Triangular vs Normal distribution I'm trying to approximate a standard normal distribution with a triangular distribution. What parameters of the triangular distribution (min, max and mode) are more suitable? Thank you
The best approximation in the $L^2$ sense is given by the value of $\alpha\in\mathbb{R}^+$ for which:

$$ \frac{d}{d\alpha}\left(\frac{1}{2\pi}\int_{|x|\geq \alpha}e^{-x^2}\,dx + \int_{-\alpha}^{\alpha}\left(\frac{1-|x/\alpha|}{\alpha}-\frac{e^{-x^2/2}}{\sqrt{2\pi}}\right)^2\,dx\right)=0,$$
i.e. by minimizing the $L^2$ norm of the difference between the pdf of a standard normal distribution, $\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$, and a distribution supported on $[-\alpha,\alpha]$ having pdf $\frac{1}{\alpha}\left(1-\left|\frac{x}{\alpha}\right|\right).$

Numerically, it is $\color{red}{\alpha\approx 2.297}$. Here there are the two distributions:

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from your Triangle Distribution you have a (min), (max), and (mode).
in a Normal Distribution you have a (mean) and (standard deviation) abbreviated as (sd) .
in the Normal Distribution the (mean) is equal to the (mode) due to the symmetry of the model.
to move from a Triangle model to a Normal model we need to estimate the (mean) and the (standard deviation) from the Triangle models parameters
for the (mean):
it is reasonable to use your (mode) of the Triangle Distribution as your (mean) for your Normal Distribution because they are equal under the Normal model.  NOTE at this point one should be aware they maybe breaking a symmetry assumption in their normal model.
a second reasonable estimate for the (mean) can come from the (midpoint) of the (min) and the (max) where (midpoint)=((max)-(min))/2.
the (mean) of the Normal distribution represents the central tendency. a (midpoint) and a (mode) are also measures of central tendency thus they are candidates that we have readily available from the Triangle Distribution.
a third method will to be use the (mean) of the Triangle Distribution and use it to estimate the mean a Normal Distribution.  the (mean) of a triangle distribution is (mean_tri)=((min)+(max)+(mode))/3
method 1 (central tendency):
(mean)=(mode)
or
method 2 (central tendency):
(midpoint)=((max)-(min))/2
(mean)=(midpoint)
or
method 3 (match moments):
(mean_tri)=((min)+(max)+(mode))/3
(mean)=(mean_tri)
for the (standard deviation):
a Triangle distribution gives us a (min) and a (max).  We can take the difference to find a range value (range)=(max)-(min).  a function of the range can provide an estimate of the standard deviation of a Normal distribution f(range)=(sd). It is up to the user to determine a reasonable function at this point.  example: four standard deviation covers 95% of a Normal Distribution's data and would result in an estimated standard devastation of (sd)=(range)/4, while a six standard deviation range covers 99% and would result in an estimate of (sd)=(range)/6.
Another method will be to take the (standard deviation) of the Triangle Distribution and use that to estimate the (standard distribution) of the Normal Distribution. the (sd) of a triangle distribution is (sd_tri)=sqrt(((min)^2+(max)^2+(mode)^2-(min)(max)-(min)(mode)-(max)(mode))/18)
method 1 (range coverage):
try setting
(range)=(max)-(min)
and using:
(sd)=(range)/2  (65% normal coverage)  (fattest of the three models)
or
(sd)=(range)/4  (95% normal coverage) (middle width of the three models)
or
(sd)=(range)/6  (99% normal coverage) (skinniest of the three models)
method 2 (match central moments):
(sd_tri)=sqrt(((min)^2+(max)^2+(mode)^2-(min)(max)-(min)(mode)-(max)(mode))/18)
(sd)=(sd_tri)
when estimating your (mean) and (standard deviation) you may want to consider how much confidence you had in your Triangle's coverage of the population and if the original distribution was skewed.  Did the (min) and (max) result from a very small sample or a very large sample?  If i am skewed, do i want to focus in on the mode or do i want to go for more coverage?