一个关于方差和平均数的数学初二问题！

已知有X1,X2,X3,X4,X5的平均数是2，方差是3分之1，那么3X1-2,3X2-2,3X3-2,3X4-2,3X5-2的平均数和方差分别是多少？请说明原因！

X平均=(3X1+3X2……+3X5)/5-2
=3(X1+X2……X5)/5-2
=3*2-2
=4
方差={[(3X1-2)-(3X3-2)]^2+……[(3X5-2)-(3X3-2)]}/5
=1/3

private sub command1_click() dim a(1 to 5) as integer dim i as integer dim ave as single dim var as single for i = 1 to 5 a(i) = inputbox("输入第" & i & "个成绩") ave = ave + a(i) next ave = ave / 5 for i = 1 to 5 var = var + (a(i) - ave) ^ 2 next var = var / 5 print "平均数:" & ave print "方差:" & var end sub

因为(X1+X2+X3+X4+X5)/5=2 (2-X1)平方+(2-X2)平方+(2-X3)平方+(2-X4)平方+(2-X5)平方=1/3 平均数:[3(X1+X2+X3+X4+X5)-2*5]/5=4 方差:[4-(3X1-2)]=6-3X1=3（2-X1） 同上 9[(2-X1)平方+(2-X2)平方+(2-X3)平方+(2-X4)平方+(2-X5)平方]=3

X平均=(X1+...+X5)/5=2. 则:3X1-2,3X2-2,3X3-2,3X4-2,3X5-2的平均数 XX平均=3*(X平均)-(2*5)/5=3*2-2=4; 方差DD=[(3X1-2 -4)^2+....+(3X5-2 -4)^2] =[(3X1-2 -4)^2+....+(3X5-2 -4)^2] =[(3X1-2)^2+....+(3X5-2)^2] -4[(3X1-2)+....+(3X5-2)]+5*4^2 =[(3X1-X平均)^2+....+(3X5-X平均)^2] -4[(3X1-2)+....+(3X5-2)]+5*4^2 =D-4*(5*XX平均)+80 D=1/3为X1,X2,X3,X4,X5的方差,则上式 =1/3-4*5*4+80 =1/3