For a dataset of 6 observations, the sum of their deviations from 10 is 18 , and the sum of the squares of their deviations from 10 is 104 . If one of the observations is 18 , then the value of + ^2 , where and ^2 are the mean and variance of the remaining 5 observations respectively, is

Step-by-step solution

Let the 6 observations be x₁, x₂, , x₆ . Let y_i = x_i - 10 . Given _ i=1 ⁶ y_i = 18 and _ i=1 ⁶ y_i^2 = 104 . One observation is 18 , so its corresponding deviation is y₆ = 18 - 10 = 8 . For the remaining 5 observations, the sum of deviations is: _ i=1 ⁵ y_i = 18 - 8 = 10 The sum of squares of deviations for the remaining 5 observations is: _ i=1 ⁵ y_i^2 = 104 - 8^2 = 104 - 64 = 40 The mean of y for these 5 observations is y = 10 5 = 2 . The mean of the original 5 observations is = y + 10 = 2 + 10 = 12 . The varia