W4_Tutorial_Q_2_8 - Tutorial 3 2. The interarrival time...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Tutorial 3 2. The interarrival time follows an exponential distribution with mean 10 or parameter 0.1: so the counting process for the arrival of events is Poisson distributed with the same parameter 0.1. Let this process be N(t). We are looking for the probability that at least two customers arrive in a 10 minute period, i.e. ( 10 2 = 1 10 = 0 10 = 1 And 10 = 0 = ? 0.110 0.1 10 0! = ? 1 10 = 1 = ? 0.110 0.1 10 1 1! = ? 1 ( 10 2 = 1 2 ? 1 = 0.2642 8. a) The number of storms N(t) up to time t depends on whether it is a good yearoranother year. If it is a good year N(t) is Poisson rate 3, but if it is an other year N(t) is Poisson rate 5 i.e. in other years there are typically more storms. Because N(t) depends on which type of year it is, let us condition on which type of year it is: = 1 ? ? ??...
View Full Document

Page1 / 2

W4_Tutorial_Q_2_8 - Tutorial 3 2. The interarrival time...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online