# The average of the averages is not equal to the average of all the numbers originally averaged. What might be the intuition behind this phenomenon?

For example: the average of {2,2,2,2,2,2,2,2,2,2,2,2,2} is 2, (N=13) and the average of {4} is 4, (N=1). The average of the averages is 3. But the average of all numbers is 30/14 ≃ 2.14

Clearly these two numbers are not same and it can be also seen through Arithmetic and Algebra. I want to know what is the intuition behind this phenomenon that they are not same. Can anyone please explain ? Thanks.

• You have to weight by sample size. In your case, the first set wins...there are a lot of $2's$.
– lulu
Oct 7, 2016 at 13:39
• Also, what makes you think that they would be the same? If I square an arbitrary number, then add $2$, that's not the same as if I squared the number after adding $2$... Oct 7, 2016 at 13:43

Here is an analogy. The small country of Ruritania is divided into three boroughs, each of which gets to send one representative to the parliament of Ruritania. The three representatives each get one vote in parliament and decide all of the country's laws.

But the first borough of Ruritania is a city of one million people, while each of the other two boroughs is a small plot of land just outside the city limits, and there is only one person living in each of those two boroughs.

So the million people elect one representative, who then goes to the parliament where she meets with the two people residing in the other two boroughs (who "elected" themselves as their own representatives). And each of these three people gets one vote on every law.

So if the representatives of the two tiny boroughs think taxes should be high, taxes will be high regardless of what the people in the city want. Two people get to tell a million others what to do.

It is not an exact analogy, but in your example you have given the $4$ in your singleton list one "vote in parliament", while all the $2$s in your much longer list received only one "vote in parliament" amongst all of them.

• Oct 23, 2016 at 17:36

Suppose you have $N_1$ numbers, $x_1,x_2,\cdots,x_{N_1}$. Their average is $$\langle x\rangle=\frac{\sum_{i=1}^{N_1}x_i}{N_1}$$

Similarly, the average of $N_2$ numbers $y_1,y_2,\cdots,y_{N_2}$ is $$\langle y\rangle=\frac{\sum_{i=1}^{N_2}y_i}{N_2}$$

On combining the observations, by definition, the average comes to be $$\langle z\rangle=\frac{\sum_{i=1}^{N_1}x_i+\sum_{i=1}^{N_2}y_i}{N_1+N_2}$$

However, the average of the averages is $$\langle z\rangle '=\frac{\frac{\sum_{i=1}^{N_1}x_i}{N_1}+\frac{\sum_{i=1}^{N_2}y_i}{N_2}}{2}$$

which is not the same as given by the definition. This discrepancy is because the number of observations that lead to each average is different. Note that if $N_1=N_2$, both become equivalent.

Instead, we need to take the number of observations into account. As the average is given by the sum of observations by the total number of observations, we need to know the sum of observations. Clearly, this can be obtained as $$\text{Sum of observations}=N_1\langle x\rangle+N_2\langle y\rangle$$ $$\text{Total number of observations}=N_1+N_2$$

Thus, $$\langle z\rangle=\frac{N_1\langle x\rangle+N_2\langle y\rangle}{N_1+N_2}$$

Note that average of $\langle x\rangle$ and $\langle y\rangle$ will be $$\langle z\rangle '=\frac{\langle x\rangle+\langle y\rangle}{2}$$

Thus, the result is similar to taking the average, except that it appears as if $\langle x\rangle$ appears $N_1$ times and $\langle y\rangle$ appears $N_2$ times. This is known as the weighted average, i.e., the contribution of $\langle x\rangle$ and $\langle y\rangle$ to the result depends on their "weight", which is the number of elements used to obtain them.

In some sense, the average of all numbers is the average of averages, with the constraint that the "weight", or contribution of each average is taken care of.