$15$ boys: mean of $47.6$, standard deviation of $11.2$. $10$ girls: mean of $49.1$, standard deviation of $15.4$. Overall standard deviation of the marks?
How would I go about accurately finding the standard deviation from this question? I already have the mean, but I have no idea how to calculate the standard deviation.
For greater generality lets say you had two samples $\{b_i\}_{i=1}^B$ and $\{g_j\}_{j=1}^G$. Then we define the statistical data $$\mu_B=\frac 1B \times \sum b_i\quad \quad \mu_G=\frac 1G \sum g_j$$
$$\sigma_B^2=\frac 1{B-1}\times \sum (b_i-\mu_B)^2\quad \quad \sigma_G^2=\frac 1{G-1}\times \sum (g_j-\mu_G)^2$$
Here I am assuming that you are using the sample variance.
The total sample then has $B+G$ elements and of course has average $$\mu = \frac 1{B+G} \times (B\mu_B+G\mu_G)$$
We want to compute $$\sigma^2=\frac 1{B+G-1}\times \left(\sum (b_i-\mu)^2+\sum (g_j-\mu)^2\right)$$
in terms of the standard statistical data for the individual samples.
Let's evaluate $\sum (b_i-\mu)^2$:
$$\sum (b_i-\mu)^2=\sum (b_i-\mu_B+(\mu_B-\mu))^2=\sum (b-\mu_B)^2+2(\mu_B-\mu)\sum (b_i-\mu_B)+B(\mu_B-\mu)^2$$
Now, $\sum (b_i-\mu_B)=0$ so we get $$\sum (b_i-\mu)^2=(B-1)\sigma_B^2+B(\mu_B-\mu)^2$$
Of course we also have $$\sum (g_j-\mu)^2=(G-1)\sigma_G^2+G(\mu_G-\mu)^2$$
Combining all this we have $$\boxed{\sigma^2=\frac 1{B+G-1}\times \left((B-1)\sigma_B^2+B(\mu_B-\mu)^2+(G-1)\sigma_G^2+G(\mu_G-\mu)^2\right)} $$
Worth noting: the "correction term" here is the sample variance of the two means.
Sanity check: note that if we had $\mu_B=\mu_G$ then this is (essentially) just the weighted average of the variances, as you'd expect.