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Chebyshev's inequality: P(|X - μ| ≥ kσ) ≤
A1/k
B1/k² (Chebyshev)
Ck/n
Dk²
Answer & Solution
Correct answer: B. 1/k² (Chebyshev)
Chebyshev: for any distribution with finite variance: P(|X-μ| ≥ kσ) ≤ 1/k². With k=2: ≤ 25% beyond 2σ. Holds without normality assumption.
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