"Additional" standard deviation? I have a train station where trains arrive at certain intervals called headways. I know the distribution of headways for a "regular" situation and a "disrupted" situation. 
Standard deviations of headways indicate reliability and I want to compute how "less" reliable the disrupted situation is compared to the regular one.
If std(disrupted)=200 seconds and std(regular)=20 seconds, can I say that the additional std of the disrupted situation compared to the regular one is 180 seconds? Or do I need to go back to the variance?
Thank you!
 A: Yes, you can accurately say that there is an additional time, with one nit-picking caveat. That is, provided that you specify absolute time, and not negative time. That sounds rather picky, but consider the physical correctness, which has to be accounted for, just like statistical correctness. Standard deviations, provided they are taken of physical quantities that have sensible units, are also physical distances, and they are never negative. Physical distances if they are measured off-axis, i.e., obliquely to an axis, and provided the measurements are orthogonal, add by vector addition. When measured parallel to an axis, a vector addition of orthogonal variables reduces to absolute value of distance. In statistical language this sounds different, but is no different.
In your case, time is orthogonal to frequency of train arrival, and a positive distance in time is perfectly consistent with the units and direction of elapsing time. Absolute values of time can be added or subtracted provided they are parallel to the time axis, and not vectors oblique to the time axis. 
In what case would the variances be in some sense "additive"? Suppose, for example, we wanted to calculate the standard deviation of a mixed schedule consisting of a fraction of both distributions so that they compose a mixture distribution, or a distribution with two bumps in it, one for the percentage of trains that are not disrupted, and one for the $1-$that percentage of trains that arrive during a disruptive event. This is done as per the moments section of the Wikipedia entry for mixture distributions:
In case of a mixture of one-dimensional distributions with weights $w_i$, means $μ_i$ and variances ${σ_i}^2$, the total mean and variance will be:
$\operatorname{E}[X] = \mu = \sum_{i = 1}^n w_i \mu_i$,
$\operatorname{E}[(X - \mu)^2] = \sigma^2 = \sum_{i=1}^n w_i((\mu_i - \mu)^{2} + \sigma_i^2)$
Please note, this is not the straight addition of variances, they have to be weighted to make any sense.
A: you have to come up with a model and then 1 or both answers are wrong! 
you are essentially assuming a model of form
$\text{disruptive time} = \text{regular time} + \text{disruptive effect}$
taking variances 
$var(\text{disruptive time}) = var(\text{regular time} + \text{disruptive effect})$
$var(\text{disruptive time}) = var(\text{regular time}) + var(\text{disruptive effect}) + 2\rho std(\text{regular time})  std(\text{disruptive effect})$
where $\rho$ is the correlation between regular time and disruptive effect.
Now without knowing too much about your problem it sounds reasonable to assume there is no correlation ($\rho=0$) between the  regular variation and disruptive effect [eg different drivers add regular variation and weather adds disruptive effects].  if on the other hand you think that regular and disruptive effects are essentially 100% correlated ( which I doubt), eg slow drivers drive slower in bad weather and fast drivers drive even faster, then $\rho=1$ and you see that 
the standard deviations add:
$var(\text{disruptive time}) = (std(\text{regular time}) + std(\text{disruptive effect}))^2$
so $std(\text{disruptive time}) = std(\text{regular time}) + std(\text{disruptive effect})$
So you have to specify which model is correct for your situation (uncorrelated or 100% correlated)
