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.
A: I'm not sure about the general answer but if the continuous distribution is the standard normal then the probability that the median lies between the min and the max of a sample of size $n$ equals 
$$ \frac{ 2^{n-1} - 1 }{2^{n-1}} $$ 
You can work this out based on the distribution of normal order statistics, which I will leave to you as an exercise since this appears to be a self study question. 
A: pick once from the distribution
then for each subsequent pick, you have a probability 0.5 that your pick is in the same half as the first pick
so your chance of getting all picks on the same side of the median is 0.5^(n-1)
so the chance of that not happening (and therefore the median being contained) is (1-that)
Eg:
n = 5
P = 1 - 0.5^4 = 0.94
