Respuesta :
Answer:
0.4866 = 48.66% probability that the engine will fail before it is inspected
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]
Exponentially distributed with mean 900 hours.
This means that [tex]m = 900, \mu = \frac{1}{900}[/tex]
If the engine is inspected every 600 hours of flight time, what is the probability that the engine will fail before it is inspected?
This is [tex]P(X \leq 600)[/tex], which is:
[tex]P(X \leq 600) = 1 - e^{-\frac{600}{900}} = 0.4866[/tex]
0.4866 = 48.66% probability that the engine will fail before it is inspected
