2. Answer the following questions. A random variable \( X \) has a mean 8 , a variance 9 , and an unknown probability distribution. Using Chebyshev's theorem, find the lower bound of \( P(|X-8| \geq 6) \).

To find the lower bound of \( P(|X-8| \geq 6) \) using Chebyshev's theorem, we first need to understand the theorem and the given information. Chebyshev's theorem states that for any random variable \( X \) with mean \( \mu \) and variance \( \sigma^2 \), the following inequality holds: \[ P(|X - \mu| \geq k) \leq \frac{\sigma^2}{k^2} \] where \( k \) is a positive real number. Given that the mean of \( X \) is 8 and the variance is 9, we can substitute these values into the inequality: \[ P(|X - 8| \geq 6) \leq \frac{9}{6^2} \] Now, we can calculate the lower bound of \( P(|X-8| \geq 6) \) using the inequality. Calculate the value by following steps: - step0: Calculate: \(\frac{9}{6^{2}}\) - step1: Rewrite the expression: \(\frac{9}{3^{2}\times 2^{2}}\) - step2: Rewrite the expression: \(\frac{3^{2}}{3^{2}\times 2^{2}}\) - step3: Reduce the fraction: \(\frac{1}{2^{2}}\) - step4: Evaluate the power: \(\frac{1}{4}\) The lower bound of \( P(|X-8| \geq 6) \) using Chebyshev's theorem is 0.25.