Is Poisson Distribution Relevant for Analyzing 19th Century Cavalry Deaths?

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To analyze the 19th-century cavalry deaths in the Prussian Army using the Poisson distribution, one must first calculate the expected mean value based on the reported deaths. The data includes 109 reports of zero deaths, 65 reports of one death, 22 reports of two deaths, 3 reports of three deaths, and 1 report of four deaths. After determining the mean, a graph can be plotted to visualize the Poisson distribution, with the number of deaths on the x-axis and the probability of each outcome on the y-axis. If the plotted data aligns with the Poisson distribution, the findings support its relevance to the analysis. This method provides a statistical framework for understanding the patterns in cavalry deaths.
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Doing Physics at University and I have never done poisson distributions.
How the hell do I do it?
The question is...
Show that the data on the number of cavalry deaths in the Prussian Army in the 19th Century are consistent with the poisson probability distribution. The date were accumulated from reports received from 10 separate cavalry corps, yearly, over a 20 year period.
There were: 109 Reports of ZERO deaths, 65 reports of 1 death, 22 reports of 2 deaths, 3 reports of 3 deaths and 1 report of 4 deaths.

Where do I begin? Does it involve drawing a graph?!
 
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Yes, it does involve drawing a graph. To begin, you need to calculate the expected mean value of the data. This can be done by multiplying the total number of cavalry deaths by the probability of each outcome. For example, 109 x 0% (probability of zero deaths) = 0. Then add up all the expected values to get the mean. Once you have the mean, you can plot the poisson distribution for the given mean on a graph using a graphing calculator or an online graphing tool. The graph should show the probability of each outcome (e.g. 0, 1, 2, 3, 4 deaths) on the y-axis and the number of reports on the x-axis. If the data points match up to the poisson distribution, then it is consistent with the distribution.
 
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