If 10 is the mean deviation of ' n ' observations x_1 ,x_2, x_3 , … ,x_n then the mean deviation of the observations 2x_1+53 , 2x_2+53 , 2x_3+53 , … . , 2x_n+53 is

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Correct answer: C. (c) Given observation are x_1 ,x_2 ,x_3 … . .x_n. x= x_1+x_2+x_3+ … .+x_nn New x= 1n( 2x_1+53+ 2x_2+53+ … . .+ 2x_n+53) New x= 2x_1+53 Mean deviation of the observations 2x_1+53 , 2x_2+53 , … … . 2x_n+53 NewM .D .= 1n 0.05em 0.05em(x_i-x) = 1n 0.05em 0.05em( 2x_i+53- 2x+53) = 1n( ( 2x_1+5-2x-53 )+ ( 2x_2+5-2x-53 )+) ( …+ ( 2x_n+5-2x-53 ))= (( 23 1n(x_1-x)+(x_2-x)+(x_3-x)+ … .+x_n-x) ) = 23 10= 203

$If 10 is the mean deviation of ' n' observations x_{1}, x_{2}, x_{3}, \dots, x_{n} then the mean deviation of the observations \frac{2 x _{1} + 5}{3}, \frac{2 x _{2} + 5}{3}, \frac{2 x _{3} + 5}{3}, \dots ., \frac{2 x _{n} + 5}{3} is$

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