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 ∪B is

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Correct answer: C. (c) Set ⁡A= (x_1 ,x_2 ,x_3 ,x_4 ,x_5 ) Mean, x= x_in_1 ⇒2= x_i5 ⇒ x_i=10 Variance, V_1= x_i^2n_1- (x )^2 ⇒4= x_i^25-4 x_i=8 5=40. Set B= (y_1 ,y_2 ,y_3 ,y_4 ,y_5 ) Mean, y= y_in_2 ⇒4= y_i5 ⇒ y_i=20 Variance, V_1^ '= y_i ^2n_2- (y )^2 ⇒5= y_i ^25-16 y_i^2=21 5=105Now, A ∪B= (x_1 ,x_2 ,x_3 ,x_4 ,x_5 ) ∪ (y_1 ,y_2 ,y_3 ,y_4 ,y_5 ) n=10 , (x_1 ,x_2 ,x_3 ,x_4 ,x_5 ,y_1 ,y_2 ,y_3 ,y_4 ,y_5 )Variance =V= z_in- (z )^2 V= x_i^2+ y_i^2n- ( x_i+ y_in )^2 = 40+10510- ( 10+2010 )^2 V=14.5-9=5.5

$For the set A = (x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) the variance is 4 and the mean is 2 . For the set B = (y_{1}, y_{2}, y_{3}, y_{4}, y_{5}) the variance is 5 and the mean is 4 . Then, the variance of A \cup B is$

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