In Exercises 9–16, use the Poisson distribution to find the indicated probabilities.

Deaths from Horse Kicks A classical example of the Poisson distribution involves the number of deaths caused by horse kicks to men in the Prussian Army between 1875 and 1894. Data for 14 corps were combined for the 20-year period, and the 280 corps-years included a total of 196 deaths. After finding the mean number of deaths per corps-year, find the probability that a randomly selected corps-year has the following numbers of deaths: (a) 0, (b) 1, (c) 2, (d) 3, (e) 4. The actual results consisted of these frequencies: 0 deaths (in 144 corps-years); 1 death (in 91 corps-years); 2 deaths (in 32 corps-years); 3 deaths (in 11 corps-years); 4 deaths (in 2 corps-years). Compare the actual results to those expected by using the Poisson probabilities. Does the Poisson distribution serve as a good tool for predicting the actual results?

The total number of deaths due to horse kicks in 280 corps-years is given to be equal to 196.

The total number of deaths due to horse kicks in 280 corps-years is given to be equal to 196 deaths

The total number of corps-years is equal to 280.

The mean number of deaths per corps-year is equal to:

\(\begin{aligned}{c}\mu = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{deaths}}}}{{{\rm{Number}}\;{\rm{of}}\;{\rm{years}}}}\\ = \frac{{196}}{{280}}\\ = 0.7\end{aligned}\)

The mean number of deaths per corps-year is equal to 0.7.

Let X be the number of deaths per corps-year.

Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 0.7\).

(a)

The probability of exactly 0 deaths per corps-year is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 0 \right) = \frac{{{{\left( {0.7} \right)}^0}{{\left( {2.71828} \right)}^{ - 0.7}}}}{{0!}}\\ = 0.496585\end{aligned}\]

Therefore,the probability of exactly 0 deaths per corps-year is equal to 0.496585.

(b)

The probability of exactly 1 death per corps-year is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 1 \right) = \frac{{{{\left( {0.7} \right)}^1}{{\left( {2.71828} \right)}^{ - 0.7}}}}{{1!}}\\ = 0.34761\end{aligned}\]

Therefore,the probability of exactly 1 death per corps-year is equal to 0.34761.

(c)

The probability of exactly 2 deaths per corps-year is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 2 \right) = \frac{{{{\left( {0.7} \right)}^2}{{\left( {2.71828} \right)}^{ - 0.7}}}}{{2!}}\\ = 0.121663\end{aligned}\]

Therefore,the probability of exactly 2deaths per corps-year is equal to 0.121663.

(d)

The probability of exactly 3 deaths per corps-year is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 3 \right) = \frac{{{{\left( {0.7} \right)}^3}{{\left( {2.71828} \right)}^{ - 0.7}}}}{{3!}}\\ = 0.028388\end{aligned}\]

Therefore,the probability of exactly 3deaths per corps-year is equal to 0.028388.

(e)

The probability of exactly 4 deaths per corps-year is computed below:

\[\begin{aligned}{c}P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\\P\left( 4 \right) = \frac{{{{\left( {0.7} \right)}^4}{{\left( {2.71828} \right)}^{ - 0.7}}}}{{4!}}\\ = 0.004968\end{aligned}\]

Therefore, the probability of exactly 4deaths per corps-year is equal to 0.004968.

The expected frequencies corresponding to different number of deaths in 280 corps years are computed below:

\(\begin{aligned}{c}{\rm{Expected}}\;{\rm{Frequenc}}{{\rm{y}}_0} = 280 \times P\left( 0 \right)\\ = 280 \times 0.496585\\ = 139.04\end{aligned}\)

\(\begin{aligned}{c}{\rm{Expected}}\;{\rm{Frequenc}}{{\rm{y}}_1} = 280 \times P\left( 1 \right)\\ = 280 \times 0.34761\\ = 97.33\end{aligned}\)

\(\begin{aligned}{c}{\rm{Expected}}\;{\rm{Frequenc}}{{\rm{y}}_2} = 280 \times P\left( 2 \right)\\ = 280 \times 0.121663\\ = 34.07\end{aligned}\)

\(\begin{aligned}{c}{\rm{Expected}}\;{\rm{Frequenc}}{{\rm{y}}_3} = 280 \times P\left( 3 \right)\\ = 280 \times 0.028388\\ = 7.95\end{aligned}\)

\(\begin{aligned}{c}{\rm{Expected}}\;{\rm{Frequenc}}{{\rm{y}}_4} = 280 \times P\left( 4 \right)\\ = 280 \times 0.004968\\ = 1.39\end{aligned}\)

The actual and expected frequencies corresponding to the number of deaths aretabulated below:

Since the actual frequencies are approximately equal to the expected frequencies, it can be said that the Poisson distribution appears to be a good tool for predicting the actual frequencies