Please explain the solution to the problem based on Poisson Distribution.

Problem:

Consumers arrive in an insurance company branch office for premium payment
according to a Poisson process with rate λ = 4 per hour. The office opens at 9:00 AM. Find the probability that “exactly one customer visits by 9:30 AM and four more visit till 11:30 AM for premium payment”.

Thanks.

 

That's a very basic stats question without any real complicating features so if that's causing you troubles then I'd suggest a bit of a refresher on the basics. ActEd do a Stats Pack that you could look at which covers Binomial, Poisson, Normal distributions etc.

http://www.acted.co.uk/Html/paper_stats_pack.htm

Good luck!

 

hi nagesh,

Here you have to use exponencial distribution caz its has waiting time property.

 

This is incorrect, based on the earlier part of the question.

It's looking for the probability that 1 person arrives between 9 and 9:30, and four arrive between 9:30 and 11:30.

That's actually the product of two Poisson probabilities, neither of which are related to waiting time.

T.

 

hi,

can you please provide the solution for this because i am not able to solve it properly even i have done similar questions...

thanx and regards
Jatin agrawal

 

India exams - Nov'2007- Question 9(b).

 

It is a Poi Process

This is a Poisson process based question.

Here you consider number of events in given time interval (= lambda L * t), instead of events per unit time (lambda L).

Using Poi(lambda L t)

Event 1 - You have 1 visitor in 30 mins ( 1/2 hrs) = prob = ( (Lt)^1 * e ^ -Lt ) / 1!
&
Event 2 - You have 4 visitors in 2 hrs = prob = ( (Lt)^4 * e^ -Lt) / 4!

Poi Dist is memoryless, so 1st visitor in 1/2 hr does not effect prob of 4 more visitors in next 2 hrs.

Also 2 events are indipendent. So we can multiply the probs.

So::: ( (Lt)^1 * e ^ -Lt ) / 1! * ( (Lt)^4 * e^ -Lt) / 4! = Answer