Respuesta :
Answer:
68% of these phones last 3.87 years.
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]
The average lifetime of a certain new cell phone is 3.4 years.
This means that [tex]m = 3.4, \mu = \frac{1}{3.4} = 0.2941[/tex]
So
[tex]P(X \leq x) = 1 - e^{-0.2941x}[/tex]
68% of these phones last how long (in years)?
This is x for which:
[tex]P(X \leq x) = 0.68[/tex]
[tex]P(X \leq x) = 1 - e^{-0.2941x}[/tex]
Then
[tex]0.68 = 1 - e^{-0.2941x}[/tex]
[tex]e{-0.2941x} = 0.32[/tex]
[tex]\ln{e{-0.2941x}} = \ln{0.32}[/tex]
[tex]-0.2941x = \ln{0.32}[/tex]
[tex]x = -\frac{\ln{0.32}}{0.2941}[/tex]
[tex]x = 3.87[/tex]
68% of these phones last 3.87 years.
