STAT-235 · Lesson 5 of 12 · 28 min

Distributions for reliability and extremes

Lesson 4 introduced the binomial, Poisson, exponential and normal distributions. They cover a great deal, but several important engineering questions need other models. How long will a bearing, a seal or a battery last, and does its failure rate rise as it ages? How long does a repair take, when most repairs are quick but a few take days? How large a flood, wind gust or snow load should a structure be designed for, when the design value is far beyond anything in the record? This lesson introduces the distributions engineers use for these questions: the Weibull distribution for component life, the lognormal distribution for quantities built up by multiplication, the gamma distribution, and the Gumbel extreme value distribution for annual maxima. Along the way it develops the language of reliability (the reliability function, the hazard rate and the bathtub curve) and shows how probability plots and maximum likelihood fit these distributions to data.

Reliability and the hazard function

For a component whose life T has density f(t) and cumulative distribution F(t), the reliability function is the probability of surviving beyond time t:

\[R(t) = P(T > t) = 1 - F(t)\]

The hazard function, or instantaneous failure rate, is the rate of failure at time t among the items still working:

\[h(t) = \frac{f(t)}{R(t)}\]

The hazard describes how aging affects failure. For the exponential distribution (Lesson 4) it is constant, \(h = \lambda\): an old component is exactly as likely to fail in the next hour as a new one, the "memoryless" property that suits random shocks and many electronic parts. Most mechanical components wear out, so their hazard rises with age; newly manufactured items often show early failures from defects, so their hazard falls at first. Combined over a product's life, these give the bathtub curve: a falling infant-mortality period, a flat useful-life period and a rising wear-out period. Burn-in testing removes the early failures before shipping; preventive replacement removes items before wear-out; and in the flat period, replacing working parts does no good at all.

Left: the bathtub curve of hazard rate against time, falling during infant mortality, flat during useful life, and rising during wear-out. Right: Weibull hazard functions for shape parameters 0.5, 1 and 3: the first decreasing, the second constant and the third increasing.
The bathtub curve and the Weibull hazard for three shape parameters.

The Weibull distribution

The Weibull distribution is the workhorse of reliability engineering because one parameter controls whether the hazard falls, stays constant or rises. With shape parameter β and scale parameter η (the characteristic life):

\[R(t) = e^{-(t/\eta)^{\beta}}, \qquad h(t) = \frac{\beta}{\eta}\left(\frac{t}{\eta}\right)^{\beta - 1}\]
  • β < 1: decreasing hazard (infant mortality, manufacturing defects).
  • β = 1: constant hazard; the Weibull reduces to the exponential distribution.
  • β > 1: increasing hazard (wear-out). Fatigue and bearing failures typically have β between about 1.5 and 4; at β ≈ 3.5 the distribution looks much like a normal one.

At t = η, the reliability is \(e^{-1} = 0.368\) for any β, so η is the life by which 63.2% of items have failed. Two other summaries are widely used: the mean time to failure \(\text{MTTF} = \eta\,\Gamma(1 + 1/\beta)\), using the gamma function, and the B10 life, the age by which 10% have failed, \(t_{10} = \eta(-\ln 0.9)^{1/\beta}\). Bearing manufacturers rate their products by B10 life.

Worked problem

Bearing life

A bearing has a Weibull life with β = 1.5 and η = 10 000 h. Find the probability that it survives 5000 h, its B10 life and its MTTF.

Show solution
  1. \(R(5000) = e^{-(0.5)^{1.5}} = e^{-0.354} = 0.702\).
  2. B10: \(t_{10} = 10\,000 \times (0.1054)^{1/1.5} = 10\,000 \times 0.223 = 2230\) h.
  3. MTTF: \(10\,000 \times \Gamma(1.667) = 10\,000 \times 0.903 = 9030\) h.

Answer: R(5000) = 0.70, B10 ≈ 2230 h and MTTF ≈ 9030 h. Ten percent of bearings fail before a quarter of the mean life, which is why maintenance plans are based on B10 or similar percentiles, not on the mean.

A useful property follows from the formula for R. For a system of n identical components in series (Lesson 3), all of which must work, the system reliability is \(R(t)^n = e^{-n(t/\eta)^\beta}\), again a Weibull distribution with the same β but with η reduced by a factor \(n^{1/\beta}\). This weakest-link behavior is why long chains, long cables and large brittle parts are weaker on average than short ones.

Fitting a Weibull distribution

Taking logarithms twice turns the Weibull distribution function into a straight line:

\[\ln\left[-\ln\left(1 - F(t)\right)\right] = \beta\ln t - \beta\ln\eta\]

So if failure times are sorted, each is given an estimated cumulative probability (the median rank \((i - 0.3)/(n + 0.4)\) for the ith of n), and the points are plotted on these axes, they should fall near a straight line whose slope is β. This Weibull probability plot is both a fitting method and a check of the model: a curved plot suggests a different distribution, a mixture of failure modes or a minimum life. The more efficient fitting method is maximum likelihood: choose the parameters that make the observed data most probable. Software does this directly, and it handles censored data, items still working when the test ended, which are common in life testing and must not simply be discarded.

