I am emailing a link of this to everyone on the class list every week. If you are not receiving these emails or want to have them sent to another email address feel free to email me at jpmccarthymaths@gmail.com and I will add you to the mailing list.

Maple Catch Up

If you want to do Maple catch up, if for example you miss a lab, please download the Maple file from your email, do the exercises in Maple (either in a new worksheet or on top of the original file with the exercises), save the worksheet and email me the worksheet you were working on. Please complete your Maple catch up before 30 April.

Summer 2011 Q. 2 (b)

This question came up last night however I said that there was an issue with interpretation but that I felt there was only one reasonable interpretation: that the question required a mix of the Poisson Distribution and the Binomial Distribution, and I have answered the question under that assumption below.

I stated however that you will not get a question like this in your exam because your exam will prompt you to use the Binomial, Poisson or indeed Normal distribution.

I have no idea where I got this question from… it wasn’t on any MATH6038 paper… a mystery!

A production department has 35 milling machines. The number of breakdowns on each machine averages 0.06 per week. Determine the probabilities of having

1. one machine breaking down in a week

2. less than three machines breaking down in a week


1. With an average rate of occurrence, \lambda=0.06 given, assume that the number of times a machine breaks down in a week, X, has a Poisson distribution, X\sim \text{Poi}[0.06]. Therefore to calculate the probability that a machine breaks down during the week we look at the probability of the machine breaking down one or more times:

\mathbb{P}[X\geq 1]=1-\mathbb{P}[\text{not-}(X=0)]=1-e^{-0.06}\approx 0.05824.

2. Now we are looking at 35 ‘trials’ where success is defined as a machine breaking down so that we have p=1-e^{-0.06} — and 1-p=1-(1-e^{-0.06})=e^{-0.06} (note that using 0.05824 introduces rounding error. Not the end of the world in an exam but perhaps in real life!).

Let Y be the number of successes from the n=35 trials so that Y has a binomial distribution; Y\sim \text{Bi}[25,1-e^{-0.06}]. We are looking at the probability of less than three machines breaking down in a week;



={35\choose 0}(1-e^{-0.06})^0(e^{-0.06})^{35}+{35\choose 1}(1-e^{-0.06})^1(e^{-0.06})^{34}+

{35\choose 2}(1-e^{-0.06})^2(e^{-0.06})^{33}



Week 10

We looked at reliability block diagrams, the binomial distribution, the Poisson distribution and we started talking about the normal distribution.

Week 11

We will look at the normal distribution in more detail and talk about sampling.

Independent Learning: Exercices

You are supposed to be working outside of class and I am supposed to help you with this. Working outside of class means doing the exercises in the notes. Any work that is handed up will be corrected by me. Also you can ask me a question here on this site and I will answer it ASAP.

Questions that you can do at this point include:

  • P. 108, Q. 1-7
  • P. 112 Q. 1-17
  • P.116, Q. 1-15

I wouldn’t necessarily advise looking at ALL of these questions: I would focus on the exam questions over the other questions.

Academic Learning Centre

Those in danger of failing need to use the Academic Learning Centre. As you can see from the timetable there is evening support. You will get best results if you come to the helpers there with specific questions.

Math.Stack Exchange

If you find yourself stuck and for some reason feel unable to ask me the question you could do worse than go to the excellent site math.stackexchange.com. If you are nice and polite, and show due deference to these principles you will find that your questions are answered promptly.

Additional Notes: E-Books

If you look in the module descriptor, you will see there is some suggested reading. Of course I think my notes are perfect but if you can look here, search for ‘Bird Higher Engineering’ you will see that the library have an E-Book resource.