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Two Wharton professors analyzed 1,613,234 putts by golfers on the Professional Golfers Association (PGA) Tour and found that 983,764 of the putts were made and 629,470 of the putts were missed. Further analysis showed that for putts that were made, \(64.0 \%\) of the time the player was attempting to make a par putt and \(18.8 \%\) of the time the player was attempting to make a birdie putt. And, for putts that were missed, \(20.3 \%\) of the time the player was attempting to make a par putt and \(73.4 \%\) of the time the player was attempting to make a birdie putt (Is Tiger Woods Loss Averse? Persistent Bias in the Face of Experience, Competition, and High Stakes, D.G. Pope and M. E. Schweitzer, June 2009, The Wharton School, University of Pennsylvania). a. What is the probability that a PGA Tour player makes a putt? b. Suppose that a PGA Tour player has a putt for par. What is the probability that the player will make the putt? c. Suppose that a PGA Tour player has a putt for birdie. What is the probability that the player will make the putt? d. Comment on the differences in the probabilities computed in parts (b) and (c).

Short Answer

Expert verified
The probability of making a putt is about 0.6095. Par putts are made about 83% of the time, while birdie putts are made about 29% of the time. Players are more successful with par putts than birdie putts.

Step by step solution

01

Calculate total number of putts

The total number of putts attempted is given in the problem. This is simply the sum of the made putts and the missed putts:\[\text{Total Putts} = 983,764 + 629,470 = 1,613,234\]
02

Determine probability of making a putt

The probability that a PGA Tour player makes a putt is the ratio of made putts to total putts:\[P(\text{Make}) = \frac{983,764}{1,613,234} \approx 0.6095\]
03

Determine probability of making a par putt

We know that 64.0% of the made putts were par putts. We compute the probability that a player makes a par putt by calculating the fraction of made par putts and using the total number of par putts attempted. Let the total number of par putts attempted be \(x\).Par Putts Made:\[P(\text{Par | Made}) = 0.64 \implies \text{Par Putts Made} = 0.64 \times 983,764 \approx 629,609.36\]We also know that 20.3% of the missed putts were par:Par Putts Missed:\[P(\text{Par | Missed}) = 0.203 \implies \text{Par Putts Missed} = 0.203 \times 629,470 \approx 127,779.41\]The total number of par putts is:\[\text{Total Par Putts} = 629,609.36 + 127,779.41 \approx 757,388.77\]Probability of making a par putt:\[P(\text{Make | Par}) = \frac{629,609.36}{757,388.77} \approx 0.8312\]
04

Determine probability of making a birdie putt

We know that 18.8% of the made putts were birdie putts.Birdie Putts Made:\[P(\text{Birdie | Made}) = 0.188 \implies \text{Birdie Putts Made} = 0.188 \times 983,764 \approx 184,942.43\]We also know that 73.4% of the missed putts were birdie:Birdie Putts Missed:\[P(\text{Birdie | Missed}) = 0.734 \implies \text{Birdie Putts Missed} = 0.734 \times 629,470 \approx 461,064.98\]The total number of birdie putts is:\[\text{Total Birdie Putts} = 184,942.43 + 461,064.98 \approx 646,007.41\]Probability of making a birdie putt:\[P(\text{Make | Birdie}) = \frac{184,942.43}{646,007.41} \approx 0.2863\]
05

