= Solution
For the Pareto model, integration gives $F(x)=1-x^{-\lambda}$ on $x\geq1$. The same plotting positions suggest $1-x_{(r)}^{-\lambda}\approx r/(n+1)$. Taking logarithms gives
$$
\log x_{(r)}\approx\frac1\lambda\left[-\log\left(1-\frac r{n+1}\right)\right].
$$
Therefore \b[plot $\log x_{(r)}$ vertically against $t_r=-\log(1-r/(n+1))$ horizontally]. A reasonable fit should be approximately a straight <line> through the origin, with slope $\boxed{1/\lambda}$. This logarithmic transformation distinguishes the Pareto tail model from the shifted exponential plot in part (i).
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