Question

Let x denote the lifetime of a mcchine component with an exponential distribution. The mean time for the component failure is 2500 hours. What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

Answer 1

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations

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 = (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 mean time for the component failure is 2500 hours.

This means that
`m = \frac{2500}, \mu = (1)/(2500) = 0.0004`

What is the probability that the lifetime exceeds the mean time by more than 1 standard deviations?

The standard deviation of the exponential distribution is the same as the mean, so this is P(X > 5000).

`P(X > x) = e^(-0.0004*5000) = 0.1353`

0.1353 = 13.53% probability that the lifetime exceeds the mean time by more than 1 standard deviations