Answer:
a) 0.00673
b)0.9596
Step-by-step 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 :
[tex]P(X=x)=\frac{e^{-( lambda).t}.(lambda.t)^{x} }{x!}[/tex]
x ≥ 0
a)
[tex]P(X=0) =\frac{e^{-5}(5)^{0} }{0!}=e^{-5}=0.00673[/tex]
b)
[tex]P(X\geq 2)=1-P(X<2)=1-[P(X=0)+P(X=1)][/tex]
[tex]P(X\geq 2)=1-[e^{-5}+\frac{e^{-5}.(5)^1}{1!}}]=1-[e^{-5}+5(e^{-5})]=1-6(e^{-5})=0.9596[/tex]
