Statistics › 🧠 Advanced & Bayesian Modeling

Weibull Reliability

One shape knob turns a lifetime model from infant-mortality to constant-risk to wear-out — the bathtub curve, in a single distribution.

AdvancedUpper-undergraduate
💡
The big idea: Real things fail at a rate that changes with age: new gadgets die from hidden defects, mature ones fail at a steady background rate, and old ones wear out. The Weibull distribution captures all three with one dial — the shape parameter k. Turn k below 1 and the failure rate falls with age (infant mortality); set k = 1 and it holds constant (the memoryless exponential); push k above 1 and it climbs (wear-out). That flexibility is why the Weibull is the standard language of reliability and lifetime modeling.
🎯 By the end, you'll be able to
  • Write the Weibull density, reliability (survival) function, and hazard rate, and see how they connect
  • Read the shape parameter k as the trend of the failure rate: falling, flat, or rising with age
  • Explain the bathtub curve as three Weibull regimes stitched together
  • Interpret the scale lambda as the characteristic life where reliability drops to about 37 percent
  • Compute reliability, hazard, and mean life for a given k and lambda
📎 You should already know
  • The exponential distribution and the memoryless property
  • Probability density and cumulative distribution functions
  • Basic calculus (exponents, the exponential function)

A single failure rate is not enough

Suppose you model the lifetime of a machine part with one average failure rate and stop there. You have quietly assumed something strong: that a brand-new part and a ten-year-old part are equally likely to fail in the next hour. For light bulbs that might be roughly true. For almost everything else it is not.

Three patterns show up again and again:

  • Infant mortality. A fresh batch carries hidden defects. The weak units die early, so the failure rate is high at first and then falls as the survivors prove themselves.
  • Random, age-independent failure. A steady background rate — voltage spikes, stray cosmic rays — that does not care how old the unit is.
  • Wear-out. Fatigue, corrosion, and erosion accumulate, so an old unit fails at a rate that rises with age.

The Weibull distribution builds all three from one family, tuned by a single shape parameter.

🔑 The Weibull, in one idea
A Weibull lifetime has two dials. The scale \(\lambda\) sets the timescale — the characteristic life. The shape \(k\) sets the trend of the failure rate: \(k<1\) makes the failure rate fall with age, \(k=1\) holds it constant, and \(k>1\) makes it rise. Change \(k\) and you change the physics of how the thing ages — not just how long it lasts.

Three views of the same distribution

Reliability work looks at a lifetime \(T\ge 0\) through three linked functions:

  • The density \(f(t)\) — how the probability of failing is spread across time.
  • The reliability (or survival) function \(R(t)=\Pr(T>t)\) — the fraction still working at time \(t\). It starts at 1 and slides down to 0.
  • The hazard rate \(h(t)\) — the instantaneous failure rate given survival so far. This is the one that tells the aging story, and it is defined as the density among the survivors, \(h(t)=f(t)/R(t)\).

Read the hazard carefully: it is conditional. A rising hazard means that the longer a unit has lasted, the more precarious its next moment becomes.

\[ f(t) = \frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1}\exp\!\left[-\left(\frac{t}{\lambda}\right)^{k}\right], \qquad t \ge 0 \]
The Weibull density. Shape k > 0 bends the curve; scale lambda > 0 stretches it along the time axis. Setting k = 1 collapses it to the exponential density.
\[ R(t) = \Pr(T>t) = \exp\!\left[-\left(\frac{t}{\lambda}\right)^{k}\right] \]
The reliability (survival) function: the fraction of units still working at time t. Clean and closed-form — one reason engineers reach for the Weibull.

The hazard rate is where k lives

Divide the density by the reliability and the exponential terms cancel, leaving a strikingly simple hazard:

\[ h(t) = \frac{f(t)}{R(t)} = \frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1} \]
The Weibull hazard rate. It is a power of t with exponent k - 1, so the shape parameter alone decides whether the failure rate falls (k<1), stays flat (k=1), or climbs (k>1).

Shape, scale, and the bathtub curve

Look at the exponent \(k-1\) in the hazard and the whole picture falls out:

  • \(k<1\Rightarrow k-1<0\): the hazard is a decreasing function of \(t\) — infant mortality.
  • \(k=1\Rightarrow k-1=0\): the hazard is constant at \(1/\lambda\) — the exponential, the only memoryless lifetime.
  • \(k>1\Rightarrow k-1>0\): the hazard increases with \(t\) — wear-out. (At \(k=2\) it rises in a straight line; this special case is the Rayleigh distribution.)

