Discuss the importance of constructing confidence intervals for the population mean., Course Project  Phase 2 help
This week you will begin working on Phase 2 of your course project. Using the same data set and variables for your selected topic, add the following information to your analysis:
 Discuss the importance of constructing confidence intervals for the population mean.
 What are confidence intervals?
 What is a point estimate?
 What is the best point estimate for the population mean? Explain.
 Why do we need confidence intervals?
 Based on your selected topic, evaluate the following:
 Find the best point estimate of the population mean.
 Construct a 95% confidence interval for the population mean. Assume that your data is normally distributed and ÃÆ’ is unknown.
 Please show your work for the construction of this confidence interval and be sure to use the Equation Editor to format your equations.
 Write a statement that correctly interprets the confidence interval in context of your selected topic.
 Based on your selected topic, evaluate the following:
 Find the best point estimate of the population mean.
 Construct a 99% confidence interval for the population mean. Assume that your data is normally distributed and ÃÆ’ is unknown.
 Please show your work for the construction of this confidence interval and be sure to use the Equation Editor to format your equations.
 Write a statement that correctly interprets the confidence interval in context of your selected topic.
 Compare and contrast your findings for the 95% and 99% confidence interval.
 Did you notice any changes in your interval estimate? Explain.
 What conclusion(s) can be drawn about your interval estimates when the confidence level is increased? Explain.
This assignment should be formatted using APA guidelines and a minimum of 2 pages in length.
Tutor Answer
Attached.
check chapter 11 it has all the information relating to confidence level.
CT3 – P C – 13
Combined Materials Pack
ActEd Study Materials: 2013 Examinations
Subject CT3
Contents
Study Guide for the 2013 exams
Course Notes
Question and Answer Bank
Series X Assignments*
*Note: The Series X Assignment Solutions should also be supplied
with this pack unless you chose not to receive them with your study material.
If you think that any pages are missing from this pack, please contact ActEd’s admin team
by email at ActEd@bpp.com.
How to use the Combined Materials Pack
Guidance on how and when to use the Combined Materials Pack is set out
in the Study Guide for the 2013 exams.
Important: Copyright Agreement
This study material is copyright and is sold for the exclusive use of the purchaser. You
may not hire out, lend, give out, sell, store or transmit electronically or photocopy any
part of it. You must take care of your material to ensure that it is not used or
copied by anybody else. By opening this pack you agree to these conditions.
The Actuarial Education Company
© IFE: 2013 Examinations
All study material produced by ActEd is copyright and is sold
for the exclusive use of the purchaser. The copyright is owned
by Institute and Faculty Education Limited, a subsidiary of
the Institute and Faculty of Actuaries.
Unless prior authority is granted by ActEd, you may not hire
out, lend, give out, sell, store or transmit electronically or
photocopy any part of the study material.
You must take care of your study material to ensure that it is
not used or copied by anybody else.
Legal action will be taken if these terms are infringed. In
addition, we may seek to take disciplinary action through the
profession or through your employer.
These conditions remain in force after you have finished using
the course.
© IFE: 2013 Examinations
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CT3: Study Guide
Page 1
2013 Study Guide
Subject CT3
Introduction
This Study Guide contains all the information that you will need before starting to study
Subject CT3 for the 2013 exams. Please read this Study Guide carefully before
reading the Course Notes, even if you have studied for some actuarial exams before.
When studying for the UK actuarial exams, you will need a copy of the Formulae and
Tables for Examinations of the Faculty of Actuaries and the Institute of Actuaries,
2nd Edition (2002). These are often referred to as simply the yellow Tables and are
available separately from the Publications shop of the Actuarial Profession. You will
also need a ‘permitted’ scientific calculator from the list published by the Profession.
Please check the list carefully since it is reviewed each year. You will find the list of
permitted calculators and a link to the Publications shop in the profession’s website at
www.actuaries.org.uk.
Contents:
Section 1
Section 2
Section 3
Section 4
Section 5
Section 6
Section 7
The Subject CT3 course structure
ActEd study support
Core Reading, the Syllabus and the Profession
Study skills
Frequently asked questions
Syllabus
Assignment deadlines
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CT3: Study Guide
The Subject CT3 course structure
There are four parts to the Subject CT3 course. This should help you plan your progress
across the study session. The parts cover related topics and have broadly equal lengths.
The parts are broken down into chapters.
The following table shows how the parts, the chapters and the syllabus items relate to
each other. The end columns show how the chapters relate to the days of the regular
tutorials.
Part
1
2
3
4
Chapter
Title
No. of
pages
33
Syllabus
items
(i)
1
Summarising data
2
Probability
37
(ii)
3
Random variables
35
(iii)
4
Probability distributions
69
(iii), (v)
5
Generating functions
46
(iv)
6
Joint distributions
58
(vi)
7
Conditional expectation
21
(xiv)
8
The Central Limit Theorem
19
(vii)
9
Sampling and statistical inference
28
(viii)
10
Point estimation
56
(ix)
11
Confidence intervals
48
(x)
12
Hypothesis testing
65
(xi)
13
Correlation and regression
62
(xii)
14
Analysis of variance
47
(xiii)
© IFE: 2013 Examinations
4 half
days
2 full
days
3 full
days
1
1
1
2
2
3
2
4
3
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ActEd study support
This section lists the study support available from ActEd for Subject CT3.
Course Notes
The Course Notes will help you develop the basic knowledge and understanding of
principles needed to pass the exam. They incorporate the complete Core Reading and
include full explanation of all the syllabus objectives, with worked examples and short
questions to test your understanding.
Each chapter includes the relevant syllabus objectives, a chapter summary and, where
appropriate, a page of important formulae or definitions.
Question and Answer Bank
The Question and Answer Bank provides a comprehensive bank of questions (including
some past exam questions) with full solutions and comments.
The Question and Answer Bank is divided into five parts. The first four parts include a
range of short and long questions to test your understanding of the corresponding part of
the Course Notes. Part five consists of 100 marks of examstyle questions.
Assignments
The four Series X Assignments (X1 to X4) cover the material in Parts 1 to 4
respectively. Assignments X1 and X2 are 80mark tests and should take you two and a
half hours to complete. Assignments X3 and X4 are 100mark tests and should take
you three hours to complete. The actual Subject CT3 examination will have a total of
100 marks.
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CT3: Study Guide
Combined Materials Pack (CMP)
The Combined Materials Pack (CMP) comprises the Course Notes, the Question and
Answer Bank and the Series X Assignments.
The CMP is available in eBook format, enabling it to be downloaded in a rightsprotected PDF format and read on your PC, laptop and ipad (but NOT currently on
Kindles). CMP eBooks can be ordered separately or as an addition to a paper CMP.
Visit www.ActEd.co.uk for more details about compatibility, printing restrictions and
software requirements.
Mock Exam
A 100mark mock exam paper (Mock Exam A) is available for students as a realistic
test of their exam preparation. The mock is based on Mock Exam A from last year but
it has been updated to reflect any changes to the Syllabus and Core Reading.
Additional Mock Pack (AMP)
The Additional Mock Pack (AMP) consists of two further 100mark mock exam papers
– Mock Exam B and Mock Exam C. This is ideal for students who are retaking and
have already sat Mock Exam A, or for those who just want some extra question
practice. If you are retaking this subject you should note that the mock exams in the
AMP use many questions from Mock Exam B for the 2011 exams and the
Y Assignments for the 2011 exams. Therefore, if you purchased Mock Exam B or the
Y Assignments that year you may wish not to purchase the AMP as many of the
questions will be duplicated.
ActEd Solutions with Exam Technique (ASET)
The ActEd Solutions with Exam Technique (ASET) contains ActEd’s solutions to the
previous four years’ exam papers, ie eight papers, plus comment and explanation. In
particular it will highlight how questions might have been analysed and interpreted so as
to produce a good solution with a wide range of relevant points. This will be valuable
in approaching questions in subsequent examinations.
A “MiniASET” will also be available in the summer session covering the April Exam
only.
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CMP Upgrade
The CMP Upgrade lists all significant changes to the Core Reading and ActEd material
so that you can manually amend last year’s study material to make it suitable for study
for this year. The Upgrade includes replacement pages and additional pages where
appropriate. If a large proportion of the material has changed significantly, making it
inappropriate to include all changes, the upgrade will still explain what has changed and
if necessary recommend that students purchase a replacement CMP or Course Notes at a
significantly reduced price. The CMP Upgrade can be downloaded free of charge from
our website at www.ActEd.co.uk.
Revision Notes
ActEd’s Revision Notes have been designed with input from students to help you revise
efficiently. They are suitable for firsttime sitters who have worked through the ActEd
Course Notes or for retakers (who should find them much more useful and challenging
than simply reading through the course again). The Revision Notes are a set of six A5
spiralbound booklets – perfect for revising on the train or tube to work. Each booklet
covers one main theme of the course and includes Core Reading (with a set of
integrated short questions to develop your bookwork knowledge), relevant past exam
questions (with concise solutions) from the last ten years, detailed analysis of key past
exam questions and other useful revision aids.
Flashcards
Flashcards are a set of A6sized cards that cover the key points of the subject that most
students want to commit to memory. Each flashcard has questions on one side and the
answers on the reverse. We recommend that you use the cards actively and test yourself
as you go.
Flashcards may be used to complement your other study and revision materials. They
are not a substitute for question practice but they should help you learn the essential
material required.
Flashcards are available in eBook format, enabling them to be downloaded in a rightsprotected PDF format and read on your PC, laptop, ipad and many smart phones (but
NOT currently on Kindles). Flashcard eBooks can be ordered separately or as an
addition to a paper copy. Visit www.ActEd.co.uk for more details about compatibility,
printing restrictions and software requirements.
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CT3: Study Guide
Marking
We are happy to mark your attempts at any of the currently available assignments, Mock
Exam A or the mock exams included within AMP. Marking is not included with the
products themselves and you need to order it separately. You can submit your scripts by
email, fax or post.
Series Marking and Mock Exam marking
Series Marking (for the Series X Assignments) and Mock Exam Marking (for Mock
Exam A) apply to a specified subject, session and student. If you purchase Series
Marking or Mock Exam Marking, you will not be able to defer the marking to a future
exam sitting or transfer it to a different subject or student.
If you order marking at the same time as you order the assignments or mock exam, you
can choose whether or not to receive a copy of the solutions in advance. If you choose
not to receive the solutions in advance, we will send the solutions to you when we
return your marked script (or following the deadline date if you don’t submit).
If you are having your attempts at the assignments marked by ActEd, you should submit
your scripts regularly throughout the session, in accordance with the schedule of
recommended dates set out in the summary at the end of this document. This will help
you to pace your study throughout the session and leave an adequate amount of time for
revision and question practice.
Any script submitted after the relevant final deadline date will not be marked. It is
your responsibility to ensure that scripts are posted in good time.
Important information
The recommended submission dates are realistic targets for the majority of students.
Your scripts will be returned more quickly if you submit them well before the final
deadline dates.
Marking Vouchers
Marking Vouchers give the holder the right to submit a script for marking at any time,
irrespective of the individual assignment deadlines, study session, subject or person.
Marking Vouchers can be used for any assignment, Mock Exam A, or the mock exams
contained within the AMP. Please note that attempts at the AMP can only be marked
using Marking Vouchers.
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Marking Vouchers are valid for four years from the date of purchase and can be refunded
at any time up to the expiry date.
Important information
Although you may submit your script with a Marking Voucher at any time, you will need
to adhere to the explicit Marking Voucher deadline dates to ensure that your script is
returned before the date of the exam. The deadline dates are given at the end of this study
guide.
If you live outside the UK you must ensure that your last script reaches the ActEd office
earlier than this to allow the extra time needed to return your marked script.
Tutorials
ActEd tutorials are specifically designed to develop the knowledge that you will acquire
from the course material into the higher level understanding that is needed to pass the
exam. We expect you to have read the relevant part of the Course Notes before
attending the tutorial so that the group can spend time on exam questions and discussion
to develop understanding rather than basic bookwork.
ActEd run a range of different tutorials at various locations. Full details are set out in
ActEd’s Tuition Bulletin, which is sent regularly to all students based in the UK, Eire
and South Africa and is also available from the ActEd website at www.ActEd.co.uk.
Regular and Block Tutorials
You can choose one of the following types of tutorial:
Regular Tutorials (usually two or three days) spread over the session.
