# 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.

$$\frac{N1 + N3 + N2 + N4 + N…}{4 + …} = y$$

• It's really hard to come up with a new math formula or theorem, because armies of mathematicians have been exploring the mathematical landscape obsessively for hundreds of years (or longer). Usually you have to study math for many years (typically by doing a PhD) before you can find something new. This way that you've described of thinking about and computing the average is well known. I like your enthusiasm for math though. This could be a good pedagogical contribution -- it's something that should perhaps be emphasized more often when teaching students to compute averages. – littleO Aug 24 at 20:12
• It true that you retain a copyright in your original form of expression when posting here, and it is not necessary to include a closing remark to that effect. On the other hand mathematical principles are not subject to copyright (or patent ownership), so one should have a narrow perspective on what is claimed. In any case you would improve your exposition by using mathematical expressions once the basics of $\LaTeX$ are learned. – hardmath Aug 24 at 20:13
• It may not be publishable as a new mathematical result, but it shows independent thinking and a good amount of effort. I applaud this. I hope you get a benefit from the responses. – David K Aug 24 at 20:49

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

• Isn't it useful on the SAT (where guess and check can be eliminated), and I've never seen it anywhere online. Can you tell me where it is online or in texts? – Yukang Jiang Aug 24 at 20:10
• @YukangJiang it's just a trivial result that nobody bothers to name it – Karn Watcharasupat Aug 24 at 20:12
• Why didn't I learn it for the SAT's then? It would speed up answering some mean questions on the SAT. – Yukang Jiang Aug 24 at 20:15
• @Berci SAT is an exam (usually includes math and english tests), the math is basic math or school math, but it has many questions(compared to the short period of time given) that have short answers or multiple choice questions. – Fareed AF Aug 24 at 20:31
• The Khan Academy treatment reminds me of math I did in 5th or 6th grade. So the idea is literally elementary. But we didn’t use any notation like in the question. So I would encourage OP to work a little more to get the notation really good. It won’t get you published (unless you can write it into a study guide) but you have to learn how to do this sort of thing if you want to be able to do anything more advanced. – David K Aug 24 at 20:44

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.

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.