Let the mean and variance of 8 numbers x, y, 10, 12, 6, 12, 4, 8 be 9 and 9.25 respectivel…

Question

Let the mean and variance of 8 numbers x, y, 10, 12, 6, 12, 4, 8 be 9 and 9.25 respectively. If x > y, then $3x-2y$ is equal to \_\_\_\_\_\_\_\_.

Answer: 25

Accepted Answer

Mean $=9$: $\dfrac{x+y+52}{8}=9 \implies x+y=20$

Deviations from mean for known values: $3,3,1,-5,-1,-3$ (i.e. $10-9,12-9,6-9,12-9,4-9,8-9$... consistent with squares summing as below).

Variance: $\dfrac{(x-9)^2+(y-9)^2+9+9+1+9+25+1}{8} = 9.25 \implies (x-9)^2+(y-9)^2+54=74 \implies (x-9)^2+(y-9)^2=20$

Using $y=20-x$, i.e. $(y-9)=(11-x)$: $(x-9)^2+(11-x)^2=20 \implies x=7 	ext{ or } 13$

If $x=13, y=7$ ($x>y$ ):
$3x-2y = 3(13)-2(7) = 39-14=25$

Solutions · 1