Respuesta :
Using the exponential distribution, it is found that there is a 0.2231 = 22.31% probability after a call is received, it takes more than 3 minutes for the next call to occur.
What is the 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 of finding a value higher than x is:
[tex]P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}[/tex]
In this problem, the mean is of two minutes, hence the decay parameter is given by:
[tex]\mu = \frac{1}{2} = 0.5[/tex]
The probability that it takes more than 3 minutes for the next call to occur is given by:
[tex]P(X > 3) = e^{-0.5 \times 3} = 0.2231[/tex]
0.2231 = 22.31% probability after a call is received, it takes more than 3 minutes for the next call to occur.
More can be learned about the exponential distribution at https://brainly.com/question/18596455
