Chapter 5, Problem 93DE

Refer to the Baseball 2012 data, which reports information on the 30 Major League Baseball teams for the 2012 season. Set up three variables:

• Divide the teams into two groups, those that had a winning season and those that did not. That is, create a variable to count the teams that won 81 games or more, and those that won 80 or less.
• Create a new variable for attendance, using three categories: attendance less than 2.0 million, attendance of 2.0 million up to 3.0 million, and attendance of 3.0 million or more.
• Create a variable that shows the teams that play in a stadium less than 15 years old versus one that is 15 years old or more.

Answer the following questions.

1. a. Create a table that shows the number of teams with a winning season versus those with a losing season by the three categories of attendance. If a team is selected at random, compute the following probabilities:
2. 1. The team had a winning season.
3. 2. The team had a winning season or attendance of more than 3.0 million.
4. 3. The team had a winning season given attendance was more than 3.0 million.
5. 4. The team has a winning season and attracted fewer than 2.0 million fans.
6. b. Create a table that shows the number of teams with a winning season versus those that play in new or old stadiums. If a team is selected at random, compute the following probabilities:
   1. 1. Selecting a team with a winning season.
   2. 2. The likelihood of selecting a team with a winning record and playing in a new stadium.
   3. 3. The team had a winning record or played in a new stadium.

To determine

Construct a table showing the number of teams with a winning season versus those with a losing season by the three categories of attendance.

1. Find the probability that the team had a winning season.

2. Find the probability that the team had a winning season or attendance of more than 3.0 million.

3. Find the probability that the team had a winning season given attendance was more than 3.0 million.

4. Find the probability that the team has a winning season and attracted fewer than 2.0 million attendance.

Answer to Problem 93DE

1. The probability that the team had a winning season is 0.5667.

2. The probability that the team had a winning season or attendance of more than 3.0 million is 0.5333.

3. The probability that the team had a winning season given attendance was more than 3.0 million is 0.7778.

4. The probability that the team has a winning season and attracted fewer than 2.0 million attendance is 0.1.

Explanation of Solution

Wins of the team are classified into two categories, namely, winning season (81 or more) and not winning season (less than 81); Attendance with three categories: low (less than 2.0 million), moderate (2.0 million to 3.0 million), and high (more than 3.0 million).

The relationship between the number of teams with a winning season versus those with a losing season by the three categories of attendance is shown in the following table:

		Attendance			Total
		Low	Moderate	High
Winning season	No	4	7	2	13
	Yes	3	7	7	17
Total		7	14	9	30

1.

The probability that the team had a winning season is calculated as follows:

$\begin{array}{c}P\left(\text{Winning season}\right)=\frac{\text{Number of teams having winning season}}{\text{Total number teams}}\\ =\frac{17}{30}\\ =0.5667\end{array}$

Thus, the probability that the team had a winning season is 0.5667.

2.

The probability that the team had a winning season or attendance of more than 3.0 million is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Winning season or}\\ \text{high attendance}\end{array}\right)=\left\{\begin{array}{l}\frac{\left(\begin{array}{l}\text{Number of teams having }\\ \text{winning season}\end{array}\right)}{\text{Total number teams}}+\frac{\left(\begin{array}{l}\text{Number of teams having }\\ \text{high attendance}\end{array}\right)}{\text{Total number teams}}\\ -\frac{\left(\begin{array}{l}\text{Number of teams having}\\ \text{winning season and }\\ \text{high attendance}\end{array}\right)}{\text{Total number teams}}\end{array}\right\}\\ =\frac{17}{30}+\frac{9}{30}-\frac{7}{30}\\ =0.5333\end{array}$

Thus, the probability that the team had a winning season or attendance of more than 3.0 million is 0.5333.

3.

The probability that the team had a winning season given attendance was more than 3.0 million is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Winning season given}\\ \text{high attendance}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of teams having winning season }\\ \text{and high attendance}\end{array}\right)}{\text{Number of teams having high attendance}}\\ =\frac{7}{9}\\ =0.7778\end{array}$

Thus, the probability that the team had a winning season given attendance was more than 3.0 million is 0.7778.

4. The probability that the team has a winning season and attracted fewer than 2.0 million attendance is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Winning season and}\\ \text{low attendance}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of teams having }\\ \text{winning season and low attendace}\end{array}\right)}{\text{Total number of teams}}\\ =\frac{3}{30}\\ =0.1\end{array}$

Thus, the probability that the team has a winning season and attracted fewer than 2.0 million attendance is 0.1.

To determine

Construct a table showing the number of teams with a winning season versus those that play in new or old stadiums.

1. Find the probability to select a team with winning season.

2. Find the probability of selecting a team with a winning record and playing in a new stadium.

3. Find the probability that the team had a winning record or played in a new stadium.

Answer to Problem 93DE

1. The probability of selecting a team with winning season is 0.5667.

2. The probability of selecting a team with a winning record and playing in a new stadium is 0.2667.

3. The probability that the team had a winning record or played in a new stadium is 0.8.

Explanation of Solution

Wins of the team are classified into two categories, namely, winning season (81 or more) and not winning season (less than 81); stadium with two categories: new (less than 15 years old) and old (15 years old or more).

The relationship between the number of teams with a winning season versus those that play in new or old stadiums is shown in the following table:

	Season		Total
	Winning	Not winning
New stadium	8	7	15
Old stadium	9	6	15
Total	17	13	30

1.

The probability of selecting a team with winning season is calculated as follows:

$\begin{array}{c}P\left(\text{Winning season}\right)=\frac{\text{Number of teams having winning season}}{\text{Total number teams}}\\ =\frac{17}{30}\\ =0.5667\end{array}$

Thus, the probability that the team had a winning season is 0.5667.

2.

The probability of selecting a team with a winning record and playing in a new stadium is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Winning season and}\\ \text{playing in new stadium}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Winning season and}\\ \text{playing in new stadium}\end{array}\right)}{\text{Total number of teams}}\\ =\frac{8}{30}\\ =0.2667\end{array}$

Thus, the probability of selecting a team with a winning record and playing in a new stadium is 0.2667.

3.

The probability that the team had a winning record or played in a new stadium is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Winning season or}\\ \text{played in new stadium}\end{array}\right)=\left\{\begin{array}{c}\frac{\text{winning season}}{\text{Total number of teams}}+\frac{\text{playing in new stadium}}{\text{Total number of teams}}-\\ \frac{\left(\begin{array}{l}\text{winning season and}\\ \text{playing in new stadium}\end{array}\right)}{\text{Total number of teams}}\end{array}\right\}\\ =\frac{17}{30}+\frac{15}{30}-\frac{8}{30}\\ =0.8\end{array}$

Thus, the probability that the team had a winning record or played in a new stadium is 0.8.