Question

**Given:** Mean $= \dfrac{7}{2}$, Variance $= \dfrac{293}{4}$, for the 8 numbers $-10, -7, -1, x, y, 9, 2, 16$. **Step 1 – Use the mean condition:** $$\frac{-10 + (-7) + (-1) + x + y + 9 + 2 + 16}{8} = \frac{7}{2}$$ $$\frac{x + y + 9}{8} = \frac{7}{2} \implies x + y + 9 = 28 \implies x + y = 19 \quad \cdots(1)$$ **Step 2 – Use the variance condition:** $$\text{Variance} = \frac{\sum z_i^2}{8} - \mu^2 = \frac{293}{4}$$ $$\frac{\sum z_i^2}{8} = \frac{293}{4} + \left(\frac{7}{2}\right)^2 = \frac{293}{4} + \frac{49}{4} = \frac{342}{4} = \frac{171}{2}$$ $$\sum z_i^2 = 684$$ Sum of squares of known numbers: $100 + 49 + 1 + 81 + 4 + 256 = 491$ $$x^2 + y^2 = 684 - 491 = 193 \quad \cdots(2)$$ **Step 3 – Solve for $x$ and $y$:** $$(x+y)^2 = x^2 + 2xy + y^2 \implies 361 = 193 + 2xy \implies xy = 84$$ So $x$ and $y$ satisfy: $t^2 - 19t + 84 = 0 \implies (t-12)(t-7) = 0$ $$\therefore\ x = 12,\ y = 7 \quad (\text{or vice versa})$$ **Step 4 – Compute the required mean:** $$x + y + 1 = 20, \quad |x - y| = 5$$ $$\text{Mean} = \frac{12 + 7 + 20 + 5}{4} = \frac{44}{4} = 11$$ **Answer: (A) $11$**