The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y where x <...

Mean $= \dfrac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$

$⇒ x + y = 12 \ldots \left(1\right)$

Variance $= \dfrac{∑ {x}_{i}^{2}}{n} - {\left(\overset{¯}{x}\right)}^{2}$

$⇒ \dfrac{9}{4} = \dfrac{{4}^{2} + {5}^{2} + {6}^{2} + {6}^{2} + {7}^{2} + {8}^{2} + {x}^{2} + {y}^{2}}{8} - {\left(6\right)}$^{2}$</p><p>$⇒ {x}^{2} + {y}^{2} = 80 \ldots \left(2\right)$</p><p>On solving equations \left(1\right)$ and \left(2\right)$, we get${x}^{2} + {\left(12 - x\right)}^{2} = 80$</p><p>$⇒ {x}^{2} + 144 + {x}^{2} - 24x = 80$</p><p>$⇒ 2{x}^{2} - 24x + 64 = 0$</p><p>$⇒ {x}^{2} - 12x + 32 = 0$</p><p>$⇒ \left(x - 4\right)\left(x - 8\right) = 0$</p><p>$⇒ x = 4$or$8$</p><p>From equation \left(2\right)$, when $x = 4, y = 8$ and when $x = 8, y = 4$.

$∴ {x}^{4} + {y}^{2} = {4}^{4} + {8}^{2} = 256 + 64 = 320$

(or) ${x}^{4} + {y}^{2} = {8}^{4} + {4}^{2} = 4096 + 16 = 4112$