Question

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

• A. 162
• B. 320
• C. 674
• D. 420

Answer 1

Answer: (B)

mean $\bar{x}=\frac{4+5+6+6+7+8+x+y}{8}=6$
$\Rightarrow x+y=48-36=12$
Variance
$=\frac{1}{8}\left(16+25+36+36+49+64+x^{2}+y^{2}\right)-36=\frac{9}{4}$
$\Rightarrow x^{2}+y^{2}=80$
$\therefore x=4 ; y=8$
$x^{4}+y^{2}=256+64=320$