Mean and variance of defective widgets

Question

A production line makes 200 widgets, each independently defective with probability p = 0.03. Let D be the number of defective widgets. What are the mean and variance of D?

Accepted Answer

The number of independent defective/good outcomes with a common defect rate is binomial: D \sim 	ext{Binomial}(n=200,\ p=0.03). Apply the binomial formulas, mean np and variance np(1-p): \mathbb{E}[D] = 200 	imes 0.03 = 6, \qquad \operatorname{Var}(D) = 200 	imes 0.03 	imes 0.97 = 5.82. Standard deviation \sqrt{5.82} \approx 2.41. Expect about 6 defects per batch, plus or minus roughly 2 to 3. Takeaway. With a small defect rate, mean np = 6 and variance np(1-p) = 5.82; the variance is just a hair under the mean because 1-p is close to 1.

Mean and variance of defective widgets

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