Consider $\frac{d}{dx}tanh(x)$ and the normal distribution of mean $0$ and std deviation $1$.
They are both bell-shaped, but the maximum height of $\frac{d}{dx}(d/dx)tanh(x)$ is $1$ and the maximum height of the standard normal distribution is $\frac{1}{\sqrt[2]{2*\pi}} \approx 0.3989423$.
Perhaps we can adjust the height and width of $\frac{d}{dx}tanh(x)$ until it approximates the standard normal distribution?
For any real numbers $h$, $w$, and for any real-valued function, $f$, let $T(h, w, f)$ denote the function $\hat{f}$ such that for all real $x$, $ \hat{f} = h*f(w*x)$
For any real $a$, $b$, with $a< b$, Let Φ$(a, b)$ denote the area under a standard normal distribution from $x= a$ to $x = b$.
When $h$ and $w$ are clear from context, let $F(a, b)$ denote the area under $T(h, w, \frac{d}{dx}tanh(x))$ from $x= a$ to $x = b$.
Find real numbers $h$ and $w$ such that the maximum of $ABS(1 - F(a, b)/Φ(a, b))$ taken over all real $a$, $b$ is minimized.
In other words, minimize the maximum relative error between area under the squished-squashed $\frac{d}{dx}tanh(x)$ and area under the normal distribution.