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San Francisco Giants hitting The table shows the probability distribution of the number of bases for a randomly selected time at bat for a San Francisco Giants player in 2010 (excluding times when the player got on base because of a walk or being hit by a pitch). In \(74.29 \%\) of the at-bats the player was out, \(17.04 \%\) of the time the player got a single (one base), \(5.17 \%\) of the time the player got a double (two bases), \(0.55 \%\) of the time the player got a triple, and \(2.95 \%\) of the time the player got a home run. a. Verify that the probabilities give a legitimate probability distribution. b. Find the mean of this probability distribution. c. Interpret the mean, explaining why it does not have to be a whole number, even though each possible value for the number of bases is a whole number. $$ \begin{array}{cc} \hline {\text { San Francisco Giants Hitting }} \\ \hline \text { Number of Bases } & \text { Probability } \\ \hline 0 & 0.7429 \\ 1 & 0.1704 \\ 2 & 0.0517 \\ 3 & 0.0055 \\ 4 & 0.0295 \\ \hline \end{array} $$

Step by step solution

01

Verify the Sum of Probabilities

To ensure the probabilities form a legitimate probability distribution, their sum must equal 1. Let's add them up: \( 0.7429 + 0.1704 + 0.0517 + 0.0055 + 0.0295 = 1.0000 \). The sum is 1, confirming it constitutes a valid probability distribution.

02

Calculate the Mean (Expected Value)

The mean of a probability distribution is calculated as the sum of each value (number of bases) times its probability. The formula is: \( E(X) = \sum (x_i \cdot p_i)\), where \(x\) is the number of bases and \(p\) is the probability. For these values: \(0 \cdot 0.7429 + 1 \cdot 0.1704 + 2 \cdot 0.0517 + 3 \cdot 0.0055 + 4 \cdot 0.0295 = 0 + 0.1704 + 0.1034 + 0.0165 + 0.1180 = 0.4083 \). Thus, the mean is approximately 0.4083 bases.

03

Interpret the Mean

The mean of approximately 0.4083 bases represents the average number of bases a player earns per at-bat. This is a statistical average and not necessarily one of the integer-valued outcomes (0, 1, 2, 3, 4). It reflects the weighted average across all at-bats, considering the probabilities of each outcome.

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Key Concepts

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

Mean of Probability Distribution

When looking at the mean of a probability distribution, we calculate the central tendency of the outcomes, accounting for their likelihood of occurring.
In this case, for the San Francisco Giants player's at-bats, we are determining the average number of bases acquired per at-bat. The mean is found using the formula: \[E(X) = \sum (x_i \cdot p_i)\] where \(x_i\) represents the number of bases, and \(p_i\) the probability of obtaining that many bases.
By multiplying each outcome by its probability and then summing these products, we achieve an expected value—in this instance, approximately 0.4083 bases. This value is vital for players and analysts as it reflects overarching performance trends that arise due to varied at-bat results.

Expected Value

Expected value is a crucial concept in probability and statistics. It provides a measure that summarizes all possible values of a random variable, weighted by their probabilities.
The expected value can also be seen as the theoretical mean of a large number of trials of a stochastic process.
For the Giants player example, the expected value of 0.4083 bases per at-bat implies that over a large number of bats, the player would average about 0.4083 bases per time.
While an individual at-bat can result in whole numbers of bases (0, 1, 2, 3, or 4), the expected value smooths these spikes over many trials, providing an essential tool for assessing performance.
It can also guide decisions when combined with other strategic considerations, such as adjustments in batting order or improvements in training.

Interpretation of Mean

Understanding the interpretation of the mean gives us insight into what the calculated value really signifies.
While the expected result isn't always an obtainable outcome in a single event, it represents the long-term average if the experiment is repeated many times.
In the context of Giants hitting statistics, the mean of 0.4083 helps illustrate the general trend: even though specific instances of batting could result in the player getting 0, 1, 2, 3, or 4 bases, the average expected performance captures the player's consistency and effectiveness over time.
It is essential to remind students that while decimals show up in these calculations, each actual at-bat results in a whole number of bases; the mean represents the broader outcome over many attempts. This difference highlights the power behind statistical averages in delivering strategic insights beyond surface-level results.