11.2 Statistical Process Control for Attributes (p, np, c, and u Charts)

Key Takeaways

  • Attribute data is bifurcated into nonconforming units (defective items governed by the binomial distribution) and nonconformities (defects per unit governed by the Poisson distribution).

  • The pp-chart monitors the fraction nonconforming and accommodates either constant or variable sample sizes (nin_i), whereas the npnp-chart tracks the absolute count of defective units and strictly requires a constant sample size across all subgroups.

  • The cc-chart tracks total defect counts over a strictly constant area of opportunity (n=1n=1), whereas the uu-chart monitors defect density per unit across variable inspection units (nin_i).

  • Because proportions, counts of defectives, and defect counts cannot be negative, calculated lower control limits that evaluate below zero must be truncated to zero (LCL=0LCL = 0).

  • Normal approximation to the binomial and Poisson distributions requires adequate expected counts (np≥5np \ge 5 or cˉ≥5\bar{c} \ge 5); small sample sizes yield severe skewness and asymmetric false alarm rates.

Last updated: October 2026

Statistical Process Control for Attributes (pp, npnp, cc, and uu Charts)

Core Principle: Attribute control charts monitor qualitative, count-based, or pass/fail quality characteristics. Attribute SPC is organized according to whether the inspection records nonconforming units (defective products modeled by the binomial distribution) or nonconformities (defects per inspection unit modeled by the Poisson distribution).

In many manufacturing, assembly, and service systems, measuring continuous numerical dimensions is neither practical nor cost-effective. Instead, quality inspectors evaluate parts using "go/no-go" functional gauges, visual optical inspection, or defect counts per assembly. Statistical Process Control for attributes provides the quantitative framework for tracking these discrete metrics. Industrial engineers must select the correct attribute chart based on the mathematical nature of the data and whether the sample inspection size remains constant or varies.


1. Fundamentals of Attribute Data: Defectives vs. Defects

A critical distinction on the PE exam is between nonconforming units (defective items) and nonconformities (defects):

Attribute Data Classification
├── Nonconforming Units (Defective Items)
│   ├── Binary outcome: Conforming (Pass) vs Nonconforming (Fail)
│   ├── Underlying distribution: Binomial Distribution B(n, p)
│   └── Appropriate Control Charts:
│       ├── p-chart: Fraction nonconforming (constant or variable n)
│       └── np-chart: Number nonconforming (strictly constant n)
└── Nonconformities (Defects per Unit)
    ├── Count of discrete imperfections, flaws, or blemishes per unit
    ├── Underlying distribution: Poisson Distribution Poisson(lambda)
    └── Appropriate Control Charts:
        ├── c-chart: Total defects across constant area of opportunity (n = 1)
        └── u-chart: Defects per inspection unit (constant or variable n_i)

The Binomial Model for Nonconforming Units

When an inspection station evaluates whether a manufactured part meets functional standards, each part represents a Bernoulli trial with two mutually exclusive outcomes: conforming (probability 1−p1 - p) or nonconforming (probability pp). In a random sample of nn independent units, the probability of observing exactly dd nonconforming units is governed by the Binomial Distribution:

P(D=d)=(nd)pd(1−p)n−dP(D = d) = \binom{n}{d} p^d (1 - p)^{n - d} Expected Count: E(D)=npVariance: Var⁡(D)=np(1−p)\text{Expected Count: } E(D) = np \qquad \text{Variance: } \operatorname{Var}(D) = np(1 - p)

For the fraction nonconforming p^=D/n\hat{p} = D / n:

E(p^)=pVar⁡(p^)=Var⁡(D)n2=p(1−p)nE(\hat{p}) = p \qquad \operatorname{Var}(\hat{p}) = \frac{\operatorname{Var}(D)}{n^2} = \frac{p(1 - p)}{n}

The Poisson Model for Nonconformities

A single complex product (e.g., an automobile door panel, a printed circuit board, or an aircraft turbine blade) can satisfy overall functional requirements while exhibiting several discrete blemishes (scratches, pinholes, solder bridges). When the opportunities for defects across a continuous area of opportunity are vast, but the probability of a defect at any specific point is extremely small, defect occurrences follow the Poisson Distribution:

