Exercise · ZW-CHEBYSHEV-INEQUALITY-002 · ★☆☆ Basic

By Chebyshev, what upper bound holds for \(P(|X-\mu|\ge3\sigma)\)?

Choose one option.

Show answer
  • B. \(\tfrac19\)

Explanation

With \(k=3\): \(1/k^2=1/9\), not \(1/3\).

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Review the concept: Chebyshev inequality

Chebyshev's inequality bounds the probability \(P\) that a variable \(X\) deviates from its mean \(\mu\) by more than \(k\) standard deviations \(\sigma\), using only its variance.

\[P\big(|X-\mu|\ge k\sigma\big)\le\dfrac{1}{k^2}\]

More practice problems: chebyshev inequality

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