Fitting a Weibull distribution to 12 seal failure times by maximum likelihood and by a probability plot.
import math
import numpy as np
from scipy import stats

# Hours to failure of 12 pump seals from a life test
t = np.array([1850, 2900, 3400, 4100, 4700, 5300, 5800, 6600, 7200, 8300, 9400, 11800])

# Maximum likelihood fit of a two-parameter Weibull (location fixed at zero)
beta, loc, eta = stats.weibull_min.fit(t, floc=0)
print(f"MLE:   shape beta = {beta:.2f}, scale eta = {eta:.0f} h")

# Median-rank regression on a Weibull probability plot
n = len(t)
F = (np.arange(1, n + 1) - 0.3) / (n + 0.4)            # median ranks
x, y = np.log(t), np.log(-np.log(1 - F))
slope, intercept = np.polyfit(x, y, 1)
print(f"Plot:  shape beta = {slope:.2f}, scale eta = {np.exp(-intercept / slope):.0f} h")

B10 = eta * (-np.log(0.9))**(1 / beta)
mttf = eta * math.gamma(1 + 1 / beta)
print(f"B10 life (10% failed): {B10:.0f} h;  mean time to failure: {mttf:.0f} h")
print(f"Probability a seal survives 3000 h: {np.exp(-(3000 / eta)**beta):.3f}")
Output
MLE:   shape beta = 2.31, scale eta = 6732 h
Plot:  shape beta = 2.14, scale eta = 6768 h
B10 life (10% failed): 2541 h;  mean time to failure: 5964 h
Probability a seal survives 3000 h: 0.857

The two methods agree closely: β ≈ 2.2 to 2.3, indicating wear-out, and η ≈ 6750 h. A seal replacement interval of about 2500 h would keep in-service failures near 10%.

Weibull probability plot with the logarithm of failure time on the horizontal axis and the transformed cumulative probability on the vertical axis. Twelve data points lie close to a straight fitted line with slope about 2.3. Horizontal reference lines mark 10% and 63.2% failed.
Weibull probability plot of the seal failure times with the fitted line.

The lognormal distribution

If ln X is normally distributed, X has a lognormal distribution. It arises when a quantity is the product of many independent factors, just as the normal distribution arises from sums (the central limit theorem, Lesson 4). Lognormal models fit repair times, particle sizes, fatigue lives at a given stress, soil permeability, pollutant concentrations and the sizes of many natural quantities. They are always positive and skewed to the right: most values are moderate, a few are very large.

Calculations are done on the logarithms. If ln X has mean μ and standard deviation σ, the median of X is \(e^{\mu}\), and its mean is \(e^{\mu + \sigma^2/2}\), larger than the median because of the long right tail.

Worked problem

Repair times

Repair times for a machine are lognormal with a median of 2.0 h and σ = 0.8 for ln(time). What fraction of repairs take more than 6 h, and what is the mean repair time?

Show solution
  1. μ = ln 2.0 = 0.693.
  2. \(P(X > 6) = P\left(Z > \dfrac{\ln 6 - 0.693}{0.8}\right) = P(Z > 1.373) = 0.085\).
  3. Mean: \(e^{0.693 + 0.32} = 2.0 \times e^{0.32} = 2.75\) h.

Answer: About 8.5% of repairs take longer than 6 h, and the mean repair time is 2.75 h, well above the typical (median) repair. Planning spare capacity on the median would underestimate total downtime.

The gamma distribution

The time until the kth event of a Poisson process with rate λ (Lesson 4) follows a gamma distribution with shape k and rate λ: it is the sum of k independent exponential waiting times, with mean k/λ. More generally the gamma distribution, with any positive shape, is a flexible model for positive, right-skewed quantities such as rainfall totals, the time to complete a multi-stage task, or the load accumulated by several independent events. For example, if a standby system can absorb three component failures and failures occur at a rate of one per 1000 h, the time to exhausting the spares is gamma with k = 3 and mean 3000 h.

Extreme values: designing for the rare event

Structures must withstand the largest flood, wind gust, snow load or earthquake they are likely to see in their lifetime, and these maxima follow their own distributions. The maximum of many independent values tends toward one of a few extreme value distributions, of which the Gumbel distribution is the most widely used for annual maximum floods, winds and loads:

\[F(x) = \exp\left[-e^{-(x - u)/a}\right], \qquad a = \frac{\sqrt{6}}{\pi}s, \quad u = \bar{x} - 0.5772\,a\]

where the parameters u and a are estimated here from the mean \(\bar{x}\) and standard deviation s of the annual maxima (the method of moments). The T-year return level, the value exceeded on average once in T years (Lesson 3), is the value with \(F(x_T) = 1 - 1/T\):

\[x_T = u - a\ln\left[-\ln\left(1 - \frac{1}{T}\right)\right]\]

Worked problem

The 100-year flood

Forty years of annual maximum flows at a river gauge have mean 300 m³/s and standard deviation 90 m³/s. Using a Gumbel distribution, estimate the 100-year flood.