P(C=c)=e−λλcc!for c=0,1,2,…P(C = c) = \frac{e^{-\lambda} \lambda^c}{c!} \quad \text{for } c = 0, 1, 2, \dots Expected Count: E(C)=λVariance: Var⁡(C)=λ\text{Expected Count: } E(C) = \lambda \qquad \text{Variance: } \operatorname{Var}(C) = \lambda

The defining characteristic of the Poisson distribution is that its variance equals its mean (σc2=μc=λ\sigma_c^2 = \mu_c = \lambda).


2. Fraction Nonconforming (pp) and Number Nonconforming (npnp) Charts

The pp-Chart (Fraction Nonconforming)

The pp-chart monitors the proportion of nonconforming items in a subgroup. It is the most versatile attribute chart because it accommodates both constant sample sizes and variable sample sizes (nin_i).

Given mm subgroups where subgroup ii contains nin_i inspected units and did_i nonconforming units, the subgroup fraction nonconforming is pi=di/nip_i = d_i / n_i. The overall weighted process fraction nonconforming is:

pˉ=∑i=1mdi∑i=1mni\bar{p} = \frac{\sum_{i=1}^m d_i}{\sum_{i=1}^m n_i}

Case A: Constant Sample Size (ni=nn_i = n)

When every subgroup contains exactly nn units, the control limits remain fixed across all intervals:

CLp=pˉCL_p = \bar{p} UCLp=pˉ+3pˉ(1−pˉ)nUCL_p = \bar{p} + 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}} LCLp=max⁡(0,pˉ−3pˉ(1−pˉ)n)LCL_p = \max\left(0, \bar{p} - 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}}\right)

Case B: Variable Sample Size (nin_i varies)

In high-mix manufacturing or shipping dock receiving inspection, the number of inspected units varies per lot. Control limits must be calculated individually for each subgroup ii:

UCLp,i=pˉ+3pˉ(1−pˉ)niUCL_{p,i} = \bar{p} + 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n_i}} LCLp,i=max⁡(0,pˉ−3pˉ(1−pˉ)ni)LCL_{p,i} = \max\left(0, \bar{p} - 3 \sqrt{\frac{\bar{p}(1 - \bar{p})}{n_i}}\right)

Notice that as nin_i increases, the standard error decreases, causing the control limits to narrow. Conversely, smaller sample sizes cause the limits to expand. Alternatively, engineers construct a Standardized pp-Chart where the plotted statistic is the standard normal variate ZiZ_i:

Zi=pi−pˉpˉ(1−pˉ)niZ_i = \frac{p_i - \bar{p}}{\sqrt{\frac{\bar{p}(1 - \bar{p})}{n_i}}} CLZ=0UCLZ=+3LCLZ=−3CL_Z = 0 \qquad UCL_Z = +3 \qquad LCL_Z = -3

The npnp-Chart (Number Nonconforming)

When operators on a shop floor inspect a fixed number of components each hour (e.g., n=50n = 50), plotting fractional percentages (such as 0.040.04 or 0.060.06) can introduce calculation errors. The npnp-chart plots the integer count of nonconforming items (did_i) directly.

Important

Strict Operational Requirement: An npnp-chart strictly requires a constant subgroup size nn across all sampling periods. If sample size varies between lots, an npnp-chart cannot be used; the engineer must use a pp-chart.

Formulation for the npnp-chart:

CLnp=npˉCL_{np} = n\bar{p} UCLnp=npˉ+3npˉ(1−pˉ)UCL_{np} = n\bar{p} + 3 \sqrt{n\bar{p}(1 - \bar{p})} LCLnp=max⁡(0,npˉ−3npˉ(1−pˉ))LCL_{np} = \max\left(0, n\bar{p} - 3 \sqrt{n\bar{p}(1 - \bar{p})}\right)

Where pˉ=∑dim⋅n\bar{p} = \frac{\sum d_i}{m \cdot n}. Note the mathematical relationship: the npnp limits are simply the pp-chart limits multiplied through by nn.


