Representation of population data mean/median (similar to BMI/age curve) I want to represent my data (n=150 samples with a curve of value over time, similar to BMI representations in figure).

Possibilities are either mean/std and median/percentiles.
The question are:

*

*Can we chose the representation we want independently of the data distributions?

*How can we justify the use of either (besides one being prettier than the other)?

*Would it be better to use confidence intervals for either ?

Thank you for your help.

TL;DR For your case, use median/percentiles.
I would decide between those two options based on the following criteria: Is your distribution generally (at any point in time) symmetric? If yes, do mean and standard deviations. If not, do median and percentiles.
Perhaps someone more knowledgeable can give a more rigorous explanation (as mine will be a bit hand-wavy), but my reasoning here is simply that standard deviation is a "symmetric" measure. It is used to describe symmetric distributions, because those are the only ones it can describe accurately. Since it is a single number expressing the "spread" of the data away from the mean, std. dev. has no way of differentiating the spread on one side of the mean from the other. So for example, the classic "~95% of the data is contained within two standard deviations of the mean" is only true if you have a normal distribution because a normal distribution is symmetric, so one std. dev. to the right of the mean has the same number of points as one std. to the left.

However, if your distribution is skewed, this breaks down:

You can see that the std. dev. left of the mean has way more points than the right one, since the distribution is skewed and non-symmetric. However, in this same image, you can see that percentiles adjust to this (by definition, as they are designed to).
So, if your example graphic is a reasonably accurate representation of your actual data, I would use the median/percentiles because that distribution is clearly skewed (the spread of points "above" the mean at any x value is much greater than the spread of points "below" the mean). Confidence intervals are based on mean/SD, so they also can't capture the asymmetric nature of your distribution (so I wouldn't use it).
I hope this explanation managed to answer all 3 of your questions!
