Respuesta :
Answer:
a) P(t>3)=0.30
b) P(t>10|t>9)=0.67
Step-by-step explanation:
We have a repair time modeled as an exponentially random variable, with mean 1/0.4=2.5 hours.
The parameter λ of the exponential distribution is the inverse of the mean, so its λ=0.4 h^-1.
The probabity that a repair time exceeds k hours can be written as:
[tex]P(t>k)=e^{-\lambda t }=e^{-0.4t}[/tex]
(a) the probability that a repair time exceeds 3 hours?
[tex]P(t>3)=e^{-0.4*3}=e^{-1.2}= 0.30[/tex]
(b) the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours?
The exponential distribution has a memoryless property, in which the probabilities of future events are not dependant of past events.
In this case, the conditional probability that a repair takes at least 10 hours, given that it takes more than 9 hours is equal to the probability that a repair takes at least (10-9)=1 hour.
[tex]P(t>10|t>9)=P(t>1)[/tex]
[tex]P(k>1)=e^{-0.4*1}=0.67[/tex]
