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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