Exercise · ZW-CHEBYSHEV-INEQUALITY-001 · ★☆☆ Basic
By Chebyshev, what upper bound holds for \(P(|X-\mu|\ge2\sigma)\)?
Choose one option.
Show answer
- A. \(\tfrac14\)
Explanation
With \(k=2\), the bound is \(1/k^2=1/4\).
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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