3. Count of Nonconformities (cc) and Nonconformities per Unit (uu) Charts

The cc-Chart (Total Defects per Constant Unit)

The cc-chart monitors the total count of nonconformities (cc) observed in a single, strictly constant area of opportunity (n=1n = 1). Typical industrial units include:

  • Number of weave defects in exactly 100 m2100\text{ m}^2 of composite cloth.
  • Number of solder bridge defects on a single complex avionics circuit board.
  • Number of surface paint imperfections on an automobile hood.

Given mm inspection intervals with defect counts c1,c2,…,cmc_1, c_2, \dots, c_m:

cˉ=1m∑i=1mci\bar{c} = \frac{1}{m} \sum_{i=1}^m c_i

Because the variance of a Poisson random variable equals its mean (σc2=cˉ  ⟹  σc=cˉ\sigma_c^2 = \bar{c} \implies \sigma_c = \sqrt{\bar{c}}), the Shewhart 3σ3\sigma control limits are:

CLc=cˉCL_c = \bar{c} UCLc=cˉ+3cˉUCL_c = \bar{c} + 3 \sqrt{\bar{c}} LCLc=max⁡(0,cˉ−3cˉ)LCL_c = \max\left(0, \bar{c} - 3 \sqrt{\bar{c}}\right)

The uu-Chart (Nonconformities per Unit)

When the size of the inspection unit varies between samples (e.g., different rolls of fabric with varying yardage, or inspecting variable batch sizes of electronic subassemblies), the raw defect count cc cannot be compared directly. The uu-chart tracks the average defect density per inspection unit (u=c/nu = c / n):

ui=ciniu_i = \frac{c_i}{n_i}

Where nin_i represents the number of inspection units in subgroup ii, and cic_i is the total count of defects observed across those nin_i units. The grand average defect density across mm subgroups is:

uˉ=∑i=1mci∑i=1mni\bar{u} = \frac{\sum_{i=1}^m c_i}{\sum_{i=1}^m n_i}

Because ui=ci/niu_i = c_i / n_i and Var⁡(ci)=niuˉ\operatorname{Var}(c_i) = n_i \bar{u}, the variance of uiu_i is Var⁡(ui)=niuˉni2=uˉni\operatorname{Var}(u_i) = \frac{n_i \bar{u}}{n_i^2} = \frac{\bar{u}}{n_i}. The standard error is σu,i=uˉni\sigma_{u,i} = \sqrt{\frac{\bar{u}}{n_i}}.

The control limits for the uu-chart are:

CLu=uˉCL_u = \bar{u} UCLu,i=uˉ+3uˉniUCL_{u,i} = \bar{u} + 3 \sqrt{\frac{\bar{u}}{n_i}} LCLu,i=max⁡(0,uˉ−3uˉni)LCL_{u,i} = \max\left(0, \bar{u} - 3 \sqrt{\frac{\bar{u}}{n_i}}\right)

If the inspection unit size is constant across all subgroups (ni=nn_i = n), the limits simplify to fixed horizontal lines with denominator uˉ/n\sqrt{\bar{u}/n}.


4. Attribute Chart Selection Decision Framework

Selecting the correct control chart is a frequent focus on the PE exam. The decision tree below structures the selection process:

Attribute Chart Selection Flowchart
                     What is being inspected?
                                |
        +-----------------------+-----------------------+
        |                                               |
   Nonconforming Units                             Nonconformities
   (Defective Items)                               (Defect Counts)
        |                                               |
   Is sample size n                                Is the area of
   strictly constant?                             opportunity constant?
        |                                               |
    +---+---+                                       +---+---+
    |       |                                       |       |
   YES      NO                                     YES      NO
    |       |                                       |       |
 np-chart  p-chart                               c-chart  u-chart
 (or p)   (variable limits)                      (n = 1)  (variable n_i)

