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No, it makes no assumptions about any distribution whatsoever. This is like Chebyshev's inequality: https://en.m.wikipedia.org/wiki/Chebyshev%27s_inequality htt

• caseysoftware 9y ago
  From that wikipedia page: > "Because it can be applied to completely arbitrary distributions provided they have a known finite mean and variance, the inequality generally gives a poor bound compared to what might be deduced if more aspects are known about the distribution involved." In the situations cited in the article, we are dealing with both an unknown mean and an unknown variance and therefore an unknowable distribution. So while Chebyshev is "weaker" than a Gaussian distribution, it's still a distribution.
  • tgb 9y ago
    Sorry, I was not saying that the article applied Chebyshev's inequality. It's not related at all. But it's an example of a result where you do not have to assume that your distribution is a member of a particular family of distributions. The result being applied here is really really general: it says that 95% of the probability mass occurs before the point at which 95% of the probability mass has occurred by. This is tautological. Like saying that 50% of people are below the median - it's true without making any assumptions about people (except that there are finite numbers of them).