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Chapter 1: Problem 3

If \(X\) is a random variable such that \(E(X)=3\) and \(E\left(X^{2}\right)=13\), use Chebyshev's inequality to determine a lower bound for the probability \(P(-2<\) \(X<8)\)

Short Answer

Expert verified
The lower bound for the probability that X lies between -2 and 8 is 0.36.

Step by step solution

01

Find variance of X.

The variance of a random variable \(X\) is given by \(\sigma^{2} = E\left(X^{2}\right) - [E(X)]^{2}\). Here, we're given that \(E(X) = 3\) and \(E\left(X^{2}\right) = 13\). Hence, \(\sigma^{2} = 13 - 3^{2} = 4 .\)
02

Standardize the range.

We're looking for the probability that \(X\) is between -2 and 8. We can standardize this by calculating how many standard deviations each of these numbers is away from the expected value \(E(X) = 3\). The standardized range is \((-2 - 3) / 4 = -5/4\) to \((8 - 3) / 4 = 5/4\). Hence, we're looking for the probability that \(X\) is less than 5/4 standard deviations away from its mean.
03

Apply Chebyshev's inequality.

By Chebyshev's inequality, the probability that a random variable \(X\) is more than \(k\) standard deviations away from its mean is at most \(1/k^2\). Hence, the probability that \(X\) is within \(k\) standard deviations of its mean \(E(X)\) is at least \(1 - 1/k^2\). Substituting \(k = 5/4\) gives the lower bound for the probability as \(1 - 1/(5/4)^2 = 1 - 16/25 = 9/25 = 0.36.\)
04

Interpret the result.

This means that the probability that \(X\) lies between -2 and 8 is at least 0.36, according to Chebyshev's inequality.

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