Let's evaluate the PDF using the formula in the documentation for normpdf at that large x value. I'll choose the smallest value in your two vectors.
x = sym(472.553);
sigma = 1;
mu = 0;
f = (1/(sigma*sqrt(2*pi)))*exp(-(x-mu)^2/(2*sigma^2))
How small is that?
vpa(f)
Yeah, that definitely underflows in double precision.
If you specified the mean (I'm going to estimate it as roughly 495) then the value of the PDF will be quite a bit larger.
sigma = 495;
vf = vpa((1/(sigma*sqrt(2*pi)))*exp(-(x-mu)^2/(2*sigma^2)))
