Please Help. I do not know what to do here...... Q: A coin is tosssed 100 times.
ID: 2950683 • Letter: P
Question
Please Help. I do not know what to do here...... Q: A coin is tosssed 100 times. Find numbers a and bthat are such that the number of heads to appear will be between aand b at least 89% of the time. Answers to choose from: a=35 & b=65, a=65 & b=35, a=55& b=45, a= 45 & b=55. Please explain how to come to the correctsolution. Please Help. I do not know what to do here...... Q: A coin is tosssed 100 times. Find numbers a and bthat are such that the number of heads to appear will be between aand b at least 89% of the time. Answers to choose from: a=35 & b=65, a=65 & b=35, a=55& b=45, a= 45 & b=55. Please explain how to come to the correctsolution.Explanation / Answer
Let's apply the central limit theorem here. The CLT basicallystates that given enough trials, certain distributions approximatea normal distribution. Here, since you have 100 trials (afairly large number) and a symmetric distribution (assuming a faircoin), then this is a classic case in which you can apply it. Let X = the number of heads flipped. sd(X) = sqrt(n*p*(1-p)) = sqrt(100 * .5 * .5) = sqrt(25) = 5 E[X] = 100 * .5 = 50 With those values calculated/defined, we can now calculate z scoresand use the ever-so-powerful z table. z = (X - 50) / 5 z(35) = (35 - 50) / 5 = -3 --> p = .9987 z(65) = (65 - 50) / 5 = 3 --> p = .0013 Thus, the probability that the number of the number of heads isbetween 35 and 65 is .9987 - .0013 = .9974 (very high!). z(45) = (45 - 50) / 5 = -1 --> p = .8413 z(55) = (55 - 50) / 5 = 1 --> p = .1587 Thus, the probability that the number of heads flipped is between45 and 55 is .8413 - .1587 = .6826 Since only the first one is greater than or equal to 89%, I'dlikely go with a = 35 and b = 65. Note: The z scores were whole numbers in this problem. Statisticians memorize the area between + or - 1, + or - 2, and +or - 3 as it provides a good benchmark to compare mentally. This rule is known as the 68, 85, 99.7 rule as that's how much ofthe distribution they cover. This problem demonstrated it for+ or - 1 and + or - 3.