I have a monthly average for a value and a standard deviation corresponding to that average. I am now computing the annual average as the sum of monthly averages, how can I represent the standard deviation for the summed average ?

For example considering output from a wind farm:

Month        MWh     StdDev
January      927     333 
February     1234    250
March        1032    301
April        876     204
May          865     165
June         750     263
July         780     280
August       690     98
September    730     76
October      821     240
November     803     178
December     850     250


We can say that in the average year the wind farm produces 10,358 MWh, but what is the standard deviation corresponding to this figure ?

Short answer: You average the variances; then you can take square root to get the average standard deviation.



Example

Month          MWh  StdDev  Variance
==========   =====  ======  ========
January        927    333     110889
February      1234    250      62500
March         1032    301      90601
April          876    204      41616
May            865    165      27225
June           750    263      69169
July           780    280      78400
August         690     98       9604
September      730     76       5776
October        821    240      57600
November       803    178      31684
December       850    250      62500
===========  =====  =======  =======
Total        10358            647564
÷12            863    232      53964


And then the average standard deviation is sqrt(53,964) = 232



From Sum of normally distributed random variables:


  If $X$ and $Y$ are independent random variables that are normally distributed (and therefore also jointly so), then their sum is also normally distributed
  
  ...the sum of two independent normally distributed random variables is normal, with its mean being the sum of the two means, and its variance being the sum of the two variances


And from Wolfram Alpha's Normal Sum Distribution:


  Amazingly, the distribution of a sum of two normally distributed independent variates $X$ and $Y$ with means and variances $(\mu_X,\sigma_X^2)$ and $(\mu_Y,\sigma_Y^2)$, respectively is another normal distribution
  
  $$
P_{X+Y}(u) = \frac{1}{\sqrt{2\pi (\sigma_X^2 + \sigma_Y^2)}} 
e^{-[u-(\mu_X+\mu_Y)]^2/[2(\sigma_X^2 + \sigma_Y^2)]}
$$
  
  which has mean
  
  $$\mu_{X+Y} = \mu_X+\mu_Y$$
  
  and variance
  
  $$ \sigma_{X+Y}^2 = \sigma_X^2 + \sigma_Y^2$$


For your data:


sum: 10,358 MWh
variance: 647,564
standard deviation: 804.71 ( sqrt(647564) )




So to answer your question: 


How to 'sum' a standard deviation?
You sum them quadratically: 

s = sqrt(s1^2 + s2^2 + ... + s12^2)



Conceptually you sum the variances, then take the square root to get the standard deviation.



Because i was curious, i wanted to know the average monthly mean power, and its standard deviation. Through induction, we need 12 normal distributions which:


sum to a mean of 10,358
sum to a variance of 647,564


That would be 12 average monthly distributions of:


mean of 10,358/12 = 863.16
variance of 647,564/12 = 53,963.6
standard deviation of sqrt(53963.6) = 232.3




We can check our monthly average distributions by adding them up 12 times, to see that they equal the yearly distribution:


Mean: 863.16*12 = 10358 = 10,358 (correct)
Variance: 53963.6*12 = 647564 = 647,564 (correct)



  Note: i'll leave it to someone with a knowledge of the esoteric Latex math to convert my formula images, and formula code into stackexchange formatted formulas.


Edit: I moved the short, to the point, answer up top. Because i needed to do this again today, but wanted to double-check that i average the variances.
