Averages and differences A student asked me a question that I couldn't answer, and I said I'd find out. 
I set a problem which was, given a set of numbers, was the mean over 50 where the  numbers are: 56, 42, 47, 59 and 48
The mean is 50.4. A student asked why it wasn't 52, and I was confused. they said they had added up the differences between 50 and the five numbers (so 50-42 = 8, 50-47=3, 48-50=2, 2+3+8=13 then 56-50=6, 59-50=9, 9+6=15) then as the difference is 2 (15-13=2) it should be 52.
This got me thinking that it was quite a clever way to work things out, and I get a feeling that something like this should work, but I can't explain why it doesn't (apart from the glib answer that "you don't do means like that" - I don't want to crush his enquiring mind!).
Thanks in advance. 
Marc
 A: The difference between each value and the mean sums to $2$, yes, but to find the mean you sum the values and divide by the count of the values -- here, that's $5$.  So the difference has to be divided by $5$ as well before it can be added, and $2/5=0.4$.
A: Setting : $\text {Sum_diff} = (a- \text {Av})+(b - \text {Av}) + \ldots + (e - \text {Av})$, the average $\text {Av}$ is the number such that $\text {Sum_diff} =0$.
Thus, from : $(a- \text {Av})+(b - \text {Av}) + \ldots + (e - \text {Av})=0$ we get :

$(a+b+c+d+e)= 5 \text {Av}$,

and thus : $\text {Av} = \dfrac {(a+b+c+d+e)}{5}$.
Now we can check that, setting $50$ as "provisional average" and computing the sum of differences, the real average will be the provisional one plus the sum of differences with respect to it divided by $5$.
Using again th equation above : $\text {Sum_diff} = (a - 50) + (50 - \text {Av}) + \ldots + (e - 50) + (50 - \text {Av})=0$, we get:

$(a-50)+ \ldots + (e -50) + 5 (50 - \text {Av})=0$.

Thus, it follows that the equation for the average computed from the sum of differences with respect to the "provisional average" will be :


$\text {Av} = 50 + \dfrac {(a-50)+ \ldots + (e -50)}{5}= 50 + \dfrac 2 5.$


A: $$\bar x=\frac{56+42+47+59+48}5=\frac{(50+6)+(50-8)+(50-3)+(50+9)+(50-2)}5=50+\frac2{\color{red}5}$$
A: The formula for the mean ($\bar{x}$) can be tranformed as follows:
$$\bar{x}=\frac{\sum_{i=1}^n x_i}{n} \iff n\bar{x}=\sum_{i=1}^n x_i \iff \\
\sum_{i=1}^nx_i-\sum_{i=1}^n\bar{x}=0 \iff \sum_{i=1}^n (x_i-\bar{x})=0$$
If the sum of differences of the numbers and assumed mean ($\bar{x}$) is equal to $k$, then the real mean ($\bar{x}_r$) is:
$$\sum_{i=1}^n (x_i-\bar{x})=k \iff \sum_{i=1}^n(x_i-\bar{x})-n\cdot \frac kn=0 \iff \\
\sum_{i=1}^n\left(x_i-\left(\bar{x}+\frac kn\right)\right)=0 \iff \bar{x}_r=\bar{x}+\frac kn$$
Hence, in your case $k=2$, the real mean is:
$$\bar{x}_r=50+\frac25=50.4.$$ 
