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I am not sure if this is the case for just four numbers, but here is my question if someone could verify. If you take the mean of four numbers, subtract the mean from each of the four numbers, and then take the sum of these differences, is the sum always zero? Here is an example if the wording is off:

w = 1;

x = 2;

y = 3;

z = 4;

mean = 10/4 = 2.5

w-mean = -1.5;

x-mean = -.5;

y-mean=.5;

z-mean=1.5;

sum of these=0

I even tried it with a random number generator but always get zero. can someone verify if this is always true or just a fluke?

Thanks!!

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    $\begingroup$ Of course, and not just for four. The mean is given by $\mu=\frac 1n \sum_{i=1}^n x_i$ so $\sum (x_i-\mu)=n\mu -n\mu =0$. $\endgroup$ – lulu Jun 12 '17 at 17:57
  • $\begingroup$ Why didn’t you do it with the four numbers $a$, $b$, $c$, and $d$? $\endgroup$ – Lubin Jun 12 '17 at 17:58
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Yes, it's always true. Good observation!

If the mean of a sequence of $n$ numbers $m_1, m_2, ..., m_n$ is $\mu$ then the sum of those numbers is $n\mu$:

$$\sum_{j=1}^n m_j = n \mu.$$

Adding up the values of the numbers subtracted by their mean:

$$\sum_{j=1}^n (m_j-\mu) = \sum_{j=1}^n m_j - \sum_{j=1}^n \mu = n \mu - n \mu = 0.$$

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In general that will be true, and it's not something special about four numbers. The mean of your four numbers $w, x, y$ and $z$ is $\mu = (w+x+y+z)/4$. So you have

$$(w-\mu) + (x-\mu) + (y-\mu) + (z-\mu) = (w+x+y+z) - 4\mu$$ and remembering the value of $\mu$ gives

$$(w+x+y+z) - {4 \times {w+x+y+z \over 4}} = (w+x+y+z) - (w+x+y+z) = 0$$

There's nothing special about $4$; if you have $n$ numbers you'll have an $n$ in the denominator and you end up multiplying by $n$, so those cancel out.

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$a_1-\frac{a_1+a_2+a_3+a_4}{4}+a_2-\frac{a_1+a_2+a_3+a_4}{4}+a_3-\frac{a_1+a_2+a_3+a_4}{4}+a_4-\frac{a_1+a_2+a_3+a_4}{4} \\=(a_1+a_2+a_3+a_4)-(a_1+a_2+a_3+a_4)=0$

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