Question 9-Part (2) While finding the missing variance I am using the formula (1/n-1)*(sum of (Xi^2)- (n*mean^2)) Through this method i am getting the value as 0.1176 But in the solutions they have used the formula (1/n-1)*(sum of (Xi-mean)^2) through which the solution is 0.1213 Doubts- Should both the formulas not give the same result 2. How to decide which formula to use 3. Will I get marks if I use the first formula Thanks for your help !!!
Both should give exactly the same answer. The values are 5.193, 4.500, 4.787 They give a sample mean of 4.827 and a sum of squares of 70.135. This gives: 0.5(70.135 - 3*4.827^2) = 0.1213
