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To approximate the mean for the given frequency distribution, we can use the following formula:

\[
\text{Mean} = \frac{\sum (f \times x)}{\sum f}
\]

where:
- \( f \) is the frequency of each class,
- \( x \) is the midpoint of each class.

### Steps:
1. **Find the midpoint** for each class interval.
2. **Multiply** the midpoint by the corresponding frequency.
3. **Sum** all the products obtained in step 2.
4. **Sum** all the frequencies.
5. **Divide** the total from step 3 by the total from step 4 to get the mean.

Let's apply these steps:

#### Table:
| Data (Class Interval) | Midpoint \( x \) | Frequency \( f \) | \( f \times x \) |
|-----------------------|------------------|-------------------|------------------|
| 60 - 64               | 62               | 1                 | 62               |
| 65 - 69               | 67               | 2                 | 134              |
| 70 - 74               | 72               | 4                 | 288              |
| 75 - 79               | 77               | 15                | 1155             |
| 80 - 84               | 82               | 20                | 1640             |
| 85 - 89               | 87               | 13                | 1131             |
| 90 - 94               | 92               | 10                | 920              |
| 95 - 99               | 97               | 5                 | 485              |
| 100 - 104             | 102              | 5                 | 510              |

### Calculations:
- **Sum of \( f \times x \)**: \( 62 + 134 + 288 + 1155 + 1640 + 1131 + 920 + 485 + 510 = 6325 \)
- **Sum of \( f \)**: \( 1 + 2 + 4 + 15 + 20 + 13 + 10 + 5 + 5 = 75 \)

### Mean:
\[
\text{Mean} = \frac{6325}{75} \approx 84.3
\]

So, the approximate mean is **84.3**.

Would you like more details or have any questions? Here are some related questions to consider:

1. How would you calculate the variance for this dataset?
2. What is the significance of using the midpoint in the calculation of the mean for grouped data?
3. How would the mean change if one of the frequencies were different?
4. Can you describe a situation where the mean might not be the best measure of central tendency?
5. How does the mean compare with the median in this distribution?

**Tip:** Always double-check that the midpoints are correctly calculated for accuracy in your mean computation.

Approximate the mean for the following frequency distribution.

Learn how to approximate the mean for a given frequency distribution using the midpoint method. This problem involves the use of a key statistical formula and is suitable for students in Grades 9-12.

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