Mathematical average formula For any mathematical average, the sum of how much larger than the average the numbers above are must equal the sum of how much smaller than the average the numbers below are.
Mathematical averages are usually calculated by adding all the numbers up and dividing the sum by how many numbers there are. The mathematical average of a group of numbers indicates the number overall the overall group is closest to.
Now an example of the theorem:
For example, if the numbers are $75$ and $79$, the average is $77$ because $79$ is 2 larger than $77$ and $75$ is 2 smaller than $77$. This is common sense.
But what if you have 3 numbers? Let's say you have $74$, $78$, and $79$. The average is $77$ because $78$ is 1 larger than $77$ and $79$ is 2 larger than $77$ and 2 + 1 = 3. $74$ is 3 smaller than $77$ so 3=3. You can use guess and check until you arrive at the average if you're not sure which number to pick.
This method can help you calculate means of numbers that are close together faster in your head without adding up the numbers or the one's digits of the numbers. It will work on any mean.
Derivation of the theorem:
$N_1 - y + N_3 - y +… = y - N_2 + y - N_4 +… $
$N_1 + N_3 + N_2 + N_4 + N… = 4y + … +y$
Sorry, I didn’t use the correct notation here, but one can see you will have all the extra $N_5$’s, $N_6$’s, and so on on the left side and the corresponding number of additional $y$’s on the right side.
$\frac{N1 + N3 + N2 + N4 + N…}{4 + …} = y$
 A: Let me just write it in proper notations for you.
Consider $X=\{x_i\}_{i = 1}^n$ and define
$$\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i\tag{1}$$
Now we denote $L=\{l_j\}=\{x_i\in X\mid x_i<\bar{x}\}$ and $h=\{h_k\}=\{x_i\in X\mid x_i>\bar{x}\}$. Let the size of $L$ be $J$ and the size of $H$ be $K$.
Clearly, $L$ and $H$ are disjoint. Now if we consider that there exist $M\ge0$ numbers in $X$ equal to $\bar{x}$, we have
\begin{align}
n\bar{x} &= \sum_i x_i\\
(J+K+M)\bar{x}&= M\bar{x}+\sum_j l_j +\sum_k h_k \\
J\bar{x} -\sum_j l_j  &= (\sum_k h_k) - K\bar{x}\\
\sum_{j} (l_j-\bar{x}) &= \sum_k (h_k-\bar{x})
\end{align}
which completes the proof.
Anyway, it's a nice observation but nowhere as novel as you may have thought. This will probably show up somewhere in Stat 101 exercises you do when you finally go to college.
A: True but not new, the idea is somehow considered trivial, but nice derivation anyway...
For example Khan Academy mentioned it here https://www.khanacademy.org/math/ap-statistics/summarizing-quantitative-data-ap/mean-median-more/a/mean-as-the-balancing-point
A: Let $X$ be a random variable. Then the expectation value of $X$ is in general denoted by $\Bbb{E}[X]$ and it represents the average of the variable $X$. If $X$ is the difference of two random variables $Y$ and $Z$, it is true that:
$$\Bbb{E}[Y-Z] = \Bbb E[Y] - \Bbb E[Z]$$
More succintly, $\Bbb E$ is linear.
So suppose $\mu = \Bbb E[X]$. Then the average of the random variable $X - \mu$ which represents the "distance from the average" is:
$$\Bbb E[X - \mu] = \Bbb E[X] - \Bbb E[\mu] = \Bbb E[X] - \mu = \Bbb E[X] - \Bbb E[X] = 0$$
Which means that, on average, a random variable will have distance zero from the average. This, in turn, implies that for every value above the average a random variable can take, there is one that balances it below the average. 
