/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 The following case study was rep... [FREE SOLUTION] | 91Ó°ÊÓ

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The following case study was reported in the article “Parking Tidkets and Missing Women," which appeared in an early edition of the book Statistics: A Guide to the Unknown. In a Swedish trial on a charge of overtime parking, a police officer testified that he had noted the position of the two air valves on the tires of a parked car: To the closest hour, one was at the one o'clock position and the other was at the six o'clock position. After the allowable time for parking in that zone had passed, the policeman returned, noted that the valves were in the same position, and ticketed the car. The owner of the car claimed that he had left the parking place in time and had returned later. The valves just happened by chance to be in the same positions. An "expert" witness computed the probability of this occurring as \((1 / 12)(1 / 12)=\) \(1 / 144\). a. What reasoning did the expert use to arrive at the probability of \(1 / 144\) ? b. Can you spot the error in the reasoning that leads to the stated probability of \(1 / 144 ?\) What effect does this error have on the probability of occurrence? Do you think that \(1 / 144\) is larger or smaller than the correct probability of occurrence?

Short Answer

Expert verified
The expert used a reasoning assuming these two instances as independent and equally likely events which lead them to the probability of \(1 / 144\). The error was considering the instances as independent and equal in probability. This error likely makes the calculated probability smaller than the true probability, \(1 / 144\) is probably an understatement of the true probability.

Step by step solution

01

Understanding the expert's reasoning

The expert calculated the probability of the car valves ending up in the same spot twice. To do this, the expert treated these two instances as independent events. Also, the expert assumed that each of the 12 positions (analog to 12 hours on a clock) on the wheel has an equal chance of being selected anytime the car parks. Thus, the chance of having the valve at a particular hour is \(1 / 12\), and the probability of this happening twice, independently, would be \((1 / 12) * (1 / 12) = 1 / 144\).
02

Identifying the error in reasoning

The mistake in the aforementioned calculation emerges from treating the instances as independent events and assuming each of the 12 positions on the wheel is equally likely. In reality, placing the car valves at a particular hour depends on the parking way and the distance travelled which makes these events not entirely independent. Additionally, the positions are not equally likely due to the distinct distances traveled in different parking situations. This misinterpretation leads to an inaccurate probability calculation.
03

Commentary on the effect of this error

This error leads to an understatement of the probability of this event occurring. Considering the factors that can influence the valve positions, the chance of the valve ending at certain hours is probably higher than calculated, hence, the correct probability is likely greater than \(1 / 144\).

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Key Concepts

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

Independent Events
Independent events in probability occur when the outcome of one event does not affect the outcome of another. This means that the probability of each event happening does not change regardless of the other events occurring. In the context of the parking ticket case, the expert assumed the positions of the air valves before and after were independent. They calculated the probability for each as if parking occurrences were unrelated.

However, it's crucial to understand that in real-life scenarios, truly independent events are rare. For instance, if the vehicle traveled a long distance, the position of the air valves might have varied greatly, making the probability truly independent. But just moving the car in or out of the parking space might not substantially change valve positions at all

This assumption of independence might not hold under conditions where small movements can lead to the same valve positions by coincidence or minimal car movement. So, when assessing real situations, ensure to check if one event actually depends on or influences another.
Probability Calculation
Calculating probability accurately is essential to evaluate real-world scenarios. In the parking case exercise, the expert calculated the probability of the valves being in the same positions again by multiplying the probabilities of each being in a specific position:
  • There are 12 possible positions (like a clock face) for each valve.
  • The probability of one valve being in any particular position is therefore \(1/12\).
  • By treating them as independent, the probability of both being in the same position again was calculated as \( (1/12) \times (1/12) = 1/144 \).
The primary issue with this calculation is it simplifies the problem to independent identical probabilities without accounting for other influencing factors. Actual movement patterns and constraints due to parking maneuver influence the probability. A simple multiplication of independent probabilities may not represent the true likelihood as it overlooks potential dependencies and positional repetition likelihood in real scenarios.
Statistical Misinterpretation
Statistical misinterpretation often arises from oversimplifying complex scenarios or neglecting real-world dependencies. In the given case study, the probability calculation was based on theoretical independence, ignoring the practical nature of parking movement and valve positioning.

This oversight can significantly distort probability estimations
  • Assuming all outcomes are equally likely can be misleading as not all events in practice occur with equal probability.
  • Disregarding potential dependencies or non-equal distribution can lead to incorrect conclusions.
Such misinterpretations cause the calculated probability \( (1/144) \) to be lower or higher than the real-world probability. It's important to remember that statistical tools should be applied with an understanding of the actual context. Thus, careful consideration of real-world dynamics is vital in making valid statistical interpretations and conclusions.

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Most popular questions from this chapter

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4 . The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9, P(2\) works \()=\) \(.9, P(3\) works \()=.9\), and \(P(4\) works \()=.9\) and that the four components work independently of one another. a. The 1-2 subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem were added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?

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Define the term chance experiment, and give an example of a chance experiment with four possible outcomes.

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