Pi by "throwing darts" at a Gaussian rather than a circle

Since the area of a Gaussian is sqrt(pi) _(or sqrt(pi)/2 if from 0 to infinity)_ you can redo the standard "dart" challenge but instead of checking if they are within a circle, you check for random (x,y) that y

Start with the JavaScript example in the issue and review the repository's existing topic or visualization conventions. The work is done when an artistic Gaussian dart simulation demonstrates the proposed approximation of pi and is suitable as a topic for the Suggestion-Box project.

Written by the indexing model from the issue text.