Exercise · ZW-POISSON-DISTRIBUTION-008 · ★★☆ Intermediate
To approximate a binomial with \(n=200\) and \(p=0.01\) by a Poisson, which value of \(\lambda\) is appropriate?
Choose one option.
Show answer
- A. \(2\)
Explanation
The means are matched: \(\lambda=np=200\cdot0.01=2\). The value \(1.98\) is the binomial variance, close but not the usual parameter.
Review the concept: Poisson distribution
A Poisson variable \(X\) counts independent events, at a constant average rate, in a fixed interval; its parameter \(\lambda\) is the mean number of events and \(P(X=k)\) the probability of observing \(k\).
\[P(X=k)=\frac{e^{-\lambda}\lambda^{k}}{k!},\quad k=0,1,2,\dots\]
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