.I the 2012 Russian presidential election russia2012, and the 2011 Canadian election canada2011 as separate data frames. Each of these data sets has the same variables, described in the table below. Name Description N Total number of voters in a precinct turnout Total number of turnout in a precinct votes Total number of votes for winner in a precinct Question 1 To analyze the 2011 Russian election results, rst compute United Russia's vote share as a proportion of the voters who turned out. Identify the 10 most frequently occurring fractions for the vote share. Create a histogram that sets the number of bins to the number of tmique fractions, with one bar created for each uniquely observed fraction, to differentiate between similar fractions Like 1 f 2 and 51/ 11K]. rI'his can be done by using the breaks argument in the hist function. Mat does this histogram look like at fractions with low numerators and denominators such as 1 ,f 2 and 2 f 3? Question 2 The mere existence of high frequencies at low fractions may not imply election fraud. Indeed, more numbers are divisible by smaller integers like 2, 3, and 4 than by larger integers like 22, 23, and 24. To investigate the possibility that the low fractions arose by chance, assume the following probability model: - Turnout for a precinct is binomially distributed, with size equal to the number of voters in the precinct and success probability equal to its observed turnout rate. . Vote counts for United Russia in a precinct is binomially distributed with size equal to the number of voters who simulated to turn out in the previous step and success probability equal to the precinct's observed vote share. Conduct a Monte Carlo simulation under these assumptions. 1000 simulated elections should be sufcient. (Note that this may be computationally intensive code. 1Write your code for a small number of simulations to