JEE Main 2022 (Online) 27th June Evening ShiftQuestion 1
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
Statistics
Solution
Known sum $=36$, mean 6 gives total $48\Rightarrow x+y=12$. $\sum x^2=8\left(\frac94+36\right)=306$; known squares $16+25+36+36+49+64=226$, so $x^2+y^2=80$. $xy=\frac{144-80}{2}=32\Rightarrow x=4,y=8$. $x^4+y^2=256+64=320$. Option B.