To solve this problem using Chebyshev's Theorem, we need to determine the range of quiz scores in which at least 88.9% of the data will reside.

Chebyshev's Theorem states that for any distribution, at least $\frac{1}{k^2}$ of the data lies within $k$ standard deviations of the mean, where $k > 1$.

To find $k$, use the fact that at least 88.9% of the data lies within this range. In decimal form, this is 0.889. Chebyshev’s Theorem tells us that:

$\frac{1}{k^2} = 1 - 0.889 = 0.111$

Now, solve for $k$:

$k^2 = \frac{1}{0.111} = 9$

$k = 3$

This means that at least 88.9% of the data lies within 3 standard deviations of the mean.

Given that the mean ($\mu$) is 7.2 and the standard deviation ($\sigma$) is 0.4, the range can be calculated as follows:

$\text{Lower bound} = \mu - 3\sigma = 7.2 - 3(0.4) = 7.2 - 1.2 = 6$ $\text{Upper bound} = \mu + 3\sigma = 7.2 + 3(0.4) = 7.2 + 1.2 = 8.4$

Thus, at least 88.9% of the quiz scores will lie within the range 6 to 8.4.

Would you like further details or clarification?

Here are some related questions:

1. How would the range change if we used 75% instead of 88.9% for Chebyshev's Theorem?
2. What is the significance of Chebyshev's Theorem in non-normal distributions?
3. How does Chebyshev’s Theorem compare with the Empirical Rule?
4. Can Chebyshev's Theorem be applied to skewed data?
5. What are other applications of Chebyshev's Theorem in statistics?

Tip: Chebyshev's Theorem applies to all data distributions, unlike the Empirical Rule, which assumes a normal distribution.