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Today we have learned about something which roughly translates to 'Markov-property' at school. I have looked it up and found this page: https://en.wikipedia.org/wiki/Markov_property, however, this is way too complicated for me and I am afraid this is not what I am looking for.

We learned that for statistical populations of units $x_1, x_2, x_3, \dots, x_n$ where all units are positive numbers (I am sorry if I use the terminology wrong, these words are completely new to me and English is not my native language), the following stands: Let $A > \bar{x}$ (Where $\bar{x}$ is the arithmetic average of the population). Then there are $\frac{n\bar{x}}{A}$ numbers which are greater or equal to $A$ (Where $n$ is the number of units in the population).

I hope I have been clear enough... Could anybody tell me what this is and where can I find more information about it?

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In English this is known as Markov's inequality.

Specifically, given a collection of samples $x_1,\ldots,x_n$ we can define an associated probability distribution by giving each sample an equal probability $\frac{1}{n}$. Then the random variable $X$ represents choosing one of the $n$ samples uniformly at random. Markov's inequality says that as long as $X$ always takes nonnegative values (i.e. all the $x_i$ are nonnegative), $$\mathbb P(X \ge a) \le \frac{\mathbb{E}(X)}{a}.$$ In this case, $\mathbb P(X \ge a)$ is the probability of choosing a sample that is larger than $a$, which is $\frac{1}{n}$ times the number of samples larger than $a$, while $\mathbb E(X)$ is the expected value of $X$, which is $\mathbb E(X) = \sum p_ix_i = \frac1n\sum x_i = \bar x$.

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    Blimey, I could have found that out by myself... Still, thank you, that's what I've been looking for, although this article seems extremely complicated to me. Never mind, I'll try to work it out somehow. – 2017-01-03
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    It's written in terms of probability theory rather than statistics, but the key point is the equation $\mathbb P(X\geq a)\leq \frac{\mathbb E(X)}{a}$: the fraction of values exceeding $a$ is less than $\bar X/a$, which is equivalent to the property in your question. – 2017-01-03
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    Isn't this $\mathbb{E}(X)$ something called the 'expected value' of $X$? Is it the same as $\bar{X}$ in this case? May I ask what that is and why is it equal to the mean? – 2017-01-03
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    @bertalanp99: Please see my edit. – 2017-01-03
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    Thank you, now it makes more sense, although I am afraid this is a bit beyond secondary school algebra. Could you also give a similar brief explanation of the corollary, 'Chebyshev's inequality'? – 2017-01-03
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    You can translate it the same way: For any collection of samples (which need not be nonnegative this time), the fraction of samples at distance $a$ or more from the mean is less than or equal to $\frac1{na^2} \sum (x_i-\bar x)^2$. – 2017-01-03