ASAP PLS SUING R CODE ###################### Part $5$: For$-$loop

y $=$ numeric$(100)$ # Create space for y to receive $100$ numbers.

for$($i in $1$:$100)\{$

y$[$i$]=2*$i # Fill y with $2,4,6,8,...,200.$

$\}$

y # Always look!

##################### Part $6$: Empirical Sampling Distribution of the Sample Mean

ntrial $=10000$

ybar $=$ numeric$($ntrial$)$ # Create space to store $10000$ sample means.

for$($i in $1$:ntrial$)\{$

s $=$ rnorm$(100,1,2)$ # Sample of size $100$ from Norm with mu$=1,$ sigma$=2$

ybar$[$i$]=$ mean$($s$)$ # Sample mean, stored in y

$\}$

ybar # $10000$ sample means.

hist$($ybar$,$ breaks$=20)$ # You are looking at the Empirical Sampling Distribution of the Sample Mean.

mean$($ybar$)$ # About $1,$ which happens to be the mu of the population.

sd$($ybar$)$ # About $0\mathrm{.}2,$ which happens to be $2/$sqrt$(100),$ i$.$e$.,$ sigma$/$sqrt$($n$)..$

Question $2$

a$)$ Revise above code to generate the empirical sampling distribution of the sample maximum.

b$)$ We have seen that the mean of the empirical sampling distribution of the sample mean is related to the population mean. Is the mean of the empirical sampling distribution of the sample maximum related to population maximum? Give some explanation for your answer.