Can a z-score be used to describe a cluster's homogeneity? As part of my thesis, we wanted to investigate the ability of using z-scores as indicators for the homogeneity of clusters (or speaking more generally of partitions of a vector). After some modelling I believe this is not entirely possible and I would like to get some opinions on this. 
Suppose I have some 2D data with one predictor c and my cluster assignment cl, which splits c into 5 groups of exactly 20 elements. There is little variation among the elements of c in every cluster and every cluster contains a distinct set of elements of c(i.e. cluster 1 contains mostly ones, cluster 2 mostly twos, a.s.o.). I then want to compute each indivdual cluster's z-score based on c as follows:
# library(data.table)
# dat<-as.data.table(dat)
meanCl<-dat[,mean(c),by=cl][,V1,]
mean0<-sapply(1:1000,function(x){
              mean(sample(dat[,c,],20,replace=T))}
sd0<-sd(mean0)
mean0<-mean(mean0)
z<-(meanCl-mean0)/sd0

... i.e. I compute the mean of c for every subgroup in cl. I go on to compute the mean of 1000 clusters containing 20 random elements from the distribution of c. I then calculate every clusters' z-score by subtracting its mean from the mean of the random clusters' means and divide by the standard deviation in those means to get:
[1] -4.96401933 -4.05057547 -0.03142247  2.89159789  5.99730703

So even though all clusters are distinctly characterised in their internal distribution of c, the z-score for cluster 3 is basically 0, as its mean tends towards the mean of the sample distribution of c. 
Typing this out, it makes sense to me to not be able to use a cluster's z-score as a predictor for its homogeneity, however, as intially stated, I would like some ideas on this.
 A: Your intuition is correct: the z-score of a cluster's mean with respect to the means of the other clusters cannot be used to determine its homogeneity. The reason for this is twofold.


*

*A cluster's homogeneity does not depend on other clusters. A cluster's homogeneity does not change depending on the location of the other clusters in the clustering; and

*A cluster's mean does not capture the "spread" of the instances assigned to it. The clusters $(3, 3, 3, 3, 3)$ and $(-1, 0, 2, 5, 9)$ both have a mean value of 3 but have ranges, for example, of $0$ and $10$.
Thinking specifically about a cluster's homogeneity I would suggest that you consider internal measures of cluster quality. These measure how well a given clustering groups similar instances together while keeping different instances apart. Internal measures such as the Davies-Bouldin index, Dunn index and Silhouette coefficient all explicitly consider the "spread" of instances assigned to a cluster.
Davies-Bouldin index: $DB = \frac{1}{n} \sum_{i=1}^n \max{j \neq i} \text{ } (\frac{\sigma_i+\sigma_j}{d(c_i,_j)})$
With $n$ the number of clusters, $c_x$ the centroid of cluster $x$, $\sigma_{x}$ is the average distance of all elements in cluster $x$ to centroid $c_{x}$, and $d(c_{i},c_{j})$ is the distance between centroids $c_i$ and $c_j$.
Dunn Index: $D =\frac{\min{i,j} d(i,j)}{\max{k} d'{k}}$
With $d(i,j)$ the distance between clusters $i$ and $j$, and $d'(k)$ the intra-cluster distance of cluster $k$.
Silhouette coefficient for instance $i$: $s(i) = \frac{b(i)-a(i)}{\max{(a(i),b(i))}}$
With  $a(i)$ the average dissimilarity of $i$ with all other data within the same cluster, and $b(i)$ the lowest average dissimilarity of $i$ to any other cluster, of which $i$ is not a member. The silhouette coefficient of the clustering is the sum of $s(i)$ for all $n$ instances clustered.
