To solve the problem based on the Chebyshev inequality given in the image, let's analyze the situation step by step.

Problem Setup:

• The mean height of students is $\mu = 168 \, \text{cm}$.
• The variance is $\sigma^2 = 145 \, \text{cm}^2$, so the standard deviation is $\sigma = \sqrt{145} \, \text{cm} \approx 12.04 \, \text{cm}$.
• We need to use Chebyshev's inequality to find an upper bound for the probability that a randomly selected student's height differs from the mean by at least 23 cm: $P(|X - 168| \geq 23)$

Chebyshev's Inequality Formula:

Chebyshev's inequality states: $P(|X - \mu| \geq k \sigma) \leq \frac{1}{k^2},$ where $k = \frac{\text{distance from the mean}}{\sigma}$.

Step 1: Calculate $k$:

The given distance from the mean is 23 cm. Using the standard deviation $\sigma = \sqrt{145} \approx 12.04$, calculate $k$: $k = \frac{23}{\sigma} = \frac{23}{12.04} \approx 1.91.$

Step 2: Apply Chebyshev's Inequality:

Plug $k = 1.91$ into the inequality: $P(|X - 168| \geq 23) \leq \frac{1}{k^2}.$ Calculate $\frac{1}{k^2}$: $k^2 = (1.91)^2 \approx 3.65, \quad \frac{1}{k^2} \approx \frac{1}{3.65} \approx 0.274.$

Final Answer:

The probability that a randomly chosen student's height differs from the mean by at least 23 cm is bounded by: $P(|X - 168| \geq 23) \leq 0.274.$

This means the probability is at most $27.4\%$.

Would you like further clarification or details on this problem?

Related Questions to Expand Understanding:

1. What is the significance of Chebyshev's inequality in probability theory?
2. How does the bound change if the variance is smaller (e.g., $\sigma^2 = 100$)?
3. How would this problem differ if we assumed a normal distribution instead of applying Chebyshev's inequality?
4. Why is Chebyshev's inequality used for any distribution, and how does it differ from distribution-specific methods?
5. What does the value $k$ represent in terms of standard deviation and distance from the mean?

Tip:

Chebyshev's inequality is very general because it does not assume a specific distribution, but its bounds can be loose compared to methods for specific distributions like the normal distribution.