Question

The mean and standard deviation of a set of $n_1$ observations are ${\overline{x}}_1$ and $s_1$, respectively while the mean and standard deviation of another set of $n_2$ observations are ${\overline{x}}_2$ and $s_2$, respectively.
Show that the standard deviation of combined set of $(n_1+n_2)$ observations is given by

$S.D.=\sqrt{\frac{n_1(s_1{)}^2+n_2(s_2{)}^2}{n_1+n_2}+\frac{n_1{n}_2({\overline{x}}_1-{\overline{x}}_2{)}^2}{(n_1+n_2{)}^2}}$

Answer 1

For two sets of observation
$x_i,i=1,2,3,.....n_1$ and
$y_j,j=1,2,3,.....n_2$

Given:

${\overline{x}}_1=\frac{1}{n_1}{\sum }_{i=1}^{n_1}{x}_i$

and

${\overline{x}}_2=\frac{1}{n_2}{\sum }_{j=1}^{n_2}{y}_j$

$\Rightarrow s_1^2=\frac{1}{n_1}{\sum }_{i=1}^{n_1}(x_i-{\overline{x}}_1{)}^2$

and

$\Rightarrow s_2^2=\frac{1}{n_2}{\sum }_{j=1}^{n_2}(y_j-{\overline{x}}_2{)}^2$

Mean, $\overline{x}=\frac{1}{n_1+n_2}[{\sum }_{i=1}^{n_1}{x}_i+{\sum }_{j=1}^{n_2}{y}_j]$

$\overline{x}=\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

$S.D^2=\frac{1}{n_1+n_2}\left[{\sum }_{i=1}^{n_1}\right(x_i-{\overline{x}}_1{)}^2+{\sum }_{j=1}^{n_2}(y_j-{\overline{x}}_2{)}^2]$

Now,

${\sum }_{i=1}^{n_1}(x_i-\overline{x}{)}^2={\sum }_{i=1}^{n_1}(x_i-{\overline{x}}_1+{\overline{x}}_1-\overline{x}{)}^2$

$={\sum }_{i=1}^{n_1}(x_i-{\overline{x}}_1{)}^2+{\sum }_{i=1}^{n_1}(x_1-\overline{x}{)}^2+2({\overline{x}}_1-\overline{x}){\sum }_{i=1}^{n_1}(x_i-{\overline{x}}_1)$

But, ${\sum }_{i=1}^{n_1}(x_i-{\overline{x}}_1)=0$

[Algebraic sum of the deviation of values of first series from their mean is zero]

Also,

${\sum }_{i=1}^{n_1}(x_i-\overline{x}{)}^2=n_1{s}_1^2+n_1({\overline{x}}_1-\overline{x}{)}^2$

${\sum }_{i=1}^{n_1}(x_i-\overline{x}{)}^2=n_1{s}_1^2+n_1{d}_1^2$

Where $d_1={\overline{x}}_1-\overline{x}$

$\therefore$ Combined standard division

$S.D=\sqrt{\frac{n_1(s_1^2+d_1^2)+n_2(s_2^2+d_2{)}^2}{n_1+n_2}}$

Where,

$d_1={\overline{x}}_1-\overline{x}={\overline{x}}_1-\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

$d_1=\frac{n_2({\overline{x}}_1-{\overline{x}}_2)}{n_1+n_2}$

And

$d_2={\overline{x}}_2-\overline{x}={\overline{x}}_2-\frac{n_1{\overline{x}}_1+n_2{\overline{x}}_2}{n_1+n_2}$

$d_2=\frac{n_1({\overline{x}}_2-{\overline{x}}_1)}{n_1+n_2}$

$\therefore S.D^2=\frac{1}{n_1+n_2}[n_1{s}_1^2+n_2{s}_2^2+\frac{n_1{n}_2^2({\overline{x}}_1-{\overline{x}}_2{)}^2}{(n_1+n_2{)}^2}+\frac{n_2{n}_1^2({\overline{x}}_2-{\overline{x}}_1{)}^2}{(n_1+n_2{)}^2}]$

$S.D.=\sqrt{\frac{n_1{s}_1^2+n_2{s}_2^2}{n_1+n_2}+\frac{n_1{n}_2({\overline{x}}_1-{\overline{x}}_2{)}^2}{(n_1+n_2{)}^2}}$

Hence proved.