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In \(1898,\) L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that the number of soldiers killed by horse kicks each year in each corps in the Prussian cavalry followed a Poisson distribution with a mean of \(0.61 .\) (a) What is the probability of more than one death in a corps in a year? (b) What is the probability of no deaths in a corps over five years?

Short Answer

Expert verified
(a) 0.126, (b) 0.047

Step by step solution

01

Identify the Poisson Parameter

The Poisson distribution is characterized by the parameter \(\lambda\), which represents the average number of events in a given interval. Here, the average number of deaths per year \(\lambda = 0.61\).
02

Calculate the Probability for More Than One Death

To find the probability of more than one death, first calculate the probabilities for 0 and 1 death, and subtract from 1. The probability mass function for the Poisson distribution is \(P(X=k) = \frac{{e^{-\lambda} \lambda^k}}{{k!}}\). \(P(X=0) = \frac{{e^{-0.61} 0.61^0}}{{0!}} = e^{-0.61}\) and \(P(X=1) = \frac{{e^{-0.61} 0.61^1}}{{1!}} = 0.61e^{-0.61}\). Hence, \(P(X > 1) = 1 - (P(X=0) + P(X=1))\).
03

Compute the Probability of Zero Deaths Over Five Years

For five years, the parameter \(\lambda\) becomes \(5 \times 0.61 = 3.05\). The probability of 0 deaths over five years is \(P(Y=0) = \frac{{e^{-3.05} 3.05^0}}{{0!}} = e^{-3.05}\).
04

Perform the Calculations

Substituting the values, we have: \(P(X=0) = e^{-0.61} \approx 0.543\) and \(P(X=1) = 0.61 \cdot e^{-0.61} \approx 0.331\). So, \(P(X > 1) = 1 - (0.543 + 0.331) = 0.126\). For zero deaths over five years, \(P(Y=0) = e^{-3.05} \approx 0.047\).

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Key Concepts

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

Probability Calculations
Probability calculations are essential for predicting the likelihood of an event occurring in a given scenario. With the Poisson distribution, these calculations help us understand the occurrence of rare events within a fixed period of time or space. To calculate probabilities using the Poisson distribution, we use the formula:\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]where:- \( e \) is the base of natural logarithms (approximately 2.71828)- \( \lambda \) is the average number of events- \( k \) is the number of occurrences for which we are calculating the probability- \( ! \) denotes factorial, which is the product of all whole numbers from 1 to \( k \).For example, calculating the probability of no deaths in a corps over five years involves replacing \( \lambda \) with the term relevant to the time frame (in this case, multiplying by 5). The ability to manipulate these calculations allows us to make informed predictions about real-world probabilities.
Probability Mass Function
The probability mass function (PMF) is a crucial tool in probability theory, especially when dealing with discrete outcomes. For the Poisson distribution, the PMF gives us the probability of observing exactly \( k \) events in a fixed interval.Utilizing the PMF lets us calculate individual probabilities based on different event counts, like zero or one death in the Prussian cavalry example. For \( X = k \):\[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \]This formula allows us to find the probability of a specified number of occurrences. It helps group similar probabilities, which can then be summed or subtracted from 1 to find the desired likelihood, such as more than one event occurring.Through PMF, complex probability scenarios become manageable, especially when several different outcomes need detailed analysis.
Average Number of Events
The average number of events, also identified by the parameter \( \lambda \) in the Poisson distribution, is a measure of the expected frequency of occurrences within a given timeframe or space.For instance, in Bortkiewicz's analysis of horse kick deaths in the Prussian cavalry, \( \lambda = 0.61 \) signifies that, on average, 0.61 deaths occur per year per corps. This average is calculated based on historical data and serves as an essential component in determining the likelihood of events using the Poisson distribution.Understanding the average number of events allows us to establish a baseline expectation. With this knowledge, it becomes easier to predict the variance and distribution of actual outcomes around this average. It gives a clearer picture of whether particular results are within typical expectations or are outliers.
Parameter Lambda
In any Poisson distribution, the parameter \( \lambda \) is of utmost importance. It provides insight into how frequent an event is expected to occur over a specific interval. In mathematical terms, \( \lambda \) is the mean or expected number of occurrences per unit. In scenarios like the Prussian cavalry study, \( \lambda \) changes based on different timeframe, such as over multiple years. For instance, if in a single year \( \lambda = 0.61 \) for deaths, over five years, \( \lambda = 5 \times 0.61 = 3.05 \).This expansion of \( \lambda \) is critical because it adjusts our characterization of risk based on the period we are examining. By comprehending how \( \lambda \) influences event probability, predictions become more nuanced and accurately reflect the expected frequency of events.

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Most popular questions from this chapter

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05 flaw per square foot of plastic panel. Assume that an automobile interior contains 10 square feet of plastic panel. (a) What is the probability that there are no surface flaws in an auto's interior? (b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws? (c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws? .

Determine the constant \(c\) so that the following function is a probability mass function: \(f(x)=c x\) for \(x=1,2,3,4\).

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