Exercise · ZW-CHEBYSHEV-INEQUALITY-004 · ★☆☆ Basic
If \(\operatorname{Var}(X)=\tfrac14\), the Chebyshev bound for \(P(|X-\mu|\ge1)\) is:
Choose one option.
Show answer
- A. \(\tfrac14\)
Explanation
The equivalent form is \(P(|X-\mu|\ge\varepsilon)\le\sigma^2/\varepsilon^2=\tfrac14/1=\tfrac14\).
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-007 What must be known about X to apply the Chebyshev inequality?
- ZW-CHEBYSHEV-INEQUALITY-008 What is the smallest k for which Chebyshev guarantees P(|X-μ|≥ kσ)≤0.04?
- ZW-CHEBYSHEV-INEQUALITY-009 If Var(X)=9, for which ε does the bound σ^2/ε^2 on P(|X-μ|≥ε) equal exactly tfrac14?