Answer:
1.83% probability that the time until the first critical-part failure is more than 2 hours.
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]
In this question, we have that:
[tex]m = 0.5, \mu = \frac{1}{0.5} = 2[/tex]
Find the probability that the time until the first critical-part failure is more than 2 hours.
[tex]P(X > 2) = e^{-2*2} = 0.0183[/tex]
1.83% probability that the time until the first critical-part failure is more than 2 hours.
