Question

Ladislaus Bortkiewicz was a Russian economist and statistician who published a book entitled "The Law of Small Numbers." In his book he showed that the number of soldiers in the Prussian cavalry killed by being kicked by a horse each year in each of 14 cavalry corps over a 20 -year period (1875-1894) followed a Poisson distribution. \({ }^{10}\) The data summary follows. $$ \begin{array}{c|r} \text { Number of deaths } & \text { Frequency } \\ \hline 0 & 144 \\ 1 & 91 \\ 2 & 32 \\ 3 & 11 \\ 4 & 2 \end{array} $$ a. Find the mean number of deaths per year per cavalry unit. [HINT: Use the grouped formula given in Exercise 21 of the "On Your Own" Exercises in Chapter \(2 .\) b. Use the result of part a and the Poisson distribution to find the probability of exactly one death per unit per year. c. Find the probability of at most two deaths per year. d. How do the probabilities in parts \(\mathrm{b}\) and \(\mathrm{c}\) compare to the observed relative frequencies in the table?

Short answer

Question: Calculate the mean number of deaths per unit/year and find the probability of exactly one death/unit/year using the given data. Answer: The mean number of deaths per unit per year is approximately 1.071. The probability of exactly one death per year per unit using the Poisson distribution is approximately 0.366 (or 36.6%).

Step-by-step solution

Calculate the mean number of deaths per year per cavalry unit.

First, let's find the mean number of deaths per year per cavalry unit using the grouped formula given in the exercise: Mean (m) = \(\sum (\text{Number of Deaths} × \text{Frequency}) / \text{Total Frequency}\) Using the given data table: $$ \begin{array}{c|r} \text { Number of Deaths } & \text { Frequency } \\\ \hline 0 & 144 \\\ 1 & 91 \\\ 2 & 32 \\\ 3 & 11 \\\ 4 & 2 \end{array} $$ We can calculate the mean as follows: Mean (m) = \((0 × 144 + 1 × 91 + 2 × 32 + 3 × 11 + 4 × 2) / (144 + 91 + 32 + 11 + 2)\) Mean (m) = \(300 / 280\) Mean (m) = \(1.071\) The mean number of deaths per year per cavalry unit is approximately 1.071.

Find the probability of exactly one death per year per unit using Poisson distribution.

Now, we will use the Poisson distribution formula to find the probability of exactly one death per unit per year: P(x; m) = \((e^{-m} × m^x) / x!\) For x=1: P(1;1.071) = \((e^{-1.071} × 1.071^1) / 1!\) P(1;1.071) ≈ \((0.342 * 1.071) / 1\) P(1;1.071) ≈ \(0.366\) The probability of exactly one death per year per cavalry unit is approximately 0.366 (or 36.6%).

Key concepts

Probability

When we delve into the world of mathematics, especially statistics, the concept of probability emerges as a crucial pillar. Probability, in a straightforward manner, can be described as the likelihood of an event occurring. It is represented numerically, ranging from 0 (signifying impossibility) to 1 (signifying certainty).

To calculate the probability of a specific event, such as the death of a soldier owing to a horse kick within the Prussian cavalry—which follows a Poisson distribution in our exercise—we use the vital Poisson probability formula: \[ P(x; \lambda) = \frac{e^{-\lambda} \times \lambda^x}{x!} \
Here, \( e \) is the base of the natural logarithm, \( x \) represents the number of occurrences (such as the number of deaths), and \( \lambda \) symbolizes the event's expected value or mean. The 'factorial' symbol (!) denotes the product of all positive integers up to \( x \). The calculated probability gives us a clear understanding of the likelihood of exactly \( x \) events happening in a given time frame.

As with our example of the cavalry unit, we can observe that the probability of exactly one death per year per cavalry unit was calculated to be approximately 36.6%. This figure is not just a number; it is a representation of predictability gleaned from historical data, providing a rational expectation of future events. Often, this is a stepping stone to further risk analysis, policy-making, or resource allocation in related fields.

Mean Calculation

Moving onto the topic of mean calculation, this constitutes a fundamental statistical measure often referred to as the 'average.' When we say 'mean,' we are typically referring to the sum of all values in a data set divided by the number of values. It represents the central value of a discrete set of numbers, which intuitively is the value that sequences of infinite random samples drawn from that distribution would average out to.

In our soldier's morbid scenario, we use the mean to establish the average number of deaths in the Prussian cavalry units due to horse kicks per year, which we calculated as approximately 1.071. To arrive at this figure, we applied the formula: \[ \text{Mean} (m) = \frac{\sum (\text{Number of Deaths} \times \text{Frequency})}{\text{Total Frequency}} \
This average is an extremely potent tool, as it lays the foundation for the Poisson distribution probability calculations. Essentially, when dealing with Poisson distribution, the mean value \( m \) is the \( \lambda \) (lambda) parameter in the Poisson probability formula, which gives us a direct link to the respective probabilities of different numbers of events occurring.

Frequency Distribution

Lastly, let’s explore the concept of frequency distribution. This is a statistical representation that displays the number of observations within a given interval. In every frequency distribution, numbers are organized into a table, and the number of occurrences of various outcomes of a study or an event are recorded.

In our textbook example, frequency distribution was used to demonstrate the number of deaths within the Prussian cavalry. The table showcased different 'Number of Deaths' as categories and their corresponding occurrence, allowing us to visually apprehend the dataset. Here's how the data appeared:

• 0 deaths: 144 occurrences
• 1 death: 91 occurrences
• 2 deaths: 32 occurrences
• 3 deaths: 11 occurrences
• 4 deaths: 2 occurrences

This table is quintessential as it directly feeds into the mean calculation by providing the 'Frequency' part of the mean formula. Moreover, the frequency distribution gives us a tangible grasp on how common or uncommon certain events were during the studied period. It can be graphically represented through histograms, bar chats, or pie charts, enhancing the comprehension of the distribution of values within a dataset and thus helping students and statisticians alike to visualize the data for comparative studies or in-depth analysis.