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 n 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), n, replace=TRUE)) ## [1] 0 1 0 1 1 0 0 0 1 1 ## [1] 1 0 1 0 0 1 0 1 1 1 ## [1] 0 0 1 0 1 1 0 0 0 1 ## (1) 0001 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) ## [1] 1 ## [1] 1 ## [1] -2 ## (1) 0 ## [1] 1 ## [1] -2 Based on the calculations, approximately what is the on the difference between the total number of tails and the expected total? (Remember: 95% of the area under a normal curve is +2SD.) In other words, approximately what is the SD of the list of numbers shown above?

Answer 1

R CODE: n = 10 trials = 6 for (i in 1: trials) { print(sample(c(0,1),n, replace=TRUE)) } for (i in 1: trials) { q[i]=sum( sample(c(0,1),n, replace=TRUE))- n/2 } sd(q) ROUTPUT: >n= 10 > trials = 6 for (i in 1: trials) { + print(sample(c(0,1),n,replace=TRUE)) +} [1]1101001101 [1]1000111111 [1]1001111000 [1]0101101110 [1100010101 01 [1]1100101100 > for (i in 1: trials) { +q[i]=sum(sample(c(0,1),n, replace=TRUE))- n/2 + } >sd(a) [1]1.94079 Interpretation: The required SD is 1.94079 i.e.the on the difference between the total number of tails and the expected total is given to be 1.94079

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