If you have a physical data set, you can compute it directly. Both the mean, and the standard deviation.
If you have the sizes of populations, say $m_1,\dots,m_n$, then the common mean is trivial to count: $$\mu=\frac{m_1\mu_1+\dots+m_n\mu_n}{m_1+\dots+m_n}$$ as the numerator is the total of all populations.
About the common variance. This is a mean of squares minus a square of mean. Then you should recreate the sums of squares. For example $\sigma_1^2+\mu_1^2$ is the mean of squares of the 1st population. Then you have also the sum of squares. Next, in this simple manner you have the joint sum of squares. Then their averge is easy to find by division by $m_1+\dots+m_n.$ Finally, subtract $\mu^2$ and we are done. :)
If $m_1=\dots=m_n=m$ (you write about the same experiment), then $$\mu=\frac{\mu_1+\dots+\mu_n}{n}.$$ The sum of squares in $i$-th experiment is $m(\sigma_i^2+\mu_i^2)$. Hence the total variance is $$\sigma^2=\frac{m(\sigma_1^2+\mu_1^2+\dots+\sigma_n^2+\mu_n^2)}{nm}-\mu^2=\frac{(\sigma_1^2+\mu_1^2+\dots+\sigma_n^2+\mu_n^2)}{n}-\mu^2.$$
About the (edited) last fragment of your question, the mean of $Z$ is $$\mu=\frac{\mu_1+\dots+\mu_n}{n},$$ while the standard deviation is $$\sigma=\sqrt{\frac{\sigma_1^2+\dots+\sigma_n^2}{n}}$$ provided that $X_1,\dots,X_n$ are independent.
Consider a graph $G$ on $\{1,\ldots,n\}$ where $\mu_i$ is assigned to node $i$, and an edge $\{i,j\}$ in $G$ indicates that the coupling 0f $\mu_i$ and $\mu_j$ is required to be optimal. If $G$ has no cycles (i.e., it is a forest) then the proof of the gluing lemma (see, e.g., [1]), applied separately to each tree of the forest, extends to show that a global coupling can be arranged to be optimal on all edges of $G$. Conversely, we next verify that if $G$ contains a cycle, then there is a choice of the measures $\mu_i$ where such a global optimal coupling does not exist.
Let us first show this for a cycle of length 3:
Suppose the variable $X$ is uniform in $\{1,2,3\}$, the variable $Y$ is uniform in $\{2,3,4\}$ and $Z$ is uniform in $\{1,3,4\}$.
Under the optimal coupling $P_1$ of $(X,Y)$, we have $P_1(Y=4|X=1)=1$.
Under the optimal coupling $P_2$ of $(Y,Z)$, we have $P_2(Z=4|Y=4)=1$.
Under the optimal coupling $P_3$ of $(X,Z)$, we have $P_3(Z=4|X=1)=0$.
This shows that there is no coupling of $(X,Y,Z)$ that projects to the optimal couplings $P_1, P_2, P_3$.
For a cycle of length $k>3$, simply assign two adjacent nodes of the cycle the laws of $X$ and $Y$, and assign all other nodes the law of $Z$.
[1] Villani, Cedric. Optimal transport: old and new. Vol. 338. Berlin: springer, 2009.
Best Answer
Let's get you started.
Suppose without loss of generality that $\mu_1 \leq \mu_2$. Then we have that $$ \mu = x_1\mu_1 + x_2\mu_2 \leq (x_1 + x_2)\mu_2 = \mu_2.$$ Each other subcomponent of the proof looks extremely similar to this line.