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

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?

• "decreasing the standard deviation of the standard normal distribution from $\sigma=0$" may involve a typographical error Commented Oct 5, 2019 at 9:52

$$\varphi(0)$$ 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