Let a dataset X consist of the first 13 terms of an arithmetic progression whose first term is 3 and common difference is 4 . A new dataset Y is formed by applying the transformation y = x + , where x X and , R . If the mean and variance of the elements of Y are 50 and 2016 respectively, then the absolute difference between all possible values of is

Step-by-step solution

The dataset X is an arithmetic progression with first term a = 3 , common difference d = 4 , and number of terms n = 13 . The mean of X is the middle term: x = a + n-1 2 d = 3 + 6(4) = 27 . The variance of an AP is given by Var ( X ) = d^2(n^2 - 1) 12 . Var ( X ) = 16(169 - 1) 12 = 16 168 12 = 16 14 = 224 . For the dataset Y , we are given y = x + . The variance of Y is Var ( Y ) = ^2 Var ( X ) . 2016 = 224 ^2 ^2 = 9 = 3 or = -3 . The mean of Y is y = x + . 50 = 27 + . Case 1: If = 3 , then 50 = 27(3) + = 50 - 81 =