Comprehensive Attribute Comparison Matrix

Attribute ChartQuality Metric MonitoredSample Size (nn)Governing DistributionCenter Line (CLCL)Control Limit Formula (UCL/LCLUCL / LCL)
pp-ChartFraction NonconformingConstant or Variable (nin_i)Binomialpˉ\bar{p}pˉ±3pˉ(1−pˉ)ni\bar{p} \pm 3\sqrt{\frac{\bar{p}(1-\bar{p})}{n_i}}
npnp-ChartNumber NonconformingStrictly Constant (nn)Binomialnpˉn\bar{p}npˉ±3npˉ(1−pˉ)n\bar{p} \pm 3\sqrt{n\bar{p}(1-\bar{p})}
cc-ChartTotal Defect CountStrictly Constant Area (n=1n=1)Poissoncˉ\bar{c}cˉ±3cˉ\bar{c} \pm 3\sqrt{\bar{c}}
uu-ChartDefects per Inspection UnitConstant or Variable (nin_i)Poissonuˉ\bar{u}uˉ±3uˉni\bar{u} \pm 3\sqrt{\frac{\bar{u}}{n_i}}

5. Lower Control Limit Truncation and Zero-Bound Mechanics

In physical manufacturing systems, the count of defects, count of defective items, and fraction nonconforming have a natural physical boundary at zero (p≥0,d≥0,c≥0p \ge 0, d \ge 0, c \ge 0). However, Shewhart formulas calculate symmetrical limits using 3σ3\sigma margins:

LCLcalculated=CL−3σLCL_{\text{calculated}} = CL - 3\sigma

When process quality is high (low pˉ\bar{p} or low cˉ\bar{c}) or sample size nn is small, 3σ3\sigma frequently exceeds the center line value, yielding a negative mathematical result (LCL<0LCL < 0).

Note

The Zero-Truncation Protocol: Whenever the calculated Lower Control Limit is negative, it must be truncated to zero: LCL=0LCL = 0. Because observed sample values can never fall below zero (pi≥0p_i \ge 0), a single point can never trigger an out-of-control alarm below the lower control limit on a zero-truncated chart.

Engineering Implications of LCL=0LCL = 0

When LCL=0LCL = 0, the control chart becomes one-sided for single-point violations. An operational breakthrough (e.g., an improved solder flux yielding zero defects) cannot register as a lower limit breach. To detect significant quality improvements on low-defect lines:

  1. Increase the subgroup size nn until the calculated LCLLCL is strictly positive (CL−3σ>0CL - 3\sigma > 0):
    • For a pp-chart: n>9(1−pˉ)pˉn > \frac{9(1 - \bar{p})}{\bar{p}}
    • For a cc-chart: cˉ>9\bar{c} > 9
  2. Apply run rules (e.g., 8 consecutive points below the center line) to confirm statistically significant quality improvements.

6. Comprehensive Worked Numerical Problem: Multi-Chart Comparison

A quality engineer at an automated electronics manufacturing facility monitors surface-mount technology (SMT) assembly on a high-density telecommunications PCB. The engineer evaluates 25 historical production shifts during stable operations. Across these 25 shifts, exactly n=100n = 100 PCBs were inspected per shift (Ntotal=25×100=2,500N_{\text{total}} = 25 \times 100 = 2,500 boards).

Inspection records reveal:

  • A cumulative total of 175175 boards were classified as nonconforming (defective) due to failure on automated optical inspection.
  • Across all inspected boards, a cumulative total of 625625 solder defects (bridging, insufficient wetting, tombstoning) were recorded. Note that some defective boards contained multiple defects.

Task A: Construct the pp-Chart and npnp-Chart Control Limits

1. Calculate baseline parameters: pˉ=∑di∑ni=1752500=0.070\bar{p} = \frac{\sum d_i}{\sum n_i} = \frac{175}{2500} = 0.070 Subgroup Size: n=100\text{Subgroup Size: } n = 100

