Respuesta :
Answer:
a) 13.53% probability that the lightbulb will function for more than 2000 days.
b) 22.31% probability that the lightbulb will function for more than 2000 days, given that it is still functional after 500 days.
Step-by-step explanation:
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]
In this problem, we have that:
[tex]m = 1000[/tex]
So
[tex]\mu = \frac{1}{1000} = 0.001[/tex]
(a) Find the probability that the lightbulb will function for more than 2000 days.
Either it will function for 2000 days or less, or it will function for more than 2000 days. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 2000) + P(X > 2000) = 1[/tex]
We want P(X > 2000). So
[tex]P(X > 2000) = 1 - P(X \leq 2000)[/tex]
In which
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
[tex]P(X \leq 2000) = 1 - e^{-0.001*2000} = 0.8647[/tex]
[tex]P(X > 2000) = 1 - P(X \leq 2000) = 1 - 0.8647 = 0.1353[/tex]
13.53% probability that the lightbulb will function for more than 2000 days.
(b) Find the probability that the lightbulb will function for more than 2000 days, given that it is still functional after 500 days.
This is:
[tex]p = \frac{P(X > 2000)}{P(X > 500)}[/tex]
From a, we have that P(X > 2000) = 0.1353.
We also have that, following the same logic as a):
[tex]P(X > 500) = 1 - P(X \leq 500)[/tex]
In which
[tex]P(X \leq x) = 1 - e^{-\mu x}[/tex]
[tex]P(X \leq 500) = 1 - e^{-0.001*500} = 0.3935[/tex]
[tex]P(X > 500) = 1 - P(X \leq 500) = 1 - 0.3935 = 0.6065[/tex]
So
[tex]p = \frac{P(X > 2000)}{P(X > 500)} = \frac{0.1353}{0.6065} = 0.2231[/tex]
22.31% probability that the lightbulb will function for more than 2000 days, given that it is still functional after 500 days.
