For the set A = {x_1, x_2, x_3, x_4, x_5},the | vedclass

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(C) For set $A$: $	ext{Mean} = \frac{\sum x_i}{5} = 2 \Rightarrow \sum x_i = 10$. Variance $= \frac{1}{5} \sum x_i^2 - (	ext{Mean})^2 = 4$ $\Rightar

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$5.5$

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(C) For set $A$: $	ext{Mean} = \frac{\sum x_i}{5} = 2 \Rightarrow \sum x_i = 10$. Variance $= \frac{1}{5} \sum x_i^2 - (	ext{Mean})^2 = 4$ $\Rightarrow \frac{1}{5} \sum x_i^2 - 4 = 4$ $\Rightarrow \sum x_i^2 = 40$. For set $B$: $	ext{Mean} = \frac{\sum y_i}{5} = 4 \Rightarrow \sum y_i = 20$. Variance $= \frac{1}{5} \sum y_i^2 - (	ext{Mean})^2 = 5$ $\Rightarrow \frac{1}{5} \sum y_i^2 - 16 = 5$ $\Rightarrow \sum y_i^2 = 105$. For $A \cup B$: Total elements $N = 10$. Combined Mean $\bar{X} = \frac{\sum x_i + \sum y_i}{10} = \frac{10 + 20}{10} = 3$. Combined Variance $= \frac{\sum x_i^2 + \sum y_i^2}{10} - (\bar{X})^2 = \frac{40 + 105}{10} - (3)^2 = 14.5 - 9 = 5.5$.

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

For the set $A = \{x_1, x_2, x_3, x_4, x_5\}$,the variance is $4$ and the mean is $2$. For the set $B = \{y_1, y_2, y_3, y_4, y_5\}$,the variance is $5$ and the mean is $4$. Then,the variance of $A \cup B$ is

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