Let the mean and variance of 8 numbers -10,-7,-1, x, y, 9,2,16 be (7)/(2) and (293)/(4), r…

Question

Let the mean and variance of 8 numbers $-10,-7,-1, x, y, 9,2,16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively.
Then the mean of 4 numbers $x, y, x+y+1,|x-y|$ is :

Accepted Answer

$\begin{gathered} =\frac{7}{2} \ \frac{-10-7-1+x+y+16+2+9}{8}=\frac{7}{2} \ \Rightarrow x+y+9=28 \ \Rightarrow x+y=19 \ =\frac{293}{4} \ \frac{100+49+1+{x}^{2}+{y}^{2}+256+4+81}{8}-\frac{49}{4}=\frac{293}{4} \ \frac{491+{x}^{2}+{y}^{2}}{8}=\frac{342}{4} \ {x}^{2}+{y}^{2}=193 \end{gathered}$

$\begin{gathered} \Rightarrow (x,y)=(7,12)(12,7) \ =\frac{x+y+x+y+1+(x-y)}{4} \ =\frac{19+20+5}{4} \ =11 \end{gathered}$

Solutions · 1