B Probability density of a normal distribution

If the normalized probability density of the normal distribution is $p(x) = \frac {1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, then if $\sigma = 0.0001$ and in the special case $x = \mu$, wouldn't the probability density at this point, $p(\mu)$, exceed 1 since it is equal to $p(\mu) = \frac {1}{\sqrt{2\pi}0.0001} > 1$? Wouldn't this mean it is not normalized?

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If the normalized probability density of the normal distribution is $p(x) = \frac {1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$, then if $\sigma = 0.0001$ and in the special case $x = \mu$, wouldn't the probability density at this point, $p(\mu)$, exceed 1 since it is equal to $p(\mu) = \frac {1}{\sqrt{2\pi}0.0001} > 1$? Wouldn't this mean it is not normalized?
No. The density function can get huge as long as its integral is equal to 1. So the density function can get very large for a short range of X.

No. The density function can get huge as long as its integral is equal to 1. So the density function can get very large for a short range of X.
Okay, just wanted to ensure I understood that. Thank you. So in general, the value for $p(x)$ always varies from 0 to $\frac{1}{\sqrt{2\pi}\sigma}$?

mfb

Mentor
Yes.

Other probability distributions can have even higher densities - as long as they are in a small range. Only the integral is important.

• Stephen Tashi

As an analogy, a 1 kg rock with a volume of 300 cc could have a density of 1.5 kg/ 300 cc at a particular location within the rock.

"Probability density of a normal distribution"

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