Question

6.81 The length of time between breakdowns of an es- sential piece of equipment is important in the decision of the use of auxiliary equipment. An engineer thinks that the best model for time between breakdowns of a generator is the exponential distribution with a mean of 15 days. (a) If the generator has just broken down, what is the probability that it will break down in the next 21 days? (b) What is the probability that the generator will op- erate for 30 days without a breakdown?

Answer 1

a) 75.34%

b) 13.53%

Explanation:

First, let us consider the cumulative exponential distribution

`F(t) = P[T\leq t] = 1 - e^(-rt)`

Where

`r=(1)/(m)` and m denotes the mean

Therefore, our cumulative distribution for this exercise is as it follows:

`P[T\leq t] = 1-e^{-(1)/(15)t }`

a) According to the distribution, the first question can be calculated as it follows:

`P[T\leq 21]=1-e^{-(1)/(15)*21 } =1-e^{-(7)/(5) }`

`P[T\leq 21]=0.7534`

Hence, the probability that the piece will break down in the next 21 days is 75.34%

b) Now, take into account that we now need the probability complement for applying the cumulative distribution like this:

`P[T\geq 30]=1-P[T<30]`

`P[T\geq 30]=1-(1-e^{-(1)/(15)*30 })=e^(-2)`

`P[T\geq 30]=0.1353`

Therefore, we have that the probability that the generator will operate for 30 days without a breakdown is 13.53%