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Distance to the average

First, consider how far each number is from the average. Suppose we have three numbers $ a$, $ b$ and $ c$, with an average of $ \frac{a+b+c}{3}$. How far is each number from the average? Well, each number ($ a,b,c$) is simply a point on the number line and the distance to the average is a simple subtraction. So, the distance from $ a$ to the average is $ a - \frac{a+b+c}{3} = \frac{2a-b-c}{3}$ Likewise, the distance of $ b$ and $ c$ to the average is $ \frac{2b-a-c}{3}$ and $ \frac{2c-a-c}{3}$, respectively. These numbers, by themselves, aren't that useful. However, let's ask a more important question: what is the total distance of $ a$, $ b$ and $ c$ to the average? The answer: $ \frac{2a-b-c}{3} + \frac{2b-a-c}{3} + \frac{2c-a-b}{3} = 0$ Now this is interesting. The sum of the distances to the average (from each point) is zero. Note that we are not talking about the absolute value of distance (such as `point a is 3 units away from the average'). We are using positive and negative distances (our convetion was $ a_i - average$). A positive distance means a point is greater than the average, and a negative distance indacates the point is less than the average. This example was just for 3 numbers. Of course, it works in general as well. Take n numbers ($ a_1$ to $ a_n$) and call the average $ a_x$ ( $ = \frac{1}{n}\sum^N{a_i}$). The sum of the distances to the average is
$\displaystyle \sum^N{(a_i - a_x)}$ $\displaystyle =$ $\displaystyle \sum^N{a_i} - \sum^N{a_x}$  
  $\displaystyle =$ $\displaystyle (n * a_x ) - (a_x * n)$  
  $\displaystyle =$ 0  

Not a hard proof, eh?
next up previous contents
Next: Interpretation Up: Mathematical Properties of the Previous: Introduction and definition   Contents
Kalid M. Azad 2002-10-31