The mean and variance of the data 4, 5,6,6,7,8, x, y where x

Question

The mean and variance of the data 4, 5,6,6,7,8, x, y where x < y are 6, and (9/4) respectively. Then x 4+ y 2 is equal to (A) 162 (B) 320 (C) 674 (D) 420

Tardigrade Admin · Accepted Answer

mean barx=(4+5+6+6+7+8+x+y/8)=6 ⇒ x+y=48-36=12 Variance =(1/8)(16+25+36+36+49+64+x2+y2)-36=(9/4) ⇒ x2+y2=80 ∴ x=4 ; y=8 x4+y2=256+64=320

Q. 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

162

320

674

420

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

Q. The mean and variance of the data 4, 5,6,6,7,8,x,y where x<y are 6, and 49​ respectively. Then x4+y2 is equal to

162

320

674

420

mean xˉ=84+5+6+6+7+8+x+y​=6 ⇒x+y=48−36=12 Variance =81​(16+25+36+36+49+64+x2+y2)−36=49​ ⇒x2+y2=80 ∴x=4;y=8 x4+y2=256+64=320

Q. 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 $\overset{x}{ˉ} = \frac{4 + 5 + 6 + 6 + 7 + 8 + x + y}{8} = 6$
$\Rightarrow x + y = 48 - 36 = 12$
Variance
$= \frac{1}{8} (16 + 25 + 36 + 36 + 49 + 64 + x^{2} + y^{2}) - 36 = \frac{9}{4}$
$\Rightarrow x^{2} + y^{2} = 80$
$∴ x = 4; y = 8$
$x^{4} + y^{2} = 256 + 64 = 320$