The mean and standard deviation of a set of n1 observations are

Solution:

Given the mean and standard deviation of a set of $n_{1}$ observations are $\bar{x}_{1}$ and $s_{1}$, respectively while the mean and standard deviation of another set of

$\mathrm{n}_{2}$ observations are $\overline{\mathrm{x}}_{2}$ and $s_{2}$, respectively

To show that the standard deviation of the combined set of $\left(n_{1}+n_{2}\right)$ observations is given by

S. D $=\sqrt{\frac{n_{1}\left(s_{1}\right)^{2}+n_{2}\left(s_{2}\right)^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$

As per given criteria, For first set

Let $x_{i}$ where $i=1,2,3,4, \ldots, n_{1}$

For second set

And $y_{i}$ where $j=1,2,3,4, \ldots, n_{2}$

And the means are

$\overline{x_{1}}=\frac{1}{n_{1}} \sum_{i=1}^{n} x_{i}, \overline{x_{2}}=\frac{1}{n_{2}} \sum_{j=1}^{n} y_{j}$

Now mean of the combined series is given by

$\bar{x}=\frac{1}{n_{1}+n_{2}}\left[\sum_{i=1}^{n} x_{j}+\sum_{j=1}^{n} y_{j}\right]=\frac{n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}}{n_{1}+n_{2}} \ldots . .(i)$

And the corresponding square of standard deviation is

$\sigma_{1}^{2}=\frac{1}{n_{1}} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}, \sigma_{2}^{2}=\frac{1}{n_{2}} \sum_{j=1}^{n}\left(y_{j}-\bar{x}\right)^{2}$

Therefore, square of standard deviation becomes,

$\sigma^{2}=\sigma_{1}^{2}+\sigma_{2}^{2}=\frac{1}{n_{1}+n_{2}}\left[\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{j=1}^{n}\left(y_{j}-\bar{x}\right)^{2}\right] \ldots . .$ (ii)

Now,

$\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}=\sum_{i=1}^{n}\left(x_{i}-\bar{x}_{j}+\bar{x}_{j}-\bar{x}\right)^{2}$

$\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}=\sum_{i=1}^{n}\left(x_{i}-\bar{x}_{j}\right)^{2}+n_{1}\left(\bar{x}_{j}-\bar{x}\right)^{2}+2\left(\bar{x}_{j}-\bar{x}\right) \sum_{i=1}^{n}\left(x_{i}-\overline{x_{j}}\right)^{2}$

But the algebraic sum of the deviation of values of first series from their mean is zero.

$\sum_{i=1}^{n}\left(x_{i}-\overline{x_{j}}\right)^{2}=0$

Also,

$\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}=n_{1} s_{1}^{2}+n_{1}\left(\bar{x}_{1}-\bar{x}\right)^{2} \ldots$(iii)

But

$\mathrm{d}_{1}=\overline{\mathrm{X}}_{1}-\overline{\mathrm{x}}$

Substituting value from equation (i), we get

$\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}=\overline{\mathrm{x}}_{1}-\frac{\mathrm{n}_{1} \overline{\mathrm{x}}_{1}+\mathrm{n}_{2} \overline{\mathrm{x}}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

$\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}=\frac{\left(\overline{\mathrm{x}}_{1}\right)\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)-\left(\mathrm{n}_{1} \overline{\mathrm{x}}_{1}+\mathrm{n}_{2} \overline{\mathrm{x}}_{2}\right)}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

$\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}=\frac{\left(\mathrm{n}_{1} \overline{\mathrm{x}}_{1}+\mathrm{n}_{2} \overline{\mathrm{x}}_{1}\right)-\left(\mathrm{n}_{1} \overline{\mathrm{x}}_{1}+\mathrm{n}_{2} \overline{\mathrm{x}}_{2}\right)}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

$\bar{x}_{1}-\bar{x}=\frac{\left(n_{2} \bar{x}_{1}\right)-\left(n_{2} \bar{x}_{2}\right)}{n_{1}+n_{2}}$

Substituting this value in equation (iii), we get

$\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}=n_{1} s_{1}^{2}+n_{1}\left(\frac{n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)}{n_{1}+n_{2}}\right)^{2}$

$\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}=n_{1} s_{1}^{2}+\frac{n_{1} n_{2}^{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}} \ldots$ (iv)

Similarly, we have

$\sum_{j=1}^{n}\left(y_{i}-\bar{x}\right)^{2}=\sum_{j=1}^{n}\left(y_{j}-\bar{x}_{i}+\bar{x}_{i}-\bar{x}\right)^{2}$

$\sum_{j=1}^{n}\left(y_{i}-\bar{x}\right)^{2}=\sum_{j=1}^{n}\left(y_{j}-\bar{x}_{j}\right)^{2}+n_{2}\left(\bar{x}_{j}-\bar{x}\right)^{2}+2\left(\bar{x}_{j}-\bar{x}\right) \sum_{j=1}^{n}\left(y_{j}-\bar{x}_{j}\right)^{2}$

But the algebraic sum of the deviation of values of second series from their mean is zero.

