Question

) During a professor’s oce hours, students arrive, on average, every ten minutes. Assume that the distribution of the time between arrivals follows an exponential distribution. Suppose that a student has just le. What is the probability that the professor has more than 25 minutes before the next student shows up?

Answer 1

`X \sim Exp (\lambda=(1)/(\mu)= (1)/(10))`

The probability density function would be given by:

`P(x)=(1)/(10) e^{-(1)/(10) x}`

And for this case we want to find this probability:

`P(X>25)`

And we can find this probability with the complement rule and the following integral:

`P(X>25) = 1-P(X<25) = 1- \int_0^(25) 0.1 e^(-0.1 x)dx`

And after solve the integral we got:

`P(X>25)= 1- [-e^(-0.1 x)] \Big|_0^(25) = 1+ [e^(-0.1x) \Big|_0^(25)]`

And using the fundamental theorem of calculus we got:

`P(X>25) =1 +[e^(-0.1*25) -e^(-0.1*0)]=0.0821`

Explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

`P(X)=\lambda e^(-\lambda x)`

Solution to the problem

For this case we define the random variable X time between arrivals and we know that the distribution is given by:

`X \sim Exp (\lambda=(1)/(\mu)= (1)/(10))`

The probability density function would be given by:

`P(x)=(1)/(10) e^{-(1)/(10) x}`

And for this case we want to find this probability:

`P(X>25)`

And we can find this probability with the complement rule and the following integral:

`P(X>25) = 1-P(X<25) = 1- \int_0^(25) 0.1 e^(-0.1 x)dx`

And after solve the integral we got:

`P(X>25)= 1- [-e^(-0.1 x)] \Big|_0^(25) = 1+ [e^(-0.1x) \Big|_0^(25)]`

And using the fundamental theorem of calculus we got:

`P(X>25) =1 +[e^(-0.1*25) -e^(-0.1*0)]=0.0821`

Answer 2

8.21% probability that the professor has more than 25 minutes before the next student shows up

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)`

During a professor’s oce hours, students arrive, on average, every ten minutes.

This means that
`m = 10, \mu = (1)/(10) = 0.1`

What is the probability that the professor has more than 25 minutes before the next student shows up?

Either he has 25 minutes or less, or he has more than 25 minutes. The sum of the probabilities of these events is decimal 1. So

`P(X \leq 25) + P(X > 25) = 1`

We want P(X > 25). So

`P(X > 25) = 1 - P(X \leq 25)`

In which

`P(X \leq 25) = 1 - e^(-0.1*25) = 0.9179`

`P(X > 25) = 1 - P(X \leq 25) = 1 - 0.9179 = 0.0821`

8.21% probability that the professor has more than 25 minutes before the next student shows up