To solve these problems, we'll use the properties of a normal distribution.

Given:

• Mean (μ): $1200
• Standard deviation (σ): $95

Each problem asks for a percentage related to specific intervals or values within this distribution. To solve, we will compute z-scores for the given values and then use standard normal distribution tables (or calculators) to find the probabilities.

1. Percentage of buyers who paid between $1105 and $1295:

• Calculate z-scores for $1105 and $1295: $z_{1105} = \frac{1105 - 1200}{95} = \frac{-95}{95} = -1$ $z_{1295} = \frac{1295 - 1200}{95} = \frac{95}{95} = 1$
• From z-tables, the cumulative percentage for z = -1 is 15.87% and for z = 1 is 84.13%.
• Percentage between $1105 and $1295 is: $84.13\% - 15.87\% = 68.26\%$

2. Percentage of buyers who paid between $915 and $1200:

• Calculate z-scores for $915 and $1200: $z_{915} = \frac{915 - 1200}{95} = \frac{-285}{95} = -3$ $z_{1200} = \frac{1200 - 1200}{95} = 0$
• From z-tables, the cumulative percentage for z = -3 is 0.13% and for z = 0 is 50%.
• Percentage between $915 and $1200 is: $50\% - 0.13\% = 49.87\%$

3. Percentage of buyers who paid more than $1485:

• Calculate the z-score for $1485: $z_{1485} = \frac{1485 - 1200}{95} = \frac{285}{95} = 3$
• From z-tables, the cumulative percentage for z = 3 is 99.87%.
• Percentage of buyers who paid more than $1485 is: $100\% - 99.87\% = 0.13\%$

4. Percentage of buyers who paid between $1200 and $1390:

• Calculate the z-score for $1390: $z_{1390} = \frac{1390 - 1200}{95} = \frac{190}{95} = 2$
• From z-tables, the cumulative percentage for z = 2 is 97.72% and for z = 0 (at $1200) is 50%.
• Percentage between $1200 and $1390 is: $97.72\% - 50\% = 47.72\%$

5. Percentage of buyers who paid between $1105 and $1200:

• We already know the z-score for $1105 is -1, and for $1200 it is 0.
• Cumulative percentage for z = 0 is 50% and for z = -1 is 15.87%.
• Percentage between $1105 and $1200 is: $50\% - 15.87\% = 34.13\%$

6. Percentage of buyers who paid more than $1390:

• We already know the z-score for $1390 is 2.
• From z-tables, the cumulative percentage for z = 2 is 97.72%.
• Percentage of buyers who paid more than $1390 is: $100\% - 97.72\% = 2.28\%$

Would you like more details on any of these? Here are five related questions you could explore:

1. How do z-scores change if the standard deviation increases?
2. What is the probability for any value less than $915?
3. How do you calculate z-scores when the mean is not a whole number?
4. How do these calculations change for non-normal distributions?
5. What percentage of buyers paid within two standard deviations of the mean?

Tip: A good way to remember the 68-95-99.7 rule is that it approximates percentages within 1, 2, and 3 standard deviations from the mean in a normal distribution.