Question

Fraction of tails Now let's look at the fraction of tails in n trials. Fraction of tails in a small number of coin flips Tails and 0 = Heads). n Suppose we Aip 10 coins and compute the fraction of tails (with 1 = - 10 trials - 6 for (i in 1: trials) { print (sun (sample(c(0, 1), n, replace=TRUE)) / n) 3 ## (1) 0.1 ## (1) 0.4 ## [1] 0.4 ## [1] 0.5 ## (1) 0.7 ## [1] 0.6 The expected fraction of tails is 0.5. Let's compute the difference between the observed fraction and the expected fraction: - 10 trials - 6 for (i in 1: trials) { print (sum (sample(c(0, 1), n, replace=TRUE)) / n 0.5) ب ## [1] -0.2 ## [1] -0.2 ## (1) -0.1 ## [1] -0.2 ## [1] o ## [1] 0.1

Answer 1

ANSWER:

• The difference between the observed fraction of tails and 0.5 does not changed with the number of trails as probablity of getting tail will not change on which trial you are tossing the coin.
• I have provided the properly commented code in both text and image format so you can easily copy the code as well as check for correct indentation.
• I have provided the output image of the code so you can easily cross-check for the correct output of the code.
• Have a nice and healthy day!!

CODE TEXT

n = 10000
trials = 6
# defining empty list to store +ve difference and -ve difference in abs form
differences = c()
for (i in 1:trials){
# difference from expected
difference = sum(sample(c(0,1),n,replace=TRUE))/n-0.5
# appending abs(difference) in differences list
differences = append(differences,abs(difference))
print(difference)
}
# finding mean difference in differences list, to get approx +- difference
app_difference = mean(differences)
# displaying
print(paste ("+- difference is: ",app_difference, sep = " "))

CODE IMAGE

1 n = 10000 2 trials = 6 3 # defining empty list to store +ve difference and -ve difference in abs form 4 differences = c() 5 for (i in 1:trials) { 6 # difference from expected 7 difference = sum(sample(c(0,1),n,replace=TRUE))/n-0.5 8 # appending abs (difference) in differences list 9 differences = append(differences,abs(difference)) 10 print(difference) 11 } 12 # finding mean difference in differences list, to get approx +- difference app_difference = mean(differences) 14 # displaying 15 print (paste ("+- difference is: ", app_difference, sep = ")) 13

OUTPUT IMAGE

• trials = 6

[1] "+- difference is: 0.00346666666666665"

• trials = 600

[1] "+- difference is: 0.00393233333333333"

• trials = 6000

[1] "+- difference is: 0.00394683333333333"