Question 1121774
See if this makes sense:
You have a population of say 100 people.  It has a mean mu and a sd sigma.
If you take samples of size 10 from this population, there are 100C10 or 1.73 x 10^13 possible samples you can choose.  The 2 quadrillion samples would be normally distributed with mean mu, the same as the population, and the sd of that distribution would be sigma/sqrt(43).  What happens is that the population distribution is now replaced by a distribution of sample size 10 distribution. 

Each of those samples has a standard deviation, but that only matters when one chooses one sample, with a mean x bar and sd s, to estimate the population mean.

It is worth doing this with 1,2,3,4, with mean 2.5 and sd (for pop) of 1.12
Samples of size 2 are taken, with replacement, and there are 16 such samples
11/12/13/14/21/22/23/24/31/32/33/34/41/42/43/44
The sums are 2/3/4/5/3/4/5/6/4/5/6/7/5/6/7/8
the means are half of that, and divided by 16, is 2.5
The sd can be shown to be 0.791, which is the original sd of 1.12, divided by sqrt(n), or sqrt (2)=1.414

The individual sample of 2 has a sd, but we don't average the individual samples. 

For the samples of 5 from 13, there are 1287 samples altogether, but the sd of the population sigma, divided by the sqrt (100), where 100 were the number of samples, would give you the sd of the new distribution, the mean of each of the 100 samples, and the overall mean of that would be very close to mu.  The sd of that distribution would be s/ sqrt(100) or s/10.