1. A fair coin is tossed 100 times. The expected number of heads is 50, and the standard deviation for the number of heads is (100 · 1/2 · 1/2)1/2 = 5. What does Chebyshev’s Inequality tell you about the probability that the number of heads that turn up deviates from the expected number 50 by three or more standard deviations (i.e., by at least 15)?

Let X by any random variable with E(X) = μ and V (X) = σ2. Then, if ε = kσ, Chebyshev’s Inequality states that

\[P(|X-μ|≥kσ)≤ \frac{σ^2}{k^2σ^2} = \frac{1}{k^2}\] Thus, for any random variable, the probability of a deviation from the mean of more than k standard deviations is ≤ 1/k2.

In our example, X = 50 and σ = 5, and k=3. Therefore…

\[P(|50-μ|≥15)≤ \frac{5^2}{3^25^2} = \frac{1}{3^2} = \frac{1}{9}\]