Jane Blaylock joined the Ladies Professional Golf Association (LPGA) in 1969 and by 1972 had become the leading money winner on the tour. During the Bluegrass Invitational Tournament, she was disqualified for an alleged rules infraction. The LPGA appointed a committee of her competitors, who suspended her from the next tournament, the Carling Open. Blaylock sued for damages and expenses under the Sherman Antitrust Act, which says that individuals cannot be prevented by their peers from working in their profession because it would lessen competition. She won but then had to come up with a method of determining how much money she might reasonably have made if she had been allowed to play in the next tournament. This is a difficult issue in most antitrust cases but was particularly problematic for a professional golfer, who might play well one day and poorly another day. Her task was challenging. She would have to use a measure that the judge and the jury would understand and that would be sufficiently convincing for a ruling in her favor. She and her legal team used a statistical procedure called the expected value, which we will study in this chapter. Using data from the nine most recent tournaments that Blaylock played in prior to the disqualification, they estimated the probability that she would achieve various scores based on her past performance. The scores for players who won money ranged from 209 (the tournament winner) to 232. To simplify things, the 24 possible scores were reduced to 8, where 210, for example, represents 209, 210, and 211. Here is a table that summarizes her possible outcomes and the probabilities calculated for each of the outcomes:

Question

Jane Blaylock joined the Ladies Professional Golf Association (LPGA) in 1969 and by 1972 had become the leading money winner on the tour. During the Bluegrass Invitational Tournament, she was disqualified for an alleged rules infraction. The LPGA appointed a committee of her competitors, who suspended her from the next tournament, the Carling Open. Blaylock sued for damages and expenses under the Sherman Antitrust Act, which says that individuals cannot be prevented by their peers from working in their profession because it would lessen competition. She won but then had to come up with a method of determining how much money she might reasonably have made if she had been allowed to play in the next tournament. This is a difficult issue in most antitrust cases but was particularly problematic for a professional golfer, who might play well one day and poorly another day. Her task was challenging. She would have to use a measure that the judge and the jury would understand and that would be sufficiently convincing for a ruling in her favor. She and her legal team used a statistical procedure called the expected value, which we will study in this chapter. Using data from the nine most recent tournaments that Blaylock played in prior to the disqualification, they estimated the probability that she would achieve various scores based on her past performance. The scores for players who won money ranged from 209 (the tournament winner) to 232. To simplify things, the 24 possible scores were reduced to 8, where 210, for example, represents 209, 210, and 211. Here is a table that summarizes her possible outcomes and the probabilities calculated for each of the outcomes:

Possible Outcomes: 210 213 216 219 222 225 228 231
Probabilities: 0.07 0.16 0.23 0.24 0.17 0.09 0.03 0.01

(outcomes match up with probabilities beneath them. they won't show lined up. 210 with 0.07, 213 with 0.16, and so on)

The probability is 0.07 that she would score 209, 210, or 211; and so forth. Using these numbers, her expected score was calculated to be approximately 218. Had she played in the tournament, her 218 would have earned her $1427.50. Not only was the jury persuaded, but they also believed that Blaylock might well have won the tournament, so they awarded her firstplace money, $4500. This amount was then tripled to $13,500 to cover legal expenses, according to the provisions of the Sherman act. Statistics to the rescue.

1. Identify the random variable of interest.
2. is this a valid probability model? support your answer.
3. verify that the probability distribution table yields an expected score of about 218.
4. determine the variance and standard deviation of X.
5. what is the probability that Baylock's score would be 218 or less? no more than 220? Between 209 and 218, inclusive?

Answer 1

The random variable is her score.

Adding up the probabilities is one way to check if it's a valid model.
Here's an example of calculating Expected Value for roll of a die:
E[X] = 1(1/6) + 2(1/6) + ... + 6(1/6) = 3 1/2
You can use a similar process on your data.
Use the Random Variable formulae for Var and σ.

Answer 2

1. The random variable of interest in this scenario is Jane Blaylock's golf score in the tournament.

2. Yes, this is a valid probability model. The probabilities given in the table sum up to 1, which is a requirement for a probability distribution to be valid. Additionally, the probabilities assigned to each possible outcome represent the likelihood of Jane Blaylock achieving those scores based on her past performance.

3. To verify the expected score of about 218, we can multiply each outcome by its corresponding probability and sum them up.

Expected Score = (210 * 0.07) + (213 * 0.16) + (216 * 0.23) + (219 * 0.24) + (222 * 0.17) + (225 * 0.09) + (228 * 0.03) + (231 * 0.01)

Expected Score = 14.7 + 34.08 + 49.68 + 52.56 + 37.74 + 20.25 + 6.84 + 2.31

Expected Score ≈ 218

Therefore, the probability distribution table yields an expected score of about 218.

4. The variance of a random variable can be calculated using the formula:

Variance (σ^2) = Σ[(x - μ)^2 * P(x)]

where x represents each outcome, μ represents the expected value, and P(x) represents the probability of each outcome.

To determine the variance of X, we can calculate it using the probability distribution table and the expected score (μ = 218):

Variance (σ^2) = [(210 - 218)^2 * 0.07] + [(213 - 218)^2 * 0.16] + [(216 - 218)^2 * 0.23] + [(219 - 218)^2 * 0.24] + [(222 - 218)^2 * 0.17] + [(225 - 218)^2 * 0.09] + [(228 - 218)^2 * 0.03] + [(231 - 218)^2 * 0.01]

Variance (σ^2) ≈ 18.06

The standard deviation can be obtained by taking the square root of the variance:

Standard Deviation (σ) ≈ √18.06 ≈ 4.25

Therefore, the variance of X is approximately 18.06 and the standard deviation is approximately 4.25.

5. To find the probability that Blaylock's score would be 218 or less, we need to sum up the probabilities of all outcomes with scores less than or equal to 218:

Probability (X ≤ 218) = Probability (X = 210) + Probability (X = 213) + Probability (X = 216) + Probability (X = 219)

Probability (X ≤ 218) = 0.07 + 0.16 + 0.23 + 0.24 ≈ 0.70

Therefore, the probability that Blaylock's score would be 218 or less is approximately 0.70.

To find the probability that Blaylock's score would be no more than 220, we need to sum up the probabilities of all outcomes with scores less than or equal to 220:

Probability (X ≤ 220) = Probability (X = 210) + Probability (X = 213) + Probability (X = 216) + Probability (X = 219) + Probability (X = 222)

Probability (X ≤ 220) = 0.07 + 0.16 + 0.23 + 0.24 + 0.17 ≈ 0.87

Therefore, the probability that Blaylock's score would be no more than 220 is approximately 0.87.

To find the probability that Blaylock's score would be between 209 and 218 (inclusive), we need to sum up the probabilities of all outcomes within that range:

Probability (209 ≤ X ≤ 218) = Probability (X = 210) + Probability (X = 213) + Probability (X = 216) + Probability (X = 219)

Probability (209 ≤ X ≤ 218) = 0.07 + 0.16 + 0.23 + 0.24 ≈ 0.70

Therefore, the probability that Blaylock's score would be between 209 and 218 (inclusive) is approximately 0.70.