The means of two groups of size 200 and 300 a | vedclass

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(C) Given: $n_1 = 200, \bar{x}_1 = 25, \sigma_1 = 3$ and $n_2 = 300, \bar{x}_2 = 10, \sigma_2 = 4$.The combined mean $\bar{x}$ is given by:$\bar{x} =

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$67.2$

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(C) Given: $n_1 = 200, \bar{x}_1 = 25, \sigma_1 = 3$ and $n_2 = 300, \bar{x}_2 = 10, \sigma_2 = 4$.The combined mean $\bar{x}$ is given by:$\bar{x} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{200 	imes 25 + 300 	imes 10}{500} = \frac{5000 + 3000}{500} = \frac{8000}{500} = 16$.Calculate the deviations $d_1$ and $d_2$:$d_1 = \bar{x}_1 - \bar{x} = 25 - 16 = 9$.$d_2 = \bar{x}_2 - \bar{x} = 10 - 16 = -6$.The combined variance $\sigma^2$ is given by:$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}$.$\sigma^2 = \frac{200(3^2 + 9^2) + 300(4^2 + (-6)^2)}{500}$.$\sigma^2 = \frac{200(9 + 81) + 300(16 + 36)}{500} = \frac{200(90) + 300(52)}{500}$.$\sigma^2 = \frac{18000 + 15600}{500} = \frac{33600}{500} = 67.2$.

The means of two groups of size $200$ and $300$ are $25$ and $10$ respectively. Their standard deviations are $3$ and $4$ respectively. What is the variance of the combined sample of size $500$?

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