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Let's say I know the following:

Let A be the average value
Let T be the number of values

Assuming that A = 4.3 and T = 5, how would I add another value (let's say 10) and update the average correctly? Is this possible?

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Yes, the new average is then

$$\frac{A\cdot T+10}{T+1}$$

Here is a more detailed proof:

Let us write $a_1,\ldots,a_T$ the numbers occurring in the average, then

$$A = \frac{1}{T}\sum_{k=1}^T a_k.$$ Now, if we want the average with one more number, say $a_{T+1}$, then the new average is $$A' = \frac{1}{T+1}\sum_{k=1}^{T+1} a_k = \frac{1}{T+1}\left(a_{T+1}+\sum_{k=1}^{T} a_k\right) = \frac{1}{T+1}\left(a_{T+1}+T\left(\frac{1}{T}\sum_{k=1}^{T} a_k\right)\right) =\frac{1}{T+1}\left(a_{T+1}+TA\right) = \frac{a_{T+1}+TA}{T+1}$$

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    $\begingroup$ Excellent answer, thank you! $\endgroup$ – Jesse Brands Feb 17 '15 at 23:49
  • $\begingroup$ @JesseBrands you're welcome :). $\endgroup$ – Surb Feb 17 '15 at 23:54
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Sum of existing values is TA Add new value V to get $TA+V$. Divide by new no. of values to get $\frac{TA+V}{T+1}$

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