Respuesta :
Answer:
[tex] P(X>2) = 1-P(X\leq 2) =1- [ 1-e^{-\frac{1}{87} x}]= e^{-\frac{1}{87}*120}=0.252[/tex]
[tex] P(X<3) = 1-e^{-\frac{1}{87} x}]= 1-e^{-\frac{1}{87}*180}=0.874[/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 that represent "The number of years a radio functions" and we know that the distribution is given by:
[tex]X \sim Exp(\lambda=\frac{1}{\mu})[/tex]
Where [tex] \lambda = \frac{1}{87}[/tex]
Solution to the problem
The cumulative distribution function for the exponential distribution is given by:
[tex]P(X\leq x) = 1-e^{-\frac{1}{87} x}[/tex]
We need to take in count that 2hours = 120 minutes and 3 hours = 180 minutes
And we want to find these probabilities:
[tex] P(X>2) = 1-P(X\leq 2) =1- [ 1-e^{-\frac{1}{87} x}]= e^{-\frac{1}{87}*120}=0.252[/tex]
[tex] P(X<3) = 1-e^{-\frac{1}{87} x}]= 1-e^{-\frac{1}{87}*180}=0.874[/tex]
