# Is the average of the averages equal to the average of all the numbers originally averaged?

I am tempted to say yes because of the following pseudo-proof (I say pseudo-proof because I am not convinced):

$$\frac{\frac{w+x}{2}+\frac{y+z}{2}}{2}=\frac{w+x}{4}+\frac{y+z}{4}=\frac{w+x+y+z}{4}$$

Is this proof enough or am I completely wrong? If I am not wrong but this is not proof, what would be a good proof?

Edit:

I guess the following proves otherwise:

$$\frac{w+x+y+z}{4} \neq \frac{\frac{w+x+y}{3}+z}{2}$$

That would be proof against my original statement by contradiction.

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Your result is for a particular case, when you have two averages of two numbers. I suggest you to state this in formal way even if it is a simple case will help you with the formality of Maths. – checkmath Mar 1 '12 at 1:45
If the smaller groups all have the same number of elements, then the result is true. – Byron Schmuland Mar 1 '12 at 1:48
A school of $100$ has very good students. The school average on a standardized set was $95$%. A not so good huge school has $3000$ students. The school average on the test was $45$%. The average of the averages is $70$%, pretty good. The true average over all students is $\frac{(95)(100)+(45)(3000)}{3100}$, about $46.6$\%. Guess which of $70$% or $46.6$% the Superintendent of Schools will quote when (s)he gives a press conference about how good the school system is. – André Nicolas Mar 1 '12 at 7:36

$1,1,1,2,2$

Their average is $\frac{7}{5}$.

But if you take it as $1,1,1$ and $2,2$, and average the averages, you get a different result.

But, what you said works if the number of numbers is a power of $2$ and you split into two equal sized sets. Interestingly, this observation was used by Cauchy to give an inductive proof of the $\text{AM} \ge \text{GM}$ inequality!

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It depends on how you group the elements, so the answer in general is no. It worked in your first example because the groups were of uniform size, while they were not in the second example.

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