?$\text {解: (1) } $?∵?$\overline{x_{\text {甲 }}}=\frac{1}{10}(12+13+···+11)=13 $?
∴?$S_{\text {甲 }}=\frac{1}{10} {[(12-13)^{2}+(13-13)^{2}+···+(11-13)^{2}]=3.6} $?
∵?$\overline {x_{乙}}=\frac {1}{10}(11+16+···+16)=13 $?
∴?$s_{乙}=\frac {1}{10} {[(11-13)^2+(16-13)^2+···+(16-13)^2]=15.8} $?
∴?$S_{\text {甲 }}^{2}<S_{乙} , $?∴ 甲的波動較小
?$(2)$?甲的平均差是?$ \frac {1}{10}(|12-13|+\mid 13-13 |+···+| 11-13 \mid )=1.6$?
乙的平均差是?$ \frac {1}{10}(|11-13|+|16-13| +···+|16-13|)=3.4$?
∵?$1.6<3.4，$?
∴ 樣本平均差能區(qū)分這兩個樣本的波動大小且甲的波動較小.
