Let $X$ be a Normal random variable of mean $\mu$ and variance $\sigma^2$. Also let $g\colon \mathbb{R}\to\mathbb{R}_{>0}$ be a positive strictly increasing bijective function.
I would like to find the symmetric 0.95 "confidence interval" of $g(X)$ centred by its mean, that is, find $a$ such that
$$ \mathbb{P}\big(z - a \leq g(X) \leq z + a\big)=0.95, $$
where $z = \mathbb{E}[g(X)]$ is the mean. Suppose that such interval exists.
However, I failed to compute the interval in closed-form. Observe that
$$ \mathbb{P}\big(z - a \leq g(X) \leq z + a\big) = \mathbb{P}\big(g^{-1}(z - a) \leq X \leq g^{-1}(z + a)\big), $$
so essentially, we want to find the $a$ by solving
$$ \mathrm{CDF}(g^{-1}(z + a)) - \mathrm{CDF}(g^{-1}(z - a)) = 0.95, $$
where $\mathrm{CDF}$ is the CDF of $X$. But how to find a closed-form solution/approximation to this equation? By "approximation", I mean an analogy of $\mu \pm 1.96\sigma$ for $\approx0.95$ of Normal.