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. …

Given: Mean

\(\mu =\frac{-18+x+y+2+9+16}{8}=\frac{7}{2}\\ \Rightarrow \frac{x+y+9}{8}=\frac{7}{2}\\ \Rightarrow x+y+9=28\\ \Rightarrow x+y=19....\left(i\right)\)

And Variance
\({\sigma }^{2}=\frac{\sum {x}_{i}^{2}}{8}-{\left(\mu \right)}^{2}=\frac{293}{4}\\ \Rightarrow \frac{{10}^{2}+{7}^{2}+{1}^{2}+{x}^{2}+{y}^{2}+{2}^{2}+{9}^{2}+{16}^{2}}{8}-{\left(\frac{7}{2}\right)}^{2}=\frac{293}{4}\)
\(\Rightarrow \frac{293}{4}+\frac{49}{4}=\frac{{x}^{2}+{y}^{2}+491}{8}\)

\(\Rightarrow \frac{171}{2}=\frac{{x}^{2}+{y}^{2}+491}{8}\)

\(\Rightarrow {x}^{2}+{y}^{2}=193...\left(ii\right)\)

After solving equation \((i)\) and \((ii)\), We get:

\(x=7\) and \(y=12\) or \(x=12\) and \(y=7\)

Hence numbers are \(7,12,20,5\)

Mean\(=\frac{20+12+7+5}{4}=\frac{44}{4}=11\)