A Block Tutorial (two or three days) held 2 to 8 weeks before the exam.
The Regular Tutorials provide an even progression through the course. Block Tutorials
cover the whole course.
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CT3: Study Guide
Revision Days
Revision Days are intensive oneday tutorials in the final runup to the exam. They are
particularly suitable for firsttime sitters who attended Regular Tutorials and would like
to spend a day close to the exam focusing on further question practice or retakers who
have already attended ActEd tutorials. Revision Days give you the opportunity to
practise interpreting and answering past exam questions and to raise any outstanding
queries with an ActEd tutor. These courses are most suitable if you have previously
attended Regular Tutorials or a Block Tutorial in the same subject.
Details of how to apply for ActEd’s tutorials are set out in our Tuition Bulletin, which is
sent regularly to all students based in the UK, Eire and South Africa and is also
available from the ActEd website at www.ActEd.co.uk.
Online Classroom
The Online Classroom is an exciting new approach to studying for the actuarial exams.
It acts as either a valuable addon to a facetoface tutorial or a great alternative to a
tutorial, particularly if you're not based in the UK or near a tutorial venue. At the heart
of the Online Classroom in each subject is a comprehensive, easilysearched collection
of over 140 tutorial units. These are a mix of:
teaching units, helping you to really get to grips with the course material, and
guided questions, enabling you to learn the most efficient ways to answer
questions and avoid common exam pitfalls.
The best way to discover the Online Classroom is to see it in action. You can watch a
sample of the Online Classroom tutorial units on the ActEd website at
www.ActEd.co.uk.
© IFE: 2013 Examinations
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Queries and feedback
From time to time you may come across something in the study material that is unclear
to you. The easiest way to solve such problems is often through discussion with friends,
colleagues and peers – they will probably have had similar experiences whilst studying.
If there’s noone at work to talk to then use ActEd’s discussion forum at
www.ActEd.co.uk/forums (or use the link from our home page at www.ActEd.co.uk).
Our online forum is dedicated to actuarial students so that you can get help from fellow
students on any aspect of your studies from technical issues to study advice. You could
also use it to get ideas for revision or for further reading around the subject that you are
studying. ActEd Tutors will visit the site from time to time to ensure that you are not
being led astray and we also post other frequently asked questions from students on the
forum as they arise.
If you are still stuck, then you can send queries by email to CT3@bpp.com or by fax to
01235 550085 (but we recommend that you try the forum first). We will endeavour to
contact you as soon as possible after receiving your query but you should be aware that
it may take some time to reply to queries, particularly when tutors are away from the
office running tutorials. At the busiest teaching times of year, it may take us more than
a week to get back to you.
If you have many queries on the course material, you should raise them at a tutorial or
book a personal tuition session with an ActEd Tutor. Information about personal tuition
is set out in our current brochure. Please email ActEd@bpp.com for more details.
If you find an error in the course, please check the corrections page of our website
(www.ActEd.co.uk/Html/paper_corrections.htm) to see if the correction has already
been dealt with. Otherwise please send details via email to CT3@bpp.com or send a
fax to 01235 550085.
Each year ActEd Tutors work hard to improve the quality of the study material and to
ensure that the courses are as clear as possible and free from errors. We are always
happy to receive feedback from students, particularly details concerning any errors,
contradictions or unclear statements in the courses. If you have any comments on this
course please email them to CT3@bpp.com or fax them to 01235 550085.
The ActEd Tutors also work with the profession to suggest developments and
improvements to the Syllabus and Core Reading. If you have any comments or
concerns about the Syllabus or Core Reading, these can be passed on via ActEd.
Alternatively, you can address them directly to the Profession’s Examination Team at
Napier House, 4 Worcester Street, Oxford, OX1 2AW or by email to
examinations@actuaries.org.uk.
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CT3: Study Guide
Core Reading, the Syllabus and the Profession
Core Reading
The Syllabus for Subject CT3 has been written by the profession to state the
requirements of the examiners. The relevant individual Syllabus Objectives are
included at the start of each course chapter and a complete copy of the Syllabus is
included in Section 6 of the Study Guide. We recommend that you use the Syllabus as
an important part of your study. The Syllabus is supplemented by Core Reading, which
has also been written by the profession. The purpose of Core Reading is to give the
examiners, tutors and students a clear, shared understanding of the depth and breadth of
treatment required by the Syllabus. In examinations students are expected to
demonstrate their understanding of the concepts in Core Reading. Examiners have the
Core Reading available when setting papers.
Core Reading deals with each Syllabus objective. Core Reading covers what is needed to
pass the exam but the tuition material that has been written by ActEd enhances it by
giving examples and further explanation of key points. The Subject CT3 Course Notes
include the Core Reading in full, integrated throughout the course. Here is an excerpt
from some ActEd Course Notes to show you how to identify Core Reading and the
ActEd material. Core Reading is shown in this bold font.
Note that in the example given above, the index will fall if the actual share price goes
below the theoretical exrights share price. Again, this is consistent with what would
happen to an underlying portfolio.
This is
ActEd
After allowing for chainlinking, the formula for the investment index then
text
becomes:
Ni ,t Pi ,t
I (t ) i
This is Core
Reading
B(t )
where Ni ,t is the number of shares issued for the ith constituent at time t;
B(t ) is the base value, or divisor, at time t.
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Core Reading accreditation
The Institute and Faculty of Actuaries would like to thank the numerous people who
have helped in the development of this material and in the previous versions of Core
Reading.
Changes to the Syllabus and Core Reading
The Syllabus and Core Reading are updated as at 31 May each year. The exams in
April and September 2013 will be based on the Syllabus and Core Reading as at
31 May 2012.
We recommend that you always use the uptodate Core Reading to prepare for the
exams.
The Profession’s Copyright
All study material produced by ActEd is copyright and is sold for the exclusive use of
the purchaser. The copyright is owned by Institute and Faculty Education Limited, a
subsidiary of the Institute and Faculty of Actuaries. Unless prior authority is granted
by ActEd, you may not hire out, lend, give out, sell, store or transmit electronically or
photocopy any part of the study material. You must take care of your study material
to ensure that it is not used or copied by anybody else.
Legal action will be taken if these terms are infringed. In addition, we may seek to
take disciplinary action through the profession or through your employer.
These conditions remain in force after you have finished using the course.
Past exam papers
You can download some past exam papers and Examiners’ Reports from the
profession’s website at www.actuaries.org.uk.
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CT3: Study Guide
Further reading
The exam will be based on the relevant Syllabus and Core Reading and the ActEd
course material will be the main source of tuition for students.
However, some students may find it useful to obtain a different viewpoint on a
particular topic covered in Subject CT3. The following list of further reading for
Subject CT3 has been prepared by the Institute and Faculty of Actuaries. This list is not
exhaustive and other useful material may be available.
Mathematical statistics. Freund, John E  7th ed.  Prentice Hall International, 2004.
614 pages. ISBN: 9780131246461.
Available from the Publications Unit.
Calculators
Please refer to the profession’s website for the latest advice on which calculators are
permitted in the exams.
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Study skills
The CT Subject exams
The Core Reading and exam papers for these subjects tend to be very technical. The
exams themselves have many calculation and manipulation questions. The emphasis in
the exam will therefore be on understanding the mathematical techniques and applying
them to various, frequently unfamiliar, situations. It is important to have a feel for what
the numerical answer should be by having a deep understanding of the material and by
doing reasonableness checks.
Subjects CT2 and CT7 are more “wordy” than the other subjects, including an “essaystyle” question in Subject CT7.
Since there will be a high level of mathematics required in the courses it is important
that your mathematical skills are extremely good. If you are a little rusty you may wish
to consider buying the Foundation ActEd Course (FAC) available from ActEd. This
covers all of the mathematical techniques that are required for the CT Subjects, some of
which are beyond ALevel (or Higher) standard. It is a reference document to which you
can refer when you need help on a particular topic.
You will have sat many exams before and will have mastered the exam and revision
techniques that suit you. However it is important to note that due to the high volume of
work involved in the CT Subjects it is not possible to leave all your revision to the last
minute. Students who prepare well in advance have a better chance of passing their
exams on the first sitting.
Unprepared students find that they are under time pressure in the exam. Therefore it is
important to find ways of maximising your score in the shortest possible time. Part of
your preparation should be to practise a large number of examstyle questions under
timed exam conditions as soon as possible. This will:
help you to develop the necessary understanding of the techniques required
highlight which are the key topics that crop up regularly in many different
contexts and questions
help you to practise the specific skills that you will need to pass the exam.
There are many sources of examstyle questions. You can use past exam papers, the
Question and Answer Bank (which includes many past exam questions), assignments,
mock exams, the Revision Notes and ASET.
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CT3: Study Guide
Overall study plan
We suggest that you develop a realistic study plan, building in time for relaxation and
allowing some time for contingencies. Be aware of busy times at work, when you may
not be able to take as much study leave as you would like. Once you have set your plan,
be determined to stick to it. You don’t have to be too prescriptive at this stage about
what precisely you do on each study day. The main thing is to be clear that you will
cover all the important activities in an appropriate manner and leave plenty of time for
revision and question practice.
Aim to manage your study so as to allow plenty of time for the concepts you meet in
this course to “bed down” in your mind. Most successful students will probably aim to
complete the course at least a month before the exam, thereby leaving a sufficient
amount of time for revision. By finishing the course as quickly as possible, you will
have a much clearer view of the big picture. It will also allow you to structure your
revision so that you can concentrate on the important and difficult areas of the course.
A sample CT subject study plan is available on our website at:
www.ActEd.co.uk/Html/help_and_advice_study_plans.htm
It includes details of useful dates, including assignment deadlines and tutorial
finalisation dates.
Study sessions
Only do activities that will increase your chance of passing. Try to avoid including
activities for the sake of it and don’t spend time reviewing material that you already
understand. You will only improve your chances of passing the exam by getting on top
of the material that you currently find difficult.
Ideally, each study session should have a specific purpose and be based on a specific
task, eg “Finish reading Chapter 3 and attempt Questions 1.4, 1.7 and 1.12 from the
Question and Answer Bank”, as opposed to a specific amount of time, eg “Three hours
studying the material in Chapter 3”.
Try to study somewhere quiet and free from distractions (eg a library or a desk at home
dedicated to study). Find out when you operate at your peak, and endeavour to study at
those times of the day. This might be between 8am and 10am or could be in the
evening. Take short breaks during your study to remain focused – it’s definitely time
for a short break if you find that your brain is tired and that your concentration has
started to drift from the information in front of you.
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Order of study
We suggest that you work through each of the chapters in turn. To get the maximum
benefit from each chapter you should proceed in the following order:
1.
Read the Syllabus Objectives. These are set out in the box on Page 1 of each
chapter.
2.
Read the Chapter Summary at the end of each chapter. This will give you a useful
overview of the material that you are about to study and help you to appreciate the
context of the ideas that you meet.
3.
Study the Course Notes in detail, annotating them and possibly making your own
notes. Try the selfassessment questions as you come to them. Our suggested
solutions are at the end of each chapter. As you study, pay particular attention to
the listing of the Syllabus Objectives and to the Core Reading.
4.
Read the Chapter Summary again carefully. If there are any ideas that you can’t
remember covering in the Course Notes, read the relevant section of the notes
again to refresh your memory.
You may like to attempt some questions from the Question and Answer Bank when you
have completed a part of the course. It’s a good idea to annotate the questions with
details of when you attempted each one. This makes it easier to ensure that you try all of
the questions as part of your revision without repeating any that you got right first time.
Once you’ve read the relevant part of the notes and tried a selection of questions from
the Question and Answer Bank (and attended a tutorial, if appropriate) you should
attempt the corresponding assignment. If you submit your assignment for marking,
spend some time looking through it carefully when it is returned. It can seem a bit
depressing to analyse the errors you made, but you will increase your chances of
passing the exam by learning from your mistakes. The markers will try their best to
provide practical comments to help you to improve.
It’s a fact that people are more likely to remember something if they review it from time
to time. So, do look over the chapters you have studied so far from time to time. It is
useful to reread the Chapter Summaries or to try the selfassessment questions again a
few days after reading the chapter itself.
To be really prepared for the exam, you should not only know and understand the Core
Reading but also be aware of what the examiners will expect. Your revision programme
should include plenty of question practice so that you are aware of the typical style,
content and marking structure of exam questions. You should attempt as many questions
as you can from the Question and Answer Bank and past exam papers.
