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?

Accepted Solution

Answer:a) 0.00673b)0.9596Step-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 tIn our exercise t = 1 yearλ : lambdaThe probability function for X is :[tex]P(X=x)=\frac{e^{-( lambda).t}.(lambda.t)^{x} }{x!}[/tex]x ≥ 0a) [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]