What is the width of a wave packet?

[RicardoMP]

I'm reading Gasiorowicz's Quantum Physics and at the beginning of chapter 2, SG introduces the concept of "wave packet" and gaussian functions associated to them. The first attached image is the 28th page of the book's 1st edition I suppose, and my question is about the paragraph inside the red box. I suppose f(x) is a normal distribution. What does SG means by the function's width? Is it its FWHM? If it were, wouldn't the width be $2\sqrt{2ln(2)}\sigma$? Or if not, why is the width of order $2\sqrt{2}$, since the function at $x=\pm 2\sqrt{2}$ falls to $\frac{1}{e}$ of its peak value?
The second attached image is the "same page" from the 3rd edition (which I found harder to understand) and another explanation (green box). How can I conclude that, since the "square falls 1/3 of its peak value when $\alpha(k-k_0)^2=1$, $\Delta k = \frac{2}{\sqrt{\alpha}}$? Is this referring to the standard deviation $\sigma$ in the normal distribution (third attached image)?
I hope I'm not missing something obvious.
Thank you for your time!

 

[MisterX]

It says in the red box it means "full width, 1/e times maximum"

 

[RicardoMP]

It says in the red box it means "full width, 1/e times maximum"

And is there a reason for using "full width, 1/e times maximum" instead of FWHM?

 

[jtbell]

For a Gaussian distribution, this gives you a value equal to the standard deviation of the distribution, which is actually the more fundamental definition. $$\Delta x = \sigma_x = \sqrt {\langle x^2 \rangle - {\langle x \rangle}^2}$$ Actually, the standard deviation is a half-width (think $\mu \pm \sigma$), so if you want to compare it to a full-width, you need a factor of 2.