Question

The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3$. If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is

• A. $5$
• B. $6$
• C. $4$
• D. $8$

Answer 1

The correct option is C $4$

$n_1=100,n_2=150$
We know that
$(n_1+n_2)\overline{X}=n_1{\overline{X}}_1+n_2{\overline{X}}_2$
where
$\overline{X}=$ mean of whole group
${\overline{X}}_1=$ mean of first sample of a group
${\overline{X}}_2=$ mean of second sample of a group
$\Rightarrow 250\times 15.6=100\times 15+150\times {\overline{X}}_2$
$\Rightarrow {\overline{X}}_2=\frac{250\times 15.6-100\times 15}{150}$
$\therefore {\overline{X}}_2=16$

Standard deviation of whole group, $\sigma =\sqrt{13.44}$
$\Rightarrow$ Standard variance of whole group ${\sigma }^2=13.44$
$\Rightarrow 13.44=\frac{\sum x^2}{n}-(\overline{X}{)}^2$
$\Rightarrow 13.44=\frac{\sum x^2}{250}-(15.6{)}^2$
$\Rightarrow \sum x^2=64200$

For first sample of a group :
Standard deviation of whole group, $\sigma =3$
$\Rightarrow$ Standard variance of whole group ${\sigma }^2=9$
$\Rightarrow 9=\frac{\sum x_1^2}{n_1}-({\overline{X}}_1{)}^2$
$\Rightarrow 9=\frac{\sum x_1^2}{100}-(15{)}^2$
$\Rightarrow \sum x_1^2=23400$
$\therefore \sum x_2^2=\sum x^2-\sum x_1^2$
$\Rightarrow \sum x_2^2=64200-23400$
$\Rightarrow \sum x_2^2=40800$

For second sample of a group :
${\sigma }^2=\frac{\sum x_2^2}{n_2}-({\overline{X}}_2{)}^2$
$\Rightarrow {\sigma }^2=\frac{40800}{150}-(16{)}^2$
$\Rightarrow \sigma =4$