Respuesta :
Answer:
11.7% probability that the time between requests is between 1 and 2 seconds.
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 = 0.5, \mu = \frac{1}{m} = 2[/tex]
Find the probability that the time between requests is between 1 and 2 seconds.
[tex]P(1 \leq X \leq 2) = P(X \leq 2) - P(X \leq 1)[/tex]
[tex]P(X \leq 2) = 1 - e^{-2*2} = 0.9817[/tex]
[tex]P(X \leq 1) = 1 - e^{-2*1} = 0.8647[/tex]
[tex]P(1 \leq X \leq 2) = P(X \leq 2) - P(X \leq 1) = 0.9817 - 0.8647 = 0.117[/tex]
11.7% probability that the time between requests is between 1 and 2 seconds.
