Question 1

What does the shape parameter k determine in the Weibull distribution?

Accepted Answer

k < 1: Hazard rate decreases over time (infant mortality / burn-in failures — items that survive early tend to last longer). k = 1: Constant hazard rate — reduces to the Exponential distribution. k = 2: Hazard rate increases linearly (Rayleigh distribution — good for metal fatigue). k = 3.44: Closely approximates the normal distribution. k > 1: Increasing hazard rate (wear-out failures — older items fail faster). k is the most important parameter; λ just scales the time axis.

Question 2

What is the hazard rate and why does it matter?

Accepted Answer

The hazard rate h(x) = f(x) / [1 − F(x)] represents the instantaneous failure rate given survival to time x. For Weibull: h(x) = (k/λ)(x/λ)^(k−1). For k > 1, h(x) increases with x — the longer an item has been running, the more likely it is to fail in the next instant. This is realistic for mechanical wear. For k = 1, h(x) = 1/λ is constant — the memoryless exponential. Plotting h(x) vs time gives the "bathtub curve" for complex systems with multiple failure modes.

Question 3

What are the mean, median, and mode of the Weibull distribution?

Accepted Answer

Mean = λ × Γ(1 + 1/k), where Γ is the gamma function. Median = λ × (ln 2)^(1/k) ≈ λ × 0.693^(1/k). Mode = λ × ((k−1)/k)^(1/k) for k > 1; Mode = 0 for k ≤ 1. Variance = λ² × [Γ(1 + 2/k) − Γ²(1 + 1/k)]. For k = 1 (exponential): Mean = Median = λ (roughly, since 0.693^1 × λ ≈ 0.693λ for median). The characteristic life λ is the value at which ~63.2% of items have failed: F(λ) = 1 − e⁻¹ ≈ 0.632.

Question 4

How do I fit a Weibull distribution to my failure data?

Accepted Answer

The most common method is Weibull probability plotting (Weibull plot). Take the CDF: F(x) = 1 − e^(−(x/λ)^k). Then ln[−ln(1−F)] = k×ln(x) − k×ln(λ). Plotting ln[−ln(1−F)] vs ln(x) should give a straight line with slope k and intercept −k×ln(λ). Maximum likelihood estimation (MLE) is the more statistically rigorous method for fitting k and λ to censored or uncensored failure data.

Question 5

What is the characteristic life λ in reliability terms?

Accepted Answer

The scale parameter λ is the characteristic life or the scale life — the time at which 63.2% of units are expected to have failed (regardless of k). This follows from F(λ) = 1 − e^(−(λ/λ)^k) = 1 − e⁻¹ ≈ 0.632. It is not the mean life (MTTF = Mean Time To Failure) unless k = 1 (exponential). MTTF = λ × Γ(1 + 1/k), which equals λ only when k = 1.

Question 6

When should I use Weibull vs exponential vs log-normal for reliability?

Accepted Answer

Exponential (k=1): random, memoryless failures — electronic components in steady state. Weibull (k<1): early-life failures — burn-in, defects. Weibull (k>1): wear-out failures — mechanical components, bearings, tyres. Log-normal: slow degradation processes — fatigue crack growth, corrosion, biological processes. The log-normal has heavier tails than Weibull and is often better for data with high variability. Use AIC/BIC or probability plots to compare fits.

k value	Hazard rate	Failure pattern	Typical cause
k < 1	Decreasing	Infant mortality / early failures	Manufacturing defects, burn-in
k = 1	Constant	Random failures (exponential)	Electronic components, accidents
k = 2	Linearly increasing	Wear-out (Rayleigh)	Metal fatigue, linear degradation
k = 3.44	Steep increase	Approximates normal distribution	Mechanical aging, many causes
k > 4	Rapid increase	Sudden wear-out failures	Brittle fracture, corrosion

Weibull Distribution Calculator

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Shape Parameter k — Failure Patterns

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