Continuous Distributions

Models for measurements and lifetimes. Learn uniform, exponential, normal (Gaussian), gamma, beta, Weibull and lognormal distributions — PDFs, parameters, mean and variance, standardising with z-scores, the exponential and Weibull in reliability, and choosing a distribution.

Resistor values scatter normally about the nominal; electronic component lifetimes are often exponential; bearing and fatigue lives follow Weibull; particle sizes and repair times are lognormal. The right distribution gives realistic tolerances and reliability predictions.

An analogy

The normal curve is the shape you get when many small independent effects add up — like the heights of people or the errors in a measurement.

The normal distribution

68–95–99.7 rule: within ±1σ, ±2σ, ±3σ of the mean. Standardise with z and use Φ tables or software. Sums of many independent effects tend to be normal (central limit theorem).

Exponential and reliability

Constant failure rate (random failures, useful-life period of the bathtub curve). Memoryless: an old component is as good as new. R(t) = e^(−t/MTBF).

Weibull

β < 1: infant mortality; β = 1: exponential; β > 1: wear-out (bearings β ≈ 1.5, fatigue β ≈ 2–4). Fitted on Weibull probability paper or by maximum likelihood.

Other families

Gamma (sums of exponentials, waiting times), beta (proportions, PERT durations on [0, 1]), lognormal (multiplicative effects, corrosion, repair times). Validate fits with probability plots and goodness-of-fit tests.

Formulas

• Normal: f = (1/(σ√2π)) e^(−(x − μ)²/(2σ²)); P(a < X < b) = Φ((b − μ)/σ) − Φ((a − μ)/σ)
• Exponential: f = λe^(−λx); R(t) = e^(−λt); MTBF = 1/λ
• Weibull: R(t) = e^(−(t/η)^β); mean = ηΓ(1 + 1/β)
• Uniform: f = 1/(b − a); μ = (a + b)/2, σ² = (b − a)²/12
• Lognormal: ln X ~ N(μ, σ²)

Worked example: Resistor yield

Given: R ~ N(100, 2²) Ω; spec 97–103 Ω

Find: Fraction within spec.

• z = ±1.5
• Φ(1.5) − Φ(−1.5) = 0.866

Answer: 86.6 % pass

Worked example: Component reliability

Given: MTBF 1000 h (exponential); Weibull β = 2, η = 1000 h

Find: R(500 h) for each.

• Exponential: e^(−0.5) = 0.607
• Weibull: e^(−0.25) = 0.779

Answer: 0.607 vs 0.779

Common mistakes

• Wrong: Assuming everything is normal. — Right: Lifetimes are skewed — use Weibull/lognormal.
• Wrong: Applying exponential to wear-out parts. — Right: Wear-out has increasing hazard (β > 1).

Applications

• Manufacturing tolerances: Yield from normal distributions.
• Reliability: Exponential and Weibull life models.
• Project planning: Beta/PERT activity durations.

Key takeaways

• Normal: z-scores and Φ; 68–95–99.7.
• Exponential: R = e^(−t/MTBF).
• Weibull shape β identifies failure mode.
• Choose distributions from physics and data.

Going further

• Maximum likelihood: Fitting distribution parameters.
• Extreme value theory: Floods, wind loads, maximum stresses.
• Stress–strength interference: Probability of failure from two distributions.