Averaging decibels

The "physically natural" quantity to average is the actual power, or energy, but it depends exponentially on the number of decibels. So if you were averaging the power or energy, the result would be pretty much equal to the power or energy of the largest (loudest) reading in decibels.

So even though it's physically less natural, you probably want to compute the average number of decibels itself. But as you said, it's abuot "preferences". Your question isn't a question about observables, it's about subjective choices, so there can't of course be any "only correct and objective" answer. For various applications, various averages may be more or less useful or representative.

There are reasons more than "preference" for the averaging. You defined it that way usually because you can get more information from that, particular for those additive quantities.

Suppose you preform a set of measurement at a particular point in space, there are two cases: (a) get the averaged value (b) take the average for the intensity itself, and then converted to decibel.

If you have the quantity in situation (b), you can know how much average energy flux passing through that point. Also, you can know the total energy flowing through that point. This information cannot be obtained from the method (a).

Similar situation for the earthquake, if you take the average for the energy, you can know the total energy released by that particular point, which is important. However, you cannot obtain this information by simply taking the average of earthquake scale.

Sure, as pointed out by Lubos, if the variation is small, these two definitions are basically the same as the $\log$ (any) function is local linear, and you can now have additive quantity again.