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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 \(P(-2< X <8)\) is 0.96 according to Chebyshev's inequality.

Step by step solution

01

Determine the mean (µ)

From the given problem, we know that the expected value \(E(X) = 3\). This expected value is also the mean, \(µ\), of the random variable, so \(µ = 3\).
02

Calculate the variance (σ²)

We are given that \(E(X^2) = 13\). To calculate the variance (\(σ²\)), we use the formula: \(σ² = E(X^2) - (E(X))^2\). Plugging in the given values: \(σ² = 13 - (3)^2 = 13 - 9 = 4\).
03

Determine the boundaries of the interval

The random variable \(X\) is bound in the interval \(-2 < X < 8\). The boundaries are \(k_1 = μ - a = 3 -(-2) = 5\) and \(k_2 = b - μ = 8 - 3 = 5\). Both boundaries are 5 units away from the mean; thus, our k is 5 in the inequality.
04

Apply Chebyshev's inequality

Chebyshev's inequality states that the probability that \(X\) deviates from its mean (expected value) by more than \(k\) standard deviations is less than or equal to \(1/k²\). However, if the deviation is less than \(k\) standard deviations, we want the complementary probability, given by: \( P( -k < X - µ < k ) ≥ 1 - 1/k² \). Substituting the values, \(1 - 1/(5)² ≥ 1 - 1/25 = 24/25 = 0.96. \)

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