Exact Gaussian chance constraint ($m+{\Phi }^{-1}(1-\epsilon )\sqrt{v}\le C$)

For $X\sim N(m,v)$ with $v>0$ and $0<\epsilon <1$, $P(X\ge C)\le \epsilon$ is equivalent to the displayed chance constraint, because the standard normal distribution function is continuous and strictly increasing. The coefficient is a standard normal quantile, which can be negative. If $v=0$, equality is an atom and the exact condition becomes $C>m$ instead.