Q:

Probability that median falls within a range? 

If I draw 5 samples from a distribution that is cont.,  what is the probability that the median of the population is between the xmin and xmax of the draw samples? 
What is the formula to apply here? 

A:

Let $m$ denote the true median of the distribution.   Since the distribution is continuous, by definition there is a probability of one-half that a given value will fall above/below the median.   Hence, you have:
$$\mathbb{P}(X_{(1)} > m) = \prod_{i=1}^n \mathbb{P}(X_i > m) = \prod_{i=1}^n \frac{1}{2} = \frac{1}{2^n},$$
$$\mathbb{P}(X_{(n)} < m) = \prod_{i=1}^n \mathbb{P}(X_i < m) = \prod_{i=1}^n \frac{1}{2} = \frac{1}{2^n}.$$
Putting this together and using the inclusion-exclusion principle you get:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X_{(1)} \leqslant m \leqslant X_{(n)})
&= 1 - \mathbb{P}(X_{(1)} > m \ \text{ or } \ X_{(n)} < m) \\[6pt]
&= 1 - \mathbb{P}(X_{(1)} > m) - \mathbb{P}(X_{(n)} < m) + \mathbb{P}(X_{(n)} < m < X_{(1)}) \\[6pt]
&= 1 - \mathbb{P}(X_{(1)} > m) - \mathbb{P}(X_{(n)} < m) + 0 \\[6pt]
&= 1 - \frac{1}{2^n} - \frac{1}{2^n} \\[6pt]
&= 1 - \frac{1}{2^{n-1}} \\[6pt]
&= \frac{2^{n-1} - 1}{2^{n-1}}.  \\[6pt]
\end{aligned} \end{equation}$$
Note that this result holds for any continuous distribution.   Unsurprisingly, as $n \rightarrow \infty$ we have $\mathbb{P}(X_{(1)} \leqslant m \leqslant X_{(n)}) \rightarrow 1$, so the range of the sample will eventually come to encompass the true median, with probability approaching one.