Answer:
[tex] P(X >2) [/tex]
And we can calculate this with the complement rule like this:
[tex] P(X>2) = 1-P(X<2)[/tex]
And using the cdf we got:
[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/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}, x>0[/tex]
And 0 for other case. Let X the random variable of interest:
[tex]X \sim Exp(\lambda=\frac{1}{2.725})[/tex]
Solution to the problem
We want to calculate this probability:
[tex] P(X >2) [/tex]
And we can calculate this with the complement rule like this:
[tex] P(X>2) = 1-P(X<2)[/tex]
And using the cdf we got:
[tex] P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480[/tex]