1. **Problem:** Find the mean of the given frequency distribution using (a) Direct method and (b) Assumed mean method with assumed mean 25. | Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | |-------------|-------|--------|--------|--------|--------|--------| | No. of students | 5 | 10 | 25 | 30 | 20 | 10 | 2. **Formula:** - Mean (Direct) = $$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$ where $f_i$ is frequency and $x_i$ is class midpoint. - Mean (Assumed) = $$\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}$$ where $A$ is assumed mean, $d_i = x_i - A$. 3. **Step 1: Calculate midpoints $x_i$:** - 0-10: $\frac{0+10}{2} = 5$ - 10-20: $15$ - 20-30: $25$ - 30-40: $35$ - 40-50: $45$ - 50-60: $55$ 4. **Step 2: Calculate $f_i x_i$ for direct method:** - $5 \times 5 = 25$ - $10 \times 15 = 150$ - $25 \times 25 = 625$ - $30 \times 35 = 1050$ - $20 \times 45 = 900$ - $10 \times 55 = 550$ Sum of frequencies $\sum f_i = 5 + 10 + 25 + 30 + 20 + 10 = 100$ Sum of $f_i x_i = 25 + 150 + 625 + 1050 + 900 + 550 = 3300$ 5. **Step 3: Calculate mean (Direct):** $$\bar{x} = \frac{3300}{100} = 33$$ 6. **Step 4: Calculate $d_i = x_i - A$ with $A=25$:** - $5 - 25 = -20$ - $15 - 25 = -10$ - $25 - 25 = 0$ - $35 - 25 = 10$ - $45 - 25 = 20$ - $55 - 25 = 30$ 7. **Step 5: Calculate $f_i d_i$:** - $5 \times (-20) = -100$ - $10 \times (-10) = -100$ - $25 \times 0 = 0$ - $30 \times 10 = 300$ - $20 \times 20 = 400$ - $10 \times 30 = 300$ Sum of $f_i d_i = -100 - 100 + 0 + 300 + 400 + 300 = 800$ 8. **Step 6: Calculate mean (Assumed):** $$\bar{x} = 25 + \frac{800}{100} = 25 + 8 = 33$$ **Final answer:** - Mean by direct method = 33 - Mean by assumed mean method = 33 Both methods give the same mean value.