Question

Supposed that over a period of several years the average number of deaths from a certain noncontagious disease has been 5. If the number of deaths from this disease follows the Poisson distribution, what is the probability that during the current year: (e −1 = 0.37, e−2 = 0.14, e−3 = 0.05, e−4 = 0.02, e−5 = 0.01) (a) There will be no deaths from the disease? (b) Two or more people will die from the disease?

Answer 1

a) 0.00673

b)0.9596

Explanation:

Let be X the random variable : ''Number of deaths from this disease''

X ~ P(λt)

Where λ is number of events per unit time and λt is number of events over time period t

In our exercise t = 1 year

λ : lambda

The probability function for X is :

`P(X=x)=(e^(-( lambda).t).(lambda.t)^(x) )/(x!)`

x ≥ 0

a)

`P(X=0) =(e^(-5)(5)^(0) )/(0!)=e^(-5)=0.00673`

b)

`P(X\geq 2)=1-P(X<2)=1-[P(X=0)+P(X=1)]`

`P(X\geq 2)=1-[e^(-5)+(e^(-5).(5)^1)/(1!)}]=1-[e^(-5)+5(e^(-5))]=1-6(e^(-5))=0.9596`