Show solution
  1. \(a = 0.7797 \times 90 = 70.2\) m³/s and \(u = 300 - 0.5772 \times 70.2 = 259.5\) m³/s.
  2. \(x_{100} = 259.5 - 70.2\ln(-\ln 0.99) = 259.5 - 70.2 \times (-4.600) = 582\) m³/s.

Answer: About 580 m³/s, nearly twice the average annual flood. Extrapolating from 40 years to 100 carries real uncertainty, often ±20% or more, and hydrologists compare several distributions and use regional data before settling on a design flow. Design codes for wind and snow loads are based on similar analyses of long weather records.

Graph of annual maximum flow against return period on a logarithmic axis from 1 to 1000 years. Forty plotted annual maxima follow the fitted Gumbel curve, which reaches about 580 cubic meters per second at 100 years. A dashed band shows the growing uncertainty of the extrapolation beyond the record length.
Gumbel fit to annual maximum flows, extrapolated to the 100-year flood.

Choosing a distribution

Physical reasoning should guide the choice: Weibull for weakest-link failure and wear-out, exponential for random failures with no aging, lognormal for multiplicative processes, gamma for sums of waiting times, Gumbel and related distributions for maxima, and normal for sums and measurement errors. Then check the choice against data with probability plots, and remember that distributions that fit equally well in the middle of the data can differ greatly in the tails, which is exactly where design decisions are made. When the tail matters, test the sensitivity of the decision to the choice of distribution.

Common mistakes

  • MistakeUsing the exponential distribution for parts that wear out.

    InsteadWear-out means a rising hazard; use a Weibull distribution with β > 1.

  • MistakePlanning maintenance around the mean life.

    InsteadUse percentiles such as B10; many items fail well before the mean.

  • MistakeDiscarding items that had not failed when a life test ended.

    InsteadTreat them as censored observations in a maximum likelihood fit.

  • MistakeAveraging lognormal data directly to find a typical value.

    InsteadWork with logarithms; the median is e^μ and the mean exceeds it.

  • MistakeTreating a 100-year return level as precise.

    InsteadExtrapolating beyond the record length carries large uncertainty; quantify it.

Summary and key terms

  • Reliability R(t) = 1 − F(t); the hazard h(t) = f(t)/R(t) is the failure rate among survivors.
  • The bathtub curve combines infant mortality, a useful-life period and wear-out.
  • The Weibull distribution has R = e^(−(t/η)^β); β < 1, = 1 and > 1 give falling, constant and rising hazard.
  • B10 = η(−ln 0.9)^(1/β) and MTTF = ηΓ(1 + 1/β).
  • Weibull probability plots and maximum likelihood fit life data, including censored items.
  • The lognormal distribution models multiplicative quantities; the gamma models sums of exponential waiting times.
  • The Gumbel distribution models annual maxima; the T-year level is u − a ln[−ln(1 − 1/T)].
Reliability function
The probability that an item survives beyond time t.
Hazard function
The instantaneous failure rate among items still working.
Bathtub curve
A hazard that falls, levels off and then rises over a product's life.
Weibull distribution
A life distribution whose shape parameter controls whether the hazard rises or falls.
Characteristic life
The Weibull scale η, by which 63.2% of items have failed.
B10 life
The age by which 10% of items have failed.
Censored data
Observations known only to exceed some value, such as items still working at the end of a test.
Maximum likelihood
Estimating parameters by making the observed data most probable.
Lognormal distribution
The distribution of a variable whose logarithm is normal.
Gumbel distribution
An extreme value distribution used for annual maxima.
Return level
The value exceeded on average once per return period.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. A Weibull life has β = 2 and η = 4000 h. Find R(2000).

    Show answer

    e^(−0.25) = 0.779.

  2. For the same item, find the B10 life.

    Show answer

    4000 × (0.1054)^0.5 = 1299 h.

  3. What does β = 0.7 suggest about a component's failures?

    Show answer

    A falling hazard: early failures from defects; burn-in may help.

  4. Find the hazard at t = η for a Weibull with β = 3.

    Show answer

    h = 3/η.

  5. Ten identical Weibull parts with β = 2, η = 8000 h are in series. What is the system η?

    Show answer

    8000/√10 = 2530 h.

  6. For a lognormal with median 50 and σ = 0.5 (logs), find P(X > 100).

    Show answer

    P(Z > ln 2/0.5 = 1.386) = 0.083.

  7. Find the mean of that lognormal.

    Show answer

    50e^(0.125) = 56.7.

  8. Failures occur at rate 0.002 per hour. What is the mean time to the third failure?

    Show answer

    3/0.002 = 1500 h (gamma with k = 3).

  9. Annual maximum wind speeds have mean 30 m/s and s = 4 m/s. Find the Gumbel u and a.

    Show answer

    a = 3.12, u = 30 − 1.80 = 28.2 m/s.

  10. Find the 50-year wind speed for that site.

    Show answer

    28.2 − 3.12 ln(−ln 0.98) = 28.2 + 3.12 × 3.902 = 40.4 m/s.

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