Respuesta :
Answer:
a) P=0.16
b) Accumulated breakdowns: 0 --> P(no special maintenance) = 0.962
Accumulated breakdowns: 1 --> P(no special maintenance) = 0.856
Accumulated breakdowns: 2 --> P(no special maintenance) = 0.641
Step-by-step explanation:
As the time between breakdowns is independent and exponential random variables, we can model the amount of breakdowns per unit of time as a Poisson process.
The rate of breakdowns is
[tex]r=1/5=0.2\,months^{-1}[/tex]
a) For a special maintenance, three consecutive breakdowns have to happen within the next five months, so we have:
[tex]P(x=k,t)=\frac{(rt)^{k}e^{-rt}}{k!}\\\\P(x=3,t=9)=\frac{(0.2*9)^3e^{-0.2*9}}{3!}\\\\P(3,9)=5.832*0.1653/6=0.16[/tex]
The probability of requiring a special maintenance within the next 9 months is P=0.16.
b) The probability of requiring a special maintenance only depends on the amounts of breakdowns since the last special maintenance. It does not depend on the amount of time without special maintenance.
In this case, we don't know if the computer has had any breakdown since the last special maintenance.
In the case the amount of breakdowns is 0 since the last special maintenance, we have to have 3 breakdowns to have a special maintenance in the next 4 months:
[tex]P(x=3,t=4)=\frac{(0.2*4)^3e^{-0.2*4}}{3!}\\\\P(3,4)=0.512*0.4493/6=0.038\\\\\\P=1-P(3,4)=1-0.038=0.962[/tex]
In the case the amount of breakdowns is 1 since the last special maintenance, we have to have 2 breakdowns to have a special maintenance in the next 4 months:
[tex]P(x=2,t=4)=\frac{(0.2*4)^2e^{-0.2*4}}{2!}\\\\P(3,4)=0.64*0.4493/2=0.144\\\\\\P=1-P(3,4)=1-0.144=0.856[/tex]
In the case the amount of breakdowns is 1 since the last special maintenance, we have to have 2 breakdowns to have a special maintenance in the next 4 months:
[tex]P(x=1,t=4)=\frac{(0.2*4)^1e^{-0.2*4}}{1!}\\\\P(3,4)=0.8*0.4493/1=0.359\\\\\\P=1-P(3,4)=1-0.359=0.641[/tex]
Using the exponential distribution, it is found that there is a:
- a) 0.8347 = 83.47% probability that a special maintenance is required within the next 9 months.
- b) 0.4493 = 44.93% probability that it will not be required within the next 4 months.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In which [tex]\mu = \frac{1}{m}[/tex] is the decay parameter.
The probability that x is lower or equal to a is given by:
[tex]P(X \leq x) = \int\limits^a_0 {f(x)} \, dx[/tex]
Which has the following solution:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
In this problem, mean of 5 months, hence:
[tex]m = 5, \mu = \frac{1}{m} = \frac{1}{5} = 0.2[/tex]
Item a:
[tex]P(X \leq 9) = 1 - e^{-0.2(9)} = 0.8347[/tex]
0.8347 = 83.47% probability that a special maintenance is required within the next 9 months.
Item b:
The times are independent, hence this is P(X > 4).
[tex]P(X > 4) = e^{-0.2(4)} = 0.4493[/tex]
0.4493 = 44.93% probability that it will not be required within the next 4 months.
To learn more about the exponential distribution, you can take a look at https://brainly.com/question/18596455
