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 $= 18.2 \text{ seconds}$,standard deviation $= 3.25 \text{ s}$.
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.

Given: $n_{1} = 25, \bar{x}_{1} = 18.2, \sigma_{1} = 3.25$.
For the first set,$\sum x_{i} = n_{1} \times \bar{x}_{1} = 25 \times 18.2 = 455$.
Using the formula $\sigma_{1}^{2} = \frac{\sum x_{i}^{2}}{n_{1}} - (\bar{x}_{1})^{2}$:
$(3.25)^{2} = \frac{\sum x_{i}^{2}}{25} - (18.2)^{2}$
$10.5625 = \frac{\sum x_{i}^{2}}{25} - 331.24$
$\frac{\sum x_{i}^{2}}{25} = 341.8025 \implies \sum x_{i}^{2} = 8545.0625$.
For the second set: $n_{2} = 15, \sum x_{i} = 279, \sum x_{i}^{2} = 5524$.
Combined sum of squares: $\sum x_{total}^{2} = 8545.0625 + 5524 = 14069.0625$.
Combined sum of observations: $\sum x_{total} = 455 + 279 = 734$.
Combined mean: $\bar{x} = \frac{734}{40} = 18.35$.
Combined standard deviation: $\sigma = \sqrt{\frac{\sum x_{total}^{2}}{n_{1} + n_{2}} - (\bar{x})^{2}}$
$\sigma = \sqrt{\frac{14069.0625}{40} - (18.35)^{2}}$
$\sigma = \sqrt{351.7265625 - 336.7225} = \sqrt{15.0040625} \approx 3.87$.