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In sports betting, Las Vegas sports books establish winning margins for a team that is favored to win a game. An individual can place a wager on the game and will win if the team bet upon wins after accounting for the spread. For example, if Team \(\mathrm{A}\) is favored by 5 points, and wins the game by 7 points, then a bet on Team \(A\) is a winning bet. However, if Team A wins the game by only 3 points, then a bet on Team \(A\) is a losing bet. In games where a team is favored by 12 or fewer points, the margin of victory for the favored team relative to the spread is approximately normally distributed with a mean of 0 points and a standard deviation of 10.9 points. Source: Justin Wolfers, "Point Shaving: Corruption in NCAA Basketball" (a) Explain the meaning of "the margin of victory relative to the spread has a mean of 0 points." Does this imply that the spreads are accurate for games in which a team is favored by 12 or fewer points? (b) In games where a team is favored by 12 or fewer points, what is the probability that the favored team wins by 5 or more points relative to the spread? (c) In games where a team is favored by 12 or fewer points, what is the probability that the favored team loses by 2 or more points relative to the spread?

Step by step solution

01

Understanding the Mean and Standard Deviation

The margin of victory relative to the spread is approximately normally distributed with a mean of 0 points and a standard deviation of 10.9 points. This means that, on average, the favored team's actual margin of victory is centered around the spread, indicating that the spreads are generally accurate. The standard deviation of 10.9 points suggests that there is variability in the actual margins of victory.

02

Calculating the Probability for Part (b)

To find the probability that the favored team wins by 5 or more points relative to the spread, we convert the value into a z-score. The z-score is calculated as \[ Z = \frac{x - \mu}{\sigma}, \] where \( x = 5 \), \( \mu = 0 \), and \( \sigma = 10.9 \). So, \[ Z = \frac{5 - 0}{10.9} \approx 0.46. \] Using standard normal distribution tables or a calculator, we find the probability corresponding to a z-score of 0.46. This gives us the probability of \( P(Z \geq 0.46) \), which is approximately 0.3228.

03

Calculating the Probability for Part (c)

To find the probability that the favored team loses by 2 or more points relative to the spread, we calculate the z-score: \[ Z = \frac{x - \mu}{\sigma}, \] where \( x = -2 \), \( \mu = 0 \), and \( \sigma = 10.9 \). So, \[ Z = \frac{-2 - 0}{10.9} \approx -0.18. \] Using standard normal distribution tables or a calculator, we find the probability corresponding to a z-score of -0.18. This gives us the probability of \( P(Z \leq -0.18) \), which is approximately 0.4286.

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Key Concepts

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

mean and standard deviation

In sports betting, especially when looking at games where a team is favored by 12 or fewer points, it's crucial to understand the concepts of mean and standard deviation. The mean, in this context, is the average margin of victory relative to the spread, which is given as 0 points.

This implies that, on average, the actual margin of victory aligns closely with the point spread established by Las Vegas sports books. In other words, the spread is generally accurate.

The standard deviation, which is 10.9 points in this case, measures the variability or spread of the actual margins of victory around the mean. A higher standard deviation would indicate more variability, while a lower standard deviation would mean the actual margins are closer to the mean. Understanding these two concepts helps in interpreting and making sense of the betting lines and the probable outcomes of the games.

z-score calculation

A z-score helps us understand how far a particular value is from the mean, in terms of the standard deviation. For sports betting, it can tell us how likely it is that a team will beat the spread by a certain margin.

The formula to calculate a z-score is:

\[ Z = \frac{x - \text{mean}}{\text{standard deviation}} \]

Let's apply this formula to an example: If we want to find the probability that the favored team wins by 5 or more points relative to the spread, we set the value of x to 5. Our mean (\text{mu}) is 0, and our standard deviation (\text{sigma}) is 10.9 points.

So, the calculation becomes:

\[ Z = \frac{5 - 0}{10.9} \approx 0.46 \]

Similarly, if we want to find out how likely it is for the favored team to lose by 2 or more points relative to the spread, we set x to -2.

The calculation is:


\[ Z = \frac{-2 - 0}{10.9} \approx -0.18 \]

These z-scores can then be used to find probabilities from standard normal distribution tables or calculators.

probability

Probability in the context of a normal distribution helps us determine the likelihood of certain outcomes. After calculating the z-score, we can use the standard normal distribution to find the probabilities.

For the z-score of 0.46, which represents a team winning by 5 or more points, we can look up this z-score in a standard normal distribution table or use a calculator. This z-score corresponds to a probability of approximately 0.6772 to the left of the z-score. However, we need the probability to the right, which is 1 - 0.6772 = 0.3228.

This means there is about a 32.28% chance that the team will win by 5 or more points.

For the z-score of -0.18, which represents a team losing by 2 or more points, the corresponding probability to the left of this z-score in the standard normal distribution table is roughly 0.4286.

This indicates that there is about a 42.86% chance that the team will lose by 2 or more points.

These probabilities help bettors gauge the risks and potential outcomes, thus making more informed betting choices.