| View Question |
|
|
 | ||
|
| Subject:
Math Question for Poisson distribution of a restaurant
Category: Science > Math Asked by: scifinut69-ga List Price: $10.00 |
Posted:
20 Oct 2004 03:29 PDT
Expires: 20 Oct 2004 13:49 PDT Question ID: 417444 |
Customers at a popular restaurant that refuses reservations arrive according to the Poisson distribution at a rate of 4 parties every 5 minutes. What is the probability that there will be more than 2 minutes between arriving parties? |
| ||
|
| There is no answer at this time. |
| ||
|
| Subject:
Re: Math Question for Poisson distribution of a restaurant
From: padmapani-ga on 20 Oct 2004 06:04 PDT |
If the arrival rate is poisson distributed then the inter-arrival time is exponentially distributed. Therefore I am hazarding a guess that the required probability is 1-e^lt where l= arrival rate = 4/5 per minute and t=2 minutes. Therefor the reqd. prob = 1 -e^(8/5)= 1-0.201896518 = 0.798. The researcher would probably confirm or deny this. |
| Subject:
Re: Math Question for Poisson distribution of a restaurant
From: pafalafa-ga on 20 Oct 2004 06:43 PDT |
It depends whether poisson was on the menu or not... |
If you feel that you have found inappropriate content, please let us know by emailing us at answers-support@google.com with the question ID listed above. Thank you. |
| Search Google Answers for |
| Google Home - Answers FAQ - Terms of Service - Privacy Policy |