A single sampling plan is specified by two numbers: the sample size \(n\) and the acceptance number \(c\). The procedure is:
Example notation: \(n = 150\), \(c = 3\) means inspect 150 items; accept if 3 or fewer are defective.
When sampling from a large lot (or with replacement), the number of defectives \(X\) in the sample follows a Binomial distribution:
The probability of acceptance is:
\[ P_a = P(X \le c) = \sum_{x=0}^{c} \binom{n}{x}\, p^x\, (1-p)^{n-x}. \]Plan: \(n = 20\), \(c = 1\). Find \(P_a\) for \(p = 0.05\).
\(P(X = 0) = (0.95)^{20} = 0.3585\).
\(P(X = 1) = \binom{20}{1}(0.05)^1(0.95)^{19} = 20 \times 0.05 \times 0.3774 = 0.3774\).
\(P_a = 0.3585 + 0.3774 = 0.7359\).
Interpretation: a lot with 5% defectives has about a 73.6% chance of being accepted under this plan.
Same plan: \(n = 20\), \(c = 1\). Compute \(P_a\) for several values of \(p\):
| \(p\) | \(P_a = P(X \le 1)\) |
|---|---|
| 0.01 | 0.983 |
| 0.03 | 0.880 |
| 0.05 | 0.736 |
| 0.10 | 0.392 |
| 0.15 | 0.176 |
| 0.20 | 0.069 |
Plotting these points gives the OC curve. It drops from near 1 to near 0 as \(p\) increases from 0 to 0.20.
When \(n\) is large (\(n \ge 50\)) and \(p\) is small (\(p \le 0.10\)), the Binomial is well approximated by the Poisson distribution with parameter \(\lambda = np\). This greatly simplifies computation, especially before computers were widely available.
Plan: \(n = 100\), \(c = 3\). Find \(P_a\) for \(p = 0.02\). \(\lambda = 100 \times 0.02 = 2.0\).
\(P_a = e^{-2}\left(\frac{2^0}{0!} + \frac{2^1}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!}\right) = e^{-2}(1 + 2 + 2 + 1.333) = 0.1353 \times 6.333 = 0.857\).
For \(p = 0.05\): \(\lambda = 5.0\). \(P_a = e^{-5}(1 + 5 + 12.5 + 20.833) = 0.0067 \times 39.333 = 0.265\).
Plan: \(n = 50\), \(c = 2\), \(p = 0.04\).
Binomial: \(P_a = \sum_{x=0}^{2}\binom{50}{x}(0.04)^x(0.96)^{50-x} = 0.130 + 0.271 + 0.276 = 0.677\).
Poisson: \(\lambda = 2.0\). \(P_a = e^{-2}(1 + 2 + 2) = 0.1353 \times 5 = 0.677\).
The Poisson approximation is excellent in this case.
Assuming rejected lots are 100% inspected and all defectives are replaced:
Plan: \(n = 80\), \(c = 2\), \(N = 3000\). Compute for \(p = 0.03\).
\(\lambda = 80 \times 0.03 = 2.4\). \(P_a = e^{-2.4}(1 + 2.4 + 2.88) = 0.0907 \times 6.28 = 0.570\).
AOQ = 0.570 × 0.03 × (3000 − 80)/3000 = 0.570 × 0.03 × 0.973 = 0.0166 = 1.66%.
ATI = 80 + (1 − 0.570)(3000 − 80) = 80 + 0.430 × 2920 = 80 + 1255.6 = 1335.6.
Plan: \(n = 80\), \(c = 2\), \(N = 3000\). AOQ ≈ \(P_a \cdot p\) (large \(N\)):
| \(p\) | \(\lambda\) | \(P_a\) | AOQ |
|---|---|---|---|
| 0.01 | 0.8 | 0.953 | 0.0095 |
| 0.02 | 1.6 | 0.783 | 0.0157 |
| 0.03 | 2.4 | 0.570 | 0.0171 |
| 0.04 | 3.2 | 0.380 | 0.0152 |
| 0.05 | 4.0 | 0.238 | 0.0119 |
| 0.06 | 4.8 | 0.143 | 0.0086 |
AOQ rises, peaks near \(p = 0.03\), then falls — the characteristic AOQ curve shape. AOQL ≈ 1.71%.
Plot AOQ (y-axis) against the lot fraction defective \(p\) (x-axis). The curve rises, reaches a peak, and then declines. The y-value at the peak is the AOQL, and the corresponding x-value is the incoming quality level at which outgoing quality is worst.
From the AOQ values in Unit 4 (reproduced), the curve peaks at approximately \(p = 0.03\) with AOQ ≈ 0.0194. Reading from the graph: AOQL ≈ 1.94%. This means that regardless of the supplier's quality, the average outgoing quality will not exceed 1.94% defective.
