Compute inverse cdf of normal distribution How can I compute inverse CDF of normal distribution using the central limit theorem on uniform distribution (u[0,1])
 A: Let $S_n:=U_1+\cdots+U_n$, where $U_1,U_2,\ldots$ are iid random variables (r.v.'s) each uniformly distributed on $[0,1]$. Then, by the central limit theorem,
\begin{equation}
    Z_n:=\frac{S_n-n/2}{\sqrt{n/12}}\to Z
\end{equation}
in distribution as $n\to\infty$, where $Z$ is a standard normal r.v. So, for each $p\in(0,1)$,
\begin{equation}
    q_n(p)\to q(p)
\end{equation}
as $n\to\infty$, where $q_n(p)$ and $q(p)$ are the $p$-quantiles of (the distributions of) $Z_n$ and $Z$, respectively, so that $q$ is the function inverse to the standard normal cdf.
Note that
\begin{equation}
    q_n(p)=\frac{X_n(p)-n/2}{\sqrt{n/12}},
\end{equation}
where $X_n(p)$ is the $p$-quantile of $S_n$, so that $X_n$ is the function inverse to the cdf of $S_n$, which let us denote by $F_n$. By the Irwin–Hall formula, $F_n$ is piecewise polynomial; more specifically,
\begin{equation}
    k\in\{1,\dots,n\}\ \&\ k-1\le x<k\implies
    F_n(x)=\frac1{n!}\sum_{j=0}^{k-1}(-1)^j\binom nj (x-j)^n. 
\end{equation}
So, the quantile/inverse functions $X_n$ and $q_n$ are piecewise algebraic.

Below is an image of a Mathematica notebook containing graphs of $q_2$ (red), $q_5$ (green), $q_{10}$ (blue), and $q$ (black). We see that $q_2$ already approximates $q$ well except near $0$ and $1$, with $q_5$ and $q_{10}$ significantly closer than $q_2$ to $q$ near $1$ (of course, the graphs of $q_n$ and $q$ are centrally symmetric about the point $(1/2,0)$).

