已知数据x1x2x3的平均数为10,方差为2.求x1的平方x2的平方x3的平方的平均数
由题意：X1+X2+X3 =30（x1-10)^2+（x2-10)^2+（x3-10)^2=6
即x1^2+x2^2+x3^2-20(x1+x2+x3)+300=6
x1^2+x2^2+x3^2=6+20x30-300=306
所以x1的平方x2的平方x3的平方的平均数为306÷3=102
绛旓細=1/3[(x1^2+x2^2+x3^2)-2*3x0*x0+3x0^2]=1/3[(x1^2+x2^2+x3^2)-3x0^2]鍘熼寰楄瘉!
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