\qquad 在日常工作学习和生产生活中，我们常常会遇到已知各层样本的均值和方差求总体均值和方差的问题，本文给出了对于此类问题的一般解法及证明。

\qquad 某高中兴趣小组想调查本校各个年级学生的平均身高和方差以及总体学生的平均身高和方差。现已知他们采用分层随机抽样的方法抽取了：

\qquad 请问全校所有学生的平均身高和方差分别为多少？

\qquad 某市有5所高中进行了一场大联考，各校的总成绩统计如下：

\qquad 求这次考试这五所学校的考生的总成绩的平均值和方差分别为多少？

在分层随机抽样中，已知：

求解：总体均值 x ‾ \overline{x} x 和总体方差 s 2 s^2 s2

x ‾ = ∑ i = 1 L ω i x i ‾ s 2 = ∑ i = 1 L ω i [ s i 2 + ( x i ‾ − x ‾ ) 2 ] \begin{align} \overline{x} &= \sum\limits_{i=1}^{L}\omega_i \overline{x_i} \nonumber \\ s^2 &= \sum\limits_{i = 1}^{L} \omega_i \left[ s^2_i + (\overline{x_i} - \overline{x})^2 \right] \nonumber \end{align} xs2​=i=1∑L​ωi​xi​​=i=1∑L​ωi​[si2​+(xi​​−x)2]​

设：

则可知：

由平均数定义，得： x ‾ = 1 N ∑ i = 1 L n i x i ‾ = ∑ i = 1 L ω i x i ‾ \overline{x} = \frac{1}{N} \sum\limits_{i=1}^{L}n_i \overline{x_i} = \sum\limits_{i=1}^{L}\omega_i \overline{x_i} x=N1​i=1∑L​ni​xi​​=i=1∑L​ωi​xi​​ 由方差定义，得： s 2 = 1 N ∑ i = 1 L ∑ j = 1 n i ( x i j − x ‾ ) 2 = 1 N ∑ i = 1 L ∑ j = 1 n i ( x i j − x i ‾ + x i ‾ − x ‾ ) 2 = 1 N ∑ i = 1 L [ ∑ j = 1 n i ( x i j − x i ‾ ) 2 + ∑ j = 1 n i 2 ( x i j − x i ‾ ) ( x i ‾ − x ‾ ) + ∑ j = 1 n i ( x i ‾ − x ‾ ) 2 ] \begin{align} s^2 &= \frac{1}{N} \sum\limits_{i=1}^{L} \sum\limits_{j=1}^{n_i} (x_{ij} - \overline{x})^2 \nonumber \\ &= \frac{1}{N} \sum\limits_{i=1}^{L} \sum\limits_{j=1}^{n_i} (x_{ij} - \overline{x_i} + \overline{x_i} - \overline{x})^2 \nonumber \\ &= \frac{1}{N} \sum\limits_{i=1}^{L} \left[ \sum\limits_{j=1}^{n_i} (x_{ij} - \overline{x_i})^2 + \sum\limits_{j=1}^{n_i} 2(x_{ij} - \overline{x_i})(\overline{x_i} - \overline{x}) + \sum\limits_{j=1}^{n_i} (\overline{x_i} - \overline{x})^2 \right] \nonumber \end{align} s2​=N1​i=1∑L​j=1∑ni​​(xij​−x)2=N1​i=1∑L​j=1∑ni​​(xij​−xi​​+xi​​−x)2=N1​i=1∑L​[j=1∑ni​​(xij​−xi​​)2+j=1∑ni​​2(xij​−xi​​)(xi​​−x)+j=1∑ni​​(xi​​−x)2]​ 由于 x i ‾ = ∑ j = 1 n i x i j n i \overline{x_i} = \frac{\sum\limits_{j=1}^{n_i} x_{ij}}{n_i} xi​​=ni​j=1∑ni​​xij​​ 则 ∑ j = 1 n i 2 ( x i j − x i ‾ ) ( x i ‾ − x ‾ ) = 2 ( x i ‾ − x ‾ ) ∑ j = 1 n i ( x i j − x i ‾ ) = 2 ( x i ‾ − x ‾ ) ( ∑ j = 1 n i x i j − n i x i ‾ ) = 0 \begin{align} &\sum\limits_{j=1}^{n_i} 2(x_{ij} - \overline{x_i})(\overline{x_i} - \overline{x}) \nonumber \\ = &2(\overline{x_i} - \overline{x})\sum\limits_{j=1}^{n_i} (x_{ij} - \overline{x_i}) \nonumber \\ = &2(\overline{x_i} - \overline{x}) \left( \sum\limits_{j=1}^{n_i} x_{ij} - n_i \overline{x_i} \right) \nonumber \\ = &0 \nonumber \end{align} ===​j=1∑ni​​2(xij​−xi​​)(xi​​−x)2(xi​​−x)j=1∑ni​​(xij​−xi​​)2(xi​​−x)(j=1∑ni​​xij​−ni​xi​​)0​ 故 s 2 = 1 N ∑ i = 1 L [ ∑ j = 1 n i ( x i j − x i ‾ ) 2 + ∑ j = 1 n i ( x i ‾ − x ‾ ) 2 ] = 1 N ∑ i = 1 L [ n i s i 2 + n i ( x i ‾ − x ‾ ) 2 ] = ∑ i = 1 L [ n i N ⋅ s i 2 + n i N ⋅ ( x i ‾ − x ‾ ) 2 ] = ∑ i = 1 L ω i [ s i 2 + ( x i ‾ − x ‾ ) 2 ] \begin{align} s^2 &= \frac{1}{N} \sum\limits_{i=1}^{L} \left[ \sum\limits_{j=1}^{n_i} (x_{ij} - \overline{x_i})^2 + \sum\limits_{j=1}^{n_i} (\overline{x_i} - \overline{x})^2 \right] \nonumber\\ &= \frac{1}{N} \sum\limits_{i=1}^{L} \left[ n_is_i^2 + n_i (\overline{x_i} - \overline{x})^2 \right] \nonumber \\ &= \sum\limits_{i=1}^{L} \left[ \frac{n_i}{N} \cdot s_i^2 + \frac{n_i}{N} \cdot (\overline{x_i} - \overline{x})^2 \right] \nonumber \\ &= \sum\limits_{i = 1}^{L} \omega_i \left[ s^2_i + (\overline{x_i} - \overline{x})^2 \right] \nonumber \end{align} s2​=N1​i=1∑L​[j=1∑ni​​(xij​−xi​​)2+j=1∑ni​​(xi​​−x)2]=N1​i=1∑L​[ni​si2​+ni​(xi​​−x)2]=i=1∑L​[Nni​​⋅si2​+Nni​​⋅(xi​​−x)2]=i=1∑L​ωi​[si2​+(xi​​−x)2]​

