Annual maxima flow rate data at Redbrook on the River Wye

$\backslash$table$[[$Water year sak flow rate $($m$3/$s$),$Log of Return Periods,$],[2006,925\mathrm{.}8,2\mathrm{.}167880759],[2007,836\mathrm{.}4,2\mathrm{.}400918682],[2008,809\mathrm{.}9,2\mathrm{.}480869537],[2009,731\mathrm{.}4,2\mathrm{.}748713627],[2010,730\mathrm{.}8,2\mathrm{.}75226946],[2011,720\mathrm{.}8,2\mathrm{.}791826309],[2012,706\mathrm{.}4,2\mathrm{.}850189812],[2013,669\mathrm{.}6,3\mathrm{.}008221684],[2014,665\mathrm{.}9,3\mathrm{.}026803801],[2015,657\mathrm{.}7,3\mathrm{.}065940419],[2016,634\mathrm{.}4,3\mathrm{.}180200644],[2017,622\mathrm{.}0,3\mathrm{.}245097251],[2018,608\mathrm{.}6,3\mathrm{.}318654068],[2019,593\mathrm{.}9,3\mathrm{.}402107193],[2020,593\mathrm{.}0,3\mathrm{.}409163016]]$Using the peak flow $($annual maxima$)$ data given in the spreadsheet $$Q$3\_$River$\_$Wye$,$ determine the empirical return period for the highest $18$ flow rates based on the Gringorten plotting position formula given by:

Return Period$=($n$+0\mathrm{.}12)/($i$-0\mathrm{.}44)$

where:

n $=$ total number of years in entire data set

i $=$ rank of data $($from the largest $($i$=1)$ to the smallest $($i$=$n$))$

Fitting a log$-$linear best fit to the data, where the return period on the x$-$axis is logged, estimate the flow rate of an event with a return period of $50$ years.