27. This problem is about probabilities and coin flipping. The goal is to compare the probability of getting 6 tails or greater when flipping a coin 10 times, versus the probability of getting 60 tails or greater when flipping a coin 100 times. In each case, the proportion of tails is the same, 2 60%, but the probabilities are different. Assume that we have a fair coin, so the odds of getting tails on any one flip are 0.5. a) First, let's calculate the odds of getting exactly 6 tails out of 10 total coin flips. Use the binomial probability formula P(x) = „Cx (p)*(1 – p)"-* to find P(6). b) You don't have to find the other probabilities by hand; they can be found from this piece 10 0 1.010 2 .044 .117 .50 of the binomial chart. Using this information, determine the probability of getting 6 tails or more. That is, find P(x > 6). Hint: it could be 6 or 7 or 8 or 9 or 10. Note that the answer to part (a) is right there next to x = 6, so you can check to see if you got part (a) correct. .001 3 .205 4 5 .246 6 .205 .117 7 8 .044 .010 10 .001 Now we would like to consider n = 100 coin flips and determine P(x 2 60). It is too difficult to add up each of the possibilities by hand from the binomial chart, so instead we will approximate the binomial distribution as a normal distribution. This is valid because n is large enough. 60 c) Using the appropriate z-score formula, determine the z-score for p = 0.60, p = .5, and n = 100. 100