Code for R Studio: the following algorithm for each of these 1000 pairs: Let the output of this algorithm be

denoted by Y.

Step 1: Generate random values jQuery22401046349511440472_1586794200440?? = ????(????) and ???? = ????(????)

Step 2: Calculate ?? = (??????)??. If ???? ??, then generate a random number ??. If ?? > ??. ??

??

accept ????as ??(that is, let ?? = ????); otherwise if ?? ??. ??, else accept ???? as ?? (that is, let ?? =

???? ).

If ???? < ??, no result is obtained, and the algorithm returns to step 1. This means that the algorithm skips the pair ???? and ???? for which ???? < ?? without generating any result and moves to the next pair ???? and????.

After repeating the above algorithm 1000 times, a number N of the Y values will be generated. Obviously ?? ????, ?????? since there will be instances when a pair ???? and ???? would not generate any result, and consequently that pair would be wasted.

Investigate the probability distribution of ?? by doing the following:

1. relative frequency histogram of ??.

2. Select a probability distribution is the best fit

3. Support your assertion above by creating a probability plot

4. Support your assertion above by performing a Chi-squared test of best fit with a 0.05 level of significance.

Results used to support a report that i write