I was teaching elementary school kids (aged 10) about the mean. The intuition I gave them is roughly as follows:

You are trying to find a value such that the sum of all the distances from the mean for the values above the mean is the same as the sum of all the distances from the mean for the values below the mean.

They were happy with this and we supplemented it by plotting sets of points and drawing the line representing the mean.

But then I told them how to compute the mean. That is add up all the values and divide by the number of values.

The method of calculation seemed completely disconnected from the intuition for what the mean is.

What is a good way to give intuition for the standard method of computing the mean?

Take 3 piles of toothpicks (different amounts in each). Shove them together and split them into 3 equal piles. Make the kids do it. Do it with marbles. Do it with money. Do it a few times: "lather, rinse, repeat". ;)

  • This is by far the best answer and it also demonstrates that the intuition that OP has tried to build up might not be optimal. – Jasper Oct 1 at 13:47
  • While this may be an effective way to lead students to an empirical notion of mean, it does little (without further development) to explain why the mean is defined the way it is defined. What is the significance of the number of toothpicks in each of the three equal piles? What more meaning has this number? – Dan Fox Oct 1 at 19:35
  • The number in the equal piles is... Exactly that. The number each pile would have to have if they were equal. I think this is an better mental image than the "sum if distances" construction. – Jasper Oct 2 at 12:29
  • Just trying to give the kids some gut feel ("intuition"). Not an all encompassing or mathematically rigorous definition. If you have 3 separate, unequal, piles of candy (and 3 kids), I am sure they will get the point of re-aligning them into equal piles. And it will mean something to them! Later, they can consider the algebraic insights. But it will build off of some gut feel. Also, this is not about experimental methods, per se, but about building gut feel, intuition. Humans are not computer programs. – guest Oct 2 at 22:17
  • I am not sure how well this works when the mean is not an integer. – Anush Oct 5 at 7:55

[Note: I interpreted the question as how to explain the relationship between the two definitions]

Let's first consider the underlying algebra before turning to real-world models.

As an example, suppose we have four reals $a_1 < a_2 < a_3 <a_4$ with mean $a$ between $a_2$ and $a_3.\,$ Here the equivalence connecting the two views of the mean is as follows
$$\overbrace{a+a+a+a = a_1 + a_2 + a_3 + a_4}^{\large\text{definition of mean } a}\iff \overbrace{a\!-a_1\, +\, a\!-a_2\, =\, a_3 - a\, +\, a_4 -a}^{\large{ a_i\ \text{are balanced around the mean } a}} $$

The direction $(\Rightarrow)$ arises as follows: in the definition, we can cancel a LHS $a$ from each $a_i \ge a$ on the RHS, and we can cancel each RHS $a_i < a$ from an $a$ on the LHS, yielding the equivalent balanced form on the right. This method can be used more generally to check an equality of sums by rewriting it more simply, e.g.

$\qquad\qquad\qquad\begin{align} &\color{#c00}{222}+\color{#0a0}{200}+1000+4119\\ =\ &\color{#c00}{100}+\color{#0a0}{328}+2113+3000\end{align}$ $\ \iff\ \begin{align} &\color{#c00}{122} + 1119\\ =\ &\color{#0a0}{128} + 1113 \end{align}$ $\iff \begin{align} &6\\ =\ &6 \end{align}$

Above we cancelled $\color{#c00}{100}$ from both sides, leaving the summand $\color{#c00}{122}$ on the new LHS.
Next, $\, $ we cancelled $\color{#0a0}{200}$ from both sides, leaving the summand $\color{#0a0}{128}$ on the new RHS, etc. The OP is just the special case when all summands are equal on one side, say $a$. This replaces each $a_i$ by its distance from $a$, doing it on the equation side that keeps all the summands nonnegative.

Conversely, direction $(\Leftarrow)$ follows by inverting the prior, i.e. by adding terms to both sides of the equation in order to move all negated terms to the opposite side of the equation, i.e. by eliminating all subtractions.

A simple real-world model is an old-fashioned balance scale. Then the prior additions and subtractions to both sides amount to adding or removing equal weights from both sides of the scale - which preserves balance. One can illustrate the above equivalence by explicitly performing the steps in the above sketched proof using weight manipulations. Of course one should use much smaller numbers than I did above (I chose those larger numbers only to highlight that the method can yield nonntrivial simplifications).

  • This seems like massive overkill for 10-year-olds. – Ben Crowell Sep 30 at 0:52
  • 1
    @Ben After your answer I see that we have interpreted the question very differently. You interpreted it simply as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it less trivially: how to explain the relationship between the two equivalent definitions. I have in fact taught this at low levels with success. Note the that above exposition is written at the level of the teacher. It should be simplified as need be for students. – Bill Dubuque Sep 30 at 1:14

For basic intuition with kids this age, I would simply draw a number line and locate 2 and 4 as dots. Ask them what number is half-way in between. They say 3. Then demonstrate that $(2+4)/2=3$. Of course this isn't a proof, but the goal is to build intuition, not to prove anything.

To amplify on this simple example, describe a situation where Alice has 2 cookies and Betty has 4, but they want to be fair and share equally. How many should each have? Describe the fact that they gain and lose equal amounts (what Betty loses equals what Alice gains), and also the fact that each one gets half the cookies.

You are trying to find a number $\mu$, the "balance point," with the property (let’s take $n=4$ to avoid summation notation)


You can solve for $\mu$ to get $\mu=\frac{x_1+x_2+x_3+x_4}{4}$

For elementary school kids, you will have to explain why at least one term $(x_j-\mu)$ will need to be negative, and another positive.

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