Answer: 1/k².
- A 1/k
- B 1/k²
- C k²
- D 1/(2k)
Correct answer: B. 1/k²
Explanation: Chebyshev's inequality (no normality needed): P(|X−μ| ≥ kσ) ≤ 1/k² for any k > 1. Proof uses Markov's inequality on (X−μ)². Equivalently, at least 1−1/k² of data lies within k standard deviations of the mean.
In a perfectly normal (bell-shaped) distribution, mean, median, and mode all coincide at the centre; about 68% of data falls within 1 standard deviation of the mean, 95% within 2, and 99.7% within 3 - the empirical rule used to judge how typical or extreme a value is.
Concept context
Mean, median, mode, standard deviation, and data interpretation