Given a sample $X_{(1)},...,X_{(8)}$ from a continuous distribution find the $P(X_{(2)} < m < X_{(7)})$ where $m$ is the population median.
My initial questions are
Do I need to make the use of $P(X \le m) \ge \frac{1}{2}$ and $\, P(X \ge m) \ge \frac{1}{2}$ where $X$ is a random variable from this distribution.
Would this be something along the lines of $\displaystyle \int_{-\infty}^m\int_{m}^\infty f_{X_{(2)},X_{(7)}}(y,z)dydz$ ?
Where $f_{X_{(2)},X_{(7)}}(y,z) = \frac{8!}{4!}f(y)f(z)[F(y)][F(z)-F(y)]^4[1-F(z)]$
If $X_{(2)} < m < X_{(7)}$ is not the case, then either $m \leq X_{(2)}$ or $X_{(7)} \leq m$.
$$P(X_{(2)} < m < X_{(7)}) = 1 - P(m \leq X_{(2)}) - P(X_{(7)} \leq m)$$
The number of draws that are below $m$ follows a binomial distribution $n=8$ and $p=1/2$, thus:
$$\begin{align} P(m \leq X_{(2)}) = P(X_{(7)} \leq m) &= 2^{-8}{8 \choose 0} + 2^{-8}{8 \choose 1} \\ &= 9\cdot 2^{-8} \end{align}$$
In conclusion: $$\begin{align} P(X_{(2)} < m < X_{(7)}) &= 1 - 9\cdot2^{-8} - 9\cdot2^{-8} \\ &= \frac{238}{256} \end{align}$$