Question: The mean score on a Statistics exam is 88 points, with a standard deviation of 4 points. Apply Chebyshev's Theorem to the data using k = 2. Interpret the results.
At least □% of the exam scores fall between □ and □.
1. **State the problem:**
We have a mean $\mu = 88$ and standard deviation $\sigma = 4$. We want to use Chebyshev's Theorem with $k = 2$ to find the minimum percentage of exam scores that fall within $k$ standard deviations of the mean.
2. **Chebyshev's Theorem formula:**
Chebyshev's Theorem states that at least $\left(1 - \frac{1}{k^2}\right) \times 100\%$ of the data lies within $k$ standard deviations of the mean.
3. **Calculate the percentage:**
$$
1 - \frac{1}{k^2} = 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75
$$
So, at least $0.75 \times 100 = 75\%$ of the scores lie within $2$ standard deviations.
4. **Calculate the interval:**
The interval is from
$$
\mu - k\sigma = 88 - 2 \times 4 = 88 - 8 = 80
$$
to
$$
\mu + k\sigma = 88 + 2 \times 4 = 88 + 8 = 96
$$
5. **Interpretation:**
At least 75% of the exam scores fall between 80 and 96 points.
**Final answer:**
At least 75% of the exam scores fall between 80 and 96.