 Solution

For the Pareto model, integration gives $F(x)=1-x^{-\lambda }$ on $x\ge 1$. The same plotting positions suggest $1-x_{(r)}^{-\lambda }\approx r/(n+1)$. Taking logarithms givesTherefore 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).