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CT3: Study Guide
Active study
Here are some techniques that may help you to study actively.
1.
Don’t believe everything you read! Good students tend to question everything
that they read. They will ask “why, how, what for, when?” when confronted
with a new concept, and they will apply their own judgement. This contrasts
with those who unquestioningly believe what they are told, learn it thoroughly,
and reproduce it (unquestioningly?) in response to exam questions.
2.
Another useful technique as you read the Course Notes is to think of possible
questions that the examiners could ask. This will help you to understand the
examiners’ point of view and should mean that there are fewer nasty surprises in
the exam room! Use the Syllabus to help you make up questions.
3.
Annotate your notes with your own ideas and questions. This will make you
study more actively and will help when you come to review and revise the
material. Do not simply copy out the notes without thinking about the issues.
4.
Attempt the questions in the notes as you work through the course. Write down
your answer before you check against the solution.
5.
Attempt other questions and assignments on a similar basis, ie write down your
answer before looking at the solution provided. Attempting the assignments
under exam conditions has some particular benefits:
6.
It forces you to think and act in a way that is similar to how you will
behave in the exam.
When you have your assignments marked it is much more useful if the
marker’s comments can show you how to improve your performance
under exam conditions than your performance when you have access to the
notes and are under no time pressure.
The knowledge that you are going to do an assignment under exam
conditions and then submit it (however good or bad) for marking can act as
a powerful incentive to make you study each part as well as possible.
It is also quicker than trying to write perfect answers.
Sit a mock exam four to six weeks before the real exam to identify your
weaknesses and work to improve them. You could use a mock exam written by
ActEd or a past exam paper.
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CT3: Study Guide
5
Page 17
Frequently asked questions
Q:
What knowledge of earlier subjects should I have?
A:
No knowledge will be assumed from other CT series courses.
Q:
What level of mathematics is required?
A:
The level of maths you need for this course is broadly Alevel standard.
However, there may be some symbols (eg the gamma function) that are not
usually included on Alevel syllabuses. You will find the course (and the
exam!) much easier if you feel comfortable with the mathematical techniques
(eg integration by parts) used in the course and you feel confident in applying
them yourself.
If you feel that you need to brush up on your mathematical skills before starting
the course, you may find it useful to study the Foundation ActEd Course (FAC)
or read an appropriate textbook. The full Syllabus for FAC, a sample of the
Course Notes and an Initial Assessment to test your mathematical skills can be
found on our website at www.acted.co.uk.
Q:
What should I do if I’ve never studied statistics before?
A:
Whilst the Core Reading does cover all that you need to know, it is our
experience that students who have not studied any statistics before can find the
pace too fast to gain a sufficient grasp. To prevent such students being
disadvantaged, we have recently developed the Stats Pack, which covers the
basics at a much slower pace with plenty of examples. Sample pages can be
found on our website at www.acted.co.uk.
Q:
What should I do if I discover an error in the course?
A:
If you find an error in the course, please check our website at:
www.acted.co.uk/Html/paper_corrections.htm
to see if the correction has already been dealt with. Otherwise please send
details via email to CT3@bpp.com or send a fax to 01235 550085.
Q:
What calculators am I allowed to use in the exam?
A:
Please refer to the Profession’s website for the latest advice.
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© IFE: 2013 Examinations
Page 18
6
CT3: Study Guide
Syllabus
The full Syllabus for Subject CT3 is given here. To the right of each objective are the
chapter numbers in which the objective is covered in the ActEd course.
Aim
The aim of the Probability and Mathematical Statistics subject is to provide a grounding
in the aspects of statistics and in particular statistical modelling that are of relevance to
actuarial work.
Links to other subjects
Subjects CT4 — Models and CT6 — Statistical Methods: use the statistical concepts
and models covered in this subject. These are then developed further in other subjects
in particular Subject ST1 — Health and Care Specialist Technical, Subject ST7 —
General Insurance — Reserving and Capital Modelling Specialist Technical and Subject
ST8 — General Insurance — Pricing Specialist Technical.
Objectives
On completion of the subject the trainee actuary will be able to:
(i)
Summarise the main features of a data set (exploratory data analysis).
(Chapter 1)
1.
Summarise a set of data using a table or frequency distribution, and
display it graphically using a line plot, a box plot, a bar chart, histogram,
stem and leaf plot, or other appropriate elementary device.
2.
Describe the level/location of a set of data using the mean, median,
mode, as appropriate.
3.
Describe the spread/variability of a set of data using the standard
deviation, range, interquartile range, as appropriate.
4.
Explain what is meant by symmetry and skewness for the distribution of
a set of data.
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CT3: Study Guide
(ii)
(iii)
Explain the concepts of probability.
Page 19
(Chapter 2)
1.
Explain what is meant by a set function, a sample space for an
experiment, and an event.
2.
Define probability as a set function on a collection of events, stating
basic axioms.
3.
Derive basic properties satisfied by the probability of occurrence of an
event, and calculate probabilities of events in simple situations.
4.
Derive the addition rule for the probability of the union of two events,
and use the rule to calculate probabilities.
5.
Define the conditional probability of one event given the occurrence of
another event, and calculate such probabilities.
6.
Derive Bayes’ Theorem for events, and use the result to calculate
probabilities.
7.
Define independence for two events, and calculate probabilities in
situations involving independence.
Explain the concepts of random variable, probability distribution, distribution
function, expected value, variance and higher moments, and calculate expected
values and probabilities associated with the distributions of random variables.
(Chapters 3 and 4)
1.
Explain what is meant by a discrete random variable, define the
distribution function and the probability function of such a variable, and
use these functions to calculate probabilities.
2.
Explain what is meant by a continuous random variable, define the
distribution function and the probability density function of such a
variable, and use these functions to calculate probabilities.
3.
Define the expected value of a function of a random variable, the mean,
the variance, the standard deviation, the coefficient of skewness and the
moments of a random variable, and calculate such quantities.
4.
Evaluate probabilities (by calculation or by referring to tables as
appropriate) associated with distributions.
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Page 20
5.
(iv)
(v)
CT3: Study Guide
Derive the distribution of a function of a random variable from the
distribution of the random variable.
Define a probability generating function, a moment generating function, a
cumulant generating function and cumulants, derive them in simple cases, and
use them to evaluate moments.
(Chapter 5)
1.
Define and determine the probability generating function of discrete,
integervalued random variables.
2.
Define and determine the moment generating function of random
variables.
3.
Define the cumulant generating function and the cumulants, and
determine them for random variables.
4.
Use generating functions to determine the moments and cumulants of
random variables, by expansion as a series or by differentiation, as
appropriate.
5.
Identify the applications for which a probability generating function, a
moment generating function, a cumulant generating function and
cumulants are used, and the reasons why they are used.
Define basic discrete and continuous distributions, be able to apply them and
simulate them in simple cases.
(Chapter 4)
1.
Define and be familiar with the discrete distributions: geometric,
binomial, negative binomial, hypergeometric, Poisson and uniform on a
finite set.
2.
Define and be familiar with the continuous distributions: normal,
lognormal, exponential, gamma, chisquare, t, F, beta and uniform on an
interval.
3.
Define a Poisson process and note the connection between Poisson
processes and the Poisson distribution, and that a Poisson process may be
equivalently characterised as: (1) the distribution of waiting times
between events, (2) the distribution of process increments and (3) the
behaviour of the process over an infinitesimal time interval.
4.
Generate basic discrete and continuous random variables using
simulation methods.
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CT3: Study Guide
(vi)
(vii)
Page 21
Explain the concepts of independence, jointly distributed random variables and
conditional distributions, and use generating functions to establish the
distribution of linear combinations of independent random variables. (Chapter 6)
1.
Explain what is meant by jointly distributed random variables, marginal
distributions and conditional distributions.
2.
Define the probability function/density function of a marginal
distribution and of a conditional distribution.
3.
Specify the conditions under which random variables are independent.
4.
Define the expected value of a function of two jointly distributed random
variables, the covariance and correlation coefficient between two
variables, and calculate such quantities.
5.
Define the probability function/density function of the sum of two
independent random variables as the convolution of two functions.
6.
Derive the mean and variance of linear combinations of random
variables.
7.
Use generating functions to establish the distribution of linear
combinations of independent random variables.
State the central limit theorem, and apply it.
(Chapter 8)
1.
State the central limit theorem for a sequence of independent, identically
distributed random variables.
2.
Apply the central limit theorem to establish normal approximations to
other distributions, and to calculate probabilities.
3.
Explain and apply a continuity correction when using a normal
approximation to a discrete distribution.
(viii) Explain the concepts of random sampling, statistical inference and sampling
distribution, and state and use basic sampling distributions.
(Chapter 9)
1.
Explain what is meant by a sample, a population and statistical inference.
2.
Define a random sample from a distribution of a random variable.
3.
Explain what is meant by a statistic and its sampling distribution.
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Page 22
(ix)
(x)
CT3: Study Guide
4.
Determine the mean and variance of a sample mean and the mean of a
sample variance in terms of the population mean, variance and sample
size.
5.
State and use the basic sampling distributions for the sample mean and
the sample variance for random samples from a normal distribution.
6.
State and use the distribution of the tstatistic for random samples from a
normal distribution.
7.
State and use the F distribution for the ratio of two sample variances
from independent samples taken from normal distributions.
Describe the main methods of estimation and the main properties of estimators,
and apply them.
(Chapter 10)
1.
Describe the method of moments for constructing estimators of
population parameters and apply it.
2.
Describe the method of maximum likelihood for constructing estimators
of population parameters and apply it.
3.
Define the terms: efficiency, bias, consistency and mean squared error.
4.
Define the property of unbiasedness of an estimator and use it.
5.
Define the mean square error of an estimator, and use it to compare
estimators.
6.
Describe the asymptotic distribution of maximum likelihood estimators
and use it.
Construct confidence intervals for unknown parameters.
(Chapter 11)
1.
Define in general terms a confidence interval for an unknown parameter
of a distribution based on a random sample.
2.
Derive a confidence interval for an unknown parameter using a given
sampling distribution.
3.
Calculate confidence intervals for the mean and the variance of a normal
distribution.
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CT3: Study Guide
(xi)
(xii)
Page 23
4.
Calculate confidence intervals for a binomial probability and a Poisson
mean, including the use of the normal approximation in both cases.
5.
Calculate confidence intervals for twosample situations involving the
normal distribution, and the binomial and Poisson distributions using the
normal approximation.
6.
Calculate confidence intervals for a difference between two means from
paired data.
Test hypotheses.
(Chapter 12)
1.
Explain what is meant by the terms null and alternative hypotheses,
simple and composite hypotheses, type I and type II errors, test statistic,
likelihood ratio, critical region, level of significance, probabilityvalue
and power of a test.
2.
Apply basic tests for the onesample and twosample situations involving
the normal, binomial and Poisson distributions, and apply basic tests for
paired data.
3.
Use a 2 test to test the hypothesis that a random sample is from a
particular distribution, including cases where parameters are unknown.
4.
Explain what is meant by a contingency (or twoway) table, and use a 2
test to test the independence of two classification criteria.
Investigate linear relationships between variables using correlation analysis and
regression analysis.
(Chapter 13)
1.
Draw scatterplots for bivariate data and comment on them.
2.
Define and calculate the correlation coefficient for bivariate data, explain
its interpretation and perform statistical inference as appropriate.
3.
Explain what is meant by response and explanatory variables.
4.
State the usual simple regression model (with a single explanatory
variable).
5.
Derive and calculate the least squares estimates of the slope and intercept
parameters in a simple linear regression model.
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© IFE: 2013 Examinations
Page 24
CT3: Study Guide
6.
Perform statistical inference on the slope parameter in simple linear
regression.
7.
Calculate R2 (coefficient of determination) and describe its use to
measure the goodness of fit of a linear regression model.
8.
Use a fitted linear relationship to predict a mean response or an
individual response with confidence limits.
9.
Use residuals to check the suitability and validity of a linear regression
model.
10.
State the usual multiple linear regression model (with several explanatory
variables).
(xiii) Explain the concepts of analysis of variance and use them.
(xiv)
(Chapter 14)
1.
Describe the circumstances in which a oneway analysis of variance can
be used.
2.
State the usual model for a oneway analysis of variance and explain
what is meant by the term treatment effects.
3.
Perform a simple oneway analysis of variance.
Explain the concepts of conditional expectation and compound distribution, and
apply them.
