Question

CASE STUDY Aces Wild on the Sixth at Oak Hill A most amazing event occurred during the second round of the 1989 U.S. Open at Oak Hill in Pittsford, New York. Four golfers- Doug Weaver, Mark Wiebe, Jerry Pate, and Nick Price made holes in one on the sixth hole. What are the chances for the occurrence of such a remarkable event? On the next day, a Boston Globe article (copyright 1989) discussed the event in detail. To quote the article. "...for perspective, consider this: This is the 89th U.S. Open, and through the thousands and thousands and thousands of rounds played in the previous 88, there had been only 17 holes in one. Yet on this dark Friday morning, there were four holes in one on the same hole in less than two hours. Four times into a cup 4 inches in diameter from 160 yards away The article also reported odds estimates obtained from several different sources. These estimates varied considerably, from about 1 in 10 million to 1 in 1,890,000,000,000,000 to 1 in 8.7 million to 1 in 332,000 After you have completed this chapter, you will be able to compute the odds for yourself. According to experts, the probability of a professional golfer making a holes L. 3709_N a) if there are 155 golfers at the event, would making a hole in one be considered a binomial event? Remember that there are multiple requirements, each of which must be verified. nonnial b) The description above stated that 4 holes in one occurred at the 89 .5. Open. What is the probablity of that happening? N-155 P( x4) 3390 .0002 = .9998 0 What is the probablity of that at most 1 golfer gets a hole in one? c) Assuming the same hole-in-one probability in 83, determine the mean and standard deviation of the number of hole in ones expected with 10,000 golfers.

Answer 1

3. (a) Since each trials are independent, n each trial there are two outcomes (making a hole or does not making a hole), probability of making a hole for each gol fer is fixed i.e. - , so it is a binomial event. Ans : Yes. Let X = no. of holes are occurred ~ Binomial(155, 1/3709) P(X = 4) = (155) (1/3709)*(1 – 1/3709)51 20. 155 (c) P(X < 1) = (1/3709)'(1 – 1/3709) 155–1 = 0.9992 I=0 4. Mean = 10000 * (1/3709) = 2.6961 SD = 10000 * (1/3709) *(1-1/3709) = 1.6418