Respuesta :
Answer:
[tex] P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048[/tex]
Step-by-step explanation:
Previous concepts
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]
Solution to the problem
For this case the time between breakdowns representing our random variable T is exponentially distirbuted [tex] T \sim Exp (\mu = 10)[/tex]
So on this case we can find the value of [tex]\lambda[/tex] like this:
[tex] \lambda = \frac{1}{\mu} = \frac{1}{10}[/tex]
So then our density function would be given by:
[tex]P(T)=\lambda e^{-\frac{t}{10}}[/tex]
The exponential distribution is useful when we want to describe the waiting time between Poisson occurrences. If we assume that the random variable T represent the waiting time between two consecutive event, we can define the probability that 0 events occurs between the start and a time t, like this:
[tex]P(T>t)= e^{-\lambda t}[/tex]
And on this case we are looking for this probability:
[tex] P(T>1) = e^{-\frac{1}{10}}= e^{-0.1}= 0.9048[/tex]
