question 14 exercise 16.02 page 341

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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 14Exercise 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…

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

Mean = 9 per hour

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.

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.

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

Answer in the back of the book0.842813

What went wrong?We ignored the cars already

waiting. (Thinking they would be INCLUDED in the Poisson

calculation).But they are EXTRA cars!

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)

Answer in the back of the book0.842813

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

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!

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)

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

more will arrive!

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)

STOP: this is getting messy!

Continuing with this approachWe drew a probability tree.

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

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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