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First, apologies for the long-winded title!

I'm helping my 10 year old son with math, and we had a set of problems based on the following scenario: Given an average of a set, a single marginal value for the set, and the resulting new average, find out the original number of items in the set.

With no further explanation, the book provides the following formula: take the difference between the original average and the new observation and then divide this difference by the difference of the old and new averages:

(X - A1)/(A2 - A1)

It is easy to answer the problems by simply plugging the values in the formula.

I have two questions. Is there any way to intuitively explain this formula or concept? All of the other problems in the book we are using have been relatively easy to explain to a 10 year old using pictures or other techniques.

The second question is how is this formula derived? In order to explain it, I tried to derive it my self. I started with the two formulas:

A1 = S/N

and

A2 = (S+X)/(N+1)

where S is the sum of the values in the set.

And was unable to solve for N (or N+1) given X, A1, and A2, though obviously my algebra skills are rusty to non-existant at this point. (As I write this, I can't remember if we were solving for N or N+1 -- I don't have the book in front of me.)

If the derivation is trivial, I'd like to know it. If it takes a fair amount of manipulation unsuitable for a 10 year old, I can probably figure it out myself eventually, so no need to provide the exact answer here.

If there is no intuitive explanation, and the only way to grasp the concept is to do the algebraic manipulation of these two equations, proving the formula holds true, then we will probably skip this section of the book. As I noted, while many problems in the book we are using can be solved doing the algebra, they all, until this one, had a much easier way to look at the problem and solve it.

Any insight would be greatly appreciated.

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Let's say with the new item, you have $n+1$ items in the set. Then, the average of the values is

$$A_{n+1} = \frac{\left(\sum_{i=1}^n x_i\right) + x_{n+1}}{n+1}.$$

Basically, the average is the sum of the first $n$ values, plus your new marginal value, divided by the total number of old and new values ($n$ old plus 1 new value).

Then, $(n+1)A_{n+1} = x_{n+1} + \sum_{i=1}^nx_i$.

Therefore, $\sum_{i=1}^n x_i = (n+1)A_{n+1}-x_{n+1}$.

Now, before we added the marginal value, we had

$$A_n = \frac{\sum_{i=1}^n x_i}{n}.$$

See how the summation term is the same as the equation above? Let's use that fact.

$$A_n = \frac{(n+1)A_{n+1}-x_{n+1}}{n}.$$

We'll expand terms out a bit:

$$\begin{align*} A_n &= \frac{nA_{n+1} + A_{n+1} -x_{n+1}}{n} \\ &= A_{n+1} + \frac{A_{n+1}}{n} - \frac{x_{n+1}}{n}. \end{align*} $$

Shifting things over a bit,

$$\begin{align*} n(A_n-A_{n+1}) &= A_{n+1}-x_{n+1}\\ n &= \frac{A_{n+1}-x_{n+1}}{A_n-A_{n+1}} \end{align*} $$

This gives you the desired result. (You can multiply the right hand side by $\frac{-1}{-1}$ to obtain the exact form that you wrote, but it's the same thing really.)

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  • $\begingroup$ I'm not sure if the algebra makes for an intuitive explanation. I'll try to come up with one. $\endgroup$ – Emily Jul 3 '13 at 17:12
  • $\begingroup$ Thanks for working out the algebra for me. If you or no one else can give an intuitive explanation (which well may not exist for mortal minds) in then next few days, I'll mark this as the correct answer. Thanks again. $\endgroup$ – Paul Mansour Jul 5 '13 at 16:30

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