Let's go through each part of the question step-by-step.

Part (a)

The question asks us to choose the correct Normal model for the tree diameters.

Given:

• Mean diameter (μ) = 10.7 inches
• Standard deviation (σ) = 4.7 inches

Using the 68-95-99.7 rule:

• 68% of data falls within one standard deviation from the mean (μ ± σ).
• 95% of data falls within two standard deviations from the mean (μ ± 2σ).
• 99.7% of data falls within three standard deviations from the mean (μ ± 3σ).

Calculations:

1. Mean ± σ (10.7 ± 4.7) = 6.0 to 15.4
2. Mean ± 2σ (10.7 ± 2×4.7) = 1.3 to 20.1
3. Mean ± 3σ (10.7 ± 3×4.7) = -3.4 to 24.8

Looking at the images provided:

• Model B is the correct one, as it has these ranges:
  • 6.0 to 15.4 for 68%
  • 1.3 to 20.1 for 95%
  • -3.4 to 24.8 for 99.7%.

Part (b)

We need to find the range for the central 95% of tree diameters using the 68-95-99.7 rule.

For the central 95%, we consider μ ± 2σ:

• Mean ± 2σ = 10.7 ± 2×4.7
• Calculation:
  • Lower bound: 10.7 - 9.4 = 1.3
  • Upper bound: 10.7 + 9.4 = 20.1

So, the central 95% of tree diameters are expected to be between 1.3 inches and 20.1 inches.

Final Answers

• (a): Model B.
• (b): The central 95% of the tree diameters are between 1.3 inches and 20.1 inches.

Would you like further clarification on any part of this solution, or additional details?

Here are 5 related questions that expand on this topic:

1. What is the 68-95-99.7 rule, and how does it apply to Normal distributions?
2. How would we calculate the probability of a tree diameter being less than 1 inch?
3. How can we determine the percentage of tree diameters that are greater than 15 inches?
4. What is the impact of a larger standard deviation on the shape of a Normal distribution?
5. If the mean diameter were to increase by 2 inches, how would that change the distribution?

Tip: When using the 68-95-99.7 rule, always confirm your calculations of ±σ, ±2σ, and ±3σ ranges carefully to match the context.