1 / 21

Question 14 Exercise 16.02 page 341

Question 14 Exercise 16.02 page 341. Carwash. This records our frustration with trying to match our answer with the back of the book. Learning did happen along the way! I uploaded it to show how tricky these problems can be…….. Personally I found the wording in this hard to understand….

aadi
Download Presentation

Question 14 Exercise 16.02 page 341

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Question 14Exercise 16.02 page 341 Carwash

  2. This records our frustration with trying to match our answer with the back of the book.Learning did happen along the way!I uploaded it to show how tricky these problems can be……..Personally I found the wording in this hard to understand…

  3. On average 9 drivers per hour pay to use a carwash.Mean = 9 per hour

  4. Each car-wash takes 5 minutes.The carwash closes at 7pm.A car leaves the carwash at 6:40 pm, when there are three cars in line.

  5. a) Assuming a Poisson distribution is an appropriate model for the number of drivers per hour that pay for a car-wash, calculate the probability that there will be one or more drivers waiting in line at closing time.

  6. The mean for the 20 minutes is 3.The three cars in line will be finished by 6:55pm allowing for one more car to arrive and be washed.So we are looking at p(x>4)p(x>4)=1-p(x<=3)= 1 – 0.64723188= 0.3528

  7. Answer in the back of the book0.842813

  8. What went wrong?We ignored the cars already waiting. (Thinking they would be INCLUDED in the Poisson calculation).But they are EXTRA cars!

  9. Lets go back and re-read the question.Perhaps we ignore the ones in line and find the probability that more than one car will come in the 20 minutes.(6:40 to 7pm = 20 mins= 4 cars)P(x>1)= 1 – p(x<=1)

  10. Answer in the back of the book0.842813(Note we were now CORRECT but did not know it!)

  11. Ok – wrong AGAIN!!!PERHAPS we need to split the time – 15min and the last 5 mins.More than one in 15 minsAND more than one in 5PLUS more than two in 15minsPLUS more than two in 5 mins!

  12. Lets look at the 15 minutes.What is the probability 2 or more cars will arrive?Mean = 9 per hour2.25 per 15 minsp(x>1)=1-p(x<=1)

  13. So whilst the 3 are being washed there is a 0.65745 chance that 2 or more will arrive!

  14. NOW Lets look at the last 5 minutes.What is the probability more than one car will arrive?Mean = 9 per hourso 9/60 per minutetimes by 5 to get per 5 minutes45/60 = 0.75p(x>1)=1-p(x<=1)

  15. STOP: this is getting messy!Continuing with this approachWe drew a probability tree.

  16. Which is the SAME answer as the p(x>2) – which was a lot quicker!

  17. Then we found a text book with the answer0.80086hand written in the back of the book.So WE WERE CORRECT!(a bit of rounding error)

More Related