Where to wait for an elevator (2010)

    import math
    import random

    B = 1000
    elevators = [-6, 0, 3]

    def time_to_enter_distribution(position):
        for _ in range(0, B):
            elevator = random.choice(elevators)
            distance = elevator - position
            # What matters is not how many people are in the elevator, but how
            # quickly they can evacuate. Approximate Poisson distribution.
            evac_time = [0, 1, 2, 3, 3, 4, 5][random.randint(0, 6)]

            # While you walk to the elevator, you give time for people to evacuate
            # and your time passes.
            evac_time -= abs(distance)
            time = abs(distance)

            # If people are still evacuating when you have arrived, you have to wait
            # for that to end by the side of the elevator, and then take the final
            # 0.5 metres to step inside.
            if evac_time > 0:
                time += evac_time + 0.5

            yield time

    def time_to_enter_summary(position):
        distr = list(time_to_enter_distribution(position))
        sum_time = sum(distr)
        sum_sq_time = sum(t**2 for t in distr)
        mean = sum_time/B
        sd = (sum_sq_time - sum_time**2/B)/(B-1)
        se = sd/math.sqrt(B)
        return (mean, se)

    if __name__ == '__main__':
        print(f'{"pos":5s}  {"t":4s} {"se":3s}')
        for p in (i/4 for i in range(-6*4, 3*4)):
            (t, se) = time_to_enter_summary(p)
            print(f'{p:5.2f}: {t:4.2f} ({se:3.2f})')

It should be obvious that the mean is the correct representative value for weighing nails. My point is that there are cases that call for the mean and cases that call for the median, and the distinction can be explained by considering the signedness (without having to consider squared distances).

Maybe the following variation on the nail example would help: Suppose the manufacturer of the nails must publish a reference value for the weight of a nail, and will be fined for every nail that is not exactly the correct weight. The fine is proportional to the absolute error. Question: Should the manufacturer publish the mean or the median in order to minimise fines? Answer: The median.

Why is the mean the correct representative value for counting nails, but not for minimising fines? I'm arguing that it's because when the nails are weighed together the errors are signed, but when the fines are calculated the errors are unsigned (in the same way that the distances to the elevators are unsigned).

> Why is the mean the correct representative value for counting nails, but not for minimising fines? I'm arguing that it's because when the nails are weighed together the errors are signed, but when the fines are calculated the errors are unsigned (in the same way that the distances to the elevators are unsigned).

And here, there are two points that bother me.

First, the mean is the correct representative for counting nails because that's how it's defined. A mean is just the answer to any question of the form "if all of these data points were equal to each other, what is the value they would all share?". That's the question you're asking about the weight of the nails.

Second, I find it difficult to believe in your stated explanation of why you might choose the mean over the median (or vice versa), because we've already observed that the mean minimizes the sum of squared errors, and squared errors are all unsigned. You're relying on the assumption that, in the above example, the fine (which should be minimized) is directly proportional to difference from the estimate. But that assumption doesn't appear in the explanation of why one metric or the other will minimize the fine.