Question

The mean and standard deviation of some data for the time taken to complete test calculated with the following results
Number of observations $=25$,
Mean $=18.2$
Standard deviation $=3.25$
Further, another set of $15$ observations $x_1, x_2, \ldots, x_n$ also in seconds, is now available and we have
$\displaystyle\sum_{i=1}^{n=15} x_i=279$ and $\displaystyle\sum_{i=1}^{15} x_i^2=5524$. The standard deviation based on all 40 observations is

• A. $2.87$
• B. $4.87$
• C. $3.87$
• D. $5.87$

Answer 1

Answer: (C)

$n=25, \bar{x}_1=18.2, \sigma_1=3.25$ and
$n=15, \displaystyle\sum_{i=1}^{15} \bar{x}_i=279, \displaystyle\sum_{i=1}^{15} x_i^2=5524$
For first set
$\Sigma x_i =25 \times 18.2=455 $
$\sigma_1^2 =\frac{\Sigma x_i^2}{25}-(18.2)^2$
$\Rightarrow (3.25)^2 =\frac{\Sigma x_i^2}{25}-331.24$
$\Rightarrow 10.5625+331.24 =\frac{\Sigma x_i^2}{25} $
$\Rightarrow \Sigma x_i^2 =25 \times 341.8025=8545.0625$
For required standard deviation, $n=40$
Combined sum of squares of two data
$\Sigma x_i^2=8545.0625+5524=14069.0625$
Combined sum of two data, $\Sigma x_i=455+279=734$
$\therefore $ Standard deviation $(\sigma) =\sqrt{\frac{14069.0625}{40}-\left(\frac{734}{40}\right)^2} $
$=\sqrt{351.726-(18.35)^2} =\sqrt{351.726-336.7225}$
$=\sqrt{15.0035}=3.87$