= Solution
{c}
For the continuous distribution in the question, $F$ maps the support monotonically from zero to one. For $0<t<1$, strict monotonicity gives
$$
\mathbb P(F(X)\leq t)=\mathbb P(X\leq F^{-1}(t))=t.
$$
Thus \b[$F(X)$ is uniform on $(0,1)$], the <probability integral transform>.
The <order statistics> are the sample values sorted increasingly, $X_{(1)}\leq\cdots\leq X_{(n)}$. Monotonicity of $F$ makes $F(X_{(i)})$ the $i$th <order statistic> of independent uniforms. Its density is $n!t^{i-1}(1-t)^{n-i}/[(i-1)!(n-i)!]$, giving
$$
\boxed{\mathbb E F(X_{(i)})=\frac{i}{n+1},\qquad i=1,\ldots,n.}
$$
This follows either by beta integrals or by the equal expected lengths of the $n+1$ uniform spacings.
For the stated log-logistic cumulative distribution,
$$
\log\frac{F(x)}{1-F(x)}=(\beta-1)\log x-(\beta-1)\log\alpha.
$$
Therefore plot $\log[p_i/(1-p_i)]$ vertically against $\log X_{(i)}$ horizontally, with plotting positions $p_i=i/(n+1)$. A plausible model gives an approximately straight line, with \b[slope $\beta-1$ and intercept $-(\beta-1)\log\alpha$]. This specifies the axes explicitly, since reversing them would invert the slope.
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