解:因为数据 ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}$的平均数是2,
所以数据 $3{x}_{1}-2,3{x}_{2}-2,3{x}_{3}-2,3{x}_{4}-2,3{x}_{5}-2$的平均数
是3×2-2=4;
因为数据 ${x}_{1},{x}_{2},{x}_{3},{x}_{4},{x}_{5}$的方差为 $\frac{1}{3}$
所以数据 $3{x}_{1},3{x}_{2},3{x}_{3},3{x}_{4},3{x}_{5}$=的方差是 $\frac{1}{3}$×3²=3,
所以数据 $3{x}_{1}-2,3{x}_{2}-2,3{x}_{3}-2,3{x}_{4}-2,3{x}_{5}-2$的方差是3.