First, let's convert the cumulative frequency table into a regular frequency distribution. It shows that 6 students scored less than 10, i.e., 6 students scored between 0â10. Also, 10 students scored less than 20, so (10â6) = 4 students scored between 10â20. Thus, the frequency table becomes: | Marks Range | 0â10 | 10â20 | 20â30 | 30â40 | 40â50 | |-------------|------|-------|-------|-------|-------| | No. of Students | 6 | 4 | 8 | 12 | 16 | Let \(a = 25\) and \(h = 10\), then: | Class Interval | Frequency (\(f_i\)) | Midpoint (\(x_i\)) | \(u_i = \frac{x_i - a}{10}\) | \(f_i u_i\) | |----------------|---------------------|--------------------|-----------------------------|-------------| | 0â10 | 6 | 5 | -2 | -12 | | 10â20 | 4 | 15 | -1 | -4 | | 20â30 | 8 | 25 = \(a\) | 0 | 0 | | 30â40 | 12 | 35 | 1 | 12 | | 40â50 | 16 | 45 | 2 | 32 | | **Total** | \(\sum f_i = 46\) | | | \(\sum f_i u_i = 28\) | \[ \therefore \text{Average marks} = a + h \times \frac{\sum f_i u_i}{\sum f_i} = 25 + 10 \times \frac{28}{46} = 25 + 6.09 = 31.09 \quad \text{(approx)} \quad \text{(Answer)} \]