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(B) Let $n_1 = 50$,$\bar{x}_1 = 10$,and $\sigma_1 = 2$ be the parameters for the first group. Let $n_2 = 50$,$\bar{x}_2$,and $\sigma_2$ be the paramet

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$3$

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(B) Let $n_1 = 50$,$\bar{x}_1 = 10$,and $\sigma_1 = 2$ be the parameters for the first group. Let $n_2 = 50$,$\bar{x}_2$,and $\sigma_2$ be the parameters for the second group. The total number of observations is $N = 100$,with mean $\bar{x} = 8$ and standard deviation $\sigma = \sqrt{10.5}$.First,find the mean of the second group:$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} \implies 8 = \frac{50(10) + 50(\bar{x}_2)}{100} \implies 800 = 500 + 50\bar{x}_2 \implies \bar{x}_2 = 6$.Next,use 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 $d_1 = \bar{x}_1 - \bar{x} = 10 - 8 = 2$ and $d_2 = \bar{x}_2 - \bar{x} = 6 - 8 = -2$.$10.5 = \frac{50(2^2 + 2^2) + 50(\sigma_2^2 + (-2)^2)}{100}$$10.5 = \frac{50(8) + 50(\sigma_2^2 + 4)}{100} = \frac{400 + 50\sigma_2^2 + 200}{100}$$1050 = 600 + 50\sigma_2^2$$50\sigma_2^2 = 450 \implies \sigma_2^2 = 9 \implies \sigma_2 = 3$.

For a group of $100$ observations,the arithmetic mean and standard deviation are $8$ and $\sqrt{10.5}$ respectively. The mean and standard deviation of $50$ items selected from these $100$ observations are $10$ and $2$ respectively. Then the standard deviation of the remaining $50$ observations is

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