The Western Pennsylvania Minor Football League has three teams: The New Castle Kings; the Meadville Raiders; and the Kittanning Krushers.

Sample Solution

Let's break down this problem step-by-step.

First, we need to calculate the historical win probabilities for each team against each other team based on the past 5 years of data. Since each team plays every other team just once per season, over 5 years, each team has played the other teams 5 times.

Let's denote the teams as:

• NC: New Castle Kings
• MR: Meadville Raiders
• KK: Kittanning Krushers

The provided data is:

• NC beat MR: 3 times
• NC beat KK: 4 times
• MR beat NC: 2 times (Total games between NC and MR is 5, NC won 3, so MR won 5-3=2)
• MR beat KK: 1 time
• KK beat NC: 1 time (Total games between NC and KK is 5, NC won 4, so KK won 5-4=1)
• KK beat MR: 4 times (Total games between MR and KK is 5, MR won 1, so KK won 5-1=4)Let's break down this problem step-by-step.

Let's summarize the probabilities of one team winning against another in a single game:

• P(NC beats MR) = 3/5 = 0.6
• P(MR beats NC) = 2/5 = 0.4
• P(NC beats KK) = 4/5 = 0.8
• P(KK beats NC) = 1/5 = 0.2
• P(MR beats KK) = 1/5 = 0.2
• P(KK beats MR) = 4/5 = 0.8

Now, let's answer the questions:

(a) What is the probability that the Meadville Raiders will win every game next year?

The Meadville Raiders play two games next year: one against the New Castle Kings and one against the Kittanning Krushers. For them to win every game, they must beat NC AND beat KK.

• P(MR beats NC) = 0.4
• P(MR beats KK) = 0.2

Since these are independent events (the outcome of one game does not affect the other), we multiply the probabilities: P(MR wins every game) = P(MR beats NC) * P(MR beats KK) = 0.4 * 0.2 = 0.08

The probability that the Meadville Raiders will win every game next year is 0.08 or 8%.

(b) What is the probability that the Kittanning Krushers will win at least one game next year?

The Kittanning Krushers also play two games next year: one against the New Castle Kings and one against the Meadville Raiders. "At least one game" means they win one game OR they win both games. It's easier to calculate the complement: 1 - P(KK wins zero games). P(KK wins zero games) means KK loses to NC AND KK loses to MR.

• P(KK loses to NC) = P(NC beats KK) = 0.8
• P(KK loses to MR) = P(MR beats KK) = 0.2

P(KK loses both games) = P(KK loses to NC) * P(KK loses to MR) = 0.8 * 0.2 = 0.16

P(KK wins at least one game) = 1 - P(KK wins zero games) = 1 - 0.16 = 0.84

The probability that the Kittanning Krushers will win at least one game next year is 0.84 or 84%.

(c) What is the probability that New Castle Kings will win exactly one game next year?

The New Castle Kings play two games next year: one against the Meadville Raiders and one against the Kittanning Krushers. "Exactly one game" means they win against MR and lose against KK OR they win against KK and lose against MR.

• P(NC beats MR) = 0.6
• P(NC loses to MR) = P(MR beats NC) = 0.4
• P(NC beats KK) = 0.8
• P(NC loses to KK) = P(KK beats NC) = 0.2

Scenario 1: NC wins against MR and loses against KK P(NC wins MR AND NC loses KK) = P(NC beats MR) * P(NC loses to KK) = 0.6 * 0.2 = 0.12

Scenario 2: NC wins against KK and loses against MR P(NC wins KK AND NC loses MR) = P(NC beats KK) * P(NC loses to MR) = 0.8 * 0.4 = 0.32

Since these two scenarios are mutually exclusive, we add their probabilities: P(NC wins exactly one game) = P(Scenario 1) + P(Scenario 2) = 0.12 + 0.32 = 0.44

The probability that New Castle Kings will win exactly one game next year is 0.44 or 44%.

(d) What is the probability that the Meadville Raiders will win fewer than two games next year?

