Badr Alghamdi
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Lessons

Acceptance sampling: a pass leaves room for defects

A sample of 200 can pass a lot with twice the reference defective fraction more often than it fails. The acceptance count explains why.

The sample plan and its curve

Taking a random sample of size 200 without replacement from 5,000 and limiting defective units to 10 passes 58.3% of lots with 5.0% defective units, so passing is more likely than failing.

Acceptance probability is the chance that a random sample contains no more defective units than a declared count, at a specified lot composition; every unit is either defective or not.

The authored lot contains 5,000 units; the two reference fractions are 2.5% and 5.0%

Acceptance probability, %

Acceptance probability across the lot's defective fractionThe shown curve falls from 100.0% acceptance at 0.0% defective units to 0.0% acceptance at 20.0% defective units. At 2.5% defective units it is 98.9%, and at 5.0% it is 58.3%. Its first point at or below 10.0% acceptance is 378 defective units out of 5,000. These values also appear in the table.Equal chancesEntered plan
Horizontal
Defective units in the lot, %
Named points on the curve
Lot compositionAcceptance probability
125 defective units, 2.5%98.9%
250 defective units, 5.0%58.3%
First count at or below 10.0% acceptance378 of 5,000 units, 7.6%

Why can the same lot be more likely to pass a sample of 200 and more likely to fail a sample of 500 when both allow at most 10 defective units?

The count leaves room

A pass is an observation about a sample. The defective fraction is a property of the lot, including units nobody inspected. Those quantities are connected by the sampling mechanism, but they are not interchangeable. At a specified defective fraction, different random samples contain different counts. A plan accepts some of those samples and rejects the rest. The curve measures the share it accepts at each possible lot composition.

The opening case fixes a lot of 5,000 units and samples 200 without replacement. A reference fraction of 2.5% means 125 defective units in the lot. A sample then contains 5 defective units on average, while this authored plan permits up to 10. That space between the average and the acceptance count explains why the reference lot passes 98.9% of the time. These are conditional chances, not a reconstruction of a delivered lot.

Double the defective fraction to 5.0% and the average sample count becomes 10. An average at the acceptance count does not mean every draw stops there. Some samples contain fewer defective units and some contain more; the accepted side includes the count of 10 itself. Summing all accepted counts gives 58.3%. The fraction doubled, yet the plan still passes the lot more often than it fails it.

Hold that same 5.0% lot composition and the acceptance count of 10, then increase the sample to 500. The average sample now contains 25 defective units, so a draw with at most 10 falls far into the lower tail. Its chance is about 0.000270, below 0.1%. This contrast changes one count, not the inspection method or the lot. It is a description of the two plans, not a reason to select either.

A small case you can count

A separate authored lot has 200 units, including 10 defective ones. Sample 20 and accept at most 1 defective unit. Exactly two disjoint outcomes contribute to a pass: no defective units and one defective unit. Their probabilities add because a sample cannot have both counts at once.

A small case you can countWith 20 drawn from 200, of which 10 are defective, no defective units has probability 34.0% and exactly one defective unit has probability 39.7%; together they give 73.7% acceptance.No defective units34.0%One defective unit39.7%
Horizontal
Acceptance probability, %
Sample outcomeAcceptance probability, %
No defective units34.0%
One defective unit39.7%
Accepted total73.7%

The first term chooses all 20 inspected units from the 190 good ones. The second chooses one defective unit from 10 and the other 19 from the good units. Both use the same denominator: every possible set of 20 units drawn from the whole lot. Counting complete samples is why the model needs no simulated run and why repeating the same inputs gives the same answer.

A larger lot need not sharpen this plan

This fixed comparison keeps the opening plan at 200 sampled and at most 10 defective units, with 5.0% of each lot defective. Growing the lot from 500 to 100,000 changes acceptance from 58.7% to 58.3%; all five cases remain more likely to pass. The lot grew two hundredfold, but the inspected count did not.

