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The Denver Post reported that, on average, a large shopping center had an incident of shoplifting caught by security once every three 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 an 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 \(10 \mathrm{A.M}\). 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
The Poisson distribution is suitable. \(\lambda = 3.67\). \(P(r \geq 1) = 1 - e^{-3.67}\), \(P(r \geq 3) = 1 - (P(r = 0) + P(r = 1) + P(r = 2))\), and \(P(r = 0) = e^{-3.67}\).

Step by step solution

01

Determine Poisson Suitability

The Poisson distribution models the number of events occurring in a fixed interval of time, especially when these events occur with a known constant rate and independently. In this case, shoplifting incidents occur independently, and their rate of occurrence is constant, making Poisson an appropriate choice for the random variable \(r\).
02

Calculate Lambda

Given that incidents occur every 3 hours on average, the rate (\(\lambda\)) for an 11-hour period is found by dividing 11 by 3, which computes to \(\lambda = \frac{11}{3} \approx 3.67\).
03

Find Probability of At Least One Incident

The probability of at least one event is given by \(P(r \geq 1) = 1 - P(r = 0)\). First, calculate the probability of zero incidents, \(P(r = 0) = \frac{\lambda^0 e^{-\lambda}}{0!} = e^{-3.67}\). Therefore, \(P(r \geq 1) = 1 - e^{-3.67}\).
04

Calculate P(r ≥ 3)

To find the probability of at least three incidents, we need to subtract the probabilities of zero, one, and two events. Calculate these using the Poisson probability formula: \(P(r = 0) = \frac{3.67^0 e^{-3.67}}{0!}, \quad P(r = 1) = \frac{3.67^1 e^{-3.67}}{1!}, \quad P(r = 2) = \frac{3.67^2 e^{-3.67}}{2!}\).Then, \(P(r \geq 3) = 1 - (P(r = 0) + P(r = 1) + P(r = 2))\).
05

Calculate P(r = 0)

The probability of no incident occurring is \(P(r = 0) = e^{-3.67}\). This uses the Poisson probability formula with \(k = 0\).

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Key Concepts

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

Probability Calculation
At the core of the Poisson distribution is its ability to calculate the likelihood of a certain number of events happening within a specified time period. In practical terms, it helps us understand and predict random occurrences, such as shoplifting incidents in a shopping center. The Poisson distribution formula is expressed as:
\[P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\]where:
  • \(P(X = k)\) is the probability of observing \(k\) events in a time period.
  • \(\lambda\) is the average number of occurrences (events) in the given time period.
  • \(k!\) is the factorial of \(k\).
To find the probability of at least one shoplifting incident, we begin by determining the chance of zero incidents:
\[P(r = 0) = \frac{3.67^0 e^{-3.67}}{0!} = e^{-3.67}\]Subtracting this from 1 gives us the probability of at least one incident: \(P(r \geq 1) = 1 - e^{-3.67}\).
Each calculation with the Poisson distribution follows this logic, and by accurately using it, we can estimate probabilities like "at least three incidents." These calculations are crucial for predictive purposes and assist in safety planning and resource allocation.
Random Variables
In statistics, a random variable is a numerical outcome of a random process. It's essential for modeling scenarios where outcomes are uncertain. The Poisson distribution uses a discrete random variable to record counts of events that occur randomly.
This shopping center scenario treats shoplifting incidents as events, and the number of such events as a random variable, \(r\). This particular random variable follows a Poisson distribution due to:
  • Events happen independently – one incident doesn't affect another.
  • The average rate of occurrence is constant over the time interval (every 3 hours, roughly 3.67 events in 11 hours).
  • Occurrences happen infrequently and can't coincide at the same instant.
The Poisson distribution perfectly models this scenario by predicting the number of shoplifting incidents that will happen in specific time periods. By understanding the behavior of random variables, students are better equipped to interpret and predict real-world situations where randomness is inherent.
Statistical Modeling
Statistical modeling involves creating mathematical representations of real-world processes. The Poisson distribution is a classic example used to model random counts of events happening over time or space.
In the shopping center problem, the Poisson model serves as a **tool** to understand and predict shoplifting patterns. It uses historical data and incident rates to foresee the future. The statistical model is built on:
  • The assumption of a consistent average rate of incidents.
  • An understanding that incidents occur independently.
After verifying Poisson suitability, the model allows us to calculate specific probabilities:
  • The probability of no incidents at all in an 11-hour window: \(P(r = 0) = e^{-3.67}\).
  • The likelihood of three or more incidents: requires finding probabilities for 0, 1, and 2 incidents using the Poisson formula, then subtracting from 1.
Statistical modeling enables businesses and researchers to systematically analyze risks and make informed decisions based on probabilities, ultimately aiding in security and operational planning.

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Most popular questions from this chapter

Poisson Approximation to Binomial: Comparisons (a) For \(n=100, p=0.02\), and \(r=2\), compute \(P(r)\) using the formula for the binomial distribution and your calculator: $$ P(r)=C_{n, t} p^{r}(1-p)^{n-r} $$ (b) For \(n=100, p=0.02\), and \(r=2\), estimate \(P(r)\) using the Poisson approximation to the binomial. (c) Compare the results of parts (a) and (b). Does it appear that the Poisson distribution with \(\lambda=n p\) provides a good approximation for \(P(r=2) ?\) (d) Repeat parts (a) to \((c)\) for \(r=3\)

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