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(B) Let $n_1 = 20$ (boys) and $n_2 = 30$ (girls). Total candidates $N = 50$.Given: $\bar{x}_b = 12$,$\sigma_b^2 = 2$,$\sigma_g^2 = 2$,and combined mea

Vedclass · Accepted Answer

$25$

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(B) Let $n_1 = 20$ (boys) and $n_2 = 30$ (girls). Total candidates $N = 50$.Given: $\bar{x}_b = 12$,$\sigma_b^2 = 2$,$\sigma_g^2 = 2$,and combined mean $\bar{x} = 15$.First,find the mean marks of girls $(\mu = \bar{x}_g)$:$N \bar{x} = n_1 \bar{x}_b + n_2 \bar{x}_g$$50 	imes 15 = 20 	imes 12 + 30 	imes \bar{x}_g$$750 = 240 + 30 \bar{x}_g$$30 \bar{x}_g = 510 \Rightarrow \bar{x}_g = 17 = \mu$.Now,calculate the combined variance $\sigma^2$:$\sigma^2 = \frac{n_1 \sigma_b^2 + n_2 \sigma_g^2}{n_1 + n_2} + \frac{n_1 n_2}{(n_1 + n_2)^2} (\bar{x}_b - \bar{x}_g)^2$$\sigma^2 = \frac{20 	imes 2 + 30 	imes 2}{50} + \frac{20 	imes 30}{50^2} (12 - 17)^2$$\sigma^2 = \frac{100}{50} + \frac{600}{2500} (-5)^2$$\sigma^2 = 2 + \frac{6}{25} 	imes 25 = 2 + 6 = 8$.Finally,$\mu + \sigma^2 = 17 + 8 = 25$.

Set	Size	Mean	Variance
Observation $I$	$10$	$2$	$2$
Observation $II$	$n$	$3$	$1$

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