Let $X_1, ... , X_n$ be a random sample from a continuous distribution F. Let $\theta$ be the median of F. Using normal approximation, calculate the probability P($X_{(3)} \leq \theta \leq X_{(n-2)}$).
Probability of Median from a continuous distribution
This one is extremely similar and I think the answer would either be $$\frac{6}{2^n}$$ or $${n \choose 6} * (\frac{1}{2})^n$$ according to the logic given in the previous question.