Respuesta :
Answer:
a) P(X > 10) = 0.6473
b) P(X > 20) = 0.4190
c) P(X < 30) = 0.7288
d) x = 68.87
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]
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]
Mean equal to 23.
This means that [tex]m = 23, \mu = \frac{1}{23} = 0.0435[/tex]
(a) P(X >10)
[tex]P(X > 10) = e^{-0.0435*10} = 0.6473[/tex]
So
P(X > 10) = 0.6473
(b) P(X >20)
[tex]P(X > 20) = e^{-0.0435*20} = 0.4190[/tex]
So
P(X > 20) = 0.4190
(c) P(X <30)
[tex]P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288[/tex]
So
P(X < 30) = 0.7288
(d) Find the value of x such that P(X > x) = 0.05
So
[tex]P(X > x) = e^{-\mu x}[/tex]
[tex]0.05 = e^{-0.0435x}[/tex]
[tex]\ln{e^{-0.0435x}} = \ln{0.05}[/tex]
[tex]-0.0435x = \ln{0.05}[/tex]
[tex]x = -\frac{\ln{0.05}}{0.0435}[/tex]
[tex]x = 68.87[/tex]
