Case Study 41 Moodle Week 41 A Bank Manager Wishes To 075450

Case Study 41 Moodle Week 41 A Bank Manager Wishes To Provide Pro

Case Study 41 Moodle Week 41 A bank manager wishes to provide prompt service for customers at the bank’s drive-up window. The bank currently can serve up to 10 customers per 15-minute period without significant delay. The average arrival rate is 7 customers per 15-minute period. Assume x, the number of customers arriving per 15-minute period, follows a Poisson distribution. Calculate the probability that exactly 10 customers arrive, the probability that 10 or fewer customers arrive, and the probability that more than 10 customers (leading to a delay) arrive during that period.

Additionally, the case involves a telephone company's service failures over a 100-mile line segment, where failures follow a Poisson distribution with an average of 2 failures per 50 miles. Determine the probability that the company meets its goal of no more than five failures, the probability that it exceeds this, and the probability of failures over 200 miles.

Furthermore, the scenario discusses an extended warranty for laptops, where the failure probabilities are 13% for minor, 8% for major, and 3% for catastrophic repairs, with associated savings if repairs are needed. Calculate the expected repair cost, the expected financial gain or loss for the consumer, and analyze risk preferences (risk-neutral, risk-averse) regarding purchasing the warranty.

Finally, the discussion covers teenage smoking trends in New York City, where the smoking rates dropped from 23% in 1997 to 8.5% in 2007. Compute the probability that at least one teen out of a group of 10 smoked in each of these years and analyze the trend in smoking behavior over the decade.

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Paper For Above instruction

Introduction

Understanding the application of probability distributions, especially the Poisson distribution, is crucial in various fields such as banking, telecommunications, consumer electronics, and public health. These applications assist in decision-making processes by quantifying risks, estimating likelihoods, and providing strategic insights. This paper explores multiple scenarios applying probability and statistical concepts, including service delays in banking, failure rates in telecommunication lines, consumer decision-making regarding extended warranties, and public health trends in adolescent smoking behavior. The analysis demonstrates the significance of probabilistic modeling in operational efficiency, risk assessment, consumer behavior, and policy evaluation.

Section 1: Poisson Distribution in Banking Service Efficiency

The first scenario involves a bank's drive-up window where customer arrivals per 15-minute interval follow a Poisson distribution with a mean (λ) of 7. The Poisson distribution models the number of events (customer arrivals) occurring within fixed intervals, based on the average rate alone, independent of previous arrivals (Ross, 2010). The probability mass function (PMF) is given by:

\[ P(x; \lambda) = \frac{e^{-\lambda} \lambda^{x}}{x!} \]

where \( x \) is the number of arrivals, and \( \lambda \) is the mean.

a) The probability that exactly 10 customers arrive (\( P(X=10) \)):

\[ P(10; 7) = \frac{e^{-7} \times 7^{10}}{10!} \]

Using a calculator or software, this yields approximately 0.083 (or 8.3%).

b) The probability of 10 or fewer arrivals:

\[ P(X \leq 10) = \sum_{x=0}^{10} P(x; 7) \]

This cumulative probability is approximately 0.762, indicating that there's about a 76.2% chance that the number of arrivals will not cause delays.

c) The probability of more than 10 arrivals (leading to delay):

\[ P(X > 10) = 1 - P(X \leq 10) \approx 0.238 \]

This suggests there's roughly a 23.8% chance of exceeding the ideal service capacity and experiencing delays.

These calculations assist the bank in understanding service levels, planning staffing, and managing customer expectations.

Section 2: Failures in Telecommunication Lines

The second scenario considers failure rates in 100 miles of telecommunication line with an average of 2 failures per 50 miles. The failure rate per 100 miles doubles to 4 failures, with \( \lambda=4 \). Applying the Poisson distribution:

a) Probability that the company meets its failure goal of no more than 5 failures:

\[ P(X \leq 5) = \sum_{x=0}^{5} \frac{e^{-4} \times 4^{x}}{x!} \]

Calculations show this probability to be approximately 0.632, indicating about 63.2% chance of meeting the goal.

b) Probability that failures exceed 5:

\[ P(X > 5) = 1 - P(X \leq 5) \approx 0.368 \]

c) For 200 miles, the failure rate doubles again to 8 failures (\( \lambda=8 \)):

\[ P(X \leq 5) = \sum_{x=0}^{5} \frac{e^{-8} \times 8^{x}}{x!} \]

This probability drops to roughly 0.189, reflecting increased risk over larger distances.

These models help telecommunication companies to schedule maintenance and allocate resources efficiently.

Section 3: Consumer Decision-Making on Extended Warranties

The third scenario presents a consumer's decision based on the probability of different repair types over three years, with associated costs and savings.

The possible outcomes are:

- No repairs (probability 76.4%)

- Minor repair (13%, savings \$80)

- Major repair (8%, savings \$320)

- Catastrophic repair (3%, savings \$500)

Expected value of repair cost:

\[ E(\text{cost}) = (0.764 \times 0) + (0.13 \times 80) + (0.08 \times 320) + (0.03 \times 500) \]

\[ = 0 + 10.4 + 25.6 + 15 = \$51 \]

The consumer’s expected gain when purchasing the warranty costing \$74:

\[ \text{Expected gain} = \text{Expected savings} - \text{Warranty cost} \]

\[ = \$51 - \$74 = -\$23 \]

This negative expected value indicates an expected loss.

Regarding risk preferences, risk-neutral consumers base decisions solely on expected monetary value. Risk-averse consumers might avoid the warranty due to potential losses, preferring to accept potential costs without paying upfront, while risk-seeking consumers might purchase the warranty for peace of mind.

Section 4: Trends in Teenage Smoking Behavior

The last scenario examines the decline in teenage smoking rates in NYC from 23% in 1997 to 8.5% in 2007. Assuming independence, calculate the probability that at least one teen out of 10 smoked:

For 2007:

\[ P(\text{at least one}) = 1 - P(\text{none}) = 1 - (1 - 0.085)^{10} \]

\[ \approx 1 - 0.915^{10} \approx 1 - 0.42 = 0.58 \]

For 2001 (smoking rate 17.6%):

\[ 1 - 0.824^{10} \approx 1 - 0.183 = 0.817 \]

For 1997 (rate 23%):

\[ 1 - 0.77^{10} \approx 1 - 0.096 = 0.904 \]

Analysis indicates a considerable reduction in the probability of smoking among the youth, reflecting the effectiveness of public health policies. The decline demonstrates a successful trend toward reduced tobacco consumption, although continued efforts are necessary to sustain and improve these results.

Conclusion

The application of probability models such as the Poisson distribution provides valuable insights across diverse fields. In banking, it aids in optimizing service levels; in telecommunications, it informs maintenance scheduling; in consumer behavior, it guides warranty offerings; and in public health, it tracks behavioral trends over time. Each scenario underscores the importance of quantitative analysis in strategic decision-making and policy development, highlighting the need for accurate data collection and model application.

References

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