2. pp-Chart Limits: CLp=pˉ=0.070CL_p = \bar{p} = 0.070 σp=pˉ(1−pˉ)n=0.070×(1−0.070)100=0.070×0.930100=0.000651=0.025515\sigma_p = \sqrt{\frac{\bar{p}(1 - \bar{p})}{n}} = \sqrt{\frac{0.070 \times (1 - 0.070)}{100}} = \sqrt{\frac{0.070 \times 0.930}{100}} = \sqrt{0.000651} = 0.025515 3σp=3×0.025515=0.076543\sigma_p = 3 \times 0.025515 = 0.07654 UCLp=pˉ+3σp=0.070+0.07654=0.14654≈0.1465UCL_p = \bar{p} + 3\sigma_p = 0.070 + 0.07654 = 0.14654 \approx 0.1465 LCLp=max⁡(0,0.070−0.07654)=max⁡(0,−0.00654)=0LCL_p = \max(0, 0.070 - 0.07654) = \max(0, -0.00654) = 0

3. npnp-Chart Limits (since n=100n = 100 is constant): CLnp=npˉ=100×0.070=7.00 defective boardsCL_{np} = n\bar{p} = 100 \times 0.070 = 7.00\text{ defective boards} σnp=npˉ(1−pˉ)=100×0.070×0.930=6.51=2.5515\sigma_{np} = \sqrt{n\bar{p}(1 - \bar{p})} = \sqrt{100 \times 0.070 \times 0.930} = \sqrt{6.51} = 2.5515 3σnp=3×2.5515=7.65453\sigma_{np} = 3 \times 2.5515 = 7.6545 UCLnp=7.00+7.6545=14.6545≈14.65 boardsUCL_{np} = 7.00 + 7.6545 = 14.6545 \approx 14.65\text{ boards} LCLnp=max⁡(0,7.00−7.6545)=0 boardsLCL_{np} = \max(0, 7.00 - 7.6545) = 0\text{ boards}

Notice that UCLnp=n×UCLp=100×0.14654=14.654UCL_{np} = n \times UCL_p = 100 \times 0.14654 = 14.654 boards.

Task B: Construct the uu-Chart Limits

Because the engineer is tracking the count of nonconformities (defects) across inspection units of n=100n = 100 boards per shift, a uu-chart monitors the average defects per board:

1. Calculate baseline defect rate per board (uˉ\bar{u}): uˉ=∑ci∑ni=6252500=0.250 defects per board\bar{u} = \frac{\sum c_i}{\sum n_i} = \frac{625}{2500} = 0.250\text{ defects per board}

2. uu-Chart Limits for standard shift (n=100n = 100 boards): CLu=uˉ=0.250 defects/boardCL_u = \bar{u} = 0.250\text{ defects/board} σu=uˉn=0.250100=0.0025=0.050 defects/board\sigma_u = \sqrt{\frac{\bar{u}}{n}} = \sqrt{\frac{0.250}{100}} = \sqrt{0.0025} = 0.050\text{ defects/board} 3σu=3×0.050=0.150 defects/board3\sigma_u = 3 \times 0.050 = 0.150\text{ defects/board} UCLu=uˉ+3σu=0.250+0.150=0.400 defects/boardUCL_u = \bar{u} + 3\sigma_u = 0.250 + 0.150 = 0.400\text{ defects/board} LCLu=uˉ−3σu=0.250−0.150=0.100 defects/boardLCL_u = \bar{u} - 3\sigma_u = 0.250 - 0.150 = 0.100\text{ defects/board}

Notice that because σu\sigma_u is small, the lower control limit is strictly positive (LCLu=0.100>0LCL_u = 0.100 > 0). If a shift exhibits an average defect rate below 0.1000.100 defects/board, it triggers an assignable cause alarm indicating a statistically significant quality improvement!

3. Evaluation of an Unusual Batch: On shift 26, line maintenance reduces production such that only n26=36n_{26} = 36 boards are assembled. Inspectors record 1212 total defects on these 36 boards. Is Shift 26 in statistical control?

  • Observed rate: u26=12/36=0.3333 defects/boardu_{26} = 12 / 36 = 0.3333\text{ defects/board}.
  • Revised standard error for n26=36n_{26} = 36: σu,26=uˉn26=0.25036=0.5006=0.08333 defects/board\sigma_{u,26} = \sqrt{\frac{\bar{u}}{n_{26}}} = \sqrt{\frac{0.250}{36}} = \frac{0.500}{6} = 0.08333\text{ defects/board} UCLu,26=0.250+3(0.08333)=0.250+0.250=0.500 defects/boardUCL_{u,26} = 0.250 + 3(0.08333) = 0.250 + 0.250 = 0.500\text{ defects/board} LCLu,26=max⁡(0,0.250−0.250)=0 defects/boardLCL_{u,26} = \max(0, 0.250 - 0.250) = 0\text{ defects/board}

Since u26=0.3333u_{26} = 0.3333 falls well within [0,0.500][0, 0.500], Shift 26 remains in statistical control despite the smaller sample size.


