10(ii)
Using short-cut method, find mean of the given frequency distribution:
| Class | Frequency |
|---|---|
| 20 - 30 | 6 |
| 30 - 40 | 9 |
| 40 - 50 | 14 |
| 50 - 60 | 10 |
| 60 - 70 | 7 |
| 70 - 80 | 4 |
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Construct the table as under, taking assumed mean as 45:
| Class | Class Mark (yi) | Deviation (di = yi - a) | Frequency(fi) | fidi |
|---|---|---|---|---|
| 20 - 30 | 25 | -20 | 6 | -120 |
| 30 - 40 | 35 | -10 | 9 | -90 |
| 40 - 50 | 45 | 0 | 14 | 0 |
| 50 - 60 | 55 | 10 | 10 | 100 |
| 60 - 70 | 65 | 20 | 7 | 140 |
| 70 - 80 | 75 | 30 | 4 | 120 |
| Total | Σfi = 50 | Σfidi = 150 |
By formula,
$\text{Mean} = a + \frac{\sum f_id_i}{\sum f_i}$
$= 45 + \frac{150}{50}$
= 45 + 3
= 48.
Hence, required mean = 48.