For the set $A=\left\{x_1, x_2, x_3, x_4, x_5\right\}$ the variance is 4 and the mean is 2 . For the set $B=0\left\{y_1, y_2, y_3, y_4, y_5\right\}$ the variance is 5 and the mean is 4 . Then, the variance of $A \cup B$ is

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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { (c) } A=\left\{x_1, x_2, x_3, x_4, x_5ight\} \ & 	ext { Mean }=\frac{\Sigma x_i}{5}=2 \Rightarrow \sum_{i=1}^5 x_i=10 \ & 	ext { Variance, } \frac{1}{5-1}\left[\Sigma x_i^2-\frac{\left(\Sigma x_iight)^2}{5}ight]=4 \ & =\frac{1}{4}\left[\Sigma x_i^2-\frac{(10)^2}{5}ight]=4 \ & \Rightarrow \ & \Rightarrow \quad \Sigma x_i^2-20=16 \ & B=\left\{y_1, y_2, y_3 y_4, y_5ight\} 	ext {, Mean }=\sum_{i=1}^5 \frac{y_i}{5}=4 \end{aligned} $ $ \Rightarrow $ $ \Sigma y_i=20 $ Variance, $\frac{1}{5-1}\left[\Sigma y_i^2-\frac{\left(\Sigma y_iight)^2}{5}ight]=5$ $ \begin{array}{rrr} \Rightarrow & \frac{1}{4}\left[\Sigma y_i^2-\frac{(20)^2}{5}ight]=5 \ \Rightarrow & \Sigma y_i^2-80=20 \ \Rightarrow & \Sigma y_i^2=100 \end{array} $ Now, $A \cup B=\left\{y_1, y_2, \ldots, y_5, x_1, x_2, \ldotsight.$, $ \left.x_5ight\}=\left\{a_i ; i=1,2, \ldots 10ight\} $ $ \begin{aligned} 	ext { Variance } & =\frac{1}{10-1}\left[\Sigma a_i^2-\frac{\left(\Sigma a_iight)^2}{10}ight] \ & =\frac{1}{9}\left[\left(x_1^2+x_2^2+\ldots+x_5^2+y_1^2+\ldots y_5^2ight)ight. \ & \left.-\frac{1}{10}\left\{x_1+\ldots+x_5+y_1+\ldots+y_5ight\}^2ight] \ & =\frac{1}{9}\left[\Sigma x_i^2+\Sigma y_i^2-\frac{1}{10}\left\{\Sigma x_i+\Sigma y_iight\}^2ight] \ & =\frac{1}{9}\left[36+100-\frac{1}{10}(10+20)^2ight] \ & =46 / 9 \sim 5.11 \end{aligned} $

For the set $A=\left\{x_1, x_2, x_3, x_4, x_5\right\}$ the variance is 4 and the mean is 2 . For the set…

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