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Probability Distributions

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FE Fundamentals of Engineering Exam

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Probability Distributions

Key Takeaways Explain This Topic AI

Binomial models successes in n independent trials: P(X=k) = C(n,k)pᵏ(1−p)ⁿ⁻ᵏ, with mean np and variance np(1−p).

Poisson models rare events at constant rate λ: P(X=k) = e⁻λλᵏ/k!, with mean = variance = λ.

The normal distribution is fully described by μ and σ; standardize with z = (x − μ)/σ before using a z-table.

Empirical rule: ~68% within ±1σ, ~95% within ±2σ, ~99.7% within ±3σ of the mean.

The exponential distribution gives time between Poisson events, mean 1/λ, and is memoryless.

Uniform on [a,b] has mean (a+b)/2 and variance (b−a)²/12.

Last updated: June 2026

Distributions assign probabilities across outcomes; the FE expects you to pick the right distribution from the problem's wording , then apply its Handbook formula. The fastest cue is the phrase: "exactly k of n trials" → binomial; "events per interval/area" → Poisson; "continuous measurement, bell-shaped" → normal; "time until/between events" → exponential.

  • Discrete Distributions
  • Binomial

Models the number of successes in n independent trials , each with success probability p: P(X=k) = C(n,k)·pᵏ·(1−p)ⁿ⁻ᵏ.

  • Parameter Formula
  • Mean μ = np
  • Variance σ² = np(1−p)
  • Std. deviation σ = √(np(1−p))

Worked example — binomial. An inspector tests 10 items, each with a 5% defect rate. P(exactly 2 defective) = C(10,2)(0.05)²(0.95)⁸ = 45 · 0.0025 · 0.6634 ≈ 0.0746 (~7.5%). The expected count is μ = np = 10(0.05) = 0.5 defectives.

Poisson

Models the count of events in a fixed interval when they occur at constant average rate λ: P(X=k) = e⁻λ·λᵏ/k!. Uniquely, its mean and variance are equal: μ = σ² = λ.

Worked example — Poisson. A call center averages 3 calls/minute. P(exactly 5 calls in a minute) = e⁻³·3⁵/5! = 0.0498·243/120 ≈ 0.1008 (~10.1%). And P(zero events) is always simply e⁻λ — for λ = 2 that is e⁻² ≈ 0.135.

The Normal (Gaussian) Distribution

The normal distribution is the most-tested continuous distribution, defined by mean μ (location) and standard deviation σ (spread). It is symmetric and bell-shaped, with mean = median = mode and total area 1.

Standardize any normal variable to the standard normal (z) using: z = x − μ σ z = \frac{x - \mu}{\sigma} z = σ x − μ ​

The z-score counts how many standard deviations x lies from the mean; you then read a cumulative probability from the standard-normal table in the Handbook.

Empirical Rule (68–95–99.7)

  • Interval Probability
  • μ ± 1σ 68.27%
  • μ ± 2σ 95.45%
  • μ ± 3σ 99.73%

Worked example — z-score. A normal variable has μ = 50 and σ = 5; find the z-score for x = 62.5. z = (62.5 − 50)/5 = 12.5/5 = 2.5 , meaning x sits 2.5 standard deviations above the mean. From a z-table, the area to the left of z = 2.5 is ≈ 0.9938, so only ~0.62% of values exceed 62.5.

Worked example — empirical rule. Bolt diameters are normal with μ = 10.0 mm, σ = 0.1 mm. What fraction lies between 9.8 and 10.2 mm? These limits are μ − 2σ and μ + 2σ, so by the empirical rule ≈ 95.45% fall in the range. Recognizing endpoints as whole multiples of σ lets you answer instantly without a table — a deliberate FE shortcut.

  • Continuous Distributions Beyond Normal
  • Uniform Distribution

Every value in [a, b] is equally likely. f(x) = 1/(b−a) on the interval.

  • Parameter Formula
  • Mean μ = (a+b)/2
  • Variance σ² = (b−a)²/12

Worked example — uniform. A random delay is uniform on [0, 6] minutes. Its mean is (0+6)/2 = 3 min and its variance is (6−0)²/12 = 36/12 = 3 min² (σ ≈ 1.73 min).

Exponential Distribution

Models the time between events in a Poisson process of rate λ: f(x) = λe⁻λˣ for x ≥ 0, with CDF P(X ≤ x) = 1 − e⁻λˣ.

  • Parameter Formula
  • Mean μ = 1/λ
  • Variance σ² = 1/λ²

It is memoryless : P(X > s + t | X > s) = P(X > t) — having waited s units does not change the future wait distribution. This makes it the standard model for the time-to-failure of components with a constant hazard rate (the flat middle of the bathtub curve).

Worked example — exponential. A component fails at rate λ = 0.5 per year, so its mean life is 1/0.5 = 2 years. P(it survives past 3 years) = e⁻λˣ = e⁻⁰·⁵·³ = e⁻¹·⁵ ≈ 0.223. Note the tight link between distributions: if failures arrive Poisson at rate λ, the waiting time between them is exponential with the same λ — a pairing the FE frequently tests in one combined reliability problem.

The Central Limit Theorem and Choosing a Distribution

The Central Limit Theorem (CLT) is why the normal distribution is everywhere: for a sufficiently large sample (commonly n ≥ 30), the distribution of the sample mean x̄ is approximately normal regardless of the population's shape, with mean μ and standard deviation σ/√n (the standard error). This is the bridge between raw data and the confidence intervals of the next section. The CLT also lets a binomial with large n be approximated by a normal with μ = np and σ = √(np(1−p)), and a Poisson with large λ by a normal with μ = σ² = λ.

Distribution selection cheat-sheet:

  • Problem wording Distribution
  • "exactly k successes in n fixed trials" binomial
  • "number of events per minute/area/length" Poisson
  • "continuous measurement near a target" normal
  • "time between / until events" exponential
  • "equally likely over a range" uniform

Worked example — CLT. A population has μ = 100 and σ = 20. For samples of n = 100, the sample mean x̄ is approximately normal with standard error σ/√n = 20/10 = 2. So about 95% of sample means fall within μ ± 2(2) = 100 ± 4, i.e., between 96 and 104 — far tighter than the spread of individual values (±40 for ±2σ). Picking the wrong distribution is the dominant error mode on these items, so always classify the scenario before reaching for a formula.