Respuesta :
Using the exponential distribution, it is found that there is a:
a) 0.2548 = 25.48% probability that the length of stay in the ICU is one day or less.
b) 0.1415 = 14.15% probability that the length of stay in the ICU is between two and three days.
c) 0.2298 = 22.98% probability that the length of stay in the ICU is more than five days.
Exponential distribution:
The exponential probability distribution, with mean m, is described by the following equation: Â
[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: Â
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
The probability of finding a value higher than x is:
[tex]P(X > x) = e^{-\mu x}[/tex]
In this problem, the mean is of 3.4 days, hence, the decay parameter is:
[tex]\mu = \frac{1}{3.4}[/tex]
Item a:
[tex]P(X \leq 1) = 1 - e^{-\frac{1}{3.4}} = 0.2548[/tex]
0.2548 = 25.48% probability that the length of stay in the ICU is one day or less.
Item b:
[tex]P(X \leq 2) = 1 - e^{-\frac{2}{3.4}} = 0.4447[/tex]
[tex]P(X \leq 3) = 1 - e^{-\frac{3}{3.4}} = 0.5862[/tex]
[tex]P(2 \leq X \leq 3) = 0.5862 - 0.4447 = 0.1415[/tex]
0.1415 = 14.15% probability that the length of stay in the ICU is between two and three days.
Item c:
[tex]P(X > 5) = e^{-\frac{5}{3.4}} = 0.2298[/tex]
0.2298 = 22.98% probability that the length of stay in the ICU is more than five days.
A similar problem is given at https://brainly.com/question/17039711
