# Can someone prove (or disprove) this assertion about the normal distribution? [closed]

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

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

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}