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
รท12 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:
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.
The standard deviation is the square root of the variance, as you might know. The variance is calculated by summing up the squared deviation from the mean, and dividing it by $n$.
$$\sigma^2 = \frac{\sum_{i}^{n} (x_i-\mu)^2 }{n}$$
Every difference $(x-\mu)$ is squared. When you take the square root of the variance, it is not the same as taking the square root of every $(x-\mu)^2$ and sum it up afterwards...
Because of the square term, the variance (and thus the standard deviation) gives more weight to more distant values and can't be negative, as positive and negative values get both positive when squared.
It is also wrong to calculate the standard deviation for all positive and negative values separately. The values lose their sign when they get squared.
Hope this helps,
Best Answer
From basic properties of variance:
$\text{Var}(X_1+X_2) = \text{Var}(X_1)+\text{Var}(X_2)+2\text{Cov}(X_1,X_2)$ and the $k-$variable case can be obtained recursively from that.