= Solution
For the shifted exponential model, $F(x)=1-e^{-\lambda(x-1)}$ for $x\geq1$. It is increasing, so the <probability integral transform> preserves order: $F(X_{(r)})=Y_{(r)}$. The expected uniform plotting positions are $r/(n+1)$ by part (b). Replacing the random transformed <order statistic> by this representative position gives
$$
1-e^{-\lambda(x_{(r)}-1)}\approx\frac r{n+1},\qquad
x_{(r)}\approx1+\frac1\lambda\left[-\log\left(1-\frac r{n+1}\right)\right].
$$
Thus \b[plotting $x_{(r)}$ against $t_r=-\log(1-r/(n+1))$ should give an approximate <line> with intercept one and slope $\boxed{1/\lambda}$].
The replacement is a quantile-plot approximation, not an assertion that <expectation> commutes exactly with the nonlinear inverse <distribution function>. The random points need not lie exactly on the <line>.
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