Respuesta :
Answer:
(a) Probability that at most 3 cars per year will experience a catastrophe is 0.2650.
(b) Probability that more than 1 car per year will experience a catastrophe is 0.9596.
Step-by-step explanation:
We are given that the distribution of the number of cars per year that will experience the catastrophe is a Poisson random variable with variance = 5.
Let X = the number of cars per year that will experience the catastrophe
SO, X ~ Poisson([tex]\lambda = 5[/tex])
The probability distribution for Poisson random variable is given by;
[tex]P(X=x) = \frac{e^{-\lambda} \times \lambda^{x} }{x!} ; \text{ where} \text{ x} = 0,1,2,3,...[/tex]
where, [tex]\lambda[/tex] = Poisson parameter = 5 {because variance of Poisson distribution is [tex]\lambda[/tex] only}
(a) Probability that at most 3 cars per year will experience a catastrophe is given by = P(X [tex]\leq[/tex] 3)
P(X [tex]\leq[/tex] 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= [tex]\frac{e^{-5} \times 5^{0} }{0!} +\frac{e^{-5} \times 5^{1} }{1!} +\frac{e^{-5} \times 5^{2} }{2!} +\frac{e^{-5} \times 5^{3} }{3!}[/tex]
= [tex]e^{-5} +(e^{-5} \times 5) +\frac{e^{-5} \times 25 }{2} +\frac{e^{-5} \times 125}{6}[/tex]
= 0.2650
(b) Probability that more than 1 car per year will experience a catastrophe is given by = P(X > 1)
P(X > 1) = 1 - P(X [tex]\leq[/tex] 1)
= 1 - P(X = 0) - P(X = 1)
= [tex]1-\frac{e^{-5} \times 5^{0} }{0!} -\frac{e^{-5} \times 5^{1} }{1!}[/tex]
= 1 - 0.00674 - 0.03369
= 0.9596
