Is it possible to plot a normal distribution with $\sigma=\frac{1}{100}$ to give some geometric interpretation?

Question

this is the PDF of the standard normal distribution.

$$\varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$

This post says

consider decreasing the standard deviation of the standard normal distribution from $\sigma=0$ to $\sigma=\frac{1}{100}$. Now $\varphi(0)=\frac{100}{\sqrt{2\pi}}$ - much more than one. Not a probability.

This description may contain some conflict, when set $\sigma=\frac{1}{100}$, the distribution is not the standard normal distribution any more, that's why I asked "In the context of normal distribution" instead of "In the context of the standard normal distribution"

I know $\varphi(0)$ is a density not a probability, the quotation has claimed this very clearly, and I agree with that. I know a density could be greater than 1.

What I want to know is

Is it possible to plot a normal distribution with $\sigma=\frac{1}{100}$ to give some geometric interpretation? If yes, what this kind of plot look like?

Answer 1

$\varphi(0)$ is a density not a probability

Densities can be greater than $1$ for part of the distribution, so long as their integral across the whole support is $1$

For example consider the density of a uniform distribution on $\left[0,\frac12\right]$

For a normal distribution with $\mu=0,\sigma=\frac1{100}$ you get a density like this. The area under the red curve is $1$ because the peak is narrow, even if it is high