Question

ANNEXURE A
Question 2.1
Class Frequency
𝑓
Cumulative
frequency
Class midpoint
𝑥
𝑖
𝑥
𝑖
· 𝑓
(0;20] 15 15
(20;40]
(40;60]
(60;80]
(80;100]
(100;120]

Answer 1

To complete the frequency distribution table for the classes provided, we need to fill in the missing frequencies, calculate the cumulative frequency, determine the class midpoints, and then compute \( x_i \cdot f \) for each class.

Let's assume we have the number of observations provided for each class:

1. **Fill in the missing frequencies**:
- Assume we have the following frequency distribution based on sample data:
- (20;40] → 20
- (40;60] → 25
- (60;80] → 30
- (80;100] → 10
- (100;120] → 5

2. **Cumulative Frequency**:
- This is computed by adding the frequency of the current class to the cumulative frequency of the previous class.

3. **Class Midpoint (\(x_i\))**:
- This is calculated as the average of the lower and upper bounds of each class interval.

4. **Calculate \(x_i \cdot f\)**:
- This is calculated by multiplying the class midpoint by the frequency of the respective class.

Let's complete the table step-by-step.

### Step 1: Frequency Distribution Table

| Class | Frequency \(f\) | Cumulative Frequency | Class Midpoint \(x_i\) | \(x_i \cdot f\) |
|--------------|-----------------|----------------------|--------------------------|------------------|
| (0;20] | 15 | 15 | 10 | 150 |
| (20;40] | 20 | 35 | 30 | 600 |
| (40;60] | 25 | 60 | 50 | 1250 |
| (60;80] | 30 | 90 | 70 | 2100 |
| (80;100] | 10 | 100 | 90 | 900 |
| (100;120] | 5 | 105 | 110 | 550 |
| **Total** | **105** | | | **4550** |

### Step 2: Explanation of Calculations
1. **Cumulative Frequency**:
- For (20;40]: 15 + 20 = 35
- For (40;60]: 35 + 25 = 60
- For (60;80]: 60 + 30 = 90
- For (80;100]: 90 + 10 = 100
- For (100;120]: 100 + 5 = 105

2. **Class Midpoint \(x_i\)**:
- (0;20]: \(x_i = \frac{0 + 20}{2} = 10\)
- (20;40]: \(x_i = \frac{20 + 40}{2} = 30\)
- (40;60]: \(x_i = \frac{40 + 60}{2} = 50\)
- (60;80]: \(x_i = \frac{60 + 80}{2} = 70\)
- (80;100]: \(x_i = \frac{80 + 100}{2} = 90\)
- (100;120]: \(x_i = \frac{100 + 120}{2} = 110\)

3. **\(x_i \cdot f\)**:
- For each class, multiply the class midpoint by the frequency for that class.

Now the frequency distribution table is complete. If you have different frequency values, replace the assumed frequency values with your actual data and recalculate the cumulative frequencies, class midpoints, and \(x_i \cdot f\).