Chapter 4 Special Distributions - 4.2 The Poisson Distribution - Questions - Page 230: 10

$\color{blue}{\begin{array}{|c|c|c|} \hline k & \rm Frequency & \rm Proportion & p_x(k) = \dfrac{e^{-0.61}0.61^k}{k!} \\ \hline 0 & 109 & 0.545 & 0.543 \\ \hline 1 & 65 & 0.325 & 0.331 \\ \hline 2 & 22 & 0.11 & 0.101 \\ \hline 3 & 3 & 0.015 & 0.021 \\ \hline 4 & 1 & 0.005 & 0.003 \\ \hline \rm Total & 200 & 1.000 & 0.999 \\ \hline \end{array}}$ Close concordance between corresponding cells of columns 3 and 4 shows that the data can be modelled by a Poisson pdf. See explanation

Let $X$ denote the number of deaths per year from horse kicks among cavalry soldiers. An estimate of the annual death rate is $\overline{k} = \dfrac{(0)(109) + (1)(65) + 2(22)+3(3)+(4)(1)}{109+65+22+3+1} = \dfrac{122}{200} \approx 0.61$ Thus, an approximate pdf for $X$ based on the data is $p_X(k) = \dfrac{e^{-0.61}0.61^k}{k!},\ k=0,1,2,\ldots. $ Columns 1 to 3 in the table below give the annual number of deaths $k$, and the observed frequencies and proportions of $k$ deaths in a year based on a total of $200$ years included in the data, while column 4 gives the corresponding proportions predicted by the Poisson model. The close concordance between the corresponding cells of columns 3 and 4 shows that the data can be modeled by a Poisson pdf. $\begin{array}{|c|c|c|} \hline k & \rm Frequency & \rm Proportion & p_X(k) = \dfrac{e^{-0.61}0.61^k}{k!} \\ \hline 0 & 109 & 0.545 & 0.543 \\ \hline 1 & 65 & 0.325 & 0.331 \\ \hline 2 & 22 & 0.11 & 0.101 \\ \hline 3 & 3 & 0.015 & 0.021 \\ \hline 4 & 1 & 0.005 & 0.003 \\ \hline \rm Total & 200 & 1.000 & 0.999 \\ \hline \end{array}$