Question

If x is the mean of n observations x₁, x₂, x₃, ... x_n, then prove that `sum_(i = 1)^n (x_i - barx) = 0`.

Answer 1

Given: `barx` is the mean of n observations x₁, x₂, x₃, ... x_n so `barx = (1/n) sum_(i = 1)^n x_i`.

To Prove: `sum_(i = 1)^n (x_i - barx) = 0`.

Proof [Step-wise]:

1. Start with the sum: `sum_(i = 1)^n (x_i - barx) = sum_(i = 1)^n x_i - sum_(i = 1)^n barx`.

2. `barx` is a constant does not depend on i, so `sum_(i = 1)^n barx = n·barx`.

3. Therefore, `sum_(i = 1)^n (x_i - barx) = sum_(i = 1)^n x_i - n·barx`.

4. By the definition of the mean, `barx = (1/n) sum_(i = 1)^n x_i`, so `n·barx = sum_(i = 1)^n x_i`.

5. Substituting gives `sum_(i = 1)^n (x_i - barx) = sum_(i = 1)^n x_i - sum_(i = 1)^n x_i = 0`.

Hence, `sum_(i = 1)^n (x_i - barx) = 0`, as required.