The mean of five observations is 4 and their variance is 5.2 . If three of these observations are 1, 2 and 6, then the other two observations are

Step-by-step solution

Let the five observations be x₁, x₂, x₃, x₄, x₅ . Given x₁ = 1, x₂ = 2, x₃ = 6 . Let the other two observations be a and b . The mean x = 1+2+6+a+b 5 = 4 . 9 + a + b = 20 a + b = 11 . The variance ^2 = x_i^2 5 - ( x )^2 = 5.2 . 1^2 + 2^2 + 6^2 + a^2 + b^2 5 - 4^2 = 5.2 . 1 + 4 + 36 + a^2 + b^2 5 - 16 = 5.2 . 41 + a^2 + b^2 5 = 21.2 . 41 + a^2 + b^2 = 106 a^2 + b^2 = 65 . Using (a+b)^2 = a^2 + b^2 + 2ab , we have 11^2 = 65 + 2ab . 121 = 65 + 2ab 2ab = 56 ab = 28 . The quadratic equation with roots a and b is t^2 - (