For two data sets,each of size 5,the variance | vedclass

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(A) Given: $\sigma_{x}^{2} = 4$ and $\sigma_{y}^{2} = 5$ for two sets of size $n_1 = 5$ and $n_2 = 5$.Means are $\bar{x} = 2$ and $\bar{y} = 4$.Using

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$\frac{11}{2}$

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(A) Given: $\sigma_{x}^{2} = 4$ and $\sigma_{y}^{2} = 5$ for two sets of size $n_1 = 5$ and $n_2 = 5$.Means are $\bar{x} = 2$ and $\bar{y} = 4$.Using the formula $\sigma^2 = \frac{\Sigma x_i^2}{n} - (\bar{x})^2$,we have:$\Sigma x_i^2 = n(\sigma_x^2 + \bar{x}^2) = 5(4 + 2^2) = 5(8) = 40$.$\Sigma y_i^2 = n(\sigma_y^2 + \bar{y}^2) = 5(5 + 4^2) = 5(21) = 105$.Combined mean $\bar{z} = \frac{n_1\bar{x} + n_2\bar{y}}{n_1 + n_2} = \frac{5(2) + 5(4)}{10} = \frac{30}{10} = 3$.Combined variance $\sigma_z^2 = \frac{\Sigma x_i^2 + \Sigma y_i^2}{n_1 + n_2} - (\bar{z})^2$.$\sigma_z^2 = \frac{40 + 105}{10} - (3)^2$.$\sigma_z^2 = \frac{145}{10} - 9 = 14.5 - 9 = 5.5 = \frac{11}{2}$.

For two data sets,each of size $5$,the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$,respectively. The variance of the combined data set is

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