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The National Safety Council (NSC) estimates that off-the-job accidents cost U.S. businesses almost \(\$ 200\) billion annually in lost productivity (National Safety Council, March 2006 ). Based on NSC estimates, companies with 50 employees are expected to average three employee off-the-job accidents per year. Answer the following questions for companies with 50 employees. a. What is the probability of no off-the-job accidents during a one-year period? b. What is the probability of at least two off-the-job accidents during a one- year period? c. What is the expected number of off-the-job accidents during six months? d. What is the probability of no off-the-job accidents during the next six months?

Short Answer

Expert verified
a. \(0.0498\); b. \(0.8009\); c. \(1.5\); d. \(0.2231\).

Step by step solution

01

Understand the Problem

The problem involves a company with 50 employees that faces, on average, 3 off-the-job accidents per year. We need to determine various probabilities and expectations related to these accidents over different time periods.
02

Define the Distribution

Since the average number of accidents is given and we need to find probabilities, we use the Poisson distribution. Here, the average rate of occurrence (\(\lambda\)) is 3 accidents per year.
03

Find the Probability of No Accidents in One Year

The probability of observing \(k\) events in a Poisson distribution is given by \(P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\). For no accidents (\(k = 0\)), the formula becomes:\[P(X = 0) = \frac{e^{-3} \times 3^0}{0!} = e^{-3} \approx 0.0498\]
04

Find the Probability of At Least Two Accidents in One Year

First, calculate the probabilities for 0 and 1 accident and then subtract from 1:\[P(X = 0) = e^{-3} \]\[P(X = 1) = \frac{e^{-3} \times 3^1}{1!} = 3 \times e^{-3} \]The probability of at least two accidents is:\[1 - (P(X = 0) + P(X = 1)) = 1 - (e^{-3} + 3e^{-3}) = 1 - 4e^{-3} \approx 0.8009\]
05

Calculate the Expected Number of Accidents in Six Months

Since we have an expected number of 3 accidents per year, for six months (half a year), the expected number is:\[\frac{3}{2} = 1.5\]
06

Find the Probability of No Accidents in Six Months

For six months, the adjusted \(\lambda\) is 1.5. The probability of no accidents (\(k = 0\)) is:\[P(X = 0) = \frac{e^{-1.5} \times 1.5^0}{0!} = e^{-1.5} \approx 0.2231\]

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Key Concepts

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

Probability Calculation
Probability is a measure of the chance that a particular event will occur. In this exercise, we are dealing with accidents, and we need to calculate the likelihood of different scenarios using the Poisson distribution. The Poisson distribution is suitable here because it applies to situations where events happen independently and at a constant average rate over a given period.

First, for calculating the probability of no accidents within one year, we identify those zero occurrences. The formula for this is:
  • \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
Where:
  • \( P(X = k) \) is the probability of observing \( k \) events,
  • \( e \) is Euler's number (approximately 2.71828),
  • \( \lambda \) is the average number of events (3 accidents per year),
  • \( k \) is the number of events (0 for no accident).
By substituting these values, we calculate the probability as approximately 0.0498. Understanding this process will help you grasp how to calculate similar probabilities where events occur repeatedly by a certain average.
Expected Value
The expected value is a crucial concept in probability and statistics that helps us determine the anticipated outcome over time for random events. It represents the mean value of a random variable and is calculated by weighing each possible outcome by its probability.

In this scenario, companies expect 3 accidents annually. To find the expected number of accidents over six months, simply adjust for the smaller time frame. Since six months is half a year, the expected value halves:
  • \[ \frac{3}{2} = 1.5 \] accidents
This calculation means that over a six-month period, one would anticipate, on average, 1.5 accidents. Therefore, the expected value is a powerful tool, providing insight into what we can reasonably anticipate about future events, which helps in resource planning and risk management.
Statistical Analysis
Statistical analysis involves the collection and exploration of data to uncover patterns and insights. Utilizing the Poisson distribution, as in our current exercise, is a method of statistical analysis.

The exercise unfolds by translating business accident data into actionable probability and expectancy data. For instance, we calculate the probability of no accidents over various periods and find probabilities for one or more accidents using the formulae. The objective is to interpret this data to aid in decision-making.

Accurate statistical analysis enables businesses to prepare for different possibilities and manage risks effectively. Such analysis provides a structured framework that transforms raw data into meaningful metrics guiding policy development and strategic planning, essential for resilience and sustainability.
Accident Statistics
Accident statistics often involve analyzing the frequency and likelihood of occurrences, which can direly impact businesses. Using such data, firms can allocate resources for preventive measures or mitigate the effects of these events.

In our case, statistical data showed a norm of 3 accidents yearly for firms with 50 employees, creating a baseline for calculating probabilities for different accident scenarios. This baseline allows organizations to:
  • Understand typical patterns of incidents,
  • Examine fluctuations compared to the average (like in different seasons or timelines),
  • Implement strategies for accident reduction and control.
By leveraging known accident statistics, companies can optimize their operations, enhance safety protocols, and reduce unexpected downtime due to off-the-job accidents. This results in not only financial savings but also a safer environment for employees.

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Most popular questions from this chapter

The National Basketball Association (NBA) records a variety of statistics for each team. Two of these statistics are the percentage of field goals made by the team and the percentage of three-point shots made by the team. For a portion of the 2004 season, the shooting records of the 29 teams in the NBA showed that the probability of scoring two points by making a field goal was \(.44,\) and the probability of scoring three points by making a threepoint shot was .34 (NBA website, January 3,2004 ). a. What is the expected value of a two-point shot for these teams? b. What is the expected value of a three-point shot for these teams? c. If the probability of making a two-point shot is greater than the probability of making a three-point shot, why do coaches allow some players to shoot the three-point shot if they have the opportunity? Use expected value to explain your answer.

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