Truncated Gaussian Law

Parameters	$$µ \in \mathbb{R} $$ $$\sigma^2>0$$ $$ -\infty < min < max < +\infty$$
Support	$$x \in [\mathbb{R}] \quad and \quad p \in \:]0,1 [$$
PDF	$$ PDF = \frac { \frac{1}{ \sqrt{2 \times \sigma^2 \pi }} e^{ \frac { (x-µ)^2 } { 2 \sigma^2 } } } { \sigma \times \left\{ \frac {1}{2} \left[ 1 + erf \left(\frac {max – µ} { \sigma \sqrt{2}} \right) \right] – \frac {1}{2} \left[ 1 + erf \left(\frac {min – µ} { \sigma \sqrt{2}} \right) \right] \right\} } $$
CDF	$$ CDF = \frac { \frac {1}{2} \left[ 1 + erf \left(\frac {x – µ} { \sigma \sqrt{2}} \right) \right] – \frac {1}{2} \left[ 1 + erf \left(\frac {min – µ} { \sigma \sqrt{2}} \right) \right] } { \frac {1}{2} \left[ 1 + erf \left(\frac {max – µ} { \sigma \sqrt{2}} \right) \right] – \frac {1}{2} \left[ 1 + erf \left(\frac {min – µ} { \sigma \sqrt{2}} \right) \right] } $$
CDF^-1	$$ CDF^{-1} = \sigma \times \sqrt{2} er f^{-1} \left( 2 \times \left\{ \frac{1}{2} \left[ 1 + er f \left ( \frac {min-µ}{ \sigma \sqrt {2}} \right ) \right ] + p \times \left\{ \frac {1}{2} \left[ 1 + er f \left( \frac{max-µ}{\sigma \sqrt{2}} \right) \right] – \frac {1}{2} \left[ 1+er f \left ( \frac {min-µ}{\sigma \sqrt {2}} \right ) \right] \right \} \right \} \right) + µ \quad $$