(Chapter 7)
1.
Define the conditional expectation of one random variable given the
value of another random variable, and calculate such a quantity.
2.
Show how the mean and variance of a random variable can be obtained
from expected values of conditional expected values, and apply this.
3.
Derive the moment generating function of the sum of a random number
of independent, identically distributed random variables (a compound
distribution), and use the result to calculate the mean and variance of
such a distribution.
© IFE: 2013 Examinations
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CT3: Study Guide
7
Page 25
Assignment Deadlines – CT Subjects
For the session leading to the April 2013 exams – CT Subjects
Marking vouchers
Subjects
Assignments
Mocks
CT1, CT2, CT5, CT8
20 March 2013
26 March 2013
CT3, CT4, CT6, CT7
26 March 2013
3 April 2013
Recommended
submission date
Final deadline
date
14 November 2012
16 January 2013
21 November 2012
23 January 2013
28 November 2012
6 February 2013
5 December 2012
13 February 2013
23 January 2013
27 February 2013
30 January 2013
6 March 2013
13 February 2013
13 March 2013
20 February 2013
20 March 2013
Subjects
Recommended
submission date
Final deadline
date
CT1, CT2, CT5, CT8
20 March 2013
26 March 2013
CT3, CT4, CT6, CT7
26 March 2013
3 April 2013
Series X Assignments
Subjects
CT1, CT2, CT5, CT8
CT3, CT4, CT6, CT7
CT1, CT2, CT5, CT8
CT3, CT4, CT6, CT7
CT1, CT2, CT5, CT8
CT3, CT4, CT6, CT7
CT1, CT2, CT5, CT8
CT3, CT4, CT6, CT7
Assignment
X1
X2
X3
X4
Mock Exams
We encourage you to work to the recommended submission dates where possible. Please remember that
the turnaround of your script is likely to be quicker if you submit it well before the final deadline date.
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© IFE: 2013 Examinations
Page 26
CT3: Study Guide
For the session leading to the September/October 2013 exams – CT Subjects
Marking vouchers
Subjects
Assignments
Mocks
CT1, CT2, CT5, CT8
28 August 2013
4 September 2013
CT3, CT4, C6, CT7
4 September 2013
11 September 2013
Recommended
submission date
Final deadline
date
12 June 2013
3 July 2013
19 June 2013
10 July 2013
3 July 2013
24 July 2013
10 July 2013
31 July 2013
24 July 2013
7 August 2013
31 July 2013
14 August 2013
7 August 2013
21 August 2013
14 August 2013
28 August 2013
Subjects
Recommended
submission date
Final deadline
date
CT1, CT2, CT5, CT8
21 August 2013
4 September 2013
CT3, CT4, C6, CT7
28 August 2013
11 September 2013
Series X Assignments
Subjects
CT1, CT2, CT5, CT8
CT3, CT4, C6, CT7
CT1, CT2, CT5, CT8
CT3, CT4, C6, CT7
CT1, CT2, CT5, CT8
CT3, CT4, C6, CT7
CT1, CT2, CT5, CT8
CT3, CT4, C6, CT7
Assignment
X1
X2
X3
X4
Mock Exams
We encourage you to work to the recommended submission dates where possible. Please remember that
the turnaround of your script is likely to be quicker if you submit it well before the final deadline date.
© IFE: 2013 Examinations
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CT3: Index
Page 1
CT3 Index
corr( X , Y ) ............................................................................................... Ch6
cov( X , Y ) ................................................................................................ Ch6
C X (t )
................................................................................................. Ch5
................................................................................................. Ch3
E (Y  X ) ................................................................................................. Ch7
Ch13
G X (t )
................................................................................................. Ch5
E( X )
p2223
p2022
p2023
p1113
p36
p3
p39, 14, 23
M X (t )
................................................................................................. Ch1
................................................................................................. Ch5
p1618
p1019, 23
o( h)
................................................................................................. Ch4
IQR
Q1, Q3
................................................................................................. Ch1
p29
p1617, 22
p1617
s2
................................................................................................. Ch1
p14
P( A  B ) ................................................................................................. Ch2
2
S
................................................................................................. Ch9
skew( X ) ................................................................................................. Ch3
................................................................................................. Ch3
................................................................................................. Ch1
p6, 89
p18
p1415
p10
................................................................................................. Ch8
Ch9
................................................................................................. Ch4
Ch9
p2
p5, 8
p2123
p8
G (a )
................................................................................................. Ch4
p14
c
................................................................................................. Ch4
p1618
m3
Ch9
................................................................................................. Ch3
................................................................................................. Ch3
p8
p1113
p18
r
................................................................................................. Ch6
p2223
s2
ti
................................................................................................. Ch3
................................................................................................. Ch14
p1415
p6
∆
«
»
Ã
Œ
................................................................................................. Ch2
................................................................................................. Ch2
................................................................................................. Ch2
................................................................................................. Ch2
................................................................................................. Ch2
p4
p4
p4
p3
p3
A¢ , A
................................................................................................. Ch2
p4
var( X )
x
X
Z
2
m
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© IFE: 2013 Examinations
Page 2
CT3: Index
Analysis of variance (ANOVA)............................................................. Ch14
Least significant difference analysis ........................................ Ch14
Approximations
Distribution of X .................................................................... Ch8
Ch9
Distribution of MLE, qˆ ........................................................... Ch10
Normal approx. of binomial..................................................... Ch8
Normal approx. of chisquare .................................................. Ch8
Normal approx. of gamma ....................................................... Ch8
Normal approx. of Poisson ...................................................... Ch8
Poisson approx. of binomial .................................................... Ch4
Asymptotic
Distribution of MLE ................................................................ Ch10
Distribution of X .................................................................... Ch8
Ch9
Bar Chart ................................................................................................ Ch1
Bayes’ Theorem ..................................................................................... Ch2
Beta
Distribution .............................................................................. Ch4
Function ................................................................................... Ch4
Bias
................................................................................................. Ch10
Box and whisker diagram ...................................................................... Ch1
Boxplot ................................................................................................. Ch1
p315
p2225
p2
p8
p26
p45
p6
p6
p56
p11
p26
p2
p8
p4
p22
p19
p19
p2122
p23
p23
Central Limit Theorem .......................................................................... Ch8
................................................................................................. Ch9
Chisquare .............................................................................................. Ch4
Ch9
Coefficient
Correlation
sample, r .................................................................... Ch13
population, r ............................................................ Ch5
p2
p8
p1618
p8
Determination .......................................................................... Ch13
Skewness
sample ........................................................................ Ch1
population .................................................................. Ch3
Compound distributions ......................................................................... Ch7
Conditional
Expectation .............................................................................. Ch7
Probabilities ............................................................................. Ch2
Probability function ................................................................. Ch6
Probability density function ..................................................... Ch6
Variance ................................................................................... Ch6
Confidence intervals
Mean
m ............................................................................... Ch11
paired data mean difference, m D ............................... Ch11
p19
p8
p2223
p22
p18
p912
p36
p1617, 22
p910
p11
p7
p9
p2627
treatment, mi ............................................................. Ch14
p19
difference between 2 means, m1  m2 ........................ Ch11
p2122
difference between 2 treatment means ....................... Ch14
p2021
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CT3: Index
Page 3
Confidence intervals (ctd)
MLEs, qˆ .................................................................................. Ch10
Poisson parameter
l ............................................................................... Ch11
difference between 2 parameters l1  l2 ................... Ch11
p1820
p25
Proportion
p ................................................................................. Ch11
difference between 2 proportions, p1  p2 ................. Ch11
p1316
p24
Slope parameter, b ................................................................. Ch13
Response
individual ................................................................... Ch13
mean........................................................................... Ch13
Variance
p26
p2223
p2526
p24, 26
s 2 .............................................................................. Ch11
p11
2
linear regression residual, s .................................... Ch13
p21
ANOVA residual, s ................................................ Ch14
p13
2
ratio of 2 variances,
s 12
s 22
...................................... Ch11
Contingency table .................................................................................. Ch12
Continuity correction ............................................................................. Ch8
Ch12
Continuous
Data
................................................................................... Ch1
Distributions............................................................................. Ch4
Random variables .................................................................... Ch3
Convolutions
................................................................................... Ch6
Ch7
Correlation ............................................................................................. Ch6
Ch13
Covariance ............................................................................................. Ch6
CramérRao Lower Bound (CRLB) ....................................................... Ch10
Cumulant
Generating Function (CGF) ..................................................... Ch5
rth, k r ..................................................................................... Ch5
Cumulative distribution function (CDF)
for discrete random variables ................................................... Ch3
for continuous random variables .............................................. Ch3
Cumulative frequency
Diagram ................................................................................... Ch1
Table ........................................................................................ Ch1
p23
p3335
p79
p13, 15
p2
p1326
p47
p2526
p9
p2223
p812
p2022
p26
p2022
p22
p67
p910
p9
p9
Data
Categorical
attribute ...................................................................... Ch1
nominal ...................................................................... Ch1
ordinal ........................................................................ Ch1
Numerical
discrete ....................................................................... Ch1
continuous .................................................................. Ch1
Sample ..................................................................................... Ch1
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p3
p3
p3
p2
p2
p2
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Page 4
CT3: Index
Discrete
Data .......................................................................................... Ch1
Distributions ............................................................................ Ch4
Random variables .................................................................... Ch3
Distribution function (DF)
for discrete random variables ................................................... Ch3
for continuous random variables .............................................. Ch3
Distributions
Discrete
Bernoulli .................................................................... Ch4
Ch6
Binomial .................................................................... Ch4
Ch5
Ch6
Geometric .................................................................. Ch4
Ch6
Hypergeometric ......................................................... Ch4
Negative binomial ...................................................... Ch4
Ch5
Ch6
Poisson ....................................................................... Ch4
Ch5
Ch6
Uniform ..................................................................... Ch4
Ch5
Distributions
Continuous
Beta ............................................................................ Ch4
Chisquare.................................................................. Ch4
Ch5
Ch6
Ch9
Exponential ................................................................ Ch4
Ch5
Ch6
Fdistribution ............................................................. Ch4
Ch9
Gamma ...................................................................... Ch4
Ch5
Ch6
Lognormal ................................................................. Ch4
Normal ....................................................................... Ch4
Ch5
Ch6
Ch9
tdistribution .............................................................. Ch4
Ch9
Uniform ..................................................................... Ch4
Dotplot ................................................................................................. Ch1
p19
p1618
p17
p35
p8
p1516
p17
p34
p26
p14
p1418
p16
p34
p24
p2023
p1819
p3536
p8, 10
p25
p1112
p13
p8
Efficiency ............................................................................................... Ch10
Element ................................................................................................. Ch2
Errors ................................................................................................. Ch12
p23
p3
p5
© IFE: 2013 Examinations
p2
p312
p47
p67
p910
p4
p30
p56
p6, 15
p30
p67
p31
p9
p89
p7, 15
p31
p1012
p8, 15
p32
p3
p6
The Actuarial Education Company
CT3: Index
Estimate ................................................................................................. Ch10
Estimation
Method of moments ................................................................. Ch10
Method of maximum likelihood (MLE) .................................. Ch10
Method of least squares ........................................................... Ch13
Ch14
Estimator ................................................................................................ Ch10
Distribution of .......................................................................... Ch10
Properties of ............................................................................. Ch10
Event ................................................................................................. Ch2
Ch3
Mutually Exclusive .................................................................. Ch2
Independent .............................................................................. Ch2
Expectation
of a distribution, E ( X ) ............................................................ Ch3
using GFs to find...................................................................... Ch5
of compound distributions, E ( S ) ............................................. Ch7
of a function of a distribution, E[ g ( X )] ................................... Ch3
of a function of two random variables E[ g ( X , Y )] ................... Ch6
of a linear function, E (aX + b) ................................................. Ch3
of linear combination, E[c1 X1 + + cn X n ] .............................. Ch6
of a sum, E[ag ( X ) + bh(Y )] ...................................................... Ch6
of a product, E[ g ( X )h(Y )] ....................................................... Ch6
Conditional............................................................................... Ch7
Fisher’s transformation .......................................................................... Ch13
Frequency distribution ........................................................................... Ch1
Grouped ................................................................................... Ch1
Page 5
p4
p38, 31
p920, 31
p14
p8
p4
p26
p2125
p3
p4
p4, 11
p18
p1113
p9, 11, 21
p10
p12
p1517
p1516
p27
p1819
p1819
p36
p11
p4
p5
Gamma
Distribution .............................................................................. Ch4
Function ................................................................................... Ch4
Geometric series..................................................................................... Ch5
Generating functions
Cumulant (CGF) ...................................................................... Ch5
Moment (MGF)........................................................................ Ch5
Ch6
Ch7
Probability (PGF) ..................................................................... Ch5
Ch6
p1418
p14
p26
p2023
p1019, 23
p3336
p11
p39, 14, 23
p2932
