Could xᵢ = x̅ ???

Question

I know it's a weird question.
But this thing is confusing me.
(x̅) : average μ

①
∵ $\frac{1}{n}\sum\limits_{i=1}^n(x_i) = \bar{x}$
∴ $\sum\limits_{i=1}^n(x_i) = {n}\bar{x}$

②
∵ $\sum\limits_{i=1}^n(C) = {n}{C}$ | C = constant
∴ $\sum\limits_{i=1}^n(\bar{x}) = {n}{\bar{x}}$

From ①, ②
$(x_i) = (\bar{x})$ ???

How could this be true?? Am I missing something?

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score 2 · Answer 1 · answered May 12 '21 at 15:02

2

$$1+3=4$$ $$2+2=4$$

We can't conclude that $1=2$.

answered May 12 '21 at 15:02

Siong Thye Goh

Ah I see. That makes sense. Thanks! – Roo Tenshi May 12 '21 at 15:09
But wait, we CAN conclude that (1+3) = (2+2). Shouldn't the same apply to xᵢ and x̅? Since the sum of both is nx̅. Right? – Roo Tenshi May 12 '21 at 15:26
we can't just extract one component $1$ and claim that it must be equal to $2$. – Siong Thye Goh May 12 '21 at 15:30
Yes but I didn't extract one component. xᵢ is all terms inside the first sum and x̅ is all the terms inside the other sum. – Roo Tenshi May 12 '21 at 16:28
what do you mean by all the terms? $x_1=1, x_2=3$. Also can I check the notation, does the braces means anything? – Siong Thye Goh May 13 '21 at 02:38
By all terms I mean the 'elements' we are adding together. x1, x2, x3, ... , xn. And the braces is just for enclosing what we are summing up. – Roo Tenshi May 25 '21 at 17:52
we have the sum are equal, but not the individual term. – Siong Thye Goh May 26 '21 at 01:09

score 0 · Answer 2 · answered May 12 '21 at 15:03

0

$\bar x$ is the average, so it's just some number that you can factor out of the summand:

$$\sum_{i=1}^n\bar x=\bar x\sum_{i=1}^n1=\bar x\left(\underbrace{1+1+\cdots+1}_{n\text{ times}}\right)=\bar x n$$

answered May 12 '21 at 15:03

user170231

this is actually very useful. but what i was looking for is whether xᵢ = x̅ or not. Because the the sum of xᵢ = the sum of x̅ = nx̅. That's why I was confused. – Roo Tenshi May 12 '21 at 15:12

2 Answers2