1. The problem statement, all variables and given/known data The number of customers visiting a store during a day is a random variable with mean EX=100and variance Var(X)=225. Using Chebyshev's inequality, find an upper bound for having more than 120 or less than 80customers in a day. That is, find an upper bound on P(X≤80 or X≥120). Using the one-sided Chebyshev inequality (Problem 21), find an upper bound for having 2. Relevant equations P(|x - Ex| ≥ b) ≤ σ2/b2 P(X ≥ a) ≤ σ2 / (σ2 + a2) 3. The attempt at a solution For part 1: P(X≤80 or X≥120) = P(80 ≥ X ≥120) = P(-20 ≥ X - 100 ≥20) = P(|X-100| ≥ 20) By Chebychev's: P(|X-100| ≥ 20) ≥ σ2/b2 thus P(|X-100| ≥ 20) ≥ 225 / 202 = 0.5625 For part 2: Using equation 2... P(X≥121) ≤ 225 / (225 + 1212) = 0.0151 This seems wrong to me because it would seem more reasonable to get a number closer to half of part 1. This answer in part 2 implies that taking just one side of Chebyshev's inequality (the upper half of part 1) and then cutting off the value for 120 reduces it by a half and then almost that entire difference over again... not to mention it imlpies that the majority of the variance isn't even probable. Did I make an arithmetic error?