Use Chebyshev's inequality: If X is a random variable with expected value μ and non-zero variance σ2, then for any real number k > 0, P(|X-µ|<k)≥1-σ2/k2.
With µ=20=σ2=k you get
P(|X-20|<20) ≥ 1-20/20² = .95
P(0 < X < 40) ≥ .95
HIma D.
asked 11/23/13
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.