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The question from the image involves a normal distribution, where the mean price paid is \$2000 and the standard deviation is \$105. Let's address the specific parts:

### Key information:
- **Mean (μ)** = \$2000
- **Standard deviation (σ)** = \$105

### Normal distribution properties:
In normal distribution, data is symmetrically distributed around the mean. Standard scores (z-scores) help us find the percentage of data within specific intervals. Z-scores are calculated as:

\[
Z = \frac{X - \mu}{\sigma}
\]

where:
- $X$ is the value in question,
- $\mu$ is the mean,
- $\sigma$ is the standard deviation.

Let's go through the questions one by one:

---

1. **Percentage of buyers who paid between \$2000 and \$2105:**
   - Calculate the Z-score for \$2105:
     \[
     Z = \frac{2105 - 2000}{105} \approx 1
     \]
   - From Z-tables, the area under the curve between $Z = 0$ and $Z = 1$ is **34.13%** (not 47.72% as provided).

2. **Percentage of buyers who paid between \$2000 and \$2210:**
   - Calculate the Z-score for \$2210:
     \[
     Z = \frac{2210 - 2000}{105} \approx 2
     \]
   - From Z-tables, the area under the curve between $Z = 0$ and $Z = 2$ is **47.72%** (as shown).

3. **Percentage of buyers who paid less than \$1790:**
   - Calculate the Z-score for \$1790:
     \[
     Z = \frac{1790 - 2000}{105} \approx -2
     \]
   - The percentage of data below $Z = -2$ is **2.28%**.

4. **Percentage of buyers who paid between \$1895 and \$2105:**
   - For \$1895:
     \[
     Z = \frac{1895 - 2000}{105} \approx -1
     \]
   - For \$2105:
     \[
     Z = 1 \quad (\text{from the first part})
     \]
   - The area between $Z = -1$ and $Z = 1$ is about **68.26%**.

5. **Percentage of buyers who paid less than \$1685:**
   - Calculate the Z-score for \$1685:
     \[
     Z = \frac{1685 - 2000}{105} \approx -3
     \]
   - The percentage of data below $Z = -3$ is **0.13%**.

6. **Percentage of buyers who paid between \$2000 and \$2315:**
   - Calculate the Z-score for \$2315:
     \[
     Z = \frac{2315 - 2000}{105} \approx 3
     \]
   - The area between $Z = 0$ and $Z = 3$ is about **49.87%** (this is correctly marked in the image).

---

Would you like more details on any part of this solution or a breakdown of the Z-scores and percentages? Here are some follow-up questions for further understanding:

1. How do Z-scores relate to percentages in a normal distribution?
2. What is the difference between a Z-table and a standard normal distribution graph?
3. Why is the area under the curve between different Z-scores important?
4. Can you calculate probabilities for values not centered around the mean?
5. How would the results change if the standard deviation were larger or smaller?

**Tip**: For normal distributions, always visualize the Z-scores to get an intuitive understanding of how far from the mean a value lies!

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $2000 and the standard deviation is $105. What are the percentages of buyers within various price ranges?

This problem demonstrates the use of normal distribution to calculate the percentage of buyers paying within specific price ranges for HD televisions, with a mean price of $2000 and standard deviation of $105.

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