Respuesta :
Using the exponential distribution, it is found that 70% of all light bulbs last at least 2.85 years.
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 problem, the mean lifetime is of 8 years, hence [tex]m = 8, \mu = \frac{1}{8} = 0.125[/tex].
Seventy percent of all light bulbs last at least x, considering that [tex]P(X > x) = 0.7[/tex].
Hence:
[tex]P(X > x) = 0.7[/tex]
[tex]e^{-0.125x} > 0.7[/tex]
[tex]\ln{e^{-0.125x}} > \ln{0.7}[/tex]
[tex]-0.125x > \ln{0.7}[/tex]
[tex]x > -\frac{\ln{0.7}}{0.125}[/tex]
[tex]x > 2.85[/tex]
70% of all light bulbs last at least 2.85 years.
To learn more about the exponential distribution, you can take a look at https://brainly.com/question/18596455
