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 $x_1,x_2,\dots ,x_n$ also in seconds, is now available and we have
$\sum _{i=1}^{n=15}{x}_i=279$ and $\sum _{i=1}^{15}{x}_i^2=5524$. 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
$n=15,\sum _{i=1}^{15}{\bar{x}}_i=279,\sum _{i=1}^{15}{x}_i^2=5524$
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 $(\sigma )=\sqrt{\frac{14069.0625}{40}-{\left(\frac{734}{40}\right)}^2}$
$=\sqrt{351.726-(18.35{)}^2}=\sqrt{351.726-336.7225}$
$=\sqrt{15.0035}=3.87$