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Question 14 Exercise 16.02 page 341

Date post: 08-Feb-2016
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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…. - PowerPoint PPT Presentation
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Question 14 Exercise 16.02 page 341 Carwash
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Page 1: Question 14 Exercise 16.02 page 341

Question 14Exercise 16.02 page 341

Carwash

Page 2: Question 14 Exercise 16.02 page 341

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…

Page 3: Question 14 Exercise 16.02 page 341

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

Mean = 9 per hour

Page 4: Question 14 Exercise 16.02 page 341

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.

Page 5: Question 14 Exercise 16.02 page 341

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.

Page 6: Question 14 Exercise 16.02 page 341

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

Page 7: Question 14 Exercise 16.02 page 341
Page 8: Question 14 Exercise 16.02 page 341

Answer in the back of the book0.842813

Page 9: Question 14 Exercise 16.02 page 341

What went wrong?We ignored the cars already

waiting. (Thinking they would be INCLUDED in the Poisson

calculation).But they are EXTRA cars!

Page 10: Question 14 Exercise 16.02 page 341

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)

Page 11: Question 14 Exercise 16.02 page 341
Page 12: Question 14 Exercise 16.02 page 341

Answer in the back of the book0.842813

(Note we were now CORRECT but did not know it!)

Page 13: Question 14 Exercise 16.02 page 341

Ok – wrong AGAIN!!!PERHAPS we need to split the time –

15min and the last 5 mins.More than one in 15 mins AND more than one in 5

PLUS more than two in 15minsPLUS more than two in 5 mins!

Page 14: Question 14 Exercise 16.02 page 341

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)

Page 15: Question 14 Exercise 16.02 page 341
Page 16: Question 14 Exercise 16.02 page 341

So whilst the 3 are being washed there is a 0.65745 chance that 2 or

more will arrive!

Page 17: Question 14 Exercise 16.02 page 341

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 minute

times by 5 to get per 5 minutes45/60 = 0.75

p(x>1)=1-p(x<=1)

Page 18: Question 14 Exercise 16.02 page 341
Page 19: Question 14 Exercise 16.02 page 341

STOP: this is getting messy!

Continuing with this approachWe drew a probability tree.

Page 20: Question 14 Exercise 16.02 page 341

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

Page 21: Question 14 Exercise 16.02 page 341

Then we found a text book with the answer

0.80086hand written in the back of the book.

So WE WERE CORRECT!(a bit of rounding error)


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