Question

The mean and standard deviation of some data for time taken to complete a test are calculated with the following results. Number of observations = 25, mean = 18.2 seconds, standard deviation = 3.25 Seconds. Further another set of observations $x_1,x_2,....,x_{15}$ 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.

Answer 1

Given : $n_1=25,\overline{x}=18.2$ & ${\sigma }_1=3.25$
$n_2=15,{\sum }_{i=1}^{15}{x}_i=279$ & ${\sum }_{i=1}^{15}{x}_i^2=5524$

Standard deviation $\sigma =\sqrt{\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2}$

Applying it for first set of data

$\therefore {\sigma }_1^2=\frac{\sum x_i^2}{n_1}-(\overline{x}{)}^2$

$\Rightarrow (3.25{)}^2=\frac{\sum x_i^2}{25}-(18.2{)}^2$

$\Rightarrow \sum x_i^2=25\left[\right(3.25{)}^2+\left(18.2{)}^2\right]$

$\Rightarrow \sum x_i^2=25\times [10.5625+331.24]$

$\Rightarrow \sum x_i^2=25\times 341.8025=8545.0625$ ...... (1)

For combined S.D. of the 40 observations n=40 and

${\sum }_{i=1}^{40}{x}_i^2=5524+8545.0625$

[From equation(1)]

$\Rightarrow {\sum }_{i=1}^{40}{x}_i^2=14069.0625$

Now,given : $\overline{x}=\frac{\sum x_i}{25}=18.2$

(For first set of Data)

$\Rightarrow \sum x_i=18.2\times 25=455$

$\Rightarrow {\sum }_{i=1}^{40}{x}_i=455+279=734$

(For both set of Data combined)

$\therefore$ Standard deviation for combined data will be

$\Rightarrow \sigma =\sqrt{\frac{14069.0625}{40}-{\left(\frac{734}{40}\right)}^2}$

$\Rightarrow \sigma =\sqrt{351.7265-(18.35{)}^2}$

$\Rightarrow \sigma =\sqrt{15.004}$

$\Rightarrow \sigma =3.87$

Hence, standard deviation based on all 40 observations is 3.87