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Let $X$ be distributed as a $Normal (\mu, \sigma^2)$. Then for a fixed $\mu$ it is always the case that:

\begin{equation} \frac{90th quantile-10th quantile}{\sigma}=constant \quad \forall \sigma>0 \end{equation}

Thanks in advance!

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    $\begingroup$ How is this "too trivial"? @Bob Jansen $\endgroup$ – Mathias Barreto Feb 20 at 18:59
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    $\begingroup$ First of all, it’s a stats question and not a quantitative finance question. Second, it follows from the definition as shown by @KeSchn. Third, you should show your own attempts at answering. $\endgroup$ – Bob Jansen Feb 20 at 21:51
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Let $p\in(0,1)$. The corresponding quantile function of $X\sim N(\mu,\sigma^2)$ is given by $$F_X^{-1}(p)=\mu+\sigma\Phi^{-1}(p)=\mu+\sqrt{2}\sigma\mathrm{erf}^{-1}(2p-1),$$ where $\Phi^{-1}$ is the inverse of the cumulative distribution function of a standard normally distributed random variable and $\mathrm{erf}^{-1}$ is the inverted error function.

Thus, \begin{align} \frac{\mathrm{Quantile}(0.9)-\mathrm{Quantile}(0.1)}{\sigma}&=\frac{\mu+\sigma\Phi^{-1}(0.9)-(\mu+\sigma\Phi^{-1}(0.1))}{\sigma} \\ &=\Phi^{-1}(0.9)-\Phi^{-1}(0.1) \\ &\approx 2.56. \end{align}

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    $\begingroup$ Impeccable, thanks! $\endgroup$ – Mathias Barreto Feb 20 at 18:09

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