Question

The average lifetime of a certain new cell phone is 3.4 years. The manufacturer will replace any cell phone failing within three years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. 68% of these phone last how long (in years)?

Answer 1

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:

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

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:

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

The probability of finding a value higher than x is:

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

The average lifetime of a certain new cell phone is 3.4 years.

This means that $m = 3.4, \mu = \frac{1}{3.4} = 0.2941$

So

$P(X \leq x) = 1 - e^{-0.2941x}$

68% of these phones last how long (in years)?

This is x for which:

$P(X \leq x) = 0.68$

$P(X \leq x) = 1 - e^{-0.2941x}$

Then

$0.68 = 1 - e^{-0.2941x}$

$e{-0.2941x} = 0.32$

$\ln{e{-0.2941x}} = \ln{0.32}$

$-0.2941x = \ln{0.32}$

$x = -\frac{\ln{0.32}}{0.2941}$

$x = 3.87$

68% of these phones last 3.87 years.