How to add up partial confidence intervals to create a total confidence interval? I have a response variable that obtains different confidence intervals (CI) when calculated with different explanatory variables. I want to add up all values of the response variable and create a CI for the sum.
I would understand how to do if someone please helped me solve the following example from a triathlon where time (minutes) is the response variable, and distance (km) and discipline are the explanatory variables:  


*

*The 95% CI for swimming 1.5 km is 40 to 50 minutes

*The 95% CI for cycling 40 km is 60 to 80 minutes      

*The 95% CI for running 10 km is 30 to 40 minutes


Q: Between how long does it take a person to complete the triathlon with a 95% CI?
[If it makes any difference, I assuming normal distribution and independence between disciplines]
Thank you
 A: In short:

*

*Take as central point of your confidence interval the sum of central points of every confidence interval (45+70+35=150 minutes).

*Take as radius of your interval the square root of the sum of the squares of the radius of every confidence interval $\sqrt{5^2+10^2+5^2}=12.25$
Therefore, a person does that triathlon in between 137.75 and 162.25 minutes with 95% probability. Anyway, beware of assumptions.
In long:
I assumed normal distribution and independence between disciplines, although first assumption may be reasonable as a rough approximation but second assumption is likely false, because I would expect that people performing well in one discipline are likely to perform well on the other ones (for example, I would expect myself to perform poorly in every discipline in a triathlon).
Assuming that times in every discipline is a normal variable, total time is just the sum of three random variables, and therefore normally distributed. Variance of the sum is also the sum of variances of the three variables, and since intervals radius is proportional to the square root of variance, you can just sum the squares of radius of every interval to get the square of the radius of the sum variable.
However, please notice that the (dubious) assumption that times for each discipline are independent narrows the resulting interval - I would say, unrealistically narrows it.
We could make the opposite assumption, that is that times for disciplines are absolutely correlated (that is, roughly, that the person who swam in 40 minutes is the same that cycled in 60 minutes and ran in 30 minutes). That assumption is probably as unrealistic as the assumption of independence was, but surely not a lot more unrealistic.
In this assumption, the radius of intervals just sum, and the triathlon is expected to be completed in between 130 and 170 minutes by 95% of athletes.
In the end, we should expect the real interval to be somewhere between [137.25,162.25] and [130,170] (both unrealistic extreme cases), but to give a more accurate result we would need to know (at least) what is the correlation between times in different discipline.
Edit after reviewing the answer a few years later: The assumption I made that results in different disciplines are likely positively correlated is reasonable if the sample includes people with different levels of fitness. However, if the sample only includes people with similar overall level in triathlon - for example, triathletes who took part in the 2020 Olympic Games - correlation between disciplines might be negative. Anyway, since assuming negative correlation yields smaller confidence intervals (or even zero length intervals), in case of lack of information about correlation I would take the conservative assumption that correlation is somewhere between 0 and 1. End of edit.
Edit about terminology
As Whuber points in his comments, it's not clear what is the meaning of the intervals given in the question. Although, the answer is valid interpreting the resulting intervals in the same way of the intervals in the question.
The two reasonable meanings of the intervals of the question are:

*

*Intervals of confidence about the mean of each sport.

*Or intervals containing the times of 95% of participants on each sport.

In spite of the wording of the question fitting better the second meaning (and hence the wording in my answer), the name "confidence interval" is usually not used with this meaning.
However, since individual times follow a normal distribution (according to assumption made in the question) and estimations of means also follow a normal distribution (if sample size is large enough or if we keep sticking to the assumption that individual times are normally distributed), the arithmetic of intervals is the same for both meanings and therefore the results given hold for both meanings.
A: Pere has given a good answer. Adding up of variances is what happens when you add distributions. However in your case, he has assumed the radius as the standard deviation. but in fact, it is 1.96 times the standard deviation. So you will need to divide the radius (pere's terminology) by 1.96 (95% conf int) and then square, sum and take sq-root.
