the Weibull distribution
/ VY-bull /
When does a machine part fail? Early on from manufacturing defects, randomly throughout its life, or later as it wears out? The exponential assumes the risk never changes; reality usually has the risk rising or falling with age. The Weibull distribution is the flexible lifetime model that lets the failure risk grow, shrink, or stay flat — it is the standard tool of reliability engineering and survival analysis.
Write T ~ Weibull(k, lambda), with a shape parameter k > 0 and a scale lambda > 0. Its power lies in the hazard rate — the instantaneous risk of failing next, given survival so far — which is proportional to t^(k-1). So k < 1 gives a DECREASING hazard (infant mortality: weak units die early, survivors are sturdier), k = 1 gives a CONSTANT hazard and reduces exactly to the exponential, and k > 1 gives an INCREASING hazard (wear-out: the longer it has run, the more likely it is to break). The scale lambda sets the typical lifespan. Its survival function is the clean P(T > t) = exp(-(t/lambda)^k).
Because one knob (k) tunes the whole aging story, the Weibull fits an enormous range of lifetimes: bearings, light bulbs, electronic components, even wind-speed records and material strength. It generalizes the exponential (k = 1) and is the natural upgrade whenever the memoryless assumption is too strong. The fair caveat: it is a phenomenological fit, not a law derived from physics — a good Weibull fit summarizes the failure pattern, it does not by itself explain the underlying mechanism.
Ball bearings tend to fail by wear-out, so their lifetime might be Weibull(k = 2, lambda = 1000 hours). Here k > 1 means the hazard rises over time — a 900-hour-old bearing is at greater risk in the next hour than a fresh one, the opposite of memorylessness.
Shape k tunes the hazard: falling (k<1), flat (k=1, exponential), or rising (k>1).
The Weibull contains the exponential as the special case k = 1; only at k = 1 is it memoryless — for any other k the failure risk changes with age.