Let (x₁, x₂, , x₁₀ ) be ten observations such that ( _ i=1 ¹⁰ (x_i-2 )=30, _ i=1 ¹⁰ (x_i- )^2=98, 2 ), and their variance is ( 4 5 ). If ( ) and ( ^2 ) are respectively the mean and the variance of (2 (x₁-1 )+4 ), (2 (x₂-1 )+4 , ., 2 (x₁₀-1 )+4 ), then ( ^2 ) is equal to :

Step-by-step solution

aligned & _ l=1 ¹⁰ (x_l-2 )=30 & _ i=1 ¹⁰ x_l=50 & Mean =5 & Variance = 4 5 = x_l^2 10 -( x )^2 & 4 5 = x_l^2 10 -25 & x_l^2=258 aligned Now, _ l=1 ¹⁰ (x_l- )^2=98 aligned & _ l=1 ¹⁰ x_l^2-2 _ l=1 ¹⁰ x_l+10 ^2=98 & 258-2 (50)+10 ^2=98 & 10 ^2-100 +160=0 & ^2-10 +16=0 & =8 as >2 aligned Now, as per the question 2 (x₁-1 )+4 , 2 (x₂-1 )+4 , 2 (x₁₀-1 )+4 Can be simplified as 2 x₁+30,2 x₂+30, , 2 x₁₀+30 aligned & =2(5)+30=40 & ^2=2^2 ( 4 5 )= 16 5 & ^2 = 8 40 16 5 =100 aligned