Exercise · ZW-CHEBYSHEV-INEQUALITY-010 · ★★★ Advanced
With \(k=1\), the Chebyshev bound for \(P(|X-\mu|\ge\sigma)\) is:
Choose one option.
Show answer
- A. \(1\), a bound that gives no information
Explanation
\(1/k^2=1\): every probability is \(\le1\). The inequality is informative only for \(k>1\).
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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- ZW-CHEBYSHEV-INEQUALITY-008 What is the smallest k for which Chebyshev guarantees P(|X-μ|≥ kσ)≤0.04?