For \(n = 100\), using AOQL ≈ \(y_c/n\) with the Dodge–Romig constants (\(y_1 = 0.840\), \(y_3 = 1.946\), \(y_5 = 3.17\)): with \(c = 1\), AOQL ≈ 0.84%; with \(c = 3\), AOQL ≈ 1.94%; with \(c = 5\), AOQL ≈ 3.17%. Increasing the acceptance number raises AOQL (worse outgoing quality guarantee) but reduces the number of lots rejected — a trade-off between quality assurance and inspection cost.
The lot quality approach designs a plan that satisfies two specified points on the OC curve:
We need \(n\) and \(c\) such that:
Given: AQL = \(p_1 = 0.01\), \(\alpha = 0.05\), LTPD = \(p_2 = 0.05\), \(\beta = 0.10\).
Required ratio \(R = p_2/p_1 = 0.05/0.01 = 5.0\).
From the Poisson unity (Cameron) table, \(\lambda_1 = np\) gives \(P_a = 0.95\) and \(\lambda_2 = np\) gives \(P_a = 0.10\):
| \(c\) | \(\lambda_1\) (\(P_a=0.95\)) | \(\lambda_2\) (\(P_a=0.10\)) | \(\lambda_2/\lambda_1\) |
|---|---|---|---|
| 2 | 0.818 | 5.322 | 6.51 |
| 3 | 1.366 | 6.681 | 4.89 |
| 4 | 1.970 | 7.994 | 4.06 |
The target ratio 5.0 falls between \(c = 2\) (6.51) and \(c = 3\) (4.89); \(c = 3\) is the closest, so take \(c = 3\).
Then fix \(n\) from the producer's point: \(n = \lambda_1/p_1 = 1.366/0.01 = 136.6 \approx 137\). (Consumer check: \(\lambda_2/p_2 = 6.681/0.05 = 133.6\), consistent.) Plan: \(n = 137\), \(c = 3\).
Given: AQL = \(p_1 = 0.02\), \(\alpha = 0.05\), LTPD = \(p_2 = 0.08\), \(\beta = 0.10\). Required ratio \(R = 0.08/0.02 = 4.0\).
From the Cameron table: \(c = 4\) gives \(\lambda_1 = 1.970\), \(\lambda_2 = 7.994\), ratio ≈ 4.06 (essentially 4.0); \(c = 5\) gives ratio ≈ 3.55 (too low). So take \(c = 4\).
\(n = \lambda_1/p_1 = 1.970/0.02 = 98.5 \approx 99\). (Consumer check: \(\lambda_2/p_2 = 7.994/0.08 = 99.9\), consistent.) Plan: \(n = 99\), \(c = 4\).
The average quality approach designs a plan that ensures the AOQL does not exceed a specified value \(p_L\). It does not require specifying both AQL and LTPD — instead, the focus is on the worst-case average outgoing quality the consumer will receive.
where \(y_c\) is a constant that depends on \(c\) (e.g., \(y_0 = 0.368\), \(y_1 = 0.840\), \(y_2 = 1.372\), \(y_3 = 1.946\)).
Desired AOQL = 2%, lot size \(N = 5000\). Choose \(c = 2\). Then \(y_2 = 1.372\).
Approximation: \(n \approx y_2 / p_L = 1.372 / 0.02 = 68.6 \approx 69\).
With \(n/N = 69/5000 = 0.0138\), the correction factor is negligible. Plan: \(n = 69\), \(c = 2\), AOQL ≈ 2%.
AOQL = 3%, \(N = 3000\).
For \(c = 0\): \(y_0 = 0.368\). \(n = 0.368/0.03 = 12.3 \approx 13\). Plan: \(n = 13\), \(c = 0\).
For \(c = 2\): \(y_2 = 1.372\). \(n = 1.372/0.03 = 45.7 \approx 46\). Plan: \(n = 46\), \(c = 2\).
The \(c = 0\) plan has a much smaller sample but a very steep OC curve (almost no lots accepted if \(p\) is even slightly above zero). The \(c = 2\) plan is more forgiving — it accepts good-quality lots with high probability but requires a larger sample. The choice depends on the cost of inspection versus the cost of rejecting good lots.
A double sampling plan allows a second sample before deciding on a lot. It is defined by \((n_1, c_1, r_1;\ n_2, c_2)\):
A clear first sample often decides the lot on a small \(n_1\), so on average fewer items are inspected than under a single plan of equal protection. This average is the Average Sample Number:
Plan \((n_1 = 50, c_1 = 1, r_1 = 4;\ n_2 = 50, c_2 = 4)\) at \(p = 0.05\) (so \(m_1 = n_1 p = 2.5\), Poisson). Probability of needing a second sample \(= P(2 \le d_1 \le 3) = 0.470\), so
\(\text{ASN} = 50 + 50(0.470) = 73.5\) items — on average below the 100 that a comparable single plan of \(n = 50 + 50\) would always inspect. A note on sequential sampling: taking this idea to the limit (deciding after each item) minimises ASN further.