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(C) Let $n_1 = 200, n_2 = 300$ be the sizes of the two samples.Let $\overline{x}_1 = 25, \overline{x}_2 = 10$ be their means.Let $\sigma_1 = 3, \sigma

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$67.2$

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(C) Let $n_1 = 200, n_2 = 300$ be the sizes of the two samples.Let $\overline{x}_1 = 25, \overline{x}_2 = 10$ be their means.Let $\sigma_1 = 3, \sigma_2 = 4$ be their standard deviations.The combined mean $\overline{x} = \frac{n_1 \overline{x}_1 + n_2 \overline{x}_2}{n_1 + n_2} = \frac{200 	imes 25 + 300 	imes 10}{500} = \frac{5000 + 3000}{500} = 16$.The variance of the combined sample is given by $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$,where $d_1 = \overline{x}_1 - \overline{x}$ and $d_2 = \overline{x}_2 - \overline{x}$.$d_1 = 25 - 16 = 9$ and $d_2 = 10 - 16 = -6$.$\sigma^2 = \frac{200(3^2 + 9^2) + 300(4^2 + (-6)^2)}{500} = \frac{200(9 + 81) + 300(16 + 36)}{500} = \frac{200(90) + 300(52)}{500} = \frac{18000 + 15600}{500} = \frac{33600}{500} = 67.2$.

The mean of two samples of size $200$ and $300$ were found to be $25$ and $10$ respectively. Their standard deviations $(S.D.)$ are $3$ and $4$ respectively. Then,the variance of the combined sample of size $500$ is:

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