Chapter 49 · Case Study
Manufacturing Defects: the Binomial 🏭
Each batch is 50 parts, and each part passes or fails inspection. The count of defects per batch is the binomial distribution. We estimate the defect rate, fit the binomial, and size the risk of a bad batch.
Statistics, Data Science and AI: A Visual Handbook · John Fisher · 2026
⚙️ Setup & data¶
In [1]:
import numpy as np, pandas as pd, matplotlib.pyplot as plt
from scipy import stats
from math import comb
AMBER="#d97706"; TEAL="#0d9488"; INK="#1a2138"; GRID="#e6e9f2"; PINK="#db2777"
plt.rcParams.update({"figure.facecolor":"white","axes.facecolor":"white","figure.dpi":110,"font.size":11,
"axes.edgecolor":GRID,"axes.grid":True,"grid.color":GRID,"axes.axisbelow":True,"axes.spines.top":False,
"axes.spines.right":False,"axes.titlesize":12,"axes.titleweight":"bold","legend.frameon":False})
BASE="https://raw.githubusercontent.com/johnfisher-ai/Statistics-Data-Science-AI-Visual-Book/main/data/"
try: df = pd.read_csv("../../data/manufacturing_qc_inspections.csv")
except FileNotFoundError: df = pd.read_csv(BASE+"manufacturing_qc_inspections.csv")
print("loaded:", df.shape)
df.head()
loaded: (1000, 5)
Out[1]:
| batch_id | inspection_date | inspector_id | batch_size | defective_count | |
|---|---|---|---|---|---|
| 0 | BATCH_50000 | 2026-01-01 | INS_04 | 50 | 1 |
| 1 | BATCH_50001 | 2026-01-01 | INS_04 | 50 | 4 |
| 2 | BATCH_50002 | 2026-01-01 | INS_04 | 50 | 3 |
| 3 | BATCH_50003 | 2026-01-01 | INS_03 | 50 | 1 |
| 4 | BATCH_50004 | 2026-01-01 | INS_01 | 50 | 1 |
BEAT 1 · n TRIALS, ONE RATE
From the part rate to the batch count
A binomial counts the successes in n independent Bernoulli trials. Here n=50 parts per batch, and a 'success' is a defect. The per-part defect rate p is the total defects over the total parts.
In [2]:
n = int(df["batch_size"].iloc[0])
p = df["defective_count"].sum() / (n*len(df))
print(f"batches : {len(df):,}")
print(f"batch size n : {n}")
print(f"per-part defect rate p = {p:.4f} ({p:.2%})")
print(f"Binomial mean = n p = {n*p:.3f}")
print(f"Binomial variance = n p(1-p) = {n*p*(1-p):.3f}")
print(f"observed mean defects/batch = {df.defective_count.mean():.3f} (matches n p)")
batches : 1,000 batch size n : 50 per-part defect rate p = 0.0385 (3.85%) Binomial mean = n p = 1.927 Binomial variance = n p(1-p) = 1.853 observed mean defects/batch = 1.927 (matches n p)
In [3]:
fig,ax=plt.subplots(figsize=(7,3.4))
kmax=int(df.defective_count.max())+1; k=np.arange(0,kmax+1)
obs=df.defective_count.value_counts(normalize=True).sort_index()
ax.bar(obs.index, obs.values, color=AMBER, alpha=0.75, label="observed batches", width=0.8)
ax.plot(k, stats.binom.pmf(k,n,p),"o-",color=TEAL,ms=5,label=f"Binomial(50, {p:.3f})")
ax.set_xlabel("defects per batch"); ax.set_ylabel("probability"); ax.set_title("Defects per batch: observed vs binomial"); ax.legend()
plt.tight_layout(); plt.show()
In [4]:
