Let x₁, x₂, , x₂₀ be twenty observations. A new series is formed as y_i = x_i - 5 2 for i = 1, 2, , 20 . It is given that _ i=1 ²⁰ y_i = 20 and _ i=1 ²⁰ y_i^2 = 100 . Furthermore, _ i=1 ²⁰ (x_i - )^2 = 640 , where < 5 . If and ^2 are respectively the mean and the variance of the observations (x₁ - ), (x₂ - ), , (x₂₀ - ) , then the value of ^2 + ^2 + ^2 is equal to :

Step-by-step solution

From the given data for y_i , the mean is _y = y_i 20 = 20 20 = 1 . The variance of y_i is Var (y) = y_i^2 20 - ( _y)^2 = 100 20 - 1^2 = 5 - 1 = 4 . Since y_i = x_i - 5 2 , we have x_i = 2y_i + 5 . The mean of x_i is _x = 2 _y + 5 = 2(1) + 5 = 7 . The variance of x_i is Var (x) = 2^2 Var (y) = 4 4 = 16 . Now, _ i=1 ²⁰ x_i = 20 7 = 140 . Also, Var (x) = x_i^2 20 - ( _x)^2 16 = x_i^2 20 - 49 x_i^2 = 20 65 = 1300 . We are given _ i=1 ²⁰ (x_i - )^2 = 640 . Expanding this, we get x_i^2 - 2 x_i + 20 ^2 = 640 . Substituti