Why is the arithmetic mean the more typical value of a data set Aristoteles said, By the mean of a thing I denote a point equally distant from either extreme. Let's call that equidistant point y. Considering two data points y1 and y2, the equidistant point y would be found by setting the equality 
y1−y = y−y2 and solving for y
y=y1+y2/2 
which is the arithmetic mean of  y1 and  y2. 
Departing from this formula of mean being an equidistant point from the extremes, how do we come to the conclusion that the general formula for mean represent the average value, or most typical value of the data set. 
 A: Let the "thing" whose mean we wish to find be the number of children born to a single woman.
Of course there are women who in their whole lives never gave birth to any children, so one extreme of the number of children is zero.
At the other extreme, The Guinness Book of World Records
says one woman gave birth to $69$ children in the 18th century.
Therefore, by Aristotle, the mean number of children that a woman might bear is $\frac{0+69}{2}=34.5,$ or at least we should say it was that much in the 18th century.
But other sources say the fertility rate in the 18th century was more like $6$ or $7$ mean births per woman.
In fact, even at that time $69$ births was an extremely unlikely number, much less likely than zero. To consider only the two extremes, zero and $69,$ and give them equal weight does not make much sense.
A: Think probabilities and expectation. Let's say you have a probability distribution over the set $A = \{x, y\}$, i.e. $\mathbb{P}(x) = \mathbb{P}(y) = \frac{1}{2}$. Let $x, y$ be real numbers. The expectation of a random variable $X$ drawn from this particular distribution over $A$ is as follows $\mathbb{E}[X] = \frac{1}{2} \cdot x + \frac{1}{2}\cdot y = \frac{x + y}{2}$. Thus, you can say that this formula is arrived by placing a uniform distribution over the set of numbers. 
Note that this generalizes readily. If $A = \{a_1, \ldots, a_n\}$, and you placed a probability mass of $\frac{1}{n}$ on each of the elements of $A$, you'd get that the expected value is $\frac{a_1 + \ldots + a_n}{n}$.
I want to repeat the main message of my answer. You are placing a uniform distribution over the elements. The uniform distribution says that each of the values is equally likely. So equally likely $\equiv$ most typical value. I hope this helps.
