Statistics and Probability for Engineers
Descriptive statistics, the normal distribution, confidence intervals, regression and reliability.
Describing data
mean x̄ = Σx ÷ n sample standard deviation s = √(Σ(x − x̄)² ÷ (n − 1)) standard error = s ÷ √n
Worked example: ten shaft diameters
Worked example: summary statistics
Ten shafts measure (mm): 24.98, 25.01, 25.03, 24.99, 25.00, 25.02, 24.97, 25.01, 25.00, 25.02.
- Mean 25.003 mm, median 25.005 mm, range 0.06 mm.
- Sum of squared deviations 0.00321; sample standard deviation s = √(0.00321 ÷ 9) = 0.0189 mm (population formula: 0.0179).
- Standard error s ÷ √10 = 0.0060 mm: the mean is much better known than any single shaft.
The normal distribution
Many measurements and manufacturing variations follow the bell-shaped normal distribution. With mean μ and standard deviation σ, about 68.3 percent of values fall within ±1σ, 95.4 percent within ±2σ and 99.73 percent within ±3σ. To find the fraction in any range, convert to z = (x − μ) ÷ σ and use the standard normal table or the function Φ(z).
Worked example: how many shafts are in tolerance?
- The process makes shafts with μ = 25.00 mm and σ = 0.02 mm; the limits are 24.95 and 25.05. z = ±0.05 ÷ 0.02 = ±2.5, so the fraction inside is Φ(2.5) − Φ(−2.5) = 98.76 percent: 12 419 parts per million (about 1.2 in 100) are out of tolerance.
- Process capability Cpk = (distance to the nearest limit) ÷ 3σ = 0.05 ÷ 0.06 = 0.83 (centered, so Cp = Cpk). A Cpk of 1.33 is the usual minimum; see Statistical Tolerancing and Process Capability.
Confidence intervals
A 95 percent confidence interval for the mean is x̄ ± t · s ÷ √n, where t comes from the Student distribution with n − 1 degrees of freedom (2.262 for 9, 2.045 for 29, approaching 1.96 for large n). For the ten shafts: 25.003 ± 2.262 × 0.0060 = 25.003 ± 0.0135 mm, i.e. 24.9895 to 25.0165 mm. It describes the uncertainty in the mean, not the range where individual shafts fall. Quadrupling the sample size halves the interval.
Regression and correlation
Worked example: a straight line through spring test data
Force (N) against compression (mm) at 2, 4, 6, 8, 10, 12 mm: 17.1, 33.5, 52.2, 66.8, 85.4, 100.9 N. Least squares gives slope m = (n Σxy − Σx Σy) ÷ (n Σx² − (Σx)²) = 8.419 N/mm and intercept 0.39 N, with R² = 0.9992. The spring rate is therefore about 8.42 N/mm, in line with the design of Springs. R² near 1 means the line explains the data; it does not prove cause, and it says nothing about whether a straight line is the right model (plot the residuals).
Probability and the binomial
P(A or B) = P(A) + P(B) − P(A and B) independent events: P(A and B) = P(A) P(B) binomial: P(k) = C(n,k) pᵏ (1 − p)ⁿ⁻ᵏ
Worked example: sampling 20 parts at 2 percent defective
P(no defects) = 0.9820 = 66.8 percent; P(exactly one) = 20 × 0.02 × 0.9819 = 27.2 percent; P(two or more) = 6.0 percent. A sample of 20 therefore finds nothing wrong about two times in three even though 2 percent of the batch is bad: small samples miss problems.
Reliability
R(t) = e−λ t MTBF = 1 ÷ λ series: R = R₁R₂R₃… parallel: R = 1 − (1 − R₁)(1 − R₂)…
- A pump with MTBF 5 000 h (λ = 0.0002 per hour) has R(1 000 h) = 81.9 percent; at 5 000 h, 36.8 percent.
- Three components in series, each 95 percent reliable: system R = 0.95³ = 85.7 percent (every added series part lowers reliability).
- Two 90 percent components in parallel (redundancy): R = 1 − 0.1² = 99 percent. Wear-out failures follow a Weibull distribution instead (shape > 1), as used for bearing life in Bearings.
Related
- Previous: Differential Equations for Engineers. Next: Engineering Economics.
- Fits and Tolerances for statistical tolerancing.
- Engineering Fundamentals: the whole guide.
FAQ
Standard deviation of the sample or of the population?
If the data are all you have (a sample of a larger process), use n − 1. For a complete population use n. For n above 30 the difference is small.
What is a p-value?
The probability of seeing data at least this extreme if there were no real effect. It is not the probability the hypothesis is true.
How many samples do I need?
It depends on the variation and on how small a difference you must detect: n ≈ (z σ ÷ E)² for a margin of error E.
References
- Montgomery, D. C., Runger, G. C., Applied Statistics and Probability for Engineers, Wiley.
- Walpole, R. E., et al., Probability and Statistics for Engineers and Scientists, Pearson.
- NIST/SEMATECH e-Handbook of Statistical Methods.
Last reviewed: October 2026. Normal-table values are computed from a standard approximation accurate to about 1e-7.