Normal Approximation to a Binomial Quality-Control Problem
A factory produces components with a 4% defect rate, independently. A batch of 500 components is inspected. A quality engineer uses a normal approximation with continuity correction to estimate the probability that at least 30 are defective. A manager instead uses the exact binomial and claims the normal approximation is 'invalid because n is only 500'. The engineer must justify the approximation's validity, compute the estimate, and compare it with the exact binomial result to assess the error.
1. Justify the use of a normal approximation here and state the mean and standard deviation.
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Let X ~ Binomial(n = 500, p = 0.04). Mean μ = np = 500 × 0.04 = 20. Variance σ² = np(1-p) = 500 × 0.04 × 0.96 = 19.2, so σ = √19.2 ≈ 4.382.
The usual rule of thumb is np ≥ 5 and n(1-p) ≥ 5; here np = 20 and n(1-p) = 480, both well above 5, so the normal approximation should be reasonable. The manager's objection based on n alone is misplaced: what matters is the number of expected successes and failures, not n by itself. However, because p is small (0.04), the distribution is right-skewed, so the approximation will be better near the mean than in the far right tail; the error at X ≥ 30 (about 2.3 standard deviations above the mean) should be modest but not negligible.
2. Compute the continuity-corrected normal estimate of P(X ≥ 30) and compare it with the exact binomial value.
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With continuity correction, P(X ≥ 30) ≈ P(Y ≥ 29.5) for Y ~ Normal(20, 19.2).
z = (29.5 - 20)/4.382 = 9.5/4.382 ≈ 2.168. Then P(Z ≥ 2.168) ≈ 0.0151.
Without continuity correction: z = (30 - 20)/4.382 ≈ 2.282, giving ≈ 0.0112 - noticeably lower, showing the correction matters at this tail.
Exact binomial: P(X ≥ 30) = 1 - P(X ≤ 29). Computing term by term (or by recursion) gives P(X ≤ 29) ≈ 0.9836, so P(X ≥ 30) ≈ 0.0164.
Comparison: the continuity-corrected estimate 0.0151 is within about 0.0013 of the exact 0.0164 (roughly 8% relative error), while the uncorrected estimate 0.0112 is off by about 0.005 (about 32% relative error). The correction substantially improves the tail estimate, confirming the approximation is usable but that tail probabilities should be treated with caution.
3. The manager wants to set a rejection threshold so that a batch is rejected only if the defect count is 'unusually high'. Using the normal approximation, find the smallest integer k such that P(X ≥ k) is at most 0.01, and discuss a limitation of relying on the approximation for this decision.
Read the worked answer
We need P(X ≥ k) ≤ 0.01, i.e. P(Y ≥ k - 0.5) ≤ 0.01 with Y ~ Normal(20, 19.2). The 99th percentile of Z is about 2.326, so k - 0.5 ≥ 20 + 2.326 × 4.382 ≈ 20 + 10.19 = 30.19, giving k ≥ 30.69, so the smallest integer is k = 31.
Check: exact binomial P(X ≥ 31) ≈ 0.0089 and P(X ≥ 30) ≈ 0.0164, so k = 31 does satisfy the 1% requirement exactly as well, which is reassuring.
Limitation: the approximation was used to locate a tail cut-off, precisely where the skew of the binomial (p = 0.04) makes the normal fit weakest; a small error in z or σ could shift k by one, changing the false-rejection rate by a factor of nearly two (0.0164 vs 0.0089). For a decision rule with a stated error rate, the exact binomial should be used to confirm the threshold, with the normal approximation serving only as a first estimate.