Lot size on a logarithmic scale; acceptance on a linear scale

A larger lot need not sharpen this planFor the fixed plan n=200 and c=10 at 5.0% defective units, lots of 500, 1,000, 5,000, 10,000 and 100,000 have acceptance probabilities 58.7%, 58.4%, 58.3%, 58.3% and 58.3%; exact values are in the adjacent table.
Horizontal
Lot size
Vertical
Acceptance probability, %
Lot sizeAcceptance probability
50058.7%
1,00058.4%
5,00058.3%
10,00058.3%
100,00058.3%

Lot size still matters when inspection consumes a large part of the lot. A census is the limiting example: inspecting all units removes sampling variation entirely. The five points above establish a narrow finding for this plan and these compositions. They do not say that lot size can always be ignored, and the model refuses a sample larger than its population rather than silently converting it into a census.

Counts behind the curve

Count the accepted samples

N is lot size, D defective units in the lot, n sampled units and c the maximum accepted defective units. k counts defective units in one sample. Each unordered sample of size n is assumed equally likely; sampling does not replace an inspected unit.

P(X≤c)=∑k=max⁡(0,n−N+D)min⁡(c,n,D)(Dk)(N−Dn−k)(Nn)P(X\le c)=\sum_{k=\max(0,n-N+D)}^{\min(c,n,D)}\frac{\binom{D}{k}\binom{N-D}{n-k}}{\binom{N}{n}}
E[X]=nD/NE[X]=nD/N
D10=min⁡{D:P(X≤c)≤0.10}D_{10}=\min\{D:P(X\le c)\le0.10\}

Three named steps at the opening values

1. Lot composition: 5,000×0.05=2505{,}000\times0.05=250 defective units.

2. Sample expectation: 200×2505,000=10\frac{200\times250}{5{,}000}=10 defective units.

3. Accepted outcomes: sum probabilities for sample counts 0 through 10 ∑k=010P(X=k)=0.583215555215\sum_{k=0}^{10}P(X=k)=0.583215555215, displayed as 58.3%.

At 377 defective units the acceptance probability is 0.100668940974; at 378 it is 0.098851640015. The latter is the first whole count at or below 0.10, so its defective fraction is 3785,000=0.0756\frac{378}{5{,}000}=0.0756.

A small case you can count

P(X=0)=(100)(19020)(20020)=0.339774376237P(X=0)=\frac{\binom{10}{0}\binom{190}{20}}{\binom{200}{20}}=0.339774376237
P(X=1)=(101)(19019)(20020)=0.397396931271P(X=1)=\frac{\binom{10}{1}\binom{190}{19}}{\binom{200}{20}}=0.397396931271
P(X≤1)=0.339774376237+0.397396931271=0.737171307508P(X\le1)=0.339774376237+0.397396931271=0.737171307508

Source basis

NIST/SEMATECH e-Handbook, section 6.2, explains operating-characteristic curves. This page derives its own hypergeometric counts and uses authored examples. ISO 2859-1 and ANSI/ASQ Z1.4 are named plan families; no plan is selected from them here.

Basis
Uniform sampling without replacement from an authored finite lot
Assumptions
Two report-held counts; lot composition is swept, never inferred
Model version
1.0.0

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This computes acceptance probabilities for the declared random sample, not a prescribed plan or a verdict on a delivered lot.

Related

Acceptance probability record

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Entered counts and authored lot

Sample size
200
Maximum accepted defective units
10
Authored lot size
5,000

Conditional result

Taking a random sample of size 200 without replacement from 5,000 and limiting defective units to 10 passes 58.3% of lots with 5.0% defective units, so passing is more likely than failing.

Numbered calculation trace

1. Lot composition: 5,000 × 0.05 = 250 defective units.

2. Sample expectation: 200 × 250 / 5,000 = 10.00 defective units.

3. Accepted outcomes: sum probabilities for possible sample counts at most 10 = 0.583215555215, displayed as 58.3%.

Model version

1.0.0

Record date

This computes acceptance probabilities for the declared random sample, not a prescribed plan or a verdict on a delivered lot.