I am trying to find $a$ that satisfies the following.
$P(X>1+a) = \frac{\alpha}{2}$ for given $\alpha \in (0,1)$ where $X \sim \mathcal{N}(1, 1.5^2)$
To solve this, I'd like to take the inverse of the Normal CDF to both sides over the following:
$CDF(1+a) = 1-\frac{\alpha}{2}$
Any idea how to apply $CDF^-1$? I can't find closed form of inverse function of CDF.