Explanation
The Weibull distribution is a positive continuous distribution widely used in reliability analysis,
maintenance engineering and lifetime modeling of components. Its versatility comes from the shape parameter k:
if k < 1 the failure rate decreases over time (early failures, typical of manufacturing defects);
if k = 1 the rate is constant and the distribution coincides with the exponential (random failures);
if k > 1 the failure rate increases over time (wear-out, aging).
The parameter λ is the characteristic life: the value of x at which the survival function equals e⁻¹ ≈ 36.8%.
Its mean is λ·Γ(1 + 1/k) and its variance is λ²·[Γ(1 + 2/k) − Γ²(1 + 1/k)].
Formula
\( f(x;k,\lambda)=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{-(x/\lambda)^k},\quad x\ge 0 \)
Parameters
- k: shape parameter (k > 0). Controls the type of failure rate: decreasing (k < 1), constant (k = 1) or increasing (k > 1).
- λ (lambda): scale parameter or characteristic life (λ > 0).
- x: lifetime or time to failure (x ≥ 0).
Worked example
Situation: The lifetime of LED bulbs is modeled with a Weibull distribution with shape parameter \(k = 2\) (increasing wear over time) and scale parameter \(\lambda = 5\) (in thousands of hours). We want to assess the reliability of the bulbs.
Question 1: What is the probability that a bulb fails before 4000 hours, \(P(X \leq 4)\)?
Solution: We apply the Weibull CDF:
\[ P(X \leq 4) = 1 - e^{-(x/\lambda)^k} = 1 - e^{-(4/5)^2} = 1 - e^{-0.64} \approx 1 - 0.5273 \approx 0.4727 \]
There is approximately a 47.27% probability that the bulb fails before 4000 hours.
Question 2: What is the expected mean lifetime of the bulbs?
Solution: The mean of the Weibull(\(k, \lambda\)) is:
\[ E[X] = \lambda \cdot \Gamma\!\left(1 + \frac{1}{k}\right) = 5 \cdot \Gamma(1.5) = 5 \times 0.8862 \approx 4.431 \text{ thousand hours} \]
The expected mean lifetime is approximately 4431 hours.
Interpretation: With \(k = 2 > 1\), the failure rate is increasing, indicating progressive wear. The 63.2nd percentile coincides exactly with \(\lambda = 5\) thousand hours: 63.2% of the bulbs will have failed before 5000 hours. This allows planning preventive replacements before that characteristic life to maximize system availability.
How to interpret the result
The calculator offers three types of output. The PDF, \( f(x) \), gives the probability density at the point \( x > 0 \): it indicates the relative concentration of failures or events at that instant. With \( k < 1 \) the curve decreases from the start (higher density early on, early failures); with \( k = 1 \) it is the exponential; with \( k > 1 \) a peak appears (the mode increases with \( k \)) and the curve tends to be more symmetric. In the chart, the green area represents the cumulative probability for the selected time range; the horizontal axis shows time or duration.
The CDF, \( P(X \leq x) = 1 - e^{-(x/\lambda)^k} \), also called the cumulative failure distribution function in reliability engineering, gives the probability that the component has failed before time \( x \). Its complement, \( R(x) = e^{-(x/\lambda)^k} \), is the reliability function: the probability that the component is still operating at instant \( x \). A value \( P(X \leq 5000) = 0.23 \) means that 23% of the components will have failed before 5000 hours of operation.
The percentile or quantile result returns the time \( x = \lambda \cdot (-\ln(1-p))^{1/k} \). The 10th percentile (called the B10 life) is the time before which 10% of components fail: it is the most common standard in mechanical engineering for warranties and preventive maintenance planning. The 63.2nd percentile always coincides with the scale parameter \( \lambda \), regardless of \( k \), because \( P(X \leq \lambda) = 1 - e^{-1} \approx 0.632 \).
Frequently asked questions
- What does the shape parameter indicate? The type of failure rate: below 1 means decreasing early-life failures, equal to 1 a constant rate (equivalent to the exponential), and above 1 wear-out failures that increase with age.
- How does it relate to the exponential? The exponential is the special case with shape k = 1. The Weibull generalises it by letting the failure risk depend on elapsed time, something the exponential cannot represent.
- Where is it commonly applied? It is the reference distribution in reliability and lifetime analysis: service life of bearings, engines or electronics, material strength, and wind speed in wind energy.
Reference: Weibull distribution — Wikipedia