证明:数据ax1-b,ax2-b,ax3-b,...axn-b的平均数是ap-b,方差a²q
1个回答

若X1+X2+X3+...+Xn的平均数为p,方差q,

则(X1+X2+X3+...+Xn)=np,

[(X1-p)²+(X2-p)²+(X3-p)²+.+(Xn-p)²]=nq

∴1/n(ax1-b+ax2-b+ax3-b+...+axn-b)

=1/n[a(x1+x2+x3+...+xn)-nb]

=1/n(anp-nb)

=ap-b

1/n{[(ax1-b)-(ap-b)]²+[(ax2-b)-(ap-b)]²+[(ax3-b)-(ap-b)]+...+[(axn-b)-(ap-b)]²}

=1/n{ [a(x1-p)]² +[a(x2-p)]²+[a(x3-p)]²+...+[a(xn-p)]²}

=1/n(a²)[(X1-p)²+(X2-p)²+(X3-p)²+.+(Xn-p)²]

=1/n*a²*nq

=a²q

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