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 Heads) Suppose we flip 10 coins and compute the fraction of tails (with 1 = Tails and 0 = = 10 trials = 6 for (i in 1: trials) { print (sum(sample(c(0, 1), n, replace-TRUE)) /n) ## [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: n = 10 trials - 6 for (i in 1: trials) { print (sum(sample(c(0, 1), n, replace=TRUE)) /n - 0.5) 3 [1] -0.2 ## (1) -0.2 ## [1] -0.1 ## [1] -0.2 *# (1) 0 ## [1] 0.1 Based on the calculations, approximately what is the on the difference between the observed fraction of tails and the expected fraction (0.5)? In other words, approximately what is the SD of the list of numbers shown above? Include your answer here.

Answer 1

Solution:

We are given the data on difference between the observed fraction and the expected fraction of tail, for 6 trials from a sample of size 10.

Using R, standard deviation is computed as:

> fraction = c(-0.2,-0.2,-0.1,-0.2,0,0.1)

 # Computing standard deviation usin function sd() > standrd_deviation = sd(fraction) > standrd_deviation [1] 0.1264911 # Computing standard deviation by variance, > variance = var(fraction) > variance [1] 0.016 > sqrt(variance) [1] 0.1264911

Hence standard deviation of the given data is 0.1265