Respuesta :
Answer:
1. P(X=0)=0.135
2. P(T>1)=0.271
Step-by-step explanation:
1. Let X be the number of accidents in the next year. Find the distribution of X and calculate P(X=0).
The rigth distribution to describe this type of event as number of accidents per unit of time is the Poisson distribution.
[tex]P(x=k)\frac{\lambda^ke^{-\lambdat}}{k!}[/tex]
In this case k=0 accidents, parameter λ=2 events/year and t=1 year:
[tex]P(X=0)=\frac{2^0e^{-2}}{0!}=\frac{1*e^{-2}}{1} = 0.135[/tex]
2. Let T be the amount of time until the next accident. Find the distribution of T and calculate P(T>1).
In this case, the time between events can be best described by an exponential distribution:
[tex]P(X>t)=e^{-\lambda t}[/tex]
In this case parameter λ=2 events/year and t=1 year:
[tex]P(X>1)= e^{-2}=0.271[/tex]
