Question

The second group of two samples has $100$ items with mean $15$ and S.D$=3$.If the whole group has $250$ items with mean $15.6$ and S.D$=\sqrt{13.44}$, then S.D of first group is

Answer 1

Number of items in 1st group$=250-100=150$ items.
let 1st group has items $y_1,y_2,\cdots ,y_{150}$ and
2nd group has items $x_1,x_2,\cdots ,x_{100}$
Now for second group
$\overline{x}=15,{\sigma }_2^2=9$
$\therefore 9=\frac{\sum {x_i}^2}{n}-{\overline{x}}^2=\frac{\sum {x_i}^2}{100}-{15}^2$
$\Rightarrow \sum {x_i}^2=23400\cdots \left(1\right)$
For whole group,
$13.44=\frac{\sum (x_i^2+y_i^2)}{250}-(15.6{)}^2$
$\Rightarrow \sum (x_i^2+y_i^2)=64200$
$\Rightarrow \sum y_i^2=64200-23400=40800\cdots \left(2\right)$ (from (1))
now,
mean of 1st group$=\frac{250\times 15.6-100\times 15}{150}=16$
Now for $1$st group,
${\sigma }_1^2=\frac{\sum y_i^2}{150}-{16}^2=\frac{40800}{150}-256$
${\sigma }^2=16$
$\therefore \sigma =4$

Alternate solution:
From the given data,
${\sigma }_2=9;\sigma =\left(\sqrt{13.44}\right)$
${\overline{x}}_2=15;\overline{x}=15.6$
$n_2=100;n=250$

So, $n_1=150$
$\overline{x}=\frac{n_1\overline{x_1}+n_2\overline{x_2}}{n_1+n_2}$
$\Rightarrow 15.6=\frac{150\left(\overline{x_1}\right)+100\left(15\right)}{250}$
$\Rightarrow \overline{x_1}=16$

Using Combined Variance formulae,
${\sigma }^2=\frac{1}{n_1+n_2}\left[n_1\right({\sigma }_1^2+d_1^2)+n_2({\sigma }_2^2+d_2^2\left)\right]$
where $d_1=\overline{x}-\overline{x_1}=15.6-16=-0.4$
$d_2=\overline{x}-\overline{x_2}=15.6-15=0.6$

$\therefore 13.44=\frac{1}{250}\left[150\right({\sigma }_1^2+0.16)+250(9+0.36\left)\right]$
$\Rightarrow {\sigma }_1^2=16$
$\Rightarrow {\sigma }_1=4$