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The Denver Post reported that, on average, a large shopping center has had an incident of shoplifting caught by security once every 3 hours. The shopping center is open from 10 A.M. to 9 P.M. (11 hours). Let \(r\) be the number of shoplifting incidents caught by security in the 11 -hour period during which the center is open. (a) Explain why the Poisson probability distribution would be a good choice for the random variable \(r\). What is \(\lambda ?\) (b) What is the probability that from 10 A.M. to 9 P.M. there will be at least one shoplifting incident caught by security? (c) What is the probability that from 10 A.M. to 9 P.M. there will be at least three shoplifting incidents caught by security? (d) What is the probability that from 10 A.M. to 9 P.M. there will be no shoplifting incidents caught by security?

Short Answer

Expert verified
(a) Poisson distribution is appropriate; \( \lambda = 3.67 \). (b) Probability is \( 0.975 \). (c) Probability is \( 0.714 \). (d) Probability is \( 0.025 \).

Step by step solution

01

Identifying the Underlying Distribution

The problem describes an average rate of occurrence for discrete events over a continuous interval: shoplifting incidents occurring with an average rate once every 3 hours. These conditions are typical for the Poisson distribution, which models rare events occurring independently over time. Thus, a Poisson distribution is appropriate for the random variable \( r \), representing the number of shoplifting incidents during the open hours.
02

Calculating the Average Rate (\( \lambda \))

The expected number of incidents in a period is represented by \( \lambda \) in a Poisson distribution. With an average rate of one incident every 3 hours, and the shopping center open for 11 hours, compute \( \lambda \) as follows: \[ \lambda = \frac{11}{3} \approx 3.67 \] incidents expected during the open hours.
03

Probability of At Least One Incident

To find the probability of at least one incident, use the complementary approach. The probability of zero incidents is \( P(r = 0) = e^{-\lambda} \frac{\lambda^0}{0!} = e^{-3.67} \). Therefore, the probability of at least one incident is: \[ P(r \geq 1) = 1 - e^{-3.67} \approx 1 - 0.025 \approx 0.975 \].
04

Probability of At Least Three Incidents

To calculate \( P(r \geq 3) \), use: \[ P(r \geq 3) = 1 - (P(r = 0) + P(r = 1) + P(r = 2)) \]. Compute each: \( P(r = 1) = e^{-3.67} \frac{3.67^1}{1!} \approx 0.092 \) and \( P(r = 2) = e^{-3.67} \frac{3.67^2}{2!} \approx 0.169 \). Therefore: \[ P(r \geq 3) = 1 - (0.025 + 0.092 + 0.169) = 0.714 \].
05

Probability of No Incidents

The probability of zero shoplifting incidents is directly: \[ P(r = 0) = e^{-3.67} \approx 0.025 \].

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Key Concepts

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

Exploring Probability and Poisson Distribution
Probability, in simple terms, is a measure of how likely an event is to occur. In the context of Poisson distribution, which is used to model the probability of a given number of events happening in a fixed interval of time or space, these events must happen independently and with a known constant mean rate.
In our shoplifting example, we're interested in the probability of shoplifting incidents over the 11-hour period during which the shopping center is open. The Poisson distribution provides a straightforward way to calculate such probabilities when incidents occur rarely or are spread out over time.
This distribution uses the rate of occurrences (\(\lambda\),) to calculate specific probabilities. For example, if we want to find out the likelihood of exactly \(r\) incidents within a given time frame, we use the Poisson probability mass function:\[ P(r) = \frac{e^{-\lambda} \cdot \lambda^r}{r!} \]where:
  • \(e\) is Euler's number (approximately 2.71828),
  • \(\lambda\) is the average rate of occurrence,
  • \(r\) is the number of incidents we are interested in.
Understanding Lambda (\(\lambda\))
In the Poisson distribution, \(\lambda\) is a crucial parameter representing the average number of occurrences within the specified interval. It can be thought of as the mean of the distribution.
For our shopping center scenario, \(\lambda\) is calculated based on the average shoplifting rate. With an incident due every 3 hours and the store being open for 11 hours, we determine \(\lambda\)as:\[ \lambda = \frac{11}{3} \approx 3.67 \]This means we expect around 3.67 incidents during the open hours.
The ability to calculate and understand \(\lambda\)is fundamental when using the Poisson distribution, as it directly influences the probability calculations for varying numbers of incidents.
Deciphering the Random Variable r
The random variable \(r\) in our scenario denotes the number of shoplifting incidents observed in the 11-hour period while the shopping center is open.
In the realm of probability distributions, a random variable is a variable that can take various values based on the outcomes of a random phenomenon. Here,\(r\)can take values like 0, 1, 2, with each representing different numbers of shoplifting incidents.
The concept of a random variable is foundational when dealing with statistical distributions, as it helps translate real-world random events into quantifiable variables that can be systematically analyzed. In Poisson distribution, the probabilities associated with different \(r\)values are calculated typically using the formula mentioned previously, and they allow us to draw insights about how common or rare different numbers of incidents are. This is crucial for understanding not just theoretical probabilities, but also for making informed predictions and decisions in real-world scenarios.

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Most popular questions from this chapter

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