Question

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

The lower bound for the probability \(P(-2 < X < 8)\) is \(21/25\) or \(0.84\).

Step-by-step solution

Determine the Variance of \(X\)

The variance of a random variable is given by \(\sigma^2 = E(X^2) - [E(X)]^2\). Using the given information, we find \(\sigma^2 = 13 - 3^2 = 4\).

Determine the Standard Deviation of \(X\)

The standard deviation is the square root of the variance. Hence, \(\sigma = \sqrt{4} = 2\).

Apply Chebyshev's Inequality

Chebyshev's inequality states that the probability of \(X\) falling more than \(k\) standard deviations from its mean is at most \(1/k^2\). In this case, we want to find a lower bound for the probability that \(X\) is within \(k=5/2\) standard deviations of the mean. Therefore, the probability that \(X\) falls outside this range is at most \(1/(5/2)^2 = 4/25\). Hence, the probability that \(X\) falls within this range is at least \(1 - 4/25 = 21/25\).