Quality control, hypergeometric vs binomial

Question

For the quality control of the example ($N=1000$, $K=20$, $n=50$): compare the hypergeometric $P(X=0)$ with the approximating binomial $B(50;\,0{,}02)$. Comment on the error.

Solution

Hypergeometric: $P_{\text{Hyp}}(X=0) = \frac{\binom{980}{50}}{\binom{1000}{50}} \approx 0{,}358.$ Binomial $B(50;\,0{,}02)$ with $p = K/N = 20/1000 = 0{,}02$: $P_{B}(X=0) = (1-0{,}02)^{50} = 0{,}98^{50} \approx 0{,}364.$ The absolute error is about $0{,}006$ (just over $1{,}5\%$ in relative terms): the approximation is excellent. The reason is the rule of thumb: the sample is $n/N = 50/1000 = 5\%$ of the population, below the $10\%$ threshold, so replacement and non-replacement give almost identical results. $\boxed{P_{\text{Hyp}}\approx 0{,}358,\ P_B\approx 0{,}364}$

Links

Topics: Probability distributions
Concepts: Binomial distribution · Hypergeometric distribution
Methods: Binomial · Hypergeometric distribution
Skills: Limiting-case analysis · Probability calculation
Exercise type: Probability problem