2012-II-45

$\begin{array}{rcl}{m}_1& =& {\displaystyle \frac{x_1+x_2+\dots +x_{100}}{100}}\\ 100m_1& =& x_1+x_2+\dots +x_{100}\end{array}$

所以，可得

$\begin{array}{rcl}{m}_2& =& {\displaystyle \frac{x_1+x_2+\dots +x_{100}+m_1}{101}}\\ & =& {\displaystyle \frac{100m_1+m_1}{101}}\\ & =& m_1\end{array}$

II 必為正確。留意平均值必定大於最小的數據，以及小於最大的數據。所以兩組數據的最大及最小數據為相等。由此，$r_1=r_2$。

III 必為不正確。留意

$\begin{array}{rcl}{v}_1& =& {\displaystyle \frac{(x_1-m_1{)}^2+\dots +(x_{100}-m_1{)}^2}{100}}\end{array}$

所以，可得

$\begin{array}{rcl}{v}_2& =& {\displaystyle \frac{1}{101}}[(x_1-m_2{)}^2+\dots +(x_{100}-m_2{)}^2+(m_1-m_2{)}^2]\\ & =& {\displaystyle \frac{(x_1-m_1{)}^2+\dots +(x_{100}-m_1{)}^2}{101}}\\ & <& {\displaystyle \frac{(x_1-m_1{)}^2+\dots +(x_{100}-m_1{)}^2}{100}}\\ & =& v_1\end{array}$