∵据x1,x2,x3,x4,x5的平均数是2, ∴=2, ∵数据x1,x2,x3,x4,x5的平均数是2,方差是, ∴[(x1-2)2+(x2-2)2+[(x3-2)2+(x4-2)2+(x5-2)2]=①; ∴3x1-2,3x2-2,3x3-2,3x4-2,3x5-2,的平均数是
(3x1-2)+(3x2-2)+(3x3-2)+(3x4-2)+(3x5-2) | 5 | =, =3×-2=4. ∴[(3x1-2-4)2+(3x2-2-4)2+(3x3-2-4)2+(3x4-2-4)2+(3x5-2-4)2] =[9(x1-2)2+9(x2-2)2+9(x3-2)2+9(x4-2)2+9(x5-2)2] =×9[(x1-2)2+(x2-2)2+(x3-2)2+(x4-2)2+(x5-2)2]② 把①代入②得,方差是:×9=3. 故答案为:3. |