Suppose that X is a random variable with E(X)=10, Var(X)=2. Using Chebyshev inequality, estimate the probability P(7

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### Example Problem: Applying Chebyshev's Inequality

#### Problem Statement:
Suppose that \( X \) is a random variable with \( E(X) = 10 \) and \( \text{Var}(X) = 2 \). Using Chebyshev's inequality, estimate the probability \( P(7 < X < 13) \).

### Explanation for Students:

Chebyshev's inequality provides a way to estimate the probability that a random variable lies within a certain number of standard deviations from the mean. The inequality states that for any random variable \( X \) with mean \( \mu \) and variance \( \sigma^2 \), and for any \( k > 0 \):

\[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \]

We are given the following information:
- Mean \( E(X) = \mu = 10 \)
- Variance \( \text{Var}(X) = \sigma^2 = 2 \), hence the standard deviation \( \sigma = \sqrt{2} \)

To find \( P(7 < X < 13) \), we rewrite this probability using Chebyshev's inequality.

First, express the interval \( 7 < X < 13 \) in terms of the deviation from the mean \( \mu \):
\[ 7 < X < 13 \]
\[ \Rightarrow -3 < X - 10 < 3 \]
\[ \Rightarrow |X - 10| < 3 \]

Thus, we want to find \( P(|X - 10| < 3) \).

Using Chebyshev's inequality:
\[ P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2} \]

Here, \( k \sigma = 3 \).
So,
\[ k = \frac{3}{\sigma} = \frac{3}{\sqrt{2}} \]

Now, apply Chebyshev's inequality:
\[ P\left(|X - 10| \geq \frac{3}{\sqrt{2}} \cdot \sqrt{2}\right) \leq \frac{\sigma^2}{(3 / \sqrt{2})^2} \leq \frac{1}{(\frac{3}{\sqrt{2}})^2
Transcribed Image Text:### Example Problem: Applying Chebyshev's Inequality #### Problem Statement: Suppose that \( X \) is a random variable with \( E(X) = 10 \) and \( \text{Var}(X) = 2 \). Using Chebyshev's inequality, estimate the probability \( P(7 < X < 13) \). ### Explanation for Students: Chebyshev's inequality provides a way to estimate the probability that a random variable lies within a certain number of standard deviations from the mean. The inequality states that for any random variable \( X \) with mean \( \mu \) and variance \( \sigma^2 \), and for any \( k > 0 \): \[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \] We are given the following information: - Mean \( E(X) = \mu = 10 \) - Variance \( \text{Var}(X) = \sigma^2 = 2 \), hence the standard deviation \( \sigma = \sqrt{2} \) To find \( P(7 < X < 13) \), we rewrite this probability using Chebyshev's inequality. First, express the interval \( 7 < X < 13 \) in terms of the deviation from the mean \( \mu \): \[ 7 < X < 13 \] \[ \Rightarrow -3 < X - 10 < 3 \] \[ \Rightarrow |X - 10| < 3 \] Thus, we want to find \( P(|X - 10| < 3) \). Using Chebyshev's inequality: \[ P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2} \] Here, \( k \sigma = 3 \). So, \[ k = \frac{3}{\sigma} = \frac{3}{\sqrt{2}} \] Now, apply Chebyshev's inequality: \[ P\left(|X - 10| \geq \frac{3}{\sqrt{2}} \cdot \sqrt{2}\right) \leq \frac{\sigma^2}{(3 / \sqrt{2})^2} \leq \frac{1}{(\frac{3}{\sqrt{2}})^2
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