7. Common PE Exam Pitfalls for Attribute Charts

  1. Using an npnp-Chart with Variable Subgroup Sizes: An npnp-chart tracks the count of defectives, which directly scales with sample size. If an exam problem states that lot sizes fluctuate from 80 to 120 units, selecting an npnp-chart is an immediate disqualifier; you must use a pp-chart.
  2. Confusing Defect Counts with Defective Unit Counts: A part with 4 solder bridges is 1 defective unit, but contributes 4 defects. Watch the problem statement terminology carefully: "fraction nonconforming" or "defective items" dictates pp or npnp; "defects per unit" or "imperfections per roll" dictates cc or uu.
  3. Failing to Truncate Negative Lower Control Limits: When calculations yield LCL=−0.024LCL = -0.024, never select an answer option displaying a negative limit. The lower limit for attribute charts is bounded by zero (LCL=0LCL = 0).
  4. Forgetting n\sqrt{n} in pp-Chart Standard Error: The standard deviation of the proportion is pˉ(1−pˉ)n\sqrt{\frac{\bar{p}(1-\bar{p})}{n}}. Examinees occasionally forget to divide by nn inside the radical or erroneously multiply by nn, confusing σp\sigma_p with σnp\sigma_{np}.
Test Your Knowledge

An industrial inspection station evaluates printed circuit boards (PCBs) for surface-mount soldering defects. Every hour, inspectors examine a sample of 25 completed boards and record the total number of nonconforming (defective) boards. Across 30 historical sampling intervals during stable production, a total of 120 nonconforming boards were identified. The quality team wishes to implement an npnp-chart to monitor the count of defective boards per sample. What are the center line (CLnpCL_{np}) and Upper Control Limit (UCLnpUCL_{np}) for this npnp-chart?

A

CLnp=4.00CL_{np} = 4.00 and UCLnp=7.84UCL_{np} = 7.84

B

CLnp=0.16CL_{np} = 0.16 and UCLnp=0.38UCL_{np} = 0.38

C

CLnp=4.00CL_{np} = 4.00 and UCLnp=9.80UCL_{np} = 9.80

D

CLnp=4.00CL_{np} = 4.00 and UCLnp=9.50UCL_{np} = 9.50

Test Your Knowledge

An aerospace composites manufacturer inspects rolls of carbon fiber fabric for surface weave imperfections (nonconformities). Because rolls vary in surface area, inspectors record the total number of imperfections cic_i and the total inspected area nin_i in square meters for each lot. Across 20 inspected lots totaling 500 m2500\text{ m}^2, inspectors found a cumulative total of 450 imperfections. For a newly produced roll measuring exactly nnew=16 m2n_{\text{new}} = 16\text{ m}^2, what are the Upper Control Limit (UCLuUCL_u) and Lower Control Limit (LCLuLCL_u) on a uu-chart?

A

UCLu=1.112 defects/m2UCL_u = 1.112\text{ defects/m}^2 and LCLu=0.688 defects/m2LCL_u = 0.688\text{ defects/m}^2

B

UCLu=1.611 defects/m2UCL_u = 1.611\text{ defects/m}^2 and LCLu=0.189 defects/m2LCL_u = 0.189\text{ defects/m}^2

C

UCLu=1.611 defects/m2UCL_u = 1.611\text{ defects/m}^2 and LCLu=0.000 defects/m2LCL_u = 0.000\text{ defects/m}^2

D

UCLu=2.809 defects/m2UCL_u = 2.809\text{ defects/m}^2 and LCLu=0.000 defects/m2LCL_u = 0.000\text{ defects/m}^2

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