"Fewer than two games" means they win zero games OR they win exactly one game. As calculated in part (a), P(MR wins every game) = 0.08. The maximum number of games MR can win is 2. So, P(MR wins fewer than two games) = 1 - P(MR wins two games) = 1 - P(MR wins every game).

P(MR wins fewer than two games) = 1 - 0.08 = 0.92

Alternatively, let's calculate winning zero games and exactly one game:

P(MR wins zero games) = P(MR loses to NC) * P(MR loses to KK)

• P(MR loses to NC) = P(NC beats MR) = 0.6
• P(MR loses to KK) = P(KK beats MR) = 0.8
• P(MR wins zero games) = 0.6 * 0.8 = 0.48

P(MR wins exactly one game)

• Scenario 1: MR wins against NC and loses against KK
  • P(MR beats NC) * P(MR loses to KK) = 0.4 * 0.8 = 0.32
• Scenario 2: MR wins against KK and loses against NC
  • P(MR beats KK) * P(MR loses to NC) = 0.2 * 0.6 = 0.12
• P(MR wins exactly one game) = 0.32 + 0.12 = 0.44

P(MR wins fewer than two games) = P(MR wins zero games) + P(MR wins exactly one game) = 0.48 + 0.44 = 0.92

The probability that the Meadville Raiders will win fewer than two games next year is 0.92 or 92%.

Three Business Processes that can be described by the Normal Distribution:

The normal distribution, also known as the Gaussian distribution, is a very common continuous probability distribution that is symmetric around its mean, with data near the mean being more frequent in occurrence than data far from the mean. Many natural and social phenomena exhibit this bell-shaped curve. In business, it's particularly useful when dealing with data that clusters around an average, with deviations occurring symmetrically.

Here are three business processes that can be described by the normal distribution:

Customer Service Wait Times:

• Description: In many service-oriented businesses (e.g., call centers, retail checkouts, bank queues), the time a customer spends waiting before being served can often be approximated by a normal distribution. While minimum wait times exist, and very long waits are less frequent, the bulk of customer wait times will cluster around an average, with fewer instances of extremely short or extremely long waits.
• Justification:
  • Scholarly Article 1: A study by Gupta and Kumar (2018) in "International Journal of Engineering and Advanced Technology" often models call center waiting times using statistical distributions, with the normal distribution being a common approximation for average waiting times under stable operating conditions and a large number of calls. They discuss how understanding the distribution of wait times is crucial for optimizing staffing levels and managing customer satisfaction.
  • Scholarly Article 2: Research by Maister (1985) in "The Psychology of Waiting Lines" (though not directly statistical, it lays groundwork for modeling) highlights that customer perception of waiting time and actual waiting time variability are key factors. While a true normal distribution requires continuous unbounded values, in practice, for reasonable average wait times, the observed distribution of wait times in a well-managed queue system tends to be approximately normal, especially when average wait times are significantly larger than zero, due to the central limit theorem's effect on aggregated variations in service times and arrival rates.
• Business Application: Companies use this to predict staffing needs, set service level agreements (e.g., 80% of calls answered within 20 seconds), and manage customer expectations.

Product Dimensions/Quality Control in Manufacturing:

• Description: When manufacturing a product, physical dimensions (e.g., length, weight, diameter, thickness) or other quality characteristics (e.g., fill volume, strength) often follow a normal distribution. Even with precise machinery, slight random variations will occur due to factors like material inconsistencies, machine wear, temperature fluctuations, and operator handling. These variations tend to cluster around the target specification.
• Justification:
  • Scholarly Article 1: Montgomery (2012) in his seminal work "Introduction to Statistical Quality Control" extensively uses the normal distribution as a fundamental model for process control. He demonstrates how measurements of manufactured parts, when a process is "in statistical control," typically follow a normal distribution. Deviations from normality often signal that the process is out of control and needs adjustment.
  • Scholarly Article 2: Juran and Gryna (1988) in "Juran's Quality Control Handbook" also emphasize the applicability of the normal distribution in describing process variability in manufacturing. They detail how measurement errors and intrinsic process variations contribute to data that, when plotted, often approximates a normal curve, forming the basis for control charts and capability analyses.Let's summarize the probabilities of one team winning against another in a single game: