Question

Let the 10 observations be $x_1, x_2, \ldots, x_9, \alpha$. **Given (original data):** $$\text{Mean} = 10 \implies \sum_{i=1}^{9} x_i + \alpha = 100 \implies \sum_{i=1}^{9} x_i = 100 - \alpha \quad \cdots (1)$$ $$\text{Variance} = 2 \implies \frac{\sum x_i^2}{10} - (10)^2 = 2 \implies \frac{\sum x_i^2}{10} = 102 \implies \sum_{i=1}^{10} x_i^2 = 1020$$ $$\therefore \sum_{i=1}^{9} x_i^2 = 1020 - \alpha^2 \quad \cdots (2)$$ **After replacing $\alpha$ with $\beta$:** $$\text{New mean} = 10.1 \implies \frac{(100 - \alpha) + \beta}{10} = 10.1$$ $$\implies \beta - \alpha = 1 \quad \cdots (3)$$ $$\text{New variance} = 1.99 \implies \frac{(1020 - \alpha^2) + \beta^2}{10} - (10.1)^2 = 1.99$$ $$\implies \frac{1020 - \alpha^2 + \beta^2}{10} = 1.99 + 102.01 = 104$$ $$\implies 1020 - \alpha^2 + \beta^2 = 1040$$ $$\implies \beta^2 - \alpha^2 = 20$$ $$\implies (\beta - \alpha)(\beta + \alpha) = 20$$ Substituting from $(3)$: $$1 \times (\alpha + \beta) = 20$$ $$\therefore \alpha + \beta = 20$$ **Answer: (D)**