The mean and variance of 10 observations are 5 and 4 respectively. If one observation x is replaced by another observation y , the new mean and new variance become 5.3 and 6.01 respectively. The value of x^2 + y^2 is equal to:

Step-by-step solution

Let the original sum of observations be x_i and the original sum of squares be x_i^2 . Given n = 10 , old mean = 5 , and old variance = 4 . x_i = 10 5 = 50 x_i^2 = 10(4 + 5^2) = 10 29 = 290 When observation x is replaced by y , the new mean is 5.3 . New sum = 50 - x + y = 10 5.3 = 53 y - x = 3 The new variance is 6.01 . New sum of squares = 10(6.01 + 5.3^2) = 10(6.01 + 28.09) = 10 34.1 = 341 Also, new sum of squares = 290 - x^2 + y^2 290 - x^2 + y^2 = 341 y^2 - x^2 = 51 Using the difference of squares: (y - x)(y +