Stitch these three regimes across a product's life — a falling hazard early, a flat stretch in the middle, a rising hazard at the end — and you get the famous bathtub curve of reliability engineering.

The scale \(\lambda\) plays a separate role: it is the characteristic life. Whatever \(k\) is, plugging \(t=\lambda\) into the reliability gives \(R(\lambda)=e^{-1}\approx 0.368\) — so by the characteristic life about 63% of units have failed. It anchors the timescale without touching the shape.

See k reshape the failure rate

Below, switch the view between the density \(f(t)\) and the hazard \(h(t)\), then drag the shape \(k\). Watch the hazard panel especially: as you slide \(k\) through 1, the failure-rate curve tips from downhill, to flat, to uphill — the three arms of the bathtub, live. The dashed marker sits at the scale \(x=\lambda\); moving \(\lambda\) slides the whole picture along the time axis without changing its shape. The readout reports the mean life \(\lambda\,\Gamma(1+1/k)\).

🎮 Interactive: Weibull density and hazard LIVE
Switch between the density f(t) and the hazard rate h(t), then vary the shape k and scale lambda. The hazard falls for k<1 (infant mortality), is flat for k=1 (exponential), and rises for k>1 (wear-out). The dashed line marks the characteristic life x = lambda.
✨ Two facts worth memorizing
1. k = 1 is the exponential. The Weibull contains the exponential as the single memoryless slice where the hazard never changes with age. Every other \(k\) has memory — the survivors' odds shift over time. 2. The scale is the 63% mark. Because \(R(\lambda)=e^{-1}\approx 0.368\) for every \(k\), the characteristic life \(\lambda\) is always the time by which about 63% of units have failed. It is a far more stable landmark than the mean, which drifts with \(k\).
📝 Worked example: A cooling-fan motor has a time-to-failure that is Weibull with shape k = 2 and scale lambda = 5000 hours. Describe how it ages, find the reliability at 2000 hours, and give the mean life.
  1. Aging pattern: \(k=2>1\), so the exponent \(k-1=1>0\) and the hazard rate rises with age — this is a wear-out component, not a random-failure one.
  2. Reliability at 2000 h: \(R(2000)=\exp[-(2000/5000)^2]=\exp[-(0.4)^2]=\exp(-0.16)\approx 0.852\).
  3. So about 85% of motors are still running at 2000 hours.
  4. Characteristic life: at \(t=\lambda=5000\) h, \(R=e^{-1}\approx 0.368\), so only ~37% remain by then.
  5. Mean life: \(\lambda\,\Gamma(1+1/k)=5000\,\Gamma(1.5)=5000\times\tfrac{\sqrt{\pi}}{2}\approx 5000\times 0.8862\approx 4431\) hours.
✓ A wear-out motor (rising hazard): about 85% survive to 2000 h, about 37% survive to the characteristic life of 5000 h, and the mean life is roughly 4431 hours.

Check your understanding

1. In a Weibull lifetime model, a shape parameter of k = 1 produces…
With k = 1 the hazard exponent k - 1 is 0, so h(t) = 1/lambda is constant. That memoryless, age-independent failure rate is exactly the exponential distribution.
2. Engineers observe a batch of parts whose failure rate DROPS the longer each part has already run (early defects dying out). Which shape fits?
A falling hazard needs a negative exponent k - 1, i.e. k < 1. That is the infant-mortality (early-failure) regime — the left arm of the bathtub curve.
3. A part has a Weibull lifetime with k = 1 and scale lambda = 200 hours. Its hazard rate is…
For k = 1, h(t) = (k/lambda)(t/lambda)^(k-1) = 1/lambda = 1/200 = 0.005 per hour, constant for all t — the memoryless exponential.
4. At the characteristic life t = lambda, the Weibull reliability R(lambda) = exp(-1) is about…
R(lambda) = exp[-(lambda/lambda)^k] = exp(-1) ≈ 0.368 for every k. So by the characteristic life about 37% remain and ~63% have failed.
✅ Key takeaways
  • The Weibull models lifetimes with two dials: scale lambda (the timescale) and shape k (the trend of the failure rate).
  • Density, reliability R(t) = exp[-(t/lambda)^k], and hazard h(t) = (k/lambda)(t/lambda)^(k-1) are three views of the same distribution, linked by h = f/R.
  • The hazard exponent is k - 1: k<1 falls (infant mortality), k=1 is constant (the exponential), k>1 rises (wear-out) — the three arms of the bathtub curve.
  • The scale lambda is the characteristic life: R(lambda) = e^-1 ≈ 0.37 for every k, so ~63% have failed by then.
  • The mean life is lambda times Gamma(1 + 1/k).