JEE Main 2022 — Mathematics Probability & Statistics

Mean $=\frac{4+5+6+6+7+8+x+y}{8}=6$=84+5+6+6+7+8+x+y​=6

$$\Rightarrow x+y=12\ldots (1)$$⇒x+y=12…(1)

Variance $\displaystyle =\frac{\sum {x}_{i}^{2}}{n}-{(\bar{x})}^{2}$=n∑xi2​​−(xˉ)2

$$\Rightarrow \frac{9}{4}=\frac{{4}^{2}+{5}^{2}+{6}^{2}+{6}^{2}+{7}^{2}+{8}^{2}+{x}^{2}+{y}^{2}}{8}-{(6)}^{2}$$⇒49​=842+52+62+62+72+82+x2+y2​−(6)2

$$\Rightarrow {x}^{2}+{y}^{2}=80\ldots (2)$$⇒x2+y2=80…(2)

On solving equations $(1)$(1) and $(2)$(2), we get ${x}^{2}+{(12-x)}^{2}=80$x2+(12−x)2=80

$$\Rightarrow {x}^{2}+144+{x}^{2}-24x=80$$⇒x2+144+x2−24x=80

$$\Rightarrow 2{x}^{2}-24x+64=0$$⇒2x2−24x+64=0

$$\Rightarrow {x}^{2}-12x+32=0$$⇒x2−12x+32=0

$$\Rightarrow (x-4)(x-8)=0$$⇒(x−4)(x−8)=0

$\Rightarrow x=4$⇒x=4 or $8$8

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

$$\therefore {x}^{4}+{y}^{2}={4}^{4}+{8}^{2}=256+64=320$$∴x4+y2=44+82=256+64=320

(or) ${x}^{4}+{y}^{2}={8}^{4}+{4}^{2}=4096+16=4112$x4+y2=84+42=4096+16=4112