To solve this problem using Chebyshev's theorem, we will work through the statements step by step.

Part (a)

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

Given:

• Mean $\mu = 0.86\%$
• Standard deviation $\sigma = 0.10\%$
• At least 56% of the data is considered.

From Chebyshev's formula:
$1 - \frac{1}{k^2} = 0.56$
Solve for $k$:
$\frac{1}{k^2} = 1 - 0.56 = 0.44$
$k^2 = \frac{1}{0.44} \approx 2.27 \quad \Rightarrow \quad k \approx \sqrt{2.27} \approx 1.51$

Thus, 1.51 standard deviations from the mean will contain at least 56% of the data.

Next, we calculate the range:
$\text{Lower bound} = \mu - k\sigma = 0.86 - 1.51(0.10) = 0.86 - 0.151 = 0.71\%$
$\text{Upper bound} = \mu + k\sigma = 0.86 + 1.51(0.10) = 0.86 + 0.151 = 1.01\%$

Final Answer for (a):

The relative increases in stock price lie between 0.71% and 1.01%.

Part (b)

Now, we are asked to choose the correct percentage of data that lies between 0.66% and 1.06%.

To calculate, find how many standard deviations this range represents:
$k = \frac{\text{distance from the mean}}{\sigma}$

For $0.66\%$ (lower bound):
$k = \frac{0.86 - 0.66}{0.10} = \frac{0.20}{0.10} = 2$

For $1.06\%$ (upper bound):
$k = \frac{1.06 - 0.86}{0.10} = \frac{0.20}{0.10} = 2$

Thus, this range covers 2 standard deviations from the mean.

Using Chebyshev's theorem:
$1 - \frac{1}{k^2} = 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = 0.75$

This means at least 75% of the data lies within 2 standard deviations of the mean.

Final Answer for (b):

The correct choice is 75%.

Summary of Answers:

• (a): 0.71% and 1.01%
• (b): 75%

Let me know if you need further clarification!

Related Questions:

1. What is Chebyshev's theorem and when can it be used?
2. How do you determine $k$ from a percentage?
3. Why does Chebyshev's theorem apply to all distributions?
4. How do you calculate bounds for other percentages (e.g., 36%, 89%)?
5. What is the difference between Chebyshev's theorem and the Empirical Rule?

Tip:

Chebyshev's theorem works for any distribution, not just normal distributions, which makes it very versatile!