Five values and the median

Idea. The median is outside the range of the sample only if all five observations fall on the same side of it. Each falls below the median with probability exactly $\frac{1}{2}$, so the distribution does not matter.

Step 1. $m$ is below the smallest observation exactly when all five are above $m$: probability ${\left(\frac{1}{2}\right)}^5=\frac{1}{32}$. Likewise, $m$ is above the largest when all five are below $m$: also $\frac{1}{32}$. (Since the distribution is continuous, no observation equals $m$.)

Step 2. $$\mathbb{P}(\min <m<\max )=1-\frac{2}{32}=\frac{15}{16}\approx 94\%.$$

Why it matters. With only five data points, the interval from the smallest to the largest is already a 94% confidence interval for the median, with no assumption on the shape of the distribution. With $n$ points the confidence is $1-2^{1-n}$.