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 $$ P(|X-8| \geq 6) $$.

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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.
The lower bound of $$ P(|X-8| \geq 6) $$ is 0.25.

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 \( P(|X-8| \geq 6) \).

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