Suppose you take a sample of size 3 with values 0, 0 and 1. (3 is the minimum size to make the sample asymmetrical and make the question non-trivial.) Ever since my first stats class, I've been vaguely confused because I didn't understand: why is it useful to calculate the mean of this sample, 1/3?
At first I naively thought that if this meant that if you assumed the background population had some standard deviation (e.g. 1), that 1/3 would be the value of the population mean that would maximize the probability of getting the observed outcome. (Of course I mean the probability of getting the observed outcome within some arbitrarily small delta, since the probability of any specific outcome is zero.) i.e. for my sample and assuming standard deviation 1, the value k that maximizes the value of f(k)*f(k)*f(1-k) where f is the standard normal distribution. But presumably that's not exactly 1/3, is it? (By the way, is there a technical term for this "observed outcome probability maximizer" value?)
[Edited to add: I was wrong. I explicitly f(k)*f(k)*f(1-k) where f is the standard normal distribution, and it has a maximum at 1/3. This is convenient, although it was far from obvious to me at the outset that this would be the case. This renders the rest of the question moot.]
But if 1/3 is not the value that maximizes the probability of the observed outcome, then what good is it?
Is it simply the case that before everyone had a computer in their pocket, calculating the "observed outcome probability maximizer" value was too hard, and calculating the mean was good enough? And now that we all have pocket computers, we should just go with the "observed outcome probability maximizer" instead, and calculating the mean is obsolete? If not, then what does "1/3" tell you about the background population?