Getting Standard Deviation of Population from s. d. of individual samples. Given mean and standard deviation of samples, with no access to data. I know mean and standard deviation of individual samples (let say $10$ data points each, and no access to the underlying data); could I get the standard deviation of the population (let say $100$ samples) from the standard deviation of the individual samples? Which is the relationship?
EDIT
Thanks to BruceET's Answer. I've applied approach-$2$ to our data, and I'm just a little puzzled on results I get when calculating $S_{all}$. I would expect that $S_{all}$ is greater than the average of the individual $S_{n}$ (since there is a spread as well of the mean of subsets); on the contrary I get an $S_{all}$ that is smaller. How this should be understood?
Processed Data

 A: Two approaches, depending on your application:
(1) If your purpose is to use SDs of several samples from
the same population to get a 'pooled' estimate of the SD of the population, then you can expand the idea of a 'pooled' variance used in a two-sample t test.
Suppose you have three random samples of sizes $n_1, n_2, n_3$ from the same
population with respective sample standard deviations $S_1, S_2, S_3.$ Then the pooled estimate of the population variance $\sigma^2$ is
$$S_p^2 = \frac{(n_1-1)S_1^2+(n_2-1)S_2^2+(n3-1)S_3^2}{n_1 +n_2+n_3-3}.$$
Take the square root to get the pooled estimate $S_p$ of $\sigma.$ The formula expands in the obvious way to pool more than three samples.
(2) If you need to find the sample standard deviation of the combined sample, then method (1) won't work. Because each sub-sample has its own mean, $S_p$ above
will not be exactly the same as the sample standard
deviation of all $n_1+n_2+n_3$ observations taken as a
grand sample.
To get the value of $S_{all},$ is a bit more tedious,
but it is possible. I will outline the procedure,
which is based on the right-hand side of the formula
below. [If you need to combine many samples, then you may want to write a brief computer program to implement this approach.]
$$S^2= \frac{1}{n-1}\sum_{i=1}^n (X_i-\bar X)^2 \\
= \frac{1}{n-1}\left(\sum_{i=1}^n X_i^2 - \frac 1n\left(\sum_{i=1}^n X_i\right)^2\right) \\
= \frac{1}{n-1}\left(Q - \frac1nT^2\right)\!.$$
a.  For the first sample, use $n_1$ and $\bar X_1$ to get
$T_1 = n_1\bar X_1.$ Similarly, from the other samples get
$T_2$ and $T_3$ and then find $T = T_1+T_2+T_3,$ the grand total for the combined sample.
b. For the first sample, use $n_1, T_1,$ and $S_1^2$ to solve
for $Q_1,$ the sum of squares for the first sample.
Similarly, from the other samples get $Q_2$ and $Q_3$
and then find $Q = Q_1+Q_2+Q_3$ for the combined sample
c. Finally, use the displayed equation once again (with $n = n_1+n_2+n_3,\,$ $Q,$ and $T)$ to get
$S_{all}^2$ for the combined sample. Take the square
root to get $S_{all}.$
A: thanks a lot for your answer, I've applied approach 2 to our data, and I'm just a little puzzled on results I get when calculating Sall. I would expect that Sall is greater than the average of the individual Sn (since there is a spread as well of the mean of subsets); on the contrary I get an Sall that is smaller. How should that be understood?
In the annex you find an immage with some of the processed data. Thanks in advance and, hopefully ..., happy 2021.
Sergio
enter image description here
