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)

320

Answer 2

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

$∴ x + y = 12 …(i)$

And variance

=$\frac{2^2+1^2+0^2+0^2+1^2+2^2+(x−6)^2+(y−6)^2}8$

=$\frac{9}{4}$

$∴ (x – 6)^2 \+ (y – 6)^2 = 8 …(ii)$

From (i) and (ii)

$x = 4$ and $y = 8$

$∴ x^4 +y^2 = 320$