Statistics Part 2

Concept:Use the definitions of mean and variance for 8 numbers to form equations in $x$ and $y$. Solve for $x$ and $y$, then compute the mean of the four required numbers.Explanation:Given mean $\frac{7}{2}$: $\frac{-10-7-1+x+y+9+2+16}{8} = \frac{7}{2}$. Sum of known numbers is $9$, so $\frac{x+y+9}{8} = \frac{7}{2} \Rightarrow x+y = 28 - 9 = 19$.Given variance $\frac{293}{4}$: $\frac{(-10)^2+(-7)^2+(-1)^2+x^2+y^2+9^2+2^2+16^2}{8} - \left(\frac{7}{2}\right)^2 = \frac{293}{4}$.Sum of squares of known numbers: $100+49+1+81+4+256 = 491$. So $\frac{491+x^2+y^2}{8} - \frac{49}{4} = \frac{293}{4} \Rightarrow \frac{491+x^2+y^2}{8} = \frac{342}{4} = \frac{171}{2}$.Cross multiply: $2(491+x^2+y^2) = 8 \cdot 171 \Rightarrow 982 + 2(x^2+y^2) = 1368 \Rightarrow 2(x^2+y^2) = 386 \Rightarrow x^2+y^2 = 193$.Now $(x+y)^2 = x^2+y^2+2xy \Rightarrow 361 = 193 + 2xy \Rightarrow xy = 84$. $x$ and $y$ are roots of $t^2 - 19t + 84 = 0 \Rightarrow (t-7)(t-12)=0$. So $(x,y) = (7,12)$ or $(12,7)$.The four numbers are $x$, $y$, $x+y+1$, and $|x-y|$. Their sum $= (x+y) + (x+y+1) + |x-y| = 19 + 20 + 5 = 44$. Mean $= \frac{44}{4} = 11$.Answer:11, Option A.