Answer:
3.97% probability that there is a waiting of more than 2 minutes between arrivals
Step-by-step explanation:
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]
In this problem, we have that:
[tex]m = 0.62, \mu = \frac{1}{0.62} = 1.6129[/tex]
What is the probability that there is a waiting of more than 2 minutes between arrivals
Either you wait 2 minutes or less, or you wait more than 2 minutes. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 2) + P(X > 2) = 1[/tex]
We want P(X > 2). So
[tex]P(X > 2) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = 1 - e^{-1.6129*2} = 0.9603[/tex]
[tex]P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9603 = 0.0397[/tex]
3.97% probability that there is a waiting of more than 2 minutes between arrivals
