Question

Flipping Coins For a fair coin, the chance of getting tails is 1/2. Counting the number of tails Let's experiment with the total number of tails in n trials. Total number of tails in a small number of coin flips ם Suppose we flip 10 coins (with 1 = Tails and 0 = Heads). Here are six samples of size 10. = 10 trials = 6 for (i in 1: trials) { print (sample(c(0, 1), 1, replace=TRUE)) مه ## [1] 0 1 0 1 1 0 0 0 1 1 ## [1] 10 10 0 1 0 1 1 1 [1] 0 0 1 0 1 1 0 0 0 1 # [1] 0 0 0 1 1 1 0 0 0 1 ## [1] 1 1 1 1 1 1 0 1 0 0 ## [1] 1 0 0 0 1 0 0 0 0 1 Let's count the number of tails then subtract the expected number of tails. The expected number of tails is half the number of coin flips. for (i in 1: trials) { print (sum(sample(c(0, 1), n, replace=TRUE)) n/2) Y ## [1] 1 ## [1] 1 ## [1] -2 ## [1] O ## [1] 1 ## [1] -2

Answer 1

Solution- R CODE → n = 10 trials = 6 for (i in 1: trials of print (sample (c(0,1),n, sieplace TRUE)) } for (in 1 trials ) { q [1] - Sum (Sample (C(0,1),n replace = TRUE 1) - 1/2 } sd (9) R OUT PUT → >n = 10 > trials 6 - for (int trials) { + point (sample CC (0,1), n, replace = TRUE )) + ven [1] 1101001101 11 [1] 100011 [!1001111000 [0] 0101101110 1010001 1100101100 OOO [!] ]

> for (i in 1: trials ) { + q [i] = sum ( sample (C(0,1),n replace = TRUE 1)-612 + S > sd (q [1] ! 94079 1.94079 interpretation - The required so is I on the defference between the olet tatal number anel espected total is given to be 1.94079 og tails