Question

On average, about 96 metal sheets are delivered to the panel construction lines in 1 hour. The engineer decides to model the arrival of the metal sheets with a Poisson process. The average waiting time between arrivals in 0.62 minute. This model assumes that the waiting time between arrivals of metal sheets are independently distributed as exponential distribution. What is the probability that there is a waiting of more than 2 minutes between arrivals

Answer 1

3.97% probability that there is a waiting of more than 2 minutes between arrivals

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)`

In this problem, we have that:

`m = 0.62, \mu = (1)/(0.62) = 1.6129`

What is the probability that there is a waiting of more than 2 minutes between arrivals

Either you wait 2 minutes or less, or you wait more than 2 minutes. The sum of the probabilities of these events is decimal 1. So

`P(X \leq 2) + P(X > 2) = 1`

We want P(X > 2). So

`P(X > 2) = 1 - P(X \leq 2)`

In which

`P(X \leq 2) = 1 - e^(-1.6129*2) = 0.9603`

`P(X > 2) = 1 - P(X \leq 2) = 1 - 0.9603 = 0.0397`

3.97% probability that there is a waiting of more than 2 minutes between arrivals