The mean and variance of the data $4,5,6,6,7,8,x,y$ where $x<y$ are $6$ and $\frac{9}{4}$ respectively. Then $x^4+y^2$ is equal to

Mean $=\frac{4+5+6+6+7+8+x+y}{8}=6$

$\Rightarrow x+y=12 \dots \left(1\right)$

Variance $=\frac{\sum x_i^2}{n}-{\left(\overline{x}\right)}^2$

$\Rightarrow \frac{9}{4}=\frac{4^2+5^2+6^2+6^2+7^2+8^2+x^2+y^2}{8}-{\left(6\right)}^2$

$\Rightarrow x^2+y^2=80 \dots \left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right)$, we get $x^2+{\left(12-x\right)}^2=80$

$\Rightarrow x^2+144+x^2-24x=80$

$\Rightarrow 2x^2-24x+64=0$

$\Rightarrow x^2-12x+32=0$

$\Rightarrow \left(x-4\right)\left(x-8\right)=0$

$\Rightarrow x=4$  or $8$

From equation $\left(2\right)$, when $x=4, y=8$ and when $x=8, y=4$.

$\therefore x^4+y^2=4^4+8^2=256+64=320$

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