The problem in the image asks to apply Chebyshev's Theorem to a Statistics exam with a mean score of 72 points and a standard deviation of 4 points, where $k = 2$.

Step-by-step solution using Chebyshev’s Theorem:

Chebyshev’s Theorem states that for any dataset (regardless of distribution), the proportion of values within $k$ standard deviations from the mean is at least: $1 - \frac{1}{k^2}$

Given:

• Mean $\mu = 72$
• Standard deviation $\sigma = 4$
• $k = 2$

The theorem guarantees that at least: $1 - \frac{1}{2^2} = 1 - \frac{1}{4} = 0.75$ So, at least 75% of the exam scores will lie within 2 standard deviations from the mean.

Range of scores:

• The range is calculated as: $\mu \pm k \sigma = 72 \pm 2(4) = 72 \pm 8$ Thus, at least 75% of the exam scores fall between 64 and 80.

Final Interpretation:

At least 75% of the exam scores will fall between 64 and 80 points.

Would you like further details or have any questions about this process?

Here are five relative questions to deepen understanding:

1. How does Chebyshev's Theorem differ from the Empirical Rule?
2. What would the percentage be if $k = 3$?
3. Can Chebyshev's Theorem be used for normally distributed data?
4. How can the theorem be applied if the standard deviation changes?
5. What are some practical situations where Chebyshev's Theorem is used?

Tip: Chebyshev's Theorem applies to any distribution, not just normal distributions, making it a versatile tool in statistics!