$\sum_{j=1}^{n}\left(y_{j}-\overline{x_{i}}\right)^{2}=0$

Also,

$\sum_{\mathrm{j}=1}^{\mathrm{n}}\left(\mathrm{y}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^{2}=\mathrm{n}_{2} \mathrm{~s}_{2}^{2}+\mathrm{n}_{2}\left(\overline{\mathrm{x}}_{2}-\overline{\mathrm{x}}\right)^{2} \ldots(\mathrm{v})$

But $d_{2}=\bar{x}_{2}-\bar{x}$

Substituting value from equation (i), we get

$\overline{\mathrm{x}}_{2}-\overline{\mathrm{x}}=\overline{\mathrm{x}}_{2}-\frac{\mathrm{n}_{1} \overline{\mathrm{x}}_{1}+\mathrm{n}_{2} \overline{\mathrm{x}}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

$\overline{\mathrm{x}}_{2}-\overline{\mathrm{x}}=\frac{\left(\overline{\mathrm{x}}_{2}\right)\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)-\left(\mathrm{n}_{1} \overline{\mathrm{x}}_{1}+\mathrm{n}_{2} \overline{\mathrm{x}}_{2}\right)}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

$\bar{x}_{2}-\bar{x}=\frac{\left(n_{1} \bar{x}_{2}+n_{2} \bar{x}_{2}\right)-\left(n_{1} \bar{x}_{1}+n_{2} \bar{x}_{2}\right)}{n_{1}+n_{2}}$

$\bar{x}_{2}-\bar{x}=\frac{\left(n_{1} \bar{x}_{2}\right)-\left(n_{1} \bar{x}_{2}\right)}{n_{1}+n_{2}}$

$\overline{\mathrm{x}}_{2}-\overline{\mathrm{x}}=\frac{\mathrm{n}_{1}\left(\overline{\mathrm{x}}_{2}-\overline{\mathrm{x}}_{1}\right)}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

Substituting this value in equation (v), we get

$\sum_{j=1}^{n}\left(y_{j}-\bar{x}\right)^{2}=n_{2} s_{2}^{2}+n_{2}\left(\frac{n_{1}\left(\bar{x}_{2}-\bar{x}_{1}\right)}{n_{1}+n_{2}}\right)^{2}$

$\sum_{j=1}^{n}\left(y_{j}-\bar{x}\right)^{2}=n_{2} s_{2}^{2}+\frac{n_{2} n_{1}^{2}\left(\bar{x}_{2}-\bar{x}_{1}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}} \ldots(v i)$

Substituting equation (iv) and (vi) in equation (ii), we get

$\sigma^{2}=\sigma_{1}^{2}+\sigma_{2}^{2}=\frac{1}{n_{1}+n_{2}}\left[\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}+\sum_{j=1}^{n}\left(y_{j}-\bar{x}\right)^{2}\right]$

$\sigma^{2}=\frac{1}{n_{1}+n_{2}}\left[n_{1} s_{1}^{2}+\frac{n_{1} n_{2}^{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}+n_{2} s_{2}^{2}+\frac{n_{2} n_{1}^{2}\left(\bar{x}_{2}-\bar{x}_{1}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}\right]$

$\sigma^{2}=\frac{1}{n_{1}+n_{2}}\left[n_{1} s_{1}^{2}+n_{2} s_{2}^{2}+\frac{n_{1} n_{2}^{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}+\frac{n_{2} n_{1}^{2}\left(-\left(\bar{x}_{1}-\bar{x}_{2}\right)\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}\right]$

$\sigma^{2}=\frac{1}{n_{1}+n_{2}}\left[n_{1} s_{1}^{2}+n_{2} s_{2}^{2}+\frac{n_{1} n_{2}^{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}+\frac{n_{2} n_{1}^{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}\right]$

$\sigma^{2}=\frac{1}{n_{1}+n_{2}}\left[n_{1} s_{1}^{2}+n_{2} s_{2}^{2}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}\left(n_{2}+n_{1}\right)\right]$

$\sigma^{2}=\left[\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}\left(n_{2}+n_{1}\right)}{n_{1}+n_{2}}\right]$

$\sigma^{2}=\left[\frac{\mathrm{n}_{1} \mathrm{~s}_{1}^{2}+\mathrm{n}_{2} \mathrm{~s}_{2}^{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\mathrm{n}_{1} \mathrm{n}_{2}\left(\overline{\mathrm{x}}_{1}-\overline{\mathrm{x}}_{2}\right)^{2}}{\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)^{2}}\right]$

So the combined standard deviation

$S . D(\sigma)=\sqrt{\frac{n_{1} s_{1}^{2}+n_{2} s_{2}^{2}}{n_{1}+n_{2}}+\frac{n_{1} n_{2}\left(\bar{x}_{1}-\bar{x}_{2}\right)^{2}}{\left(n_{1}+n_{2}\right)^{2}}}$

Hence proved