.2\sqrt{1.2} \approx 1.2\times 1.0954 \approx 1.3145\)

\[ P(X>6000) \approx e^{-1.3145} \approx \color{#1a237e}{\mathbf{0.2686}} \] Answer: \(P(X>6000)\approx 0.269\)

Excel

=1 - WEIBULL.DIST(6000, 1.5, 5000, TRUE) ➜ 0.2686

Interpretation: With these parameters, ~26.9% of bearings last beyond 6000 hours.

🧰 Worked Example 2 — Warranty Percentile

Scenario. Smartphone battery life \(X\) (months) follows \(\text{Weibull}(\beta=2.2,\ \delta=30)\). The company wants a warranty so that only the bottom 5% fail before the warranty end. Find the cutoff \(x_{0.05}\) and also compute the mean lifetime.

Part A — 5th percentile \(x_{0.05}\)

Use the quantile formula \(x_p=\delta\,[-\ln(1-p)]^{1/\beta}\):

\[ x_{0.05} = 30 \times \big[-\ln(1-0.05)\big]^{1/2.2} = 30 \times [ -\ln(0.95) ]^{0.4545} \]

\(-\ln(0.95) \approx 0.051293\) so \([0.051293]^{0.4545} \approx 0.3052\).

\[ x_{0.05} \approx 30 \times 0.3052 \approx \color{#1a237e}{\mathbf{9.16\ \text{months}}} \] Answer: Warranty cutoff ≈ 9.16 months

Excel

=WEIBULL.INV(0.05, 2.2, 30) ➜ 9.16

Part B — Mean lifetime

Mean \(=\delta \Gamma(1+1/\beta)=30\times \Gamma(1+1/2.2)=30\times \Gamma(1.4545)\).

Using a calculator or Excel’s GAMMA, \(\Gamma(1.4545)\approx 0.8850\).

\[ \mathbb{E}[X]\approx 30\times 0.8850 \approx \color{#1a237e}{\mathbf{26.55\ \text{months}}} \] Answer: Mean ≈ 26.6 months

Excel

=30 * GAMMA(1 + 1/2.2) ➜ 26.55

🧾 Excel Quick Reference (Weibull)

• Survival \(P(X>x)\): =1 - WEIBULL.DIST(x, alpha, beta, TRUE)
• CDF \(P(X\le x)\): =WEIBULL.DIST(x, alpha, beta, TRUE)
• PDF \(f(x)\): =WEIBULL.DIST(x, alpha, beta, FALSE)
• Percentile \(x_p\): =WEIBULL.INV(p, alpha, beta)
• Mean: =beta * GAMMA(1 + 1/alpha) (remember: Excel’s beta is the scale \(\delta\))
• Variance: =beta^2 * ( GAMMA(1 + 2/alpha) - GAMMA(1 + 1/alpha)^2 )
• Excel mapping: alpha = shape = \(\beta\), beta = scale = \(\delta\)

🧪 Weibull Explorer (try it yourself)