Quality Control with Binomial and Normal Approximation
A factory produces circuit boards. Historically 4% are defective. A new production line is installed and a quality inspector tests a random sample of 200 boards, finding 14 defective. The manager wants to know whether this is evidence that the defect rate has changed. The inspector proposes using a binomial model with n = 200 and p = 0.04, and also considers a normal approximation. A second concern is that boards are packed in boxes of 25; the inspector wants the probability that a box contains at least 2 defective boards under the historical rate.
1. Using the binomial model, determine whether 14 defective boards in 200 is unusual under the historical rate, and state a conclusion.
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Under H0, X ~ Binomial(200, 0.04). Mean = np = 8, variance = np(1 - p) = 200 × 0.04 × 0.96 = 7.68, standard deviation ≈ 2.771. The observed 14 is (14 - 8)/2.771 ≈ 2.16 standard deviations above the mean. Using a normal approximation with continuity correction, P(X ≥ 14) ≈ P(Z ≥ (13.5 - 8)/2.771) = P(Z ≥ 1.984) ≈ 0.024. Since this is below 0.05, there is moderate evidence at the 5% level that the defect rate has increased. However, the binomial is skewed for small p, so the normal approximation is only approximate; an exact binomial calculation would give a similar order of magnitude. The conclusion should be cautious: the result is unusual but not extreme, and a single sample of 200 could produce such a result by chance about 2–3% of the time. A larger sample or repeat sampling would strengthen the conclusion.
2. Calculate the probability that a box of 25 boards contains at least 2 defective boards under the historical rate, and explain why a normal approximation may be unreliable here.
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Let Y ~ Binomial(25, 0.04). P(Y ≥ 2) = 1 - P(Y = 0) - P(Y = 1). P(Y = 0) = 0.96^25. ln(0.96) ≈ -0.040822, times 25 = -1.02055, e^-1.02055 ≈ 0.3604. P(Y = 1) = 25 × 0.04 × 0.96^24 = 1 × 0.96^24 ≈ 0.3754. So P(Y ≥ 2) ≈ 1 - 0.3604 - 0.3754 = 0.2642, about 26%. A normal approximation with n = 25 and p = 0.04 gives mean 1 and standard deviation √(25 × 0.04 × 0.96) = √0.96 ≈ 0.98; the rule of thumb np ≥ 5 and n(1 - p) ≥ 5 fails badly (np = 1), so the normal approximation is unreliable. The binomial distribution is strongly right-skewed and discrete, so exact binomial probabilities should be used.