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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.

The graph illustrates a normal distribution for the prices paid for a particular model of HD television. The mean price paid is $1200 and the standard deviation is $95. What are the approximate percentages for buyers in the following price ranges?

Explore the normal distribution for HD television prices with a mean of $1200 and a standard deviation of $95. Calculate the approximate percentage of buyers within various price ranges using Z-scores and the standard normal distribution.

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