Exercise · ZW-BINOMIAL-DISTRIBUTION-008 · ★★☆ Intermediate

A binomial variable has mean 12 and variance 3. What are \(n\) and \(p\)?

Choose one option.

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  • A. \(n=16,\ p=\tfrac34\)

Explanation

From \(np=12\) and \(np(1-p)=3\) we get \(1-p=\tfrac14\), so \(p=\tfrac34\) and \(n=16\). With \(p=\tfrac14\) the mean would be 4, not 12.

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Review the concept: Binomial distribution

A binomial variable \(X\) counts the successes \(k\) in \(n\) independent Bernoulli trials with the same success probability \(p\); \(P(X=k)\) is the probability of getting \(k\) successes.

\[P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k},\quad k=0,1,\dots,n\]

More practice problems: binomial distribution

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