I have a random variable $X$ that is binomially distributed $B(n_B, p)$ with a mean $\mu_B$ and standard deviation $\sigma_B$. Lets say I take a sample of this distribution of size $n_s$.
Thus, the sample mean, $\bar{Y}$, of this sample has an expected value of:
$$E[\bar{Y}] = \mu_{\bar{Y}} = \frac{1}{n_s}(E[X_1] + ...E[X_{n_s}]) = n_B*p = \mu_B$$
And the variance of $\bar{Y}$ is $$Var(\bar{Y}) = \sigma_{\bar{Y}}^2 = \frac{\sigma_B}{n_s}$$
Using the Central Limit Theorem, if $n_S$ is large enough, I can approximate the distribution of $\bar{Y}$ using $N(\mu_{\bar{Y}}, \sigma_{\bar{Y}})$. However, I am not interested in an approximation. I am curious if $\bar{Y}$ is continuous and how to find its exact PDF (if it has one).
Suppose you have an i.i.d sample $X_1, \dots, X_n \sim f(x; \theta)$. Of course, your sample mean $\bar X$ will follow some distribution, since it is a random variable. The question is, can we write a closed-form expression for that distribution? The sample mean is $\frac{1}{n}Y$ where $Y = \sum_iX_i$. If we know the distribution of $Y$, it's easy to get $Y/n$ if you know how to find distributions of transformations of random variables. It would just be $Y/n \sim n*f(ny)$. The important consideration is whether we can get the distribution of $Y$. This is not always possible.
As an example, if your sample follows $Exponenial(\theta)$, a continuous distribution, then $\bar Y \sim Gamma(n, \theta)$. This is an example where you can find the exact PDF. On the other hand, see here for an example where there is no closed form for the distribution of the mean.
If your sampling distribution is discrete (e.g. binomial distributions), it is not possible for your sample mean to have a continuous distribution.
Edit: for your particular example of binomially distributed random variables, the sum of binomials is itself binomial. You can find the mean, then, by applying the scaling transformation I mentioned above.