Normal Distribution Engineering Example

A worked example applying the normal distribution to dimensional tolerance analysis — computing the probability of a manufactured dimension falling outside the tolerance band.

Probabilistic & Reliability5 min read

worked examplenormal distributiontolerancemanufacturingprobability

Problem

A machined bore diameter has a nominal value of 50.00 mm and a standard deviation of 0.02 mm based on historical process data. The tolerance band is 50.00 +/- 0.05 mm. What is the probability that a randomly selected bore falls outside the tolerance band?

Given

Nominal (mean): mu = 50.00 mm
Standard deviation: sigma = 0.02 mm
Lower tolerance limit: 49.95 mm
Upper tolerance limit: 50.05 mm

Method

Assume the bore diameter is normally distributed. Compute the z-scores for the tolerance limits and find the tail probabilities.

Solution

z_lower = (49.95 - 50.00) / 0.02 = -2.50
z_upper = (50.05 - 50.00) / 0.02 = +2.50

P(X < 49.95) = Phi(-2.50) = 0.0062
P(X > 50.05) = 1 - Phi(2.50) = 0.0062

P(outside tolerance) = 2 * 0.0062 = 0.0124 = 1.24%

Interpretation

Approximately 1.24% of bores are expected to fall outside the tolerance band. This is an illustrative example using representative values — actual process capability must be based on manufacturing data.

Assumptions

Normal distribution is valid for this process
Process is in statistical control
No systematic drift or tool wear