Histogram............................................................................................... Ch1
Hypothesis
Alternative, H1 ........................................................................ Ch12
p6
p3
Composite ................................................................................ Ch12
Null, H 0 .................................................................................. Ch12
p3
p3
Simple ...................................................................................... Ch12 p3
Test .......................................................................................... see Tests
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CT3: Index
Independent
Events ...................................................................................... Ch2
p18
Functions of random variables ................................................. Ch6
p13
Random variables .................................................................... Ch6
p1114, 22
Interpolation ........................................................................................... Ch4
p21
Stats Pack
Interquartile range .................................................................................. Ch1
p16
Ch3
p18
Invariance property of MLEs ................................................................. Ch10 p11
Least significant difference analysis of treatment means ....................... Ch14
Least squares estimation
ANOVA model ........................................................................ Ch14
Linear regression model........................................................... Ch13
Level of a test......................................................................................... Ch12
Likelihood .............................................................................................. Ch10
Method of maximum (MLE) ................................................... Ch10
Ratio test .................................................................................. Ch12
Linear regression analysis
Full model ................................................................................ Ch13
Multiple model......................................................................... Ch13
Simple model ........................................................................... Ch13
Lineplot ................................................................................................. Ch1
Location, measures of ............................................................................ Ch1
p2124
p8
p14
p5
p9
p920, 31
p67
p21
p30
p1316
p8
p1012
Maximum likelihood, method of ........................................................... Ch10 p920, 31
Mean
p10
Sample, x ................................................................................ Ch1
Ch8
p2
Ch9
p5, 8
Population, m .......................................................................... Ch3
p11
see also expectation
Mean square error (MSE) ...................................................................... Ch10 p2325
Median
Sample ..................................................................................... Ch1
p1112, 23
Population ................................................................................ Ch3
p19
Ch4
p15
Memoryless property ............................................................................. Ch4
p7, 16
Mode
Sample ..................................................................................... Ch1
p12
Population ................................................................................ Q&A 1.9
X
1.4
Moments (see also mean, expectation, variance and skewness)
Generating function (MGF) ..................................................... Ch5
p1019, 23
Method of ................................................................................ Ch10 p38, 31
Population ................................................................................ Ch3
p1718
using GFs to find ....................................................... Ch5
p8, 1112
of compound distributions ......................................... Ch7
p10
Sample ..................................................................................... Ch1
p15
NeymanPearson Lemma ....................................................................... Ch12
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p6
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CT3: Index
Page 7
Oneway analysis of variance (ANOVA) .............................................. see analysis of variance
Outliers ................................................................................................. Ch1
p23
Partition
Sample space............................................................................ Ch2
Variance in ANOVA model..................................................... Ch14
Variance in linear regression model......................................... Ch13
p21
p11
p11
Pivotal
method of constructing confidence intervals ........................... Ch11 p46
quantity .................................................................................... Ch11 p4
Point estimation ..................................................................................... see estimation
Poisson
Distribution .............................................................................. Ch4
p1012
Process ..................................................................................... Ch4
p12, 2735
Waiting time between events ................................................... Ch4
p15, 3435
Power ................................................................................................. Ch12 p5
Probabilities ........................................................................................... Ch2
p911
Addition rule ............................................................................ Ch2
p12
Conditional............................................................................... Ch2
p1617, 22
Law of total .............................................................................. Ch2
p21
Multiplication rule ................................................................... Ch2
p18
Probability density function (PDF) ........................................................ Ch3
p8
Conditional............................................................................... Ch6
p11
Joint (bivariate) ........................................................................ Ch6
p56
Marginal ................................................................................... Ch6
p89
Probability function (PF) ....................................................................... Ch3
p5
Conditional............................................................................... Ch6
p910
Joint (bivariate) ........................................................................ Ch6
p34
Marginal ................................................................................... Ch6
p78
Process
Poisson ..................................................................................... Ch4
p12, 2735
Waiting time ............................................................................ Ch4
p15, 3435
pvalue ................................................................................................. Ch12 p78
Quartiles
Lower, Q1 ................................................................................ Ch1
p1617, 23
Ch3
Upper, Q3 ................................................................................ Ch1
p19
p1617, 23
Ch3
Random sample, X ............................................................................... Ch9
p19
p3
Random variables................................................................................... Ch3
p3
Continuous ............................................................................... Ch3
p810
Discrete .................................................................................... Ch3
p47
Distributions of ........................................................................ see distributions
Functions of ............................................................................. Ch3
p2024
Simulation ................................................................................ Ch4
p3640
Range ................................................................................................. Ch1
p16
Residuals
ANOVA model ........................................................................ Ch14 p1718
Linear regression model ........................................................... Ch13 p2728
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CT3: Index
Sample
Correlation coefficient, r ........................................................ Ch13
Data .......................................................................................... Ch1
Mean, x ................................................................................... Ch1
Ch8
Ch9
Median ..................................................................................... Ch1
Mode ........................................................................................ Ch1
Moments .................................................................................. Ch1
Random, X ............................................................................. Ch9
p8
p2
p10
p2
p5, 8
p11, 23
p12
p15
p3
Size, for confidence intervals................................................... Ch11
Space........................................................................................ Ch2
Ch3
Standard deviation, s .............................................................. Ch1
p78
p3
p3
p1314
Variance, s 2 ............................................................................ Ch1
Ch9
Scatterplot .............................................................................................. Ch13
Sets
................................................................................................. Ch2
Complement, A¢, A .................................................................. Ch2
p1314
p6, 89
p3
p3
p4
Intersection of, « .................................................................... Ch2
Mutually exclusive................................................................... Ch2
Null, ∆ .................................................................................... Ch2
Subset....................................................................................... Ch2
Ch3
Union of, » ............................................................................. Ch2
Simulation .............................................................................................. Ch4
Size of a test ........................................................................................... Ch12
Skewness
Coefficient
Sample ....................................................................... Ch1
Population .................................................................. Ch3
Formula
Sample ....................................................................... Ch1
Population m3 = skew( X ) ........................................... Ch3
p4
p4, 11
p4
p3
p4
p4
p3640
p5
Ch5
Types ....................................................................................... Ch1
Spread, measures of ............................................................................... Ch1
Standard deviation
Population, s .......................................................................... Ch3
Sample, s ................................................................................ Ch1
Standard error ........................................................................................ Ch9
Statistic Ch9........................................................................................... p4
Ch12
Stem and leaf diagrams .......................................................................... Ch1
Subset ................................................................................................. Ch2
Ch3
Symmetry ............................................................................................... Ch1
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p22
p18
p22
p18
p21
p1921
p1318
p14
p1314
p5
p4
p7
p3
p4
p1819
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CT3: Index
Page 9
Test
ANOVA ................................................................................... Ch14
Best .......................................................................................... Ch12
p13
p67
c 2 goodness of fit ................................................................... Ch12
p29
Correlation coefficient, r ....................................................... Ch13
p11
Mean
m ............................................................................... Ch12
paired data mean difference, m D ............................... Ch12
p9
p25
difference between 2 means, m1  m2 ........................ Ch12
p17
Onesided ................................................................................. Ch12
Poisson parameter
l ............................................................................... Ch12
difference between 2 parameters l1  l2 ................... Ch12
p15
p23
Proportion
p ................................................................................. Ch12
difference between 2 proportions, p1  p2 ................. Ch12
p13
p21
Slope parameter, b ................................................................. Ch13
Twosided ................................................................................ Ch12
Variance
s 2 .............................................................................. Ch12
ratio of 2 variances,
s 12
s 22
...................................... Ch12
Unbiasedness.......................................................................................... Ch10
Uncorrelated........................................................................................... Ch6
Uniqueness property of GFs .................................................................. Ch5
p4
p2223
p4
p12
p20
p2122
p23
p4, 13
Variance
Sample, s 2 ............................................................................... Ch1
Ch9
p1314
p6, 89
Pooled sample, sP2 ................................................................... Ch11
p22
Ch12
p17
Population, s 2 = var( X ) .......................................................... Ch3
Ch5
of a compound distribution ........................................ Ch7
of a linear function, var( aX + b) ................................. Ch3
of linear combination, var[c1 X1 + + cn X n ] .............. Ch6
p1415
p9, 21
p10
p1516
p27
of a sum, var( X + Y ) .................................................. Ch6
of a difference, var( X  Y ) ......................................... Ch6
Conditional ................................................................ Ch7
Venn Diagrams ................................................................................... Ch2
p2324
p24
p7
p57, 1314
Waiting times
p15, 3435
................................................................................... Ch4
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CT301: Summarising data
Page 1
Chapter 1
Summarising data
Syllabus objectives
(i)
0
1.
Summarise a set of data using a table or frequency distribution, and
display it graphically using a line plot, a bar chart, histogram, stem and
leaf plot, or other appropriate elementary device.
2.
Describe the level/location of a set of data using the mean, median,
mode, as appropriate.
3.
Describe the spread/variability of a set of data using the standard
deviation, range, interquartile range, as appropriate.
4.
Explain what is meant by symmetry and skewness for the distribution of a
set of data.
Introduction
This chapter deals with descriptive statistics, that is, the methodology for
describing or summarising a set of data using tables, diagrams and numerical
measures.
Presenting the data in a descriptive form is usually the first stage in any statistical
analysis, as it allows us to spot any patterns in the data. The numerical measures
mentioned are the “average” of the data (ie mean, median and mode) and the “spread”
of the data (ie range, interquartile range (IQR) and variance).
If you have done any statistics before then this chapter will be very straightforward. For
students who have not met this material before and would like more examples and
careful explanations of where these results actually come from, the Stats Pack has been
developed to help. See http://www.acted.co.uk/Html/paper_stats_pack.htm or contact
StatsPack@bpp.com for further details.
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1
Tabular and graphical methods
1.1
Types of data
CT301: Summarising data
Batch data are a set of related observations, such as the current inflation rates of
EU countries.
Sample data are a set of observations selected from a population and designed
to be representative of that population, such as the sums assured for a sample
of 100 policies selected from a company’s wholelife business.
The “sum assured” for a life insurance policy is the amount of benefit that is paid when
the policyholder dies or the policy matures. “Wholelife business” refers to a particular
type of policy. These terms are not important here.
The objectives of an analysis involving batch data will usually be to extract the
important features by summarising the data. For an analysis involving sample
data the objectives will be the same plus the main objective of making inferences
about the population.
In other words, we can use sample data to determine certain properties of the underlying
population from which the sample was taken. This might be useful for example, in
setting up a model which could then be used to predict future behaviour, eg estimating
the number of claims that will be made on a certain type of policy in a given time
period.
Data involve the values of a variable and there are several types of variable:
NUMERICAL
Numerical data can be classified into two types: discrete and continuous. The
distinction between discrete and continuous data is that discrete data can only take one
of a set of particular values, whereas continuous data can take any value within a
specified range (or the possible values are so close together that they can be considered
to occupy a continuous range).
Discrete data arise from counting, eg numbers of actuaries, numbers of claims.
It is also possible to have discrete data that take negative or fractional values. To take a
nonactuarial example, data obtained from measuring the spins of subatomic particles,
which can take any “halfinteger” value ( , 1½, 1, ½,0, ½, 1, 1½, ), would
also be discrete.
Continuous data arise from measuring, eg height, amount, age.
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CT301: Summarising data
Page 3
In actuarial work an “amount” often refers to an amount of money, eg an employee’s
annual salary. “Age” here refers to a person’s “exact age”, not age last birthday. For
example your age might be 21.85 years.
CATEGORICAL
Attribute (or dichotomous) data have only two categories, eg yes/no,
male/female, claim/no claim.
Nominal data have several unordered categories, eg type of policy, nature of
claim.
Ordinal data have several ordered categories, eg questionnaire responses such
as “strongly in favour / … / strongly against”.