以 引例2 为例： x ‾ = 15 % × 443 + 23 % × 450 + 22 % × 470 + 16 % × 453 + 24 % × 466 = 457.67 s 2 = 15 % × [ 28 + ( 443 − 457.67 ) 2 ] + 23 % × [ 33 + ( 450 − 457.67 ) 2 ] + 22 % × [ 23 + ( 470 − 457.67 ) 2 ] + 16 % × [ 40 + ( 453 − 457.67 ) 2 ] + 24 % × [ 43 + ( 466 − 457.67 ) 2 ] = 132.971 \begin{align} \overline{x} =& 15\% \times 443 + 23\% \times 450 + 22\% \times 470 + 16\% \times 453 + 24\% \times 466 \nonumber \\ =& 457.67 \nonumber \\ s^2 = &15\% \times [28 + (443 - 457.67)^2] + \nonumber \\ &23\% \times [33 + (450 - 457.67)^2] + \nonumber \\ &22\% \times [23 + (470 - 457.67)^2] + \nonumber \\ &16\% \times [40 + (453 - 457.67)^2] + \nonumber \\ &24\% \times [43 + (466 - 457.67)^2] \nonumber \\ =& 132.971 \nonumber \end{align} x==s2==​15%×443+23%×450+22%×470+16%×453+24%×466457.6715%×[28+(443−457.67)2]+23%×[33+(450−457.67)2]+22%×[23+(470−457.67)2]+16%×[40+(453−457.67)2]+24%×[43+(466−457.67)2]132.971​ 因此，这五所学校成绩的总体均值为457.67，方差为132.971.