Is 95% confidance interval always included in minimum-maximum interval ?
I have a set of altitudes (Z). I calculate the mean Z and the 95% confidance interval. Will the bounds of this confidance interval be included in the [Zmin - Zmax] interval, or can they be out of it ?
No, not always.
If you take the usual definition of a $95\%$ confidence interval for the mean as
$$\frac1n \sum x_i \pm t_{n-1:0.975}\,\sqrt{\frac1{n(n-1)}\sum\left(x_i -\frac1n \sum x_j\right)^2}$$
then I think this confidence interval will never enclose $\big(\min(x_i),\max(x_i)\big)$ when $n\in\{2,3,4\}$, while for larger $n$ it can. An example of the best case is $x_1=-n, x_n=n$ and the other $x_i=0$, in which case $\bar{x}=0$ and the confidence interval for the mean when $n=4$ is about $[-5.2,5.2]$ so not contained in $[-4,4]$; when $n=5$ the confidence interval is about $[-4.4,4.4]$ and is contained in $[-5,5]$, and similarly for large $n$ with a confidence interval heading towards about $[-2.77,2.77]$ as $n$ increases, well within $[-n,n]$. Note that $2.77 \approx 1.96 \sqrt{2}$.
For $n\ge 7$ drawn from a normally distributed random variable, the range probably will contain the confidence interval for the mean, but it does not have to do so even for extremely large $n$. Consider as an example of a worst case the observations $x_n=n$ and the other $x_i=0$. Then $\bar{x}=1$ and the confidence interval for the mean heads towards about $[-0.96, 2.96]$ (wider for smaller $n$) and clearly overlaps the minimum observation of $0$.
What is true is that for all $n\gt 1$ the $50\%$ confidence interval for the mean is always contained in the range of observations (for $n=2$ the two intervals coincide), and that for large $n>90$ the $68\%$ confidence interval for the mean is always contained in the range of observations (note $2\Phi(1)-1\approx 0.6827$).