Question 1.1
Answer the following dating agency questionnaire and state what type of data is
required in each question:
1. How old are you? (Give your age last birthday.)
2. How tall are you? (State as accurately as you can.)
3. What sex are you?
4. What colour are your eyes?
5. Do you smoke?
6. How would you rate your looks? (10 =Dropdead gorgeous, 1= Seen better days)
Most of our work in this course will relate to numerical data.
The distinction between discrete and continuous data is important because we will be
using different statistical models to deal with each type.
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1.2
CT301: Summarising data
Frequency distributions
The data from a discrete distribution can be summarised using a frequency
distribution, that is, by counting the number of 0’s, 1’s, 2’s, etc. For example, the
number of children in a sample of 80 families might be summarised as follows:
Number of children under 16, x
Number of families in sample, f
0
8
1
12
2
28
3
19
4
7
5
4
6
1
7
1
8 or more
0
Question 1.2
How would you calculate how many children there were altogether?
To represent the data graphically, a bar chart is used.
B ar ch art o f n u mb e r o f ch ild re n in familie s
30
25
20
f
15
10
5
0
0
1
2
3
4
5
6
7
x
(The above graph is part of Core Reading.)
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CT301: Summarising data
1.3
Page 5
Histograms and grouped frequency distributions
The next example involves a typical continuous variable and will introduce the
idea of a grouped frequency distribution and a histogram. The data involve cash
amounts given to the nearest £1. Cash amounts are actually discrete, being
measured in whole numbers of pence, but here the sums are so large that they
can be considered as continuous. In practice, all variables are discrete due to
the fact that they will be rounded to a certain accuracy and nothing can be
measured to infinitesimal accuracy.
For example, a sample of 100 claims for damage due to water leakage on an
insurance company’s household contents policies might be as follows:
243
425
127
293
221
176
293
359
330
291
306
324
74
464
355
342
236
352
238
299
271
228
523
200
324
443
223
273
248
265
396
113
164
392
374
239
371
267
419
318
287
226
366
265
347
302
287
277
330
415
399
176
343
403
261
483
400
184
319
372
466
320
330
372
278
231
314
286
440
238
269
230
436
259
113
292
468
214
427
323
295
404
141
426
135
373
337
351
343
411
330
487
388
262
291
346
308
270
414
494
These data might be summarised in the following grouped frequency
distribution:
Group
Frequency
50–99
1
100–149
5
150–199
4
200–249
14
250–299
22
300–349
20
350–399
14
400–449
13
450–499
6
500–549
1
This is a grouped frequency distribution with equal group sizes.
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CT301: Summarising data
A histogram is similar to a bar chart, but with a continuous scale.
In other words, a bar chart is called a histogram when it is used to present continuous
numerical data. So the responses to any of the six questions on the dating agency
questionnaire could be represented as a bar chart, but only Question 2 could be shown
as a histogram.
Histograms are usually presented with vertical rectangles, but the one given
below is unusual as it is in horizontal form.
In this histogram the continuous scale has been broken down into categories by dividing
the data values into £50 bands. The bar labelled “125”, for example, represents the 5
data values (113, 127, 141, 113 and 135) that were in the range 100 to 149 (inclusive).
am ounts
N = 100
1
75
5
125
175
4
Midpoint
225
14
22
275
20
325
14
375
425
13
475
6
1
525
0
4
8
12
16
20
24
C ount
The midpoints on this diagram have been rounded to the nearest whole number for
49.5 99.5
99.5 149.5
convenience. The actual values should be
74.5,
124.5,
2
2
The above grouped frequency distribution and histogram have equal group
widths. In some situations it may be convenient to have one or two wider groups
at the extremes of the distributions. For such cases it should be noted that it is
the areas of the rectangles that are proportional to the frequencies not their
heights.
This is an important point to note and you may wonder why this is the case. Think of a
histogram as a “fairer” bar chart: if a group has twice the width, it should only be half
the height for a given frequency.
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CT301: Summarising data
Page 7
Question 1.3
If we had had a single group containing all values greater than or equal to 350, how
would you have drawn the bar representing this group?
Question 1.4
Write down a formula to calculate the correct height for any bar in a histogram.
1.4
Stem and leaf diagrams
An alternative to the histogram is the stem and leaf display. It gives a visual
representation similar to the histogram but does not lose the detail of the
individual data points in the grouping. Here is a stem and leaf display for the
water leakage data claim amounts:
0
1
1
2
2
3
3
4
4
5
7
11344
6888
0122333344444
566677777778899999999
00011122222233334444
55556677777799
000001122333444
677899
2
The stems are on the left with units of £100 and the leaves are on the right with
units of £10. So the individual data points can be represented, although they are
rounded to the nearest £10. It is usual to put a key on the diagram to show this.
These diagrams are useful to observe the general shape of the distribution of data and
can be used to calculate values such as the median or interquartile range.
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1.5
CT301: Summarising data
Lineplots
For smaller data sets another alternative diagram is the dotplot or lineplot in
which the data points are plotted as “dots” or “crosses” along a line with a
scale.
Here is a computer generated dotplot for the water leakage claim amount data.
This computer package hasn’t lined things up perfectly. The notches and numbers on
the horizontal scale both need to be moved slightly to the left.
Here is a lineplot for the first row of 10 amounts from the claims data:
200
300
400
amounts
500
If there are repeats of any data points (eg two values of 300) in a lineplot, the crosses are
placed above each other as in a dotplot diagram.
The dotplot/lineplot is used frequently throughout the CT3 course:
●
To check the normality of a data set. This is required for some confidence
intervals (Chapter 11), hypothesis tests (Chapter 12), analysis of variance
(Chapter 14) and as a test for the appropriateness of the linear regression model
(Chapter 13).
●
To check whether two data sets have a common variance. This is required for
the twosample ttest (Chapter 12) and for analysis of variance (Chapter 14).
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CT301: Summarising data
Cumulative frequency tables
Cumulative frequencies are obtained by accumulating the frequencies to give the
total number of observations up to and including the value or group in question.
For grouped data it is natural to relate the cumulative frequency to the upper
boundaries of the groups.
The following table shows the cumulative frequencies for 200 motor insurance claims
received by an office in a month.
Claim size
Cumulative frequency
up to £1,000
up to £2,000
up to £3,000
up to £4,000
up to £5,000
up to £6,000
24
75
136
175
195
200
A cumulative frequency diagram can be drawn from these data as follows:
Cumulative Frequency
1.6
Page 9
200
180
160
140
120
100
80
60
40
20
0
0
2000
4000
6000
Claim (£)
Notice on the diagram that the cumulative frequency is plotted against the highest claim
size in the group.
The cumulative frequency table or diagram is commonly used to find the median or
interquartile range.
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Page 10
2
CT301: Summarising data
Measures of location
There are a number of different quantities, which can be used to estimate the
central point of a sample. These are called measures of central tendency, or
measures of location.
The quantities described in this section are just different ways of calculating the
“average” value for the data set.
2.1
The sample mean
By far the most common measure for describing the location of a set of data is
the mean.
For a set of observations denoted by x1, x 2 , , x n or x i , i = 1,2, , n the mean is
defined by x =
For
a
n
1
x i (read as “x bar”).
n i =1
Â
frequency
distribution
with
possible
values
x1, x 2 , , x k
with
corresponding frequencies f1, f2 , , fk , where Â fi = n , the mean is given by
1
x=
n
k
Â fi xi .
i =1
For example, for the family size distribution data the mean number of children in
the sample is, from the frequency distribution:
x=
Â fi x i 186
=
= 2.325
Â fi
80
Question 1.5
Confirm the above figure obtained for the mean number of children using the family size
data in Section 1.2.
For grouped data the midpoint of each group would normally be used in the
frequency distribution to determine the mean.
Question 1.6
Calculate x for the water leakage data from the frequency distribution table in
Section 1.3 using 74.5, 124.5, etc as midpoints for the bands.
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CT301: Summarising data
2.2
Page 11
The median
Another useful measure of location is the median. Consider placing the n
observations in order of magnitude. The median is a value, which splits the data
set into two equal halves, so that half the observations are less than the median
and half are greater than the median.
This definition says “a value”, rather than “the value”, because there will often be a
range of possible values satisfying this condition.
If n is odd, then the median is the middle observation. If n is even, then the
median is the midpoint of the middle two observations. This can be conveniently
expressed as the (n + 1) / 2 th observation.
One of the potential advantages of the median for certain data sets is that it is
robust or resistant to the effects of extreme observations.
Extreme observations (outliers) can cause problems because they can have a
disproportionate effect on the calculated value of some of the quantities we have been
looking at. The word “robust” is used with this special meaning in statistics. There is a
whole area of statistical theory called “robust statistics”, which deals with this issue.
For the family size distribution data, the 40th and 41st observations are both 2.
So the median number of children is 2.
Question 1.7
The ages of seven policyholders in a portfolio of insurance policies are as follows:
39
34
26
41
70
34
28
(i)
Find the median age of the policyholders in this portfolio.
(ii)
Another policyholder aged 41 years is added to the portfolio. Find the median age
of policyholders in the portfolio.
(iii)
Why would the mean be a poor measure of central tendency for these data?
If we were asked to calculate the median of grouped data we would use interpolation.
The median is that value corresponding to a cumulative frequency of 50% and it
can be read from a plot of the cumulative frequency distribution.
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CT301: Summarising data
Question 1.8
Calculate the median for the water leakage data given that, when the data values are
arranged in ascending order, the 50th and 51st values are £314 and £318.
Question 1.9
Estimate the median for the motor insurance claims from the cumulative frequency
table in Section 1.6.
2.3
The mode
A third measure of location is the mode. It is defined as the value which occurs
with the greatest frequency or the most typical value. Its use in practice is
limited but there are occasions when, for example, a company is interested in the
most typical policyholder.
For the family size distribution data the largest frequency is 28, so the modal
number of children in the sample is 2.
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CT301: Summarising data
3
Page 13
Measures of spread
The location of a data set is usually the main feature of interest. Another feature
of interest is the spread (or variability or dispersion or scatter): how widely
spread the data are about the mean (or other measure of location).
3.1
The sample standard deviation
The most commonly used measure of spread is the standard deviation.
Essentially it is a measure of how far on average the observations are from the
mean.
For a data set x i , i = 1,2, , n with mean x , consider an individual observation
x i . Then ( x i  x ) is the distance of x i from the mean, also called the deviation
of x i from the mean.
Here we are thinking of x i and x as points plotted on a line. In mathematical
language, xi x (which can be a positive or a negative number) is the “displacement”
and  x i x  is the “distance” between the points, which is always a positive number.
(The vertical lines indicate the “absolute value”.)
The sum of the squares of these deviations is:
Â( x i  x )2
Because each of the terms in this sum is a squared quantity (and hence a positive
number) the total must be a positive number.
The variance is this total divided by n 1.
You might think it would be more logical to divide by n , rather than n 1 . We will see
later in Chapter 10 that dividing by n 1 makes the sample variance an unbiased
estimator of the population variance. That is to say, on average, the sample variance
gives us the correct value for the real variance of the whole population.
You might also think it would be more logical to use  x i x  instead of ( x i x )2
in this calculation. The reason this is not done is simply because the absolute function
is not a nice function to deal with mathematically – in particular, the graph of x has a
nasty kink when x 0 , whereas the graph of x 2 is always a nice “smooth” curve.
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CT301: Summarising data
The standard deviation is the positive square root of the variance. The symbol
s 2 is used to denote the variance:
s2 =
1
Â( x i  x )2
n 1
The standard deviation and variance can be calculated more easily using the
alternative formula:
s2 =
1 È
Â x i2  nx 2 ˘
˚
n 1Î
For the family size distribution data the variance of the number of children for
the sample of families is given by:
s2 =
2
1 È
1 È
Ê 186 ˆ ˘
Í 592  80 ¥ Á
˙ = 2.02
Â fi x i2  nx 2 ˘ =
˚ 79 Í
Ë 80 ˜¯ ˙
n 1Î
Î
˚
And the standard deviation is s 2.02 142
. .
Note that where the original data has “units”, eg if the values were amounts measured in
Euros, the variance (being based on a sum of squares) would be measured in “square
Euros”. But the standard deviation (being the square root of the variance) would be
measured in the original units – Euros in this case. For this reason, standard deviations
are often more “friendly” to work with in numerical comparisons.
Question 1.10
Show that the two formulae s 2
1
1
( x i x )2 and s 2
x i2 nx 2
n 1
n 1
are
mathematically equivalent.
