91Ó°ÊÓ

To start understanding the Poisson distribution, we need to find the mean number of events (in this case, deaths) that occur in a fixed interval of time or space. For our example involving the Prussian army, the mean number of deaths per corps-year provides a foundation for further calculations. We calculate this mean by dividing the total number of deaths (196) by the total number of years (280). Mathematically, this is expressed as: \[ \text{Mean} (\text{λ}) = \frac{\text{Total Deaths}}{\text{Total Corps-Years}} = \frac{196}{280} \ \text{Mean} (\text{λ}) \text{ is approximately } 0.7 \ \text{deaths per corps-year.} \] This calculated mean will be used as the 'lambda' (\text{λ}) parameter in the Poisson distribution formula.

Once we have the mean number of deaths (\text{λ}), we can use it to find the probability of a certain number of deaths occurring in a single corps-year. The Poisson probability formula is: \[ P(X = k) = \frac{e^{-\text{λ}} \text{λ}^{k}}{k!} \] This formula helps us calculate the probability of observing exactly k deaths in a specified time period. By substituting \text{λ} = 0.7 (our mean), we can calculate specific probabilities: (a) For 0 deaths: \[P(X=0) = \frac{e^{-0.7} (0.7)^{0}}{0!} = e^{-0.7} \approx 0.4966 \](b) For 1 death: \[P(X=1) = \frac{e^{-0.7} (0.7)^{1}}{1!} = 0.7 e^{-0.7} \approx 0.3476 \]. (c) For 2 deaths: \[P(X=2) = \frac{e^{-0.7} (0.7)^{2}}{2!} = \frac{0.49 e^{-0.7}}{2} \approx 0.1217 \](d) For 3 deaths: \[P(X=3) = \frac{e^{-0.7} (0.7)^{3}}{3!} = \frac{0.343 e^{-0.7}}{6} \approx 0.0284 \](e) For 4 deaths: \[P(X=4) = \frac{e^{-0.7} (0.7)^{4}}{4!} = \frac{0.2401 e^{-0.7}}{24} \approx 0.005 \]. To make these calculations directly relevant, you can apply them to your data set to find the likelihood of different numbers of deaths per year.

Now that we have the expected probabilities, it's time to compare these values to the actual historical data. For our Prussian army example, we need to look at the actual observed frequencies of deaths and see how well they match our Poisson distribution predictions. Here are the comparisons:

• 0 deaths: Predicted: 0.4966 x 280 ≈ 139.048 | Actual: 144
• 1 death: Predicted: 0.3476 x 280 ≈ 97.328 | Actual: 91
• 2 deaths: Predicted: 0.1217 x 280 ≈ 34.076 | Actual: 32
• 3 deaths: Predicted: 0.0284 x 280 ≈ 7.952 | Actual: 11
• 4 deaths: Predicted: 0.005 x 280 ≈ 1.400 | Actual: 2

While there are slight variations, the predicted frequencies are reasonably close to the actual counts. This suggests the Poisson distribution serves as a useful tool for modeling the given data.

Historically, the data from the Prussian army between 1875 and 1894 provides an intriguing and classical example of the Poisson distribution in action. Over 20 years, deaths in each of the 14 corps were documented, totaling 196 deaths over 280 corps-years. These detailed records allow statisticians to test the applicability of statistical models like the Poisson distribution. In this situation, the Poisson distribution helps us predict and understand the occurrences of rare events (like deaths from horse kicks). By examining a statistical distribution of rare events, one can make informed decisions and assessments about similar real-world phenomena. This rich historical data set, therefore, serves as a cornerstone case study in applied statistics, illustrating how theoretical models can be applied to analyze, predict, and understand real-life events.