Neither the suggestion in a previous (now deleted) Answer nor the suggestion in the following Comment is correct for the sample standard deviation of the combined sample.

**Known data for reference.:** First, it is helpful to have actual data at hand to verify results, so I simulated samples of sizes $n_1 = 137$ and $n_2 = 112$ that are roughly the same as the ones in the question.

**Combined sample mean:** You say 'the mean is easy' so let's look at that first. The sample mean $\bar X_c$ of the combined sample can be expressed in terms of the means
$\bar X_1$ and $\bar X_2$ of the first and second
samples, respectively, as follows. Let $n_c = n_1 + n_2$ be the sample size of the combined sample, and let
the notation using brackets in subscripts denote the
indices of the respective samples.

$$ \bar X_c = \frac{\sum_{[c]} X_i}{n} =
\frac{\sum_{[1]} X_i + \sum_{[2]} X_i}{n_1 + n_1}
= \frac{n_1\bar X_1 + n_2\bar X_2}{n_1+n_2}.$$

Let's verify that much in R, using my simulated dataset (for now, ignore the standard deviations):

```
set.seed(2025); n1 = 137; n2 = 112
x1 = rnorm(n1, 35, 45); x2 = rnorm(n2, 31, 11)
x = c(x1,x2) # combined dataset
mean(x1); sd(x1)
[1] 31.19363 # sample mean of sample 1
[1] 44.96014
mean(x2); sd(x2)
[1] 31.57042 # sample mean of sample 2
[1] 10.47946
mean(x); sd(x)
[1] 31.36311 # sample mean of combined sample
[1] 34.02507
(n1*mean(x1)+n2*mean(x2))/(n1+n2) # displayed formula above
[1] 31.36311 # matches mean of comb samp
```

**Suggested formulas give incorrect combined SD:** Here is a demonstration that neither of the proposed formulas finds $S_c = 34.025$ the combined sample:

According to the first formula $S_a = \sqrt{S_1^2 + S_2^2} = 46.165 \ne 34.025.$ One reason this formula is wrong is that it does not
take account of the different sample sizes $n_1$ and $n_2.$

According to the second formula we have $S_b = \sqrt{(n_1-1)S_1^2 + (n_2 -1)S_2^2} = 535.82 \ne 34.025.$

To be fair, the formula $S_b^\prime= \sqrt{\frac{(n_1-1)S_1^2 + (n_2 -1)S_2^2}{n_1 + n_2 - 2}} = 34.093 \ne 34.029$ is more reasonable. This is the formula for the 'pooled standard deviation' in a pooled 2-sample t test.
If we may have two samples from populations with different means, this is a reasonable *estimate* of the
(assumed) common population standard deviation $\sigma$ of the two samples. However, it is not a correct
formula for the standard deviation $S_c$ of the combined sample.

```
sd.a = sqrt(sd(x1)^2 + sd(x2)^2); sd.a
[1] 46.16528
sd.b = sqrt((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2); sd.b
[1] 535.8193
sd.b1 = sqrt(((n1-1)*sd(x1)^2 + (n2-1)*sd(x2)^2)/(n1+n2-2))
sd.b1
[1] 34.09336
```

**Method for correct combined SD:** It is possible to find $S_c$ from $n_1, n_2, \bar X_1, \bar X_2, S_1,$ and $S_2.$ I will give an indication how this can be done. For now, let's
look at sample variances in order to avoid square root signs.

$$S_c^2 = \frac{\sum_{[c]}(X_i - \bar X_c)^2}{n_c - 1} = \frac{\sum_{[c]} X_i^2 - n\bar X_c^2}{n_c - 1}$$

We have everything we need on the right-hand side
except for $\sum_{[c]} X_i^2 = \sum_{[1]} X_i^2 + \sum_{[2]} X_i^2.$ The two terms in this sum
can be obtained for $i = 1,2$ from $n_i, \bar X_i$ and $S_c^2$
by solving for $\sum_{[i]} X_i^2$ in a formula
analogous to the last displayed equation.
[In the code below we abbreviate this sum as
$Q_c = \sum_{[c]} X_i^2 = Q_1 + Q_2.$]

Although somewhat messy, this process of obtaining combined sample variances (and thus combined sample SDs) is used
in many statistical programs, especially when
updating archival information with a subsequent sample.

**Numerical verification of correct method:** The code below verifies that the this formula
gives $S_c = 34.02507,$ which is the result we
obtained above, directly from the combined sample.

```
q1 = (n1-1)*var(x1) + n1*mean(x1)^2; q1
[1] 408219.2
q2 = (n2-1)*var(x2) + n2*mean(x2)^2; q1
[1] 123819.4
qc = q1 + q2
sc = sqrt( (qc - (n1+n2)*mean(x)^2)/(n1+n2-1) ); sc
[1] 34.02507
```