Given just a mean and standard deviation and proportion of the data that is over X can we determine if it is a normal distribution? I want to say no, because even if the we take X - mean / SD and take that Z-score and use it to find the proportion, even if that proportion is equal to the proportion given, I don't think that is still enough information to determine if it's a normal distribution because we don't know how the data below X is distributed. But I could be missing something?
2026-04-01 23:53:29.1775087609
Given a mean, standard deviation and proportion can we determine if it is a normal distribution?
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Say you were given $E(X)=0$, $\text{Var}(X)=1$ and $P(X\ge0)=0.5$. Then you cannot just assume that $X\sim N(0,1)$ because any symmetrical distribution with variance $1$ satisfies these conditions. For example the uniform continuous distribution - $X\sim\text{unif}(-\sqrt{3},\sqrt{3})$ - is such a distribution.