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CT301: Summarising data
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Question 1.11
Given that for the water leakage data:
n 100
x 31,353 x2 10, 687, 041
Calculate the:
3.2
(i)
sample mean
(ii)
sample standard deviation.
Moments
The mean and variance are special cases of a set of summary measures called
the moments of a set of data.
In general the k thorder moment about the value a is defined by:
n
1
( x  a )k
n i =1 i
Â
Here is any fixed number.
So the mean is the firstorder moment about the origin, and the variance is
essentially the secondorder moment about the mean with a divisor of n 1
rather than n .
Again, we are thinking of points plotted along a line, so that the “origin” just refers to
the zero position, ie 0 . When equals the mean of the distribution, we get a
central moment.
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3.3
CT301: Summarising data
The range
The range is a very simple measure of spread defined, as its name suggests, by
the difference between the largest and smallest observations in the data set.
R max( x i ) min( x i )
i
i
For the family size distribution data, R 7 0 7 .
The range is a poor measure of the spread of the data as it relies on the extreme values,
which aren’t necessarily representative of the data as a whole.
Question 1.12
What is the range of the water leakage data?
3.4
The interquartile range
The interquartile range (IQR) is another measure of spread which is like the
range but which is not affected by the data extremes.
First we must define the quartiles of a set of data:
Just as the median divides a set of data into two halves, the quartiles divide a set
of data into four quarters. They are denoted by Q1 , Q2 and Q3 .
Thus a quarter of the data have values less than Q1 , a quarter have values
between Q1 and Q2 , a quarter have values between Q2 and Q3 , and the last
quarter have values greater than Q3 . Note that Q2 is just the median, while Q1 is
called the lower quartile and Q3 the upper quartile.
When dealing with grouped data, the quartiles can be read from the cumulative
frequency diagram or calculated using interpolation.
When dealing with discrete ungrouped data, we determine the appropriate value in a
similar way to that used to calculate the median.
Recall that the median was specified as the (n + 1) / 2 th observation. In a similar
way Q1 can be defined to be the (n + 2) / 4 th observation counting from below
and Q3 as the same counting from above, with relevant interpolation if needed.
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CT301: Summarising data
Page 17
The lower quartile, the median and the upper quartile are also referred to as the 25th,
50th and 75th percentiles. The p th percentile corresponds to the
(
p
100
)
¥ n + 12 th data
value.
The interquartile range is defined as Q3  Q1 .
So, for the family size distribution data, the interquartile range is calculated as
follows:
Since n 80 , Q1 is based on the (80 2) / 4 “20½th” observation.
From the frequency distribution, the 20th observation is 1 and the 21st is 2 and
so Q1 = 1.5 . From the other end, the 20th and 21st observations are both 3 and
so Q3 = 3 .
So the IQR = 3  1.5 = 1.5 .
Note: An alternative to the definitions of Q1 and Q3 given above is:
Q1 is the
n +1
th observation counting from below;
4
Q3 is the
n +1
th observation counting from above.
4
Using this alternative, the IQR = 3  1.25 = 1.75 .
Either pair of definitions is acceptable.
Many students express concern that the alternative definitions can give a different
answer for the IQR. As both are in the Core Reading, either answer will be given full
credit in the exam.
Question 1.13
Calculate the interquartile range (using the first definition) for the water leakage data
given that, when the data values are arranged in ascending order:
–
the 25th and 26th values are 259 and 261
–
the 75th and 76th values are 373 and 374.
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CT301: Summarising data
Question 1.14
Estimate the interquartile range for the motor insurance claims from the cumulative
frequency table in Section 1.6.
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CT301: Summarising data
4
Page 19
Symmetry and skewness
The next feature of interest is the shape of the distribution of a data set, that is,
whether it is symmetric or skewed to one side or the other.
The approximate shape of a distribution can be determined by looking at a
histogram, stem and leaf display or dotplot.
Here are histograms for three data sets each of 200 observations:
–
the first is fairly symmetrical
–
the second is positively skewed
–
the third is negatively skewed.
(The histograms are shown on the next page.)
It should be noted that for sample data, a small data set of 10 observations is
unlikely to show up the skewness of a population unless its skewness is very
severe, whereas a large data set of 200 observations will reflect the shape of the
population quite well.
In the more usual format, a positively skewed, a symmetrical and a negatively skewed
distribution would look like this:
positively skewed
symmetrical
negatively skewed
The reason for describing these as positively or negatively skewed is that the skewness
can be measured numerically by calculating the third central moment. For a positively
(negatively) skewed distribution, this works out to a positive (negative) answer. For a
symmetrical distribution it equals zero.
You many wonder why the skewness is called positive when the ‘hump’ is on the left.
This is because there are more data points to the right (on the positive side) of the ‘hump’
than the left and vice versa for negative skewness.
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CT301: Summarising data
Positively skew distributions are most common in actuarial work because we are often
dealing with quantities, such as claim amounts, which must be positive but have no
upper limit. Hence, most of the probability distributions we will meet in Subjects CT3
and CT6 are positively skewed (eg the Poisson, exponential, gamma and lognormal
distributions).
Symmetric
data
N = 200
3
38
40
6
42
7
17
44
22
Midpoint
46
33
48
50
31
52
30
54
21
56
18
6
58
5
60
62
1
0
6
© IFE: 2013 Examinations
12
18
Count
24
30
36
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CT301: Summarising data
Page 21
Positively skew
data
N = 200
22
0
55
20
40
57
31
Midpoint
60
13
80
6
100
120
5
140
5
2
160
4
180
0
10
20
30
40
50
60
Count
Negatively skew
data
N = 200
5
4
2
10
3
15
5
20
25
10
7
Midpoint
30
11
35
40
21
45
21
50
31
28
55
60
27
65
21
70
8
1
75
0
6
12
18
24
30
36
Count
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4.1
CT301: Summarising data
Measuring skewness for data
One particular measure of skewness is based on the third moment about the
mean:
n
1
( x i  x )3
n i =1
Â
The cubic power in this formula gives a positive or negative value depending on which
side of the mean the value xi is. Consequently, positively skewed distributions with a
long tail on the right give a positive value, and negatively skewed distribution with a
long tail on the left give a negative value.
The coefficient of skewness is a scaled version of this moment obtained by
dividing it by the second moment about the mean raised to the power 3/2.
Recall that the second moment was almost our sample variance formula (but dividing
by n instead of n  1 ). Denoting this by sn2 , we have:
coeff of skew =
1 n
( xi  x )3
Â
n i =1
1.5
È1 n
˘
2
Í Â ( xi  x ) ˙
ÎÍ n i =1
˚˙
=
skewness
( )
1.5
sn2
=
skewness
sn3
We can see that we are dividing by a cubic measure. Since the skewness is a cubic
measure, we obtain a dimensionless measure when we divide by another cubic measure.
Note that this does not depend on the units of the data and is such that its sign
reflects the skewness with a value of zero corresponding to a symmetric set of
data.
Question 1.15
Given that for the water leakage data:
n 100
( xi x )2 856,934.91 ( xi x )3 11,949,848.3946
Calculate the:
(a)
skewness
© IFE: 2013 Examinations
(b)
coefficient of skewness.
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CT301: Summarising data
4.2
Page 23
Boxplots
The quartiles together with the extremes provide a useful way of displaying a
data set in summary form, called a boxplot. It consists of a box or rectangle with
ends at Q1 and Q3 divided with a line at the median Q2 . Then lines are drawn
from Q1 and Q3 out to the extremes.
These diagrams are also called “box and whiskers” diagrams (the whiskers referring to
the lines protruding at each end).
Here is a boxplot of the water leakage claim amounts data. This boxplot was
generated by a computer package which labels potential “outliers”, here by an
asterisk. Outliers are observations which are in some way detached from the
main bulk of the observations.
Outliers are just untrustworthy (“rogue”) values present in the data. In actuarial work
outliers may sometimes be values that have been recorded incorrectly, eg a pensioner’s
age given as 27 instead of 72, or an amount of money typed with an extra or missing 0
on the end. However, they may also be a genuine part of the data.
The median can be read off as a measure of location, the interquartile range (and
the range) as measures of spread, and the relative symmetry can also be
observed.
Question 1.16
Sketch a possible boxplot for each of the following data sets:
(i)
the annual salaries of the employees of a mediumsized company
(ii)
the scores obtained by a class of students in an easy exam
The precise numerical values are not important, but your sketches should show clearly
the likely “shape” of the distributions.
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5
CT301: Summarising data
Exam Question
Past Exam Question (Subject C1, April 1998, Q10)
In the claims department of an insurance office various quantities are computed at the
end of each day’s business. On Monday, 10 claims are received for a particular class of
policy. The mean claim amount is calculated to be £426 and the standard deviation to
be £112.
On Tuesday it is found that one of Monday’s claims for £545, was classified wrongly
and it is removed from the set of 10 claims. Calculate the resulting mean and standard
deviation of the reduced set of 9 claims.
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CT301: Summarising data
Page 25
Chapter 1 Summary
Numerical data can be discrete or continuous.
Categorical data can be dichotomous (attribute), nominal or ordinal.
Data can be presented either in tabular form (using a frequency table, a cumulative
frequency table or a stem and leaf diagram) or in graphical form (using a lineplot, a
dotplot, a boxplot, a bar chart or a histogram).
The location of a data set can be summarised using the mean, the median or the mode.
The spread of a data set can be summarised using the standard deviation, the range or
the interquartile range.
The variance measures the spread squared.
Third moments can be used to summarise the skewness (ie the degree of asymmetry) of
a data set.
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CT301: Summarising data
Chapter 1 Formulae
Measures of location
x=
1 n
Â xi
n i =1
M =
( 12 n + 12 )
th
Â fi xi
Â fi
x=
or
value
Measures of spread
R = max( xi )  min( xi )
i
i
IQR = Q3  Q1 , Q3 =
( 34 n + 12 )
th
value Q1 =
( 14 n + 12 )
value
Q3 =
( 34 n + 34 )
th
value Q1 =
( 14 n + 14 )
value
alternatively
s2 =
or
th
th
˘
1 n
1 Èn 2
2
2
=
x
x
(
)
ÍÂ xi  nx ˙
Â
i
n  1 i =1
n  1 ÍÎ i =1
˙˚
Â
n
1
fi ( xi  x ) 2 =
Â
fi  1 i =1
Â
1
ÈÂ fi xi 2  nx 2 ˘
˚
fi  1 Î
Measures of skewness
skewness =
1 n
Â ( xi  x )3
n i =1
coeff of skew =
skewness
sn3
where sn2 =
1 n
Â ( xi  x )2
n i =1
Sample moments
kth moment =
1 n k
Â xi
n i =1
kth moment about a =
kth central moment =
© IFE: 2013 Examinations
1 n
Â ( xi  a )k
n i =1
1 n
Â ( xi  x )k
n i =1
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CT301: Summarising data
Page 27
Chapter 1 Solutions
Solution 1.1
1.
Discrete
2.
Continuous
3.
Dichotomous
4.
Nominal
5.
Dichotomous
6.
Ordinal
Solution 1.2
The total number of children is n = Sfi xi = 8 ¥ 0 + 12 ¥ 1 + 28 ¥ 2 + + 1 ¥ 7 = 186 .
Solution 1.3
The total area of the last four groups must remain the same. So these groups would be
34
replace by a single “thick” group with a count of
8.5 .
4
Solution 1.4
Since the area ( height width ) is proportional to the frequency:
height
frequency
width
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CT301: Summarising data
Solution 1.5
The mean is calculated as follows:
x=
=
1
(8 ¥ 0 + 12 ¥ 1 + 28 ¥ 2 + 19 ¥ 3 + 7 ¥ 4 + 4 ¥ 5 + 1 ¥ 6 + 1 ¥ 7)
80
1
¥ 186 = 2.325
80
Solution 1.6
The mean is:
x=
S fi xi 1 ¥ 74.5 + 5 ¥ 124.5 + + 1 ¥ 524.5 31, 200
=
=
= £312
S fi
1+ 5 + +1
100
This figure is an approximation because we have used the midpoints and have assumed
that the values in each group are uniformly distributed. For comparison, the value of x
calculated from the original data values is £313.53.
Solution 1.7
(i)
First placing the ages in increasing order of magnitude:
26
28
34
34
39
41
70
The median is the (7 + 1) / 2 = 4 th observation, which is 34.
