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. Homework Equations
P(|x - Ex| ≥ b) ≤ σ2/b2
P(X ≥ a) ≤ σ2 / (σ2 + a2)
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)
P(|X-100| ≥ 20) ≥ σ2/b2
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?