1. **State the problem:** We have test marks $x$ of 15 students with $\sum x = 820$ and standard deviation $\sigma = 6.3$. We need to find: (i) $\sum (x - 53)$ (ii) $\sum (x - 53)^2$ 2. **Calculate the mean $\bar{x}$:** $$\bar{x} = \frac{\sum x}{n} = \frac{820}{15} = 54.67$$ 3. **Calculate $\sum (x - 53)$:** Using the property of summation: $$\sum (x - 53) = \sum x - \sum 53 = 820 - 15 \times 53 = 820 - 795 = 25$$ 4. **Calculate $\sum (x - 53)^2$:** Recall variance formula: $$\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}$$ Given $\sigma = 6.3$, so: $$\sum (x - \bar{x})^2 = n \sigma^2 = 15 \times 6.3^2 = 15 \times 39.69 = 595.35$$ Use the formula to relate $\sum (x - 53)^2$ to $\sum (x - \bar{x})^2$: $$\sum (x - 53)^2 = \sum \big[(x - \bar{x}) + (\bar{x} - 53)\big]^2$$ $$= \sum (x - \bar{x})^2 + 2(\bar{x} - 53) \sum (x - \bar{x}) + n(\bar{x} - 53)^2$$ Since $\sum (x - \bar{x}) = 0$, the middle term is zero: $$\sum (x - 53)^2 = \sum (x - \bar{x})^2 + n(\bar{x} - 53)^2$$ Calculate: $$n(\bar{x} - 53)^2 = 15 \times (54.67 - 53)^2 = 15 \times 1.67^2 = 15 \times 2.7889 = 41.83$$ Therefore: $$\sum (x - 53)^2 = 595.35 + 41.83 = 637.18$$ **Final answers:** (i) $\sum (x - 53) = 25$ (ii) $\sum (x - 53)^2 = 637.18$