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I see that you have a frequency distribution table in the image with the goal of finding the mean, median, and modal class. Let's go step by step to fill the table and calculate what is required.

The table has the following incomplete columns:
- **Ages** (Class intervals)
- **Frequency** (F)
- **Class Mark (X)**: This is the midpoint of each class interval and can be calculated by averaging the lower and upper boundaries of the class.
- **F × X**: This is the product of the frequency and class mark, used to calculate the mean.

### Step 1: Fill the Class Mark (X)
Class mark = \(\frac{\text{Lower boundary} + \text{Upper boundary}}{2}\).

1. \(7 - 3: \frac{3 + 7}{2} = 5\)
2. \(4 - 6: \frac{4 + 6}{2} = 5\)
3. \(7 - 9: \frac{7 + 9}{2} = 8\)
4. \(10 - 12: \frac{10 + 12}{2} = 11\)
5. \(13 - 15: \frac{13 + 15}{2} = 14\)

### Step 2: Fill F × X for each row
1. \(2 \times 5 = 10\)
2. \(5 \times 5 = 25\)
3. \(10 \times 8 = 80\)
4. \(5 \times 11 = 55\)
5. \(3 \times 14 = 42\)

### Step 3: Calculate the Summations
- Total frequency \(\sum F = 2 + 5 + 10 + 5 + 3 = 25\)
- Total of \(F \times X\): \(\sum F \times X = 10 + 25 + 80 + 55 + 42 = 212\)

### Step 4: Find the Mean
Mean = \(\frac{\sum F \times X}{\sum F} = \frac{212}{25} = 8.48\)

### Step 5: Median Class
- The cumulative frequency (CF) is used to find the median.
  - Cumulative frequency for each class:
    - For \(7 - 3\), CF = 2
    - For \(4 - 6\), CF = 2 + 5 = 7
    - For \(7 - 9\), CF = 7 + 10 = 17
    - For \(10 - 12\), CF = 17 + 5 = 22
    - For \(13 - 15\), CF = 22 + 3 = 25

- Median class is the class where \(\frac{N}{2} = \frac{25}{2} = 12.5\). The class where the cumulative frequency exceeds 12.5 is \(7 - 9\).

### Step 6: Modal Class
The modal class is the one with the highest frequency, which is \(7 - 9\) (frequency = 10).

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### Summary:
- **Mean**: 8.48
- **Median Class**: \(7 - 9\)
- **Modal Class**: \(7 - 9\)

Let me know if you'd like more details or have any questions. 

Here are five questions related to the above concepts:
1. What does the class mark represent in a frequency distribution?
2. How would you calculate the median if the class intervals were different sizes?
3. Why is the modal class important in data analysis?
4. How does the cumulative frequency help in identifying the median class?
5. What is the relationship between mean, median, and mode in a normal distribution?

**Tip**: Always ensure to list class intervals and frequencies accurately to avoid errors in calculating the mean, median, and mode.

Copy and fill the frequency distribution table and use it to find the mean, median, and modal class of the data.

Learn how to find the mean, median, and modal class from a frequency distribution table using statistical concepts. This example includes a step-by-step solution for calculating these values for Grades 8-10 students.

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