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Fit a Poisson distribution to following data
No. of deaths 0 1 2 3 4
frequencies 123 59 14 3 1
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Solution :

$$ \begin{array}{l}\text { Let, } m=\frac{\Sigma f_{i} x_{i}}{\sum f_{i}}=\frac{(0 \times 123)+(1 \times 59)+(2 \times 14)+(3 \times 3)+(4 \times 1)}{123+59+14+3+1} \\ \therefore m=0.5 \text { and } N=200 \end{array} $$ By poisson distribution formulae, $$ \begin{array}{l} P(X=x)=\frac{e^{-m} m^{x}}{x !} \\ P(X=0)=\frac{e^{-m} m^{0}}{0 !}=e^{-0.5} \times 1=0.6065 \end{array} $$ Now, By poisson recurrance distribution $$ \begin{aligned} f(0)=N P(0) &=200 \times 0.6065\\ &=121.3 \\ &\approx 121 \\ f(1)=N P(1) &=200 \times e^{-0.5} \times 0.5 \\ &=60.65 \\ &\approx 61 \end{aligned} $$ $$ \begin{aligned} f(2)=N P(2) &=200 \times \frac{e^{-0.5}(0.5)^{2}}{2 !} \\ &=15.16 \\ & \approx 15\\ f(3)=N P(3) &=200 \times \frac{e^{-0.5}(0.5)^{3}}{3 !} \\ &=2.5272 \\ & \approx 3 \\ f(4)=N P(4) &=200 \times \frac{e^{-0.5}(0.5)^{4}}{4 !} \\ &=0.3159 \\ & \approx 0 \end{aligned} $$

$$ \begin{array}{|c|c|c|c|c|c|} \hline \text { No of deaths } & 0 & 1 & 2 & 3 & 4 \\ \hline \text { frequencies } & 121 & 61 & 15 & 3 & 0 \\ \hline \end{array} $$

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