Question

The length of life (in days) of an alkaline battery has an exponential distribution with an average life of 340 days. (Round your answers to three decimal places.) (a) What is the 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.)

Answer 1

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:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

In this problem, we have that:

$m = 340$. So

$\mu = \frac{1}{340} = 0.0029$

$P(X \leq x) = 1 - e^{-0.0029x}$

(a) What is the probability that an alkaline battery will fail before 185 days?

$P(X \leq x) = 1 - e^{-0.0029x}$

$P(X \leq 185) = 1 - e^{-0.0029*185} = 0.4196 = 0.42$

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

$P(X \leq 730) + P(X > 730) = 1$

We want P(X > 730). So

$P(X > 370) = 1 - P(X \leq 730) = 1 - (1 - e^{-0.0029*730}) = 0.12$

0.12 = 12% probability that an alkaline battery will last beyond 2 years