Respuesta :
Answer:
a) [tex]\lambda = 0.0666[/tex]
b) 48.62% probability of a waiting time less than 10 minutes between successive speeders
c) 18.92% probability of a waiting time in excess of 25 minutes between successive speeders
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]
(a) Calculate the rate parameter λ.
Mean of 15 minutes. So
[tex]\lambda = \mu = \frac{1}{15} = 0.0666[/tex]
(b) What is the probability of a waiting time less than 10 minutes between successive speeders?
[tex]P(X \leq 10) = 1 - e^{-0.0666*10} = 0.4862[/tex]
48.62% probability of a waiting time less than 10 minutes between successive speeders
(c) What is the probability of a waiting time in excess of 25 minutes between successive speeders?
Either the waiting time is 25 or less, or it is in excess of 25 minutes. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 25) + P(X > 25) = 1[/tex]
We want P(X > 25). So
[tex]P(X > 25) = 1 - P(X \leq 25) = 1 - (1 - e^{-0.0666*25}) = 0.1892[/tex]
18.92% probability of a waiting time in excess of 25 minutes between successive speeders
