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 one death in a corps in a year? b. What is the probability of no deaths in a corps over five years?

Answer 1

A) 0.1252

B) 0.0474

Explanation:

A) First of all let's call the number of soldiers killed by horse kicks each year "X"

Thus, T = 1 and so X will have a Poisson distribution of;

Mean EX = λT = 0.61 x 1 = 0.61

Thus, the probability of more than one death in a corps in a year is given as;

P(X>1) = 1 - P( X = 0) - P(X=1)

= 1 - e^(-0.61)(0.61^(0)/0!) - e^(-0.61)(0.61^(1)/1!)

= 1 - 0.5434 - 0.3314 = 0.1252

B) let X1 represent the number of soldiers killed by horse kicks in 5 years.

Thus, T = 5 and so X1 will have a Poisson distribution of;

Mean EX1 = λT = 0.61 x 5 = 3.05

probability of no deaths in a corps over five years is given as;

P(X=0) = e^(-3.05)(3.05^(0)/0!)

= 0.0474

Answer 2

The answers are;

a. The probability of more than one death in a corps in a year is 0.12521.

b. The probability of no deaths in a corps over five years is 4.736 × 10⁻².

Explanation:

For Poisson Distribution we have

Pₓ(k) =
`((\lambda t)^k e^(-(\lambda t)) )/(k!)`

Where:

λ = Mean per unit time

k = Specified data point

t = time

e = Euler constant

a. The probability of more than one death in a corps in a year is given by

The mean of the Poisson distribution for one year is given as

λ·t = 0.61 × 1 = 0.61

Therefore by using complement principle, we have

P (X >1) = 1 - P(X = 0) - - P(X = 1)

=1-
`((0.61)^0 e^(-(0.61)) )/(0!)` -
`((0.61)^1 e^(-(0.61)) )/(1!)`

= 1 - 0.543 - 0.3314

= 0.12521

b. Here we have t = 5

Therefore the mean = λ·t = 0.61×5 = 3.05

The probability of there being no deaths in a corps for over five years is

P (X =0) =
`((3.05)^0 e^(-(3.05)) )/(0!)` = 4.736 × 10⁻²

The probability is 4.736 × 10⁻² .