Respuesta :
Using the exponential distribution, it is found that the proportion of the population that will visit an emergency room in the next six months is of 0.1813 = 18.13%.
Exponential distribution
The exponential probability distribution, with mean m, is described by the following mass function:
[tex]f(x) = \mu e^{-\mu x}[/tex]
In this function, [tex]\mu = \frac{1}{m}[/tex] represents 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]
The solution of the definite integral gives the cumulative mass function as follows:
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
In this problem, the mean is of 2.5 years, hence the decay parameter is calculated as follows:
[tex]\mu = \frac{1}{2.5} = 0.4[/tex]
The proportion of the population that will visit the emergency room in the next six months is P(X < 0.5), as 6 months = 0.5 years, hence:
[tex]P(X \leq 0.5) = 1 - e^{-0.4 \times 0.5} = 0.1813.[/tex]
Missing Information
The problem asks for the proportion of the population that will visit an emergency room in the next six months.
More can be learned about the exponential distribution at https://brainly.com/question/14634921
#SPJ1
