3. Calculate the mean of the following data :
| Class | 4–7 | 8–11 | 12–15 | 16–19 |
|---|---|---|---|---|
| Frequency | 5 | 4 | 9 | 10 |
\(\displaystyle 12.93\) (approx.)
Step 1: Find the class marks (mid-points)
The class mark of each class is the middle value. It is found by the formula:
\( \text{Class mark} = \dfrac{\text{Lower limit + Upper limit}}{2} \)
For 4–7: \( (4+7)/2 = 5.5 \)
For 8–11: \( (8+11)/2 = 9.5 \)
For 12–15: \( (12+15)/2 = 13.5 \)
For 16–19: \( (16+19)/2 = 17.5 \)
Class marks: 5.5, 9.5, 13.5, 17.5
Step 2: Multiply frequency (f) with class mark (x)
| Class | Frequency (f) | Class mark (x) | f × x |
|---|---|---|---|
| 4–7 | 5 | 5.5 | 27.5 |
| 8–11 | 4 | 9.5 | 38.0 |
| 12–15 | 9 | 13.5 | 121.5 |
| 16–19 | 10 | 17.5 | 175.0 |
Now add the totals:
\( \sum f = 5+4+9+10 = 28 \)
\( \sum f x = 27.5+38+121.5+175 = 362 \)
Step 3: Apply the formula for mean
Formula: \( \bar{x} = \dfrac{\sum f x}{\sum f} \)
\( \bar{x} = \dfrac{362}{28} \)
Step 4: Simplify
\( \bar{x} = 12.93 \) (approx.)
Final Answer: Mean ≈ 12.93