1. **State the problem:** We are given a frequency distribution with class intervals, midpoints, and frequencies (f). We need to fill in the missing columns d and fd for each class interval. 2. **Identify the variables:** - Midpoints ($x$): 5, 15, 25, 35, 45 - Frequencies ($f$): 11, 29, 36, 22, 8 3. **Calculate the mean midpoint ($ar{x}$):** $$\bar{x} = \frac{\sum f x}{\sum f} = \frac{(11)(5)+(29)(15)+(36)(25)+(22)(35)+(8)(45)}{11+29+36+22+8}$$ Calculate numerator: $$(11 \times 5) = 55$$ $$(29 \times 15) = 435$$ $$(36 \times 25) = 900$$ $$(22 \times 35) = 770$$ $$(8 \times 45) = 360$$ Sum numerator $= 55 + 435 + 900 + 770 + 360 = 2520$ Calculate denominator $= 11 + 29 + 36 + 22 + 8 = 106$ So, $$\bar{x} = \frac{2520}{106} \approx 23.77$$ 4. **Calculate $d = x - \bar{x}$ for each midpoint:** - For $5$: $d = 5 - 23.77 = -18.77$ - For $15$: $d = 15 - 23.77 = -8.77$ - For $25$: $d = 25 - 23.77 = 1.23$ - For $35$: $d = 35 - 23.77 = 11.23$ - For $45$: $d = 45 - 23.77 = 21.23$ 5. **Calculate $fd = f \times d$ for each class:** - For $5$: $fd = 11 \times (-18.77) = -206.47$ - For $15$: $fd = 29 \times (-8.77) = -254.33$ - For $25$: $fd = 36 \times 1.23 = 44.28$ - For $35$: $fd = 22 \times 11.23 = 247.06$ - For $45$: $fd = 8 \times 21.23 = 169.84$ 6. **Summary table:** | Class Interval | Midpoint ($x$) | Frequency ($f$) | $d = x - \bar{x}$ | $fd$ | |---|---|---|---|---| | 0-10 | 5 | 11 | -18.77 | -206.47 | | 10-20 | 15 | 29 | -8.77 | -254.33 | | 20-30 | 25 | 36 | 1.23 | 44.28 | | 30-40 | 35 | 22 | 11.23 | 247.06 | | 40-50 | 45 | 8 | 21.23 | 169.84 | 7. **Check that the sum of $fd$ is (approximately) zero:** $$\sum fd = -206.47 -254.33 + 44.28 + 247.06 + 169.84 \approx 0.38 \approx 0,$$ which is expected because $d$ is deviations from the mean. The filled-in columns give insight into deviations of midpoints from the overall mean weighted by frequency.