If E(X) = 17 and E(X
2
) = 298. Use Chebyshev’s theorem to determine;
(i) a lower bound for P(10 < X < 24) (ii) an upper bound for P(|X − 17| ≥ 16).

To solve the problem using Chebyshev's theorem, we begin by finding the variance of the random variable \( X \), denoted as \( \text{Var}(X) \).

### Given:
- \( E(X) = 17 \)
- \( E(X^2) = 298 \)

We know that:
\[
\text{Var}(X) = E(X^2) - (E(X))^2
\]

Substituting the given values:
\[
\text{Var}(X) = 298 - (17)^2 = 298 - 289 = 9
\]

Thus, the standard deviation \( \sigma = \sqrt{\text{Var}(X)} = \sqrt{9} = 3 \).

### (i) Lower bound for \( P(10 < X < 24) \)

Chebyshev’s theorem states that for any random variable \( X \) with mean \( \mu \) and standard deviation \( \sigma \), the probability that \( X \) lies within \( k \) standard deviations of the mean is at least:

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

Here, the mean \( \mu = 17 \) and the standard deviation \( \sigma = 3 \). The interval \( (10, 24) \) corresponds to \( 17 - 7 = 10 \) and \( 17 + 7 = 24 \), meaning we are looking for the probability within 7 units of the mean.

Thus, \( k = \frac{7}{3} \).

Now applying Chebyshev’s theorem:
\[
P(|X - 17| < 7) \geq 1 - \frac{1}{\left( \frac{7}{3} \right)^2} = 1 - \frac{1}{\frac{49}{9}} = 1 - \frac{9}{49} = \frac{40}{49}
\]

So, the lower bound for \( P(10 < X < 24) \) is \( \frac{40}{49} \) or approximately 0.816.

### (ii) Upper bound for \( P(|X - 17| \geq 16) \)

Again using Chebyshev’s theorem, we know that:

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

Here, \( |X - 17| \geq 16 \) corresponds to \( k = \frac{16}{3} \).

Thus, applying Chebyshev’s inequality:
\[
P(|X - 17| \geq 16) \leq \frac{1}{\left( \frac{16}{3} \right)^2} = \frac{1}{\frac{256}{9}} = \frac{9}{256}
\]

So, the upper bound for \( P(|X - 17| \geq 16) \) is \( \frac{9}{256} \) or approximately 0.035.

---

### Summary:
(i) The lower bound for \( P(10 < X < 24) \) is \( \frac{40}{49} \) or approximately 0.816.  
(ii) The upper bound for \( P(|X - 17| \geq 16) \) is \( \frac{9}{256} \) or approximately 0.035.

Would you like more details or clarifications?

### Related Questions:
1. How is Chebyshev's theorem useful in statistics?
2. Can Chebyshev’s theorem give an exact probability for normal distributions?
3. How does Chebyshev’s inequality compare with the Empirical Rule?
4. What is the significance of the variance in Chebyshev’s theorem?
5. Can Chebyshev’s theorem be used for non-symmetric distributions?

**Tip:** Chebyshev’s theorem is especially useful when you do not know the distribution of the data—it provides a conservative estimate based purely on variance.

If E(X) = 17 and E(X^2) = 298. Use Chebyshev’s theorem to determine; (i) a lower bound for P(10 < X < 24) (ii) an upper bound for P(|X − 17| ≥ 16).

Learn how to apply Chebyshev's theorem to find the probability bounds for P(10 < X < 24) and P(|X − 17| ≥ 16), given E(X) = 17 and E(X^2) = 298. Explore key concepts such as variance and standard deviation, and see detailed step-by-step calculations for these bounds.

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