What is the probability of a value to be greater than the mean plus three times the standard deviation (assuming a normal distribution)? What is this probability for an arbitrary distribution?
@Kodiologist "For an arbitrary distribution" has any answer consistent with the standard inequalities (Chebyshev, etc.). It basically asks how to compute the CDF for any distribution.
The first part of the question has been answered countless times already, so I ignored that part.
I can see how one might nevertheless make a good answer out of such a vague question, so if you're eager to do so, we can vote to reopen.
@whuber Rather than discussing how to compute the CDF, I think I would just have written "Compute $1 - F(μ + 3σ)$ where $F$ is the CDF, $μ$ the mean, and $σ$ the SD". My guess is that the asker thought there'd be an easier answer that didn't require a CDF.
