Respuesta :
Answer:
a) 0.42 = 42% probability that an alkaline battery will fail before 185 days
b) 0.12 = 12% probability that an alkaline battery will last beyond 2 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]
In this problem, we have that:
[tex]m = 340[/tex]. So
[tex]\mu = \frac{1}{340} = 0.0029[/tex]
[tex]P(X \leq x) = 1 - e^{-0.0029x}[/tex]
(a) What is the probability that an alkaline battery will fail before 185 days?
[tex]P(X \leq x) = 1 - e^{-0.0029x}[/tex]
[tex]P(X \leq 185) = 1 - e^{-0.0029*185} = 0.4196 = 0.42[/tex]
0.42 = 42% probability that an alkaline battery will fail before 185 days
(b) What is the probability that an alkaline battery will last beyond 2 years? (Use the fact that there are 365 days in one year.)
2 years = 2*365 = 730 days.
Either it lasts 730 or less days, or it last more. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 730) + P(X > 730) = 1[/tex]
We want P(X > 730). So
[tex]P(X > 370) = 1 - P(X \leq 730) = 1 - (1 - e^{-0.0029*730}) = 0.12[/tex]
0.12 = 12% probability that an alkaline battery will last beyond 2 years
