= Solution
By <continuity> of the <cumulative distribution function> and nonnegativity, $F(0)=0$; strict increase and $F(x)\to1$ allow inversion for every $0<t<1$. The <probability integral transform> follows from
$$
\mathbb P(F(X)\leq t)=\mathbb P(X\leq F^{-1}(t))=t.
$$
Thus $F(X)$ is <uniformly distributed> on $(0,1)$.
The <order statistics> arrange the sample increasingly: $X_{(1)}\leq\cdots\leq X_{(n)}$. Strict monotonicity means $F(X_{(i)})$ is the $i$th <order statistic> of $n$ independent uniform variables. Its <density> is
$$
\frac{n!}{(i-1)!(n-i)!}\,t^{i-1}(1-t)^{n-i},\qquad0<t<1,
$$
obtained by choosing the $i-1$ observations below $t$ and the $n-i$ above it. This is the <beta distribution> with parameters $i,n+1-i$, so
$$
\boxed{\mathbb E F(X_{(i)})=\frac i{n+1}}.
$$
For the given <log-logistic distribution>, rearrangement gives
$$
\log\frac{F(x)}{1-F(x)}=(\beta-1)\log x-(\beta-1)\log\alpha.
$$
Consequently plot $y_i=\log(i/(n+1-i))$ vertically against $x_i^*=\log X_{(i)}$ horizontally. Under a good fit, replacing $F(X_{(i)})$ by its plotting position $i/(n+1)$ gives an approximate straight line with
$$
\boxed{\text{slope}=\beta-1,\qquad\text{intercept}=-(\beta-1)\log\alpha}.
$$
This is a diagnostic <probability plot>, rather than an assertion that the <expectation> of the nonlinear transformed <order statistic> equals that transformed plotting position. Positive observations occur almost surely under the model.
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