Compare probabilities for par and birdie

The probability of making a par putt (approximately 0.8312) is much higher than the probability of making a birdie putt (approximately 0.2863). This indicates players are more successful with par putts than with birdie putts, possibly due to the difference in the putt's difficulty or pressure conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
In golf statistics, understanding conditional probability allows us to assess the likelihood of one event occurring, given another event has already occurred. For example, calculating the probability of a golfer making a par putt, given they have already attempted the putt, involves conditional probability. This is calculated by comparing the number of successful par putts made to the total number of par putts attempted, using the formula: \[ P( ext{Make | Par}) = \frac{\text{Par Putts Made}}{\text{Total Par Putts}} \] By knowing both the number of attempts and the success rate for specific types of putts, like par or birdie putts, we can derive more detailed insights into player performance based on varying conditions.
Statistical Analysis in Sports
Statistical analysis is crucial in understanding performance patterns in sports, including golf. By analyzing a massive dataset of over 1.6 million putts, we can identify trends and outcomes based on various scenarios. In the provided exercise, statistical analysis is used to determine how frequently golfers succeed in putts overall, and specifically in par and birdie putts. The analysis shows that successful putts are spread differently across par and birdie attempts, providing insights into players' performance under pressure. This type of analysis aids in - recognizing strengths and weaknesses in players - aiding coaches to tailor training programs - enabling players to maximize their performance.
Data Interpretation
Effective data interpretation is essential for turning mere numbers into actionable insights. With the data from the golf exercise, we can draw conclusions about the patterns of success and failure in putts. For instance, the data shows - 60.95% of all putts are successful - Par putts have a higher success rate compared to birdie putts Interpreting the data involves understanding these percentages and what they imply. The higher success rate of par putts compared to birdies might suggest that either par putts are easier, or that players feel less pressure with them. These interpretations help inform changes or adaptations in techniques and strategies.
Experimental Analysis
Experimental analysis involves using the gathered data to test hypotheses about player behavior and game dynamics in golf. In this scenario, the vast sample size allows for significant experimental observations about player tendencies. For example, - we assess that players are more likely to miss birdie putts compared to par putts. This could lead to an exploration of why players are less successful with birdies—perhaps it involves factors such as distance, angle, or psychological pressure. - Identifying specific areas for improvement can lead to initiatives to increase putt success rates in competitive scenarios. By experimenting with different variables, like altering training focus or psychological preparedness, players can aim to overcome identified weaknesses.

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Most popular questions from this chapter

The American Council of Education reported that \(47 \%\) of college freshmen earn a degree and graduate within five years. Assume that graduation records show women make up \(50 \%\) of the students who graduated within five years, but only \(45 \%\) of the students who did not graduate within five years. The students who had not graduated within five years either dropped out or were still working on their degrees. a. \(\quad\) Let \(A_{1}=\) the student graduated within five years \(A_{2}=\) the student did not graduate within five years \(W=\) the student is a female student Using the given information, what are the values for \(P\left(A_{1}\right), P\left(A_{2}\right), P\left(W | A_{1}\right),\) and \\[ P\left(W | A_{2}\right) ? \\] b. What is the probability that a female student will graduate within five years? c. What is the probability that a male student will graduate within five years? d. Given the preceding results, what are the percentage of women and the percentage of men in the entering freshman class?

High school seniors with strong academic records apply to the nation's most selective colleges in greater numbers each year. Because the number of slots remains relatively stable, some colleges reject more early applicants. Suppose that for a recent admissions class, an Ivy League college received 2851 applications for early admission. Of this group, it admitted 1033 students early, rejected 854 outright, and deferred 964 to the regular admission pool for further consideration. In the past, this school has admitted \(18 \%\) of the deferred early admission applicants during the regular admission process. Counting the students admitted early and the students admitted during the regular admission process, the total class size was \(2375 .\) Let \(E, R,\) and \(D\) represent the events that a student who applies for early admission is admitted early, rejected outright, or deferred to the regular admissions pool. a. Use the data to estimate \(P(E), P(R),\) and \(P(D)\). b. Are events \(E\) and \(D\) mutually exclusive? Find \(P(E \cap D)\). c. For the 2375 students who were admitted, what is the probability that a randomly selected student was accepted during early admission? d. Suppose a student applies for early admission. What is the probability that the student will be admitted for early admission or be deferred and later admitted during the regular admission process?

Suppose that we have two events, \(A\) and \(B,\) with \(P(A)=.50, P(B)=.60,\) and \(P(A \cap B)=.40\). a. \(\quad\) Find \(P(A | B)\). b. Find \(P(B | A)\). c. Are \(A\) and \(B\) independent? Why or why not?

Consider the experiment of tossing a coin three times. a. Develop a tree diagram for the experiment. b. List the experimental outcomes. c. What is the probability for each experimental outcome?

An experiment has four equally likely outcomes: \(E_{1}, E_{2}, E_{3},\) and \(E_{4}\). a. What is the probability that \(E_{2}\) occurs? b. What is the probability that any two of the outcomes occur (e.g., \(E_{1}\) or \(E_{3}\) )? c. What is the probability that any three of the outcomes occur (e.g., \(E_{1}\) or \(E_{2}\) or \(E_{4}\) )?

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