$\begin {array} {1 1} (A)\;5 & \quad (B)\;4 \\ (C)\;6 & \quad (D)\;3.52 \end {array}$

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- $ \sigma^2= \large\frac{n_1( \sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$

Use $ \sigma^2= \large\frac{n_1( \sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}$

where $d_1=m_1-a$

$d_2=m_2-a$

$a$ = mean of the whole group

Let $ m_2$ = mean of the second group

$ \therefore 15.6 = \large\frac{100 \times 15+150 \times m_2}{250}$

$\Rightarrow\: 250 \times 15.6 = 100 \times 15 + 150 \times m_2$

$\Rightarrow\: \large\frac{250 \times15.6 - 100 \times 15}{150} $$= m_2$

$\Rightarrow\: \large\frac{3900-1500}{150}$$=m_2$

$\Rightarrow\: \large\frac{2400}{150}$$=m_2$

$\Rightarrow\:m_2=16$

$ 13.44=\large\frac{(100 \times 9 + 150 \times \sigma^2)+100 \times (0.6)^2+150 \times (0.4)^2}{250}$

$ \Rightarrow \sigma = 4$

Hence Ans : (B)

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