140k views
0 votes
The time between customers is exponentially distributed with a mean of 0.5 minutes. Given that 60 minutes have elapsed since the last customer, find the probability that we must wait an additional 5 mins for the arrival of the next customer

User Jalperin
by
8.5k points

1 Answer

5 votes

Answer:

The probability that we must wait 5 more minutes is 0.0000454

Explanation:

Let X be the amount of minutes that passed between the two customers. X has emponential distribution with parameter
\lambda = 1/0.5 = 2 and its density is


f_X(x) = 2 \, e^(-2 \, x)

We want to compute is P(X> 65 | X > 60), which is equal to P(X>65)/P(X>60) (Because the conditional event is contained in the other event). Lets do some computations


P(X>60) = \int\limits_(60)^(+\,\infty) {2e^(-2 \, x) } \, dx = - e^(-2x) \, |^(+\infty)_(\, 60) =e^(-120)


P(X>65) = \int\limits_(65)^(+\,\infty) {2e^(-2 \, x) } \, dx = - e^(-2x) \, |^(+\infty)_(\, 65) = e^(-130)

As a consequence


P(X > 65 | X > 60) = (P(X > 65))/(P(X>60)) = (e^(-130))/(e^(-120)) = e^(-10) = 0.0000454

Note that this is the same result than just calculate P(X > 5), and 5 is, by the way, the difference between 60 and 65. This is commonly referred as the memoryless property: The past doesnt have influence in the future.

We conclude that the probability that we must wait 5 more minutes is 0.0000454.

User Yegor Babarykin
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories