Question

In 1898, L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that the number of soldiers killed by horse kicks each year in each corps in the Prussian cavalry followed a Poisson distribution with a mean of 0.61. (a) What is the probability of more than 1 death in a corps in a year? (b) What is the probability of no deaths in a corps over 7 years? Round your answers to four decimal places (e.g. 98.7654).

Answer 1

Probability of more than 1 death in a corps in a year

12.52%

Probability of no deaths in a corps over 7 years

1.40%

Explanation:

If the data collected follow a Poisson distribution with mean 0.61, then

Probability of k deaths in a year is

`P(k\;deaths\;in\;one\;year)=(0.61^ke^(-0.61))/(k!)`

The probability of more than 1 death in a corps in a year would be

(1)
`P(2)+P(3)+...=\sum_(k=2)^(\infty)P(k)=\sum_(k=2)^(\infty)(0.61^ke^(-0.61))/(k!)=e^(-0.61)\sum_(k=2)^(\infty)(0.61^k)/(k!)`

But the Taylor series around 0 for the exponential function is

`e^x=\sum_(k=0)^(\infty)(x^k)/(k!)`

So,

`e^(0.61)=(0.61^0)/(0!)+(0.61^1)/(1!)+\sum_(k=2)^(\infty)(0.61^k)/(k!)`

and

`\sum_(k=2)^(\infty)(0.61^k)/(k!)=e^(0.61)-1.61`

Replacing in (1), we obtain

`\sum_(k=2)^(\infty)P(k)=e^(-0.61)(e^(0.61)-1.61)=1-1.61e^(-0.61)=1-0.874794\approx 0.1252`

or 12.52%

b)

The probability of 0 deaths in one year is

`P(0)=e^(-0.61)`

As the events are independent, the probability of 0 deaths over 7 years is the product of P(0) by itself 7 times:

`P(0))^7=(e^(-0.61))^7=e^(-4.27)\approx 0.0140`

or 1.40%