$X_1, X_2, X_3, ..., X_n$ are iid distributed according to $U[0,1]$. Computing the mean of the expected order statistics is easy:
$$\frac{1}{n}\sum_{i=1}^n\mathbb{E}[X_{(i)}] = \sum_{i=1}^n \int_0^1 {n-1\choose i-1}X^i(1-X)^{n-i}dX = \frac{1}{2}.$$
Instead I would like to multiply each order statistic by the probability that it lies below some value $z$.
$$\sum_{i=1}^n F_i(z) \int_0^1 {n-1\choose i-1}X^i(1-X)^{n-i}dX = \sum_{i=1}^n \sum_{j=i}^n {n \choose j}z^j(1-z)^{n-j} \int_0^1 {n-1\choose i-1}X^i(1-X)^{n-i}dX $$ Suddenly the computation is not so easy anymore.
Is there a closed form?