Solution
\[
\begin{aligned}
& 2,3,5,10,11,13,15,21, a \& b \\
& \text { Mean }=\frac{\sum x_{i}}{10}=\frac{2+3+5+10+11+13+15+21+a+b}{10}=9
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow 80+a+b=90 \Rightarrow a+b=10 \\
& \text { Variance }=\frac{1}{n} \sum x_{i}^{2}-(\text { Meau })^{2}=34.2 \\
& \Rightarrow \frac{1}{10}\left(4+9+25+100+121+169+225+441+a^{2}+b^{2}\right)-81=34.2 \\
& \Rightarrow a^{2}+b^{2}=58 \Rightarrow a^{2}+(10-a)^{2}=58 \\
& \Rightarrow 2 a^{2}-20 a+100=58 \Rightarrow a^{2}-10 a+21=0 \\
& \Rightarrow(a-3)(a-7)=0 \\
& a=3 \quad b=7 \\
& b=7 \quad b=3
\end{aligned}
\]
All number 2,3,3,5,7,10,11,13,15,21
\[
\text { Median }=\frac{\mathrm{T}_{5}+\mathrm{T}_{6}}{2}=\frac{7+10}{2}=8.5
\]
Mean derivation about median $=\frac{\sum\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{n}}$
\[
\begin{aligned}
& =\frac{6.5+5.5+5.5+3.5+1.5+1.5+2.5+4.5+6.5+12.5}{10} \\
& =5 \text { Answer }(1)
\end{aligned}
\]
.
