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Maybe I am not well-versed with the actual definition of mean, but I have a doubt. On most resources, people say that arithmetic mean is the sum of $n$ observations divided by $n$. So my first question:

How does this formula work? Is there any derivation to it? If not, then while creating this definition, what was the creator thinking?

Okay, so using my intuition, I thought that it is the value that lies in the centre. And it worked for some cases, like the mean of $1$ , $2$ and $3$ is $2$ , which is the central value. But, lets imagine a number line from numbers $0$ to $9$. Now, I choose $3$ numbers, say $1$, $8$ and $9$. By the formula, I get the mean is equal to $6$. But, if mean really is a central value, shouldn't it be $5$(I know we call $5$ the median in this case)? But it seems like the mean is getting closer to $8$ and $9$, which means it is not central? So my final question?

Have I imagined mean incorrectly? What kind of central value really mean is?

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  • $\begingroup$ The thread here might be of some use to you; it discusses the geometric mean, but many of the answers start with why the arithmetic mean is the way it is (and proceed by analogy). $\endgroup$ Jan 3, 2016 at 17:18
  • $\begingroup$ So you know the median is a different kind of "average" or central value, and you know the definition of the arithmetic mean. Depending on the application, one notion of central value may be more useful than the other, and in many important distributions the median is the arithmetic mean. Given that you have the definitions correct, are you interested in the history of these concepts? $\endgroup$
    – hardmath
    Jan 3, 2016 at 17:19

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The arithmetic mean formula is a definition of the term, so there is no derivation. The mean does not involve the range that the numbers might be over, like your $1$ to $9$ example. It only involves the actual numbers. You can define the term central value to take a set of numbers and return half the sum of the max and the min. That seems to be what you are doing with your example of $1,8,9$. That is a fine definition. Whether it is useful or not is yet to be seen. Given $1,8,9$, the mean of $6$ reflects the fact that two of the numbers are high. It is also the expected value-if you draw many times with replacement from $1,8,9$, the sum will be close to $6$ times the number of draws.

There are several statistics that can be used similarly. You are trying to take a distribution of numbers and report one number that "gives a feel" for the set. The arithmetic mean is one. Others are the median, the mode, the geometric mean, etc. You need to look at your purpose and choose the one that suits the need, then be careful to say which you have used.

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I really don't think that people came up with the formula for the arithmetic mean by starting with a nice property and deriving the formula to obtain that property. However, one useful characterization of the mean is that it is the unique minimizer of the function $f(m)=\sum_{i=1}^n (x_i-m)^2$. That is, it is the number closest to the sample vector in the sense of the ordinary Euclidean distance. From this property it follows that $\sum_{i=1}^n (x_i-m)=0$, so that the signed deviations add up to zero.

The median has a similar characterization: it is a minimizer (in general not the only one) of the function $g(m)=\sum_{i=1}^n |x_i-m|$. So this is the closest to the sample vector in the sense of absolute distance. From this property we get $\sum_{i=1}^n \operatorname{sign}(x_i-m)=0$, so that there are an equal number of samples larger than the median and smaller than the median.

By the way, the metric defining the median is bad for a number of reasons. One emerges if you consider the linear regression problem for minimizing the sum of $p$th powers of the residuals where $p \geq 1$. When $p=1$ you have the metric which defines the median. In this case, if you have $N-1$ points on the line and $1$ point not on the line, one of the lines minimizing the sum of the absolute residuals is the line through the $N-1$ points regardless of how far away the outlier point is from the line. This is a bad property, the system should respond to that outlier at least a little bit.

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You are talking about median, given some sequence of ordered numbers

$$1,1,2,2,3,3,\color{red}{3},6,6,7,8,9,9\tag{1}$$

The number at the center is called a median. When there are even number of elements,

$$1,1,2,2,3,\color{red}{3},\color{red}{6},6,7,8,9,9$$

Median is defined as a mean of those two numbers, namely $4.5$.

Different thing goes on with arithmetic mean. This is the sum of elements divided by number of them. Arithmetic mean of $(1)$ is $$\frac{1+1+2+2+3+3+3+6+6+7+8+9+9}{13}=\frac{60}{13}\approx 4.6153$$

which clearly differ from its median. The median cares only about numbers that are in half of the sequence, arithmetic mean cares about all numbers.

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    $\begingroup$ @kamilo9875, I think that didnt answer my question. I know what median is... I was asking some more info., like how does the mean formula work.. and what kind of central tendency it is.. $\endgroup$
    – codetalker
    Jan 3, 2016 at 17:16
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Let me give the bureaucratic answer first:

No, there is no "derivation" (supposing that you mean something like a "theorem") that concludes that $E(X)$ is what it is. $E(X)$ is defined that way.

Now to properly answer your question:

At a very naive level, you can think of the mean as "given a collection of values coming from $n$ persons, the value you should distribute to each person in order that all of them receive the same value". It is a nice intuition, but you may feel it is a bit circular...

Well, in my opinion, if you don't understand the intuition (or think it is not justified) of the definition, the proper way to get a feeling of what "mean" is is by taking a look at the Law of Large Numbers.

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I suppose how you would come up with mean:

Suppose we have played some soccer games, and here is the list of the goals in the soccer games we played:

$${1,1,1,1,1,9}$$

And we want to get a sense of how we did as a whole or how much we scored per game as a whole.

Well looking at the list we see that we played for $6$ games and in that $6$ games we scored $14$ goals, so our rate as a whole would be about:

$$2.33 goals/game$$

Which we see is a pretty good measure of how we did as a whole. But of course in this case median is a better measure of how we would do on average, as it is not influenced by outliers, meaning that that $9$ was very unusual so it doesn't influence the median. We look at many tools to find an average, like mean, median,... and use what we find best in the situation.

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