# Formal goodness-of-fit: chi-square on observed vs expected counts.
# Rule of thumb: merge adjacent bins so every EXPECTED count is at least 5, then
# chi-square = sum((obs-exp)**2 / exp), with df = (bins - 1 - number of fitted parameters).
def chisq_gof(counts, pmf, n_params):
N = len(counts); kmax = int(counts.max())
obs = np.array([(counts == k).sum() for k in range(kmax + 1)], float)
exp = np.array([N * pmf(k) for k in range(kmax + 1)], float)
obs = np.append(obs, 0.0); exp = np.append(exp, N - exp.sum()) # tail bin for k > kmax
mo, me, co, ce = [], [], 0.0, 0.0
for o, e in zip(obs, exp):
co += o; ce += e
if ce >= 5:
mo.append(co); me.append(ce); co = ce = 0.0
if ce > 0 and mo:
mo[-1] += co; me[-1] += ce
mo, me = np.array(mo), np.array(me)
chi2 = ((mo - me) ** 2 / me).sum(); dof = len(mo) - 1 - n_params
return chi2, dof, stats.chi2.sf(chi2, dof)
chi2, dof, pval = chisq_gof(df.defective_count.values, lambda k: stats.binom.pmf(k, n, p), 1)
print(f"Chi-square goodness-of-fit: chi2 = {chi2:.2f}, df = {dof}, p = {pval:.3f} -> "
+ ("cannot reject the fit (Binomial is a good model)" if pval > 0.05 else "reject: not Binomial"))
Chi-square goodness-of-fit: chi2 = 5.18, df = 5, p = 0.394 -> cannot reject the fit (Binomial is a good model)
The bars line up, but validate it: a chi-square goodness-of-fit test compares the observed count of batches with 0, 1, 2, ... defects against the counts a Binomial(n, p) predicts (merging rare tail bins so every expected count is at least 5). Here chi-square = 5.18 on 5 df, p = 0.39, so we cannot reject the Binomial, the model is validated, not just drawn.
The binomial PMF, built from C(50, k) pᵏ(1−p)⁷⁻ᵏ, lands almost perfectly on the observed batch counts. The combinatorial term C(50, k) counts the ways k defects can fall among 50 parts, the counting from Chapter 32 at work.
BEAT 2 · THE RISK OF A BAD BATCH
CDF for cumulative boundaries
Quality control cares about thresholds. The CDF P(X <= 2) gives the chance a batch passes a 'no more than 2 defects' rule; its complement is the reject risk.
In [5]:
for thr in [0,1,2,3,5]:
print(f"P(X <= {thr}) = {stats.binom.cdf(thr,n,p):.4f} P(X > {thr}) = {stats.binom.sf(thr,n,p):.4f}")
print()
# exact hand calculation of P(X=2) via the formula, to confirm scipy
k=2; manual = comb(n,k)*p**k*(1-p)**(n-k)
print(f"manual P(X=2) = C(50,2) p^2 (1-p)^48 = {manual:.4f} (scipy {stats.binom.pmf(2,n,p):.4f})")
P(X <= 0) = 0.1401 P(X > 0) = 0.8599 P(X <= 1) = 0.4210 P(X > 1) = 0.5790 P(X <= 2) = 0.6969 P(X > 2) = 0.3031 P(X <= 3) = 0.8738 P(X > 3) = 0.1262 P(X <= 5) = 0.9878 P(X > 5) = 0.0122 manual P(X=2) = C(50,2) p^2 (1-p)^48 = 0.2758 (scipy 0.2758)
About 70% of batches have 2 or fewer defects, so a "reject if more than 2" rule would scrap roughly 30% of production, expensive. The hand-computed P(X=2) using C(50,2) matches SciPy exactly, confirming the model. Tuning the threshold trades scrap rate against quality, a decision the CDF makes precise.
BEAT 3 · SIMULATE A PRODUCTION RUN
Long-run defect totals
Stack many batches and the daily defect total is a sum of binomials, itself approximately normal by the CLT. We simulate a production run to see the spread of outcomes.
In [6]:
rng = np.random.default_rng(49)
sim = rng.binomial(n, p, size=200_000)
print(f"simulated mean defects/batch = {sim.mean():.3f} (theory {n*p:.3f})")
print(f"P(a batch is defect-free) = {stats.binom.pmf(0,n,p):.4f} ({(sim==0).mean():.4f} simulated)")
run=30; daily=rng.binomial(n,p,size=(100_000,run)).sum(axis=1)
print(f"over {run} batches/day: expected total = {run*n*p:.1f}, P(>80 defects) = {(daily>80).mean():.4f}")
simulated mean defects/batch = 1.924 (theory 1.927) P(a batch is defect-free) = 0.1401 (0.1410 simulated) over 30 batches/day: expected total = 57.8, P(>80 defects) = 0.0023
A defect-free batch happens about 14% of the time, P(X=0) = (1−p)⁵⁰. Across a 30-batch day the total defects center near 58 with predictable spread, letting the plant set staffing and rework budgets from the distribution rather than from a guess.
✅ Case closed
Defects per fixed batch are binomial. We estimated the per-part rate p, fit Binomial(50, p) to the data, used the CDF to size the reject risk (P(X≤2) = 0.70), and simulated production totals. The combinatorial C(n,k) is what turns a per-part rate into a batch-level forecast.
Statistics, Data Science and AI: A Visual Handbook · © 2026 John Fisher