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(C) Given $n_{1} = 100$,$\bar{x}_{1} = 15$,$\sigma_{1} = 3$.Total items $n = 250$,combined mean $\bar{x} = 15.6$,combined variance $\sigma^{2} = 13.44

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$4$

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(C) Given $n_{1} = 100$,$\bar{x}_{1} = 15$,$\sigma_{1} = 3$.Total items $n = 250$,combined mean $\bar{x} = 15.6$,combined variance $\sigma^{2} = 13.44$.Since $n = n_{1} + n_{2}$,we have $n_{2} = 250 - 100 = 150$.Using the combined mean formula $\bar{x} = \frac{n_{1}\bar{x}_{1} + n_{2}\bar{x}_{2}}{n_{1} + n_{2}}$:$15.6 = \frac{100(15) + 150(\bar{x}_{2})}{250}$ $\Rightarrow 3900 = 1500 + 150\bar{x}_{2}$ $\Rightarrow 150\bar{x}_{2} = 2400$ $\Rightarrow \bar{x}_{2} = 16$.Using the combined variance formula $\sigma^{2} = \frac{n_{1}(\sigma_{1}^{2} + d_{1}^{2}) + n_{2}(\sigma_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}}$,where $d_{1} = \bar{x}_{1} - \bar{x} = 15 - 15.6 = -0.6$ and $d_{2} = \bar{x}_{2} - \bar{x} = 16 - 15.6 = 0.4$:$13.44 = \frac{100(3^{2} + (-0.6)^{2}) + 150(\sigma_{2}^{2} + (0.4)^{2})}{250}$.$13.44 	imes 250 = 100(9 + 0.36) + 150(\sigma_{2}^{2} + 0.16)$.$3360 = 936 + 150\sigma_{2}^{2} + 24$.$3360 = 960 + 150\sigma_{2}^{2}$ $\Rightarrow 2400 = 150\sigma_{2}^{2}$ $\Rightarrow \sigma_{2}^{2} = 16$.Thus,$\sigma_{2} = 4$.

$x_i$	$2$	$4$	$6$	$8$	$10$	$12$	$14$	$16$
$f_i$	$4$	$4$	$\alpha$	$15$	$8$	$\beta$	$4$	$5$

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:

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