To determine the percentage of students with grades between 65 and 91, we can use Chebyshev’s Theorem since the shape of the distribution is not symmetrical. Chebyshev’s Theorem provides a minimum percentage of data that falls within $k$ standard deviations from the mean for any distribution.

Step 1: Calculate $k$

The mean ($\mu$) is 78, and the standard deviation ($\sigma$) is 12.4. We calculate $k$ as the number of standard deviations for the interval [65, 91]:

1. Distance from 65 to the mean:
   $|65 - 78| = 13$

2. Distance from 91 to the mean:
   $|91 - 78| = 13$

Thus, the interval [65, 91] is 13 units away from the mean on either side. To find $k$: $k = \frac{13}{12.4} \approx 1.048$

Step 2: Apply Chebyshev’s Theorem

Chebyshev’s Theorem states that at least $1 - \frac{1}{k^2}$ of the data lies within $k$ standard deviations of the mean. Substituting $k \approx 1.048$:

$\text{Proportion} = 1 - \frac{1}{(1.048)^2} \approx 1 - \frac{1}{1.098} \approx 1 - 0.91 = 0.09$

This means at least 9% of the data lies outside the interval. Therefore, at least 91% of the grades are within the interval [65, 91].

Final Answer:

At least 91% of students' grades fall between 65 and 91.

Would you like a detailed breakdown of Chebyshev’s Theorem, or to apply this concept to a different question?

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Relative Questions:

1. What is Chebyshev’s Theorem, and why is it useful for non-symmetric distributions?
2. How does the calculation of $k$ vary with different intervals?
3. What happens to the percentage when $k$ increases or decreases?
4. Can this problem be solved using the Empirical Rule if the shape were symmetric?
5. How do you interpret results outside the interval in this context?

Tip:

For symmetric distributions like normal distributions, the Empirical Rule is often quicker and more precise than Chebyshev’s Theorem.