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## Homework Statement

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

2. Homework Equations

P(|x - Ex| ≥ b) ≤ σ

^{2}/b

^{2}

P(X ≥ a) ≤ σ

^{2}/ (σ

^{2}+ a

^{2})

## 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) ≥ σ

^{2}/b

^{2}

thus

P(|X-100| ≥ 20) ≥ 225 / 20

^{2}= 0.5625

For part 2:

Using equation 2...

P(X≥121) ≤ 225 / (225 + 121

^{2}) = 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?