Solution
Chebyshev's Guarantee
Show the problem again
For any distribution with finite mean and variance, what is the minimum fraction of observations guaranteed to lie within 2 standard deviations of the mean?
Worked solution
The answer is at least 75%. Chebyshev's inequality states P(|X − μ| ≥ kσ) ≤ 1/k², so at most 1/4 of the probability mass lies beyond 2σ, leaving at least 3/4 within. The bound holds for every distribution (heavy tails, skew, anything), which is why it is much weaker than the normal-specific 95%; it's the price of universality.
Source: Chebyshev's inequality, after Pafnuty Chebyshev; textbook material in probability. Statement written for AxiomIQ.