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The mean and standard deviation of some data for the time taken to complete . a test are calculated with the following results:
Number of observations $=25,$ mean $=18.2$ seconds, standard deviation $=3.25 s$
Further, another set of 15 observations $x_{1}, x_{2}, \ldots, x_{15},$ also in seconds, is now available and we have $\sum_{i=1}^{15} x_{i}=279$ and $\sum_{i=1}^{15} x_{i}^{2}=5524 .$ Calculate the standard deviation based on all 40 observations.
Solution
Given, $n_{1}=25, \bar{x}_{i}=18.2, \sigma_{1}=3.25$
$n_{2}=15, \sum_{i=1}^{15} x_{i}=279$ and $\sum_{i=1}^{15} x_{i}^{2}=5524$
For first set $\Sigma x_{i}=25 \times 18.2=455$
$\therefore$
$\sigma_{1}^{2}=\frac{\Sigma x_{i}^{2}}{25}-(18.2)^{2}$
$\Rightarrow \quad(3.25)^{2}=\frac{\Sigma x_{i}^{2}}{25}-(18.2)^{2} \Rightarrow 10.5625+331.24=\frac{\Sigma x_{i}^{2}}{25}$
$\Rightarrow \quad \Sigma x_{i}^{2}=25 \times(10.5625+331.24)=25 \times 341.8025=8545.0625$
For combined SD of the 40 observations, $n=40$.
Now $\quad \sum_{i=1}^{40} x_{i}^{2}=5524+8545.0625=14069.0625$
and $\quad \sum_{i=1}^{40} x_{i}=455+279=734$
$\therefore \quad SD =\sqrt{\frac{14069.0625}{40}-\left(\frac{734}{40}\right)^{2}}=\sqrt{351.1726-(18.35)^{2}}$
$=\sqrt{351.726-336.7225}=\sqrt{15.0035}=3.87$
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