The mean of 6 distinct observations is 6 . 5 and their variance is 10 . 25 . If 4 out of 6 observations are 2 , 4 , 5 and 7 , then the remaining two observations are:

Step-by-step solution

Let the other two observations be a and b , then a + b + 2 + 4 + 5 + 7 6 = 6 . 5 ⇒ a + b = 21 ⇒ b = 21 - a Then, other two numbers are a and 21 - a Now, variance is 10 . 25 , therefore 10 . 25 = 4 + 16 + 25 + 49 + a 2 + 21 - a 2 6 - 6 . 5 2 ⇒ 6 10 . 25 + 6 6 . 5 2 = 94 + a 2 + 21 - a 2 ⇒ a 2 + 21 - a 2 = 221 ∴   a = 10 and 21 - a = 21 - 10 = 11 So, remaining two observations are 10 , 11 .