When to use Mean(X/Y) versus Mean(X)/Mean(Y)? So, I understand that for two vectors of numbers X and Y the mean of X/Y is different from the mean of X divided by the mean of Y. But I'm not sure when it is appropriate to use each. Specifically, I need to calculate the average consumption per days active for a set of individuals, and I don't know which approach to use.
To illustrate, say there are 3 individuals: A, B, and C. Their consumption values are (1,3,7) respectively, and their days active are (3, 1, 6) respectively. If I individually calculate their consumption per days active I get 1/3, 3, and 7/6, which I can then take the mean of to get 1.5. But if I sum their consumption (11) and their active days (10) and then work out consumption per active days, I get 1.1. Which approach is 'correct'?
I hope I've expressed this ok, I wasn't really sure what the title should be which is why I've struggled to find answers because I'm not totally sure what to search. If anyone can explain the different properties of the two quantities and when using each might be appropriate I'd really appreciate it!
EDIT: I have found this question which seems to touch on a very similar issue, but I'm not totally satisfied with the answer. I think the second answer comes closest to what I'm looking for, but I'd just like to know when it's appropriate to use each (not specific to some simulation context like in the question) Usages of Mean(X/Y) vs. Mean(X) / Mean(Y)

I first consider  $mean(x/y)$ versus  $mean(x)/mean(y)$.
As Eoin  and dariober suggest, the more practically relevant quantity should be used, which is often the former, although   the latter is not that outlandish.
To illustrate both, let's modify the OP's example and imagine opening a bakery with bakers A, B and C.

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*A usecase for $mean(x/y)$. On the  initial (test) day, A, B and C  worked $X =(3, 1, 7)$ hrs and made $Y=(1, 3, 6)$ cakes, respectively; later on, each baker will work the same hours per month. The hourly productivity of worker $i$ is $z_i=y_i/x_i$, and the expected productivity  $\sum z_i/n = 1.5$ cakes per hour/worker. If, say, A worked for 6 hrs and made 2 cakes on the test day, this expectation won't change.

*A usecase for $mean(x)/mean(y)$. Suppose a single   baker can work at a  time, and the bakers do $X =(3, 1, 7)$ hrs per day. Then the bakery's expected hourly output, $\sum y_i/\sum x_i=  1.1$, is meaningful. Since $\sum y_i/\sum x_i= \sum w_i z_i$, where $w_1 = x_1/\sum x_i$ is the "timeshare" of worker 1, this is like the previous case, but weighted. Similar weighting could be based for example on  customer expenditure.

*Using $mean(x)/mean(y)$  when X and Y are not (fully) matched. Suppose we pay $X =(3, 1, 7)$ pounds per hour to $(A, B, C)$ and want to assess competing wages. This assessment is easy if  the bakers moonlight for  $Y=(1, 3, 6)$ pounds per hour at a competitor's. It is less so if only A and C do; then we have unequal samples or "missing" data. The sizes are equal but samples are still unmatched if $Y=(1, 3, 6)$ are randomly drawn IT wages. (Appropriate matching could be possible, but not always.)

The second ("robustness") aspect, implied by  dariober and several commenters, is whether means are suitable for ratios in the first place. As suggested by @Nick Cox under the  question, the geometric mean(s) might be better; see also. Alternatives include removal of  outliers; the use  of trimmed means; the use   of $median(x)/median(y)$ or  $median(x/y)$. Also, bootstrap estimation could be appropriate for ratios.
