1. **Stating the problem:** We have an incomplete frequency distribution table of marks and the total number of students is 50. The mean of the distribution is given as 30.2. | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | Total | |--------------|-------|--------|--------|--------|--------|-------| | No. of students | 4 | x | 10 | y | 10 | 50 | We need to find the missing frequencies $x$ and $y$. 2. **Formula used:** The mean of grouped data is given by $$\text{Mean} = \frac{\sum f_i m_i}{\sum f_i}$$ where $f_i$ is the frequency and $m_i$ is the midpoint of each class interval. 3. **Calculate midpoints:** - For 0-10: $m_1 = \frac{0+10}{2} = 5$ - For 10-20: $m_2 = 15$ - For 20-30: $m_3 = 25$ - For 30-40: $m_4 = 35$ - For 40-50: $m_5 = 45$ 4. **Set up equations:** Let $x$ be the frequency for 10-20 and $y$ for 30-40. Total frequency: $4 + x + 10 + y + 10 = 50 \implies x + y = 26$ Sum of $f_i m_i$: $$4 \times 5 + x \times 15 + 10 \times 25 + y \times 35 + 10 \times 45 = 50 \times 30.2 = 1510$$ Calculate known terms: $$20 + 15x + 250 + 35y + 450 = 1510$$ Simplify: $$15x + 35y + 720 = 1510 \implies 15x + 35y = 790$$ 5. **Solve the system:** From $x + y = 26$, we get $y = 26 - x$. Substitute into the second equation: $$15x + 35(26 - x) = 790$$ $$15x + 910 - 35x = 790$$ $$-20x = 790 - 910 = -120$$ $$x = \frac{120}{20} = 6$$ Then, $$y = 26 - 6 = 20$$ 6. **Final answer:** The missing frequencies are $x = 6$ and $y = 20$. Thus, the complete frequency distribution is: | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | Total | |--------------|-------|--------|--------|--------|--------|-------| | No. of students | 4 | 6 | 10 | 20 | 10 | 50 |