Why the value of Gaussian curve drop to 1/19 at 2 standard deviation?

Question

Taken from Guide to DSP where it says:

... at two ... standard deviations from the mean, the value of the Gaussian curve has dropped to about 1/19 ...

It seems to be a straight forward calculation but my math just didn't work out to 1/19. The Gaussian probability distribution function is given by:

$$ P(x) = {1 \over \sqrt{2\pi} \sigma } e^{-{{(x - \mu)^2} \over {2 \sigma^2}}} $$

At the mean (where maximum probability occurs),

$$ \begin{align} P(\mu) &= {1 \over \sqrt{2\pi} \sigma } e^{-{{(\mu - \mu)^2} \over {2 \sigma^2}}} \\ &= {1 \over \sqrt{2\pi} \sigma } e^{0} = {1 \over \sqrt{2\pi} \sigma} \end{align} $$

And at 2 standard deviation, or $ x = \mu +2\sigma $

$$ \begin{align} P(\mu + 2\sigma) &= {1 \over \sqrt{2\pi} \sigma } e^{-{{(\mu + 2\sigma - \mu)^2} \over {2 \sigma^2}}} \\ &= {1 \over \sqrt{2\pi} \sigma } e^{-{{4\sigma^2} \over {2 \sigma^2}}} = {1 \over \sqrt{2\pi} \sigma } e^{-2} \end{align} $$

To see how much probability has dropped from max probability at $ \mu $ to probability at $ 2\sigma $, I can simply take the ratio:

$$ { P(\mu + 2 \sigma) \over P(\mu) } = { {{1 \over {\sqrt{2\pi}\sigma}} e^{-2}} \over {{1 \over {\sqrt{2\pi}\sigma}}}} = e^{-2} = 0.135 \approx {2 \over 15} \neq {1 \over 19} $$

Answer 1

You've computed the most general case and have shown it's always 2/15, thus 1/19 is incorrect, at least if interpreting "1/x-th of value" as $f(\text{value})/f_\text{max}$. This simulation confirms it.

Edit: I took a guess, 1/19 is the value of the Gaussian at two standard deviations for $(\mu, \sigma) = (0, 1)$, which does qualify per exact wording of "the value drops to 1/19". But I agree your measure is more meaningful as it's independent of $(\mu, \sigma)$ and interprets as "the value drops to 2/15 of its peak value".

Answer 2

As noted by AlexTP and OverLordGoldDragon, the confusion is that the actual value of the Gaussian probability density curve is 1/19, while the value compared to the 0.4 peak value is 2/15.

When talking about probability distributions, the middle or peak value does not have a particular significance. The absolute value of PDF is more useful in practice than the ratio to peak value.

For example, a distribution that has a wide flat peak with sharp edges would have a lower peak probability, for example 0.2. If it has density of 1/19 at 2 sigma, it has the same probability of having outliers that far as the Gaussian distribution would. But if you used the comparison to peak value, you would think it was less sharp drop than the Gaussian, just because the middle is more flat.

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As an example, let's compare the Laplace distribution to Gaussian (normal) distribution:

^{(Image source)}

The ratio of value at 2 standard deviations compared to peak value is 0.1 for both distributions. The actual probability of events outside 2 sigma is lower for Gaussian distribution, while Laplace distribution does have steeper descent at the middle.

The discussion in the original article was that in Gaussian distribution the "tails drop toward zero very rapidly", where tails means the long portion further away from middle.