(ii)
The ages in increasing order of magnitude are now:
26
28
34
34
39
41
41
70
The median is the (8 + 1) / 2 = 4.5 th observation, which is (34 + 39) / 2 = 36.5 .
(iii)
The mean for the original seven policyholders is 38.86. This figure has been
affected by the extreme value of 70, whereas the median is robust (resistant to the
effects of extreme observations).
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Page 29
Solution 1.8
The median is the (100 + 1) / 2 = 50.5 th observation, which is (314 + 318) / 2 = £316 .
Solution 1.9
The median value splits the data set into two equal halves, so that half of the
observations (100) lie below the median and half lie above the median.
25 36
64
75
2000
3000
claim amount
m
We can see that 75 observations are equal to or below £2,000 and that 64 observations
are above £3,000, so the median lies in the group 2, 000 < X £ 3, 000 . If we assume the
values are distributed uniformly and apply linear interpolation, the median, m , is given
by:
100  75
m  2, 000
Ê 25 ˆ
=
ﬁ m = 2, 000 + Á ˜ ¥ 1, 000 = £2, 410
Ë 61 ¯
3, 000  2, 000 136  75
We’ve rounded this to the nearest £ because our calculation is not likely to be very
accurate. (We have assumed the values are distributed uniformly within (very few)
groups).
Solution 1.10
We need to show that Â( xi  x ) 2 is equivalent to Â xi2  nx 2 .
Expanding the first expression and taking any constant factors outside the sums gives:
n
n
n
n
n
i =1
i =1
i =1
i =1
i =1
Â ( xi  x )2 = Â ( xi2  2 xxi + x 2 ) = Â xi2  2 x Â xi + x 2 Â1
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CT301: Summarising data
n
Since
Â xi = nx and
i =1
n
Â ( xi  x )
i =1
2
n
Â1 = n , we get:
i =1
n
=Â
i =1
xi2
n
 2nx + nx = Â xi2  nx 2
2
2
i =1
So the two methods of calculating s 2 are equivalent.
Solution 1.11
Now we have:
Â x = 31,353 Â x 2 = 10, 687, 041
31,353
= £313.53
100
(i)
x=
(ii)
s2 =
1 È
10, 687, 041  100 ¥ 313.532 ˘˚ = 8655.91
Î
99
So the sample standard deviation is:
s = 8655.91 = £93.04
Solution 1.12
After a bit of searching in the original data table, we find that the smallest and largest
values are 74 and 523, giving a range of £449.
Solution 1.13
The quartiles correspond to the 25½th and 75½th values. So the IQR is:
IQR = 373½  260 = £113.50
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CT301: Summarising data
Page 31
Solution 1.14
The quartiles split the data set into four equal quarters, so that a quarter of the
observations (50) lie below the lower quartile (l) and a quarter lie above the upper
quartile (u).
26 25
24
14 25
125
1000 l 2000
136
claim amount
3000 u
25
4000
claim amount
If we assume the values are distributed uniformly and apply linear interpolation, the
lower quartile, l, is given by:
Ê 26 ˆ
l = 1, 000 + Á ˜ ¥ 1, 000 = £1,510
Ë 51 ¯
and the upper quartile, u, is given by:
Ê 14 ˆ
u = 3, 000 + Á ˜ ¥ 1, 000 = £3,359
Ë 39 ¯
The estimated interquartile range is therefore:
IQR = £3,359  £1,510 = £1,849
Solution 1.15
(a)
The skewness is given by:
Â ( xi  x )3 =  11,949,848.3946 = 119, 498.483946
n
100
The negative value tells us the water leakage claims are negatively skewed.
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(b)
CT301: Summarising data
The second central moment is:
Â ( xi  x )2 = 856,934.91 = 8,569.3491
n
100
Hence the coefficient of skewness is given by:

119, 498.483946
8,569.34911.5
= 0.15064
Now the skewness is “standardised” we can see that it is only slightly negatively
skewed (as it is close to a value of zero which means the distribution is
symmetrical).
Solution 1.16
(i)
Annual salaries
This data set is likely to have a positively skew distribution. A boxplot might look like
this:
0
10
20
30
40
50
60
70
80
90
100
salary (£000's)
(ii)
Exam scores
This data set is likely to have a negatively skew distribution. A boxplot might look like
this:
0
10
20
30
40
50
60
70
80
90
100
exam score (%)
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CT301: Summarising data
Page 33
Past Exam Question (Subject C1, April 1998, Q10)
This means that we can calculate the total of the claims:
Â x = 426 ¥ 10 = 4, 260
Removing the unwanted claim means that n = 9 and
This gives a new mean of
1122 =
1
9
(Â x
2
Â x = 4, 260  545 = 3, 715 .
3, 715
= £412.78 .
9
 10 ¥ 4262
)
Removing the unwanted claim gives
ﬁ
Â x2 = (9 ¥ 1122 + 10 ¥ 4262 ) = 1,927, 656
Â x2 = 1,927, 656  5452 = 1, 630, 631 .
The new variance is:
2
1Ê
Ê 3715 ˆ ˆ
= 12,145.19
S = Á1, 630, 631  9 ¥ Á
Ë 9 ˜¯ ˜¯
8Ë
2
ie the standard deviation is £110.21.
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CT302: Probability
Page 1
Chapter 2
Probability
Syllabus objectives
(ii)
1.
Explain what is meant by a set function, a sample space for an
experiment, and an event.
2.
Define probability as a set function on a collection of events, stating
basic axioms.
3.
Derive basic properties satisfied by the probability of occurrence of an
event, and calculate probabilities of events in simple situations.
4.
Derive the addition rule for the probability of the union of two events,
and use the rule to calculate probabilities.
5.
Define the conditional probability of one event given the occurrence of
another event, and calculate such probabilities.
6.
Derive Bayes’ Theorem for events, and use the result to calculate
probabilities.
7.
Define independence for two events, and calculate probabilities in
situations involving independence.
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0
CT302: Probability
Introduction
Probability is a numerical way of describing how likely something is to happen. This
chapter looks at calculating probabilities using set theory. We also look at various rules
for calculating “or”, “and” and “conditional” probabilities, including Bayes’ Theorem.
Much of this material will be familiar to anyone who has studied statistics before. For
students who have not met probability (and in particular probability tree diagrams)
before and would like more examples and careful explanations, the Stats Pack has been
recently developed to help. See http://www.acted.co.uk/Html/paper_stats_pack.htm or
contact StatsPack@bpp.com for further details. Alternatively, refer to an Alevel (or
equivalent) statistics textbook.
Please be aware that permutations and combinations are assumed knowledge for CT3.
Therefore probability questions in CT3 can (and do) involve their use. Again, if you are
unfamiliar with these please refer to Stats Pack or an Alevel (or equivalent) textbook.
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CT302: Probability
1
Sets
1.1
Basic terminology
Page 3
A set is defined as a collection of objects and each individual object is called an
element of that set.
For example, the days of the week form a set:
D = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}
Tuesday is an element of the set D. We write this as:
Tuesday Œ D
However, we are interested in using sets for probabilities:
Supposing an experiment is defined as any sort of operation whose outcome
cannot be predicted in advance with certainty, the sample space S for such an
experiment is the set of all possible outcomes that might be observed.
For example, supposing the experiment is one roll of a normal sixsided dice, the
sample space would be defined as S = {1, 2, 3, 4, 5, 6} , ie all the possible numbers
that can be rolled.
Question 2.1
Write down the sample space for an experiment where the scores of two dice are added.
An event A is defined as a subset of the sample space S, which contains any
element of S. A is only a subset of S (written as A Ã S ) if every element that
belongs to A also belongs to S.
For example, C = {Tuesday, Thursday} is a subset of the days of the week set D given
above. We write C Ã D . It might be helpful to think of the subset symbol as a bit like
a “less than” symbol.
In the example above, we could define throwing an even number as an event,
therefore A = {2, 4, 6} . The event is said to have occurred if any one of its
elements is the outcome observed, for example if a 2 was rolled.
Think of an event as anything we might wish to observe from our experiment.
The null set ∆ is the set with no elements.
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CT302: Probability
Therefore: ∆ = {}
An event B is said to be complementary to another A in a sample space, S, if B
contains all the elements of S that are not in A. The complement of A is normally
written as A¢ or A .
Considering the example of rolling a dice, where S = {1, 2, 3, 4, 5, 6} . If A = {2, 4, 6} ,
then A¢ = {1, 3, 5} .
We can think of the complement as “not”, ie all the elements not in A.
1.2
Set operations
The union of two sets, A and B, written as A » B , is the set that consists of all
the elements that belong to A or B or both, for example if A = {1, 2} and B = {1, 3}
then A » B = {1, 2, 3} .
We can think of the union as “or”, ie all the elements in A or B (or both).
The intersection of two sets, A and B, written as A « B is the set that consists of
all elements that belong to both A and B so in the example above, A « B = {1} .
We can think of the intersection as “and”, ie all the elements in A and B.
If there are no elements common to both sets, they are known as mutually
exclusive, for example, if A = {1, 2} and B = {3, 4} then A « B = ∆ , the null set.
Question 2.2
Write down the sets A « B and A » B where:
A={2, 8, 11, 16, 121}
B={the square numbers less than 144}
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2
Venn diagrams
2.1
Basic diagrams
Page 5
A convenient way to represent sets is by drawing Venn diagrams.
S
A
3
2
4
1
6
5
The rectangle shows the complete sample space for our dice roll, S = {1, 2, 3, 4, 5, 6} .
The circle shows the event “rolling an even number”, A = {2, 4, 6} . Note how A is
contained within S, because A Ã S .
A
B
S
5
2
3
1
6
4
The Venn diagram above for a dice roll shows the sets A = {1, 2} and B = {1, 3} . Since
1 is in both sets it is placed in the overlap (the intersection) between A and B.
B
A
S
3
1
2
6
4
5
The Venn diagram above for a dice roll shows the sets A = {1, 2} and B = {3, 4} . These
sets are mutually exclusive – they have no elements in common, so they are drawn with
no overlap.
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CT302: Probability
Question 2.3
Draw a separate Venn diagram for each of the following sample spaces:
2.2
(i)
A = {1} , B = {1, 2, 3} from S = {1, 2, 3, 4, 5, 6}
(ii)
A = {1, 4, 7} , B = {3, 7, 8, 9} , C = {4, 7, 8, 9} from S = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Set operations on Venn diagrams
We can use shading to show the appropriate part represented by an operation:
S
A
The diagram above shows A¢ (the complement of A), ie everything that is not in A.
A
B
S
The diagram above shows A « B (the intersection of A and B), ie everything in A and
B.
A
B
S
The diagram above shows A » B (the union), ie everything in A or B (or both).
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Question 2.4
For the Venn diagram below, use shading to identify the following regions:
A
(i)
2.3
A¢ « B
B
A¢ « B ¢
(ii)
S
( A « B)¢
(iii)
(iv)
( A » B)¢
Using Venn diagrams to solve problems
Instead of writing the elements on the diagram we could simply write the total number
of elements in each region (or the probability of each region).
Example
In a group of 25 people, 18 have a mortgage, 13 own some shares and 2 people have
neither a mortgage nor any shares. How many people have both?
Solution
Adding up the number of people who have a mortgage and own shares, we should get
23, since there are two people who have neither. The actual figure is 31, but this
includes the people who have a mortgage and own shares twice. Therefore there must
be 8 people in the intersection.
A
B
10
8
S
5
2
There are 8 people with both a mortgage and some shares.
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CT302: Probability
Question 2.5
In a group of students 60% have passed CT1, 45% have passed CT3 and 25% have
passed both. What percentage of students have passed neither?
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3
Page 9
Probabilities
In the previous section, we used sets to define events. A probability is a numerical way
of describing how likely (or not) an event is to happen.
If each of the elements in the sample space are equally likely, then we can define the
probability of event A as:
P ( A) =
number of elements in A
number of elements in S
Example
Find the probability of rolling an even number on an ordinary dice.
Solution
We have a sample space of S = {1, 2, 3, 4, 5, 6} .
Defining “throwing an even number” as event A, we have A = {2, 4, 6} .
So the probability of throwing an even number is given by:
P ( A) =
3 1
=
6 2
Question 2.6
One card is picked from an ordinary pack of 52 playing cards. What is the probability
of obtaining:
(i)
a diamond
(ii)
an ace
(iii)
the ace of diamonds
(iv)
a jack, queen or king.
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3.1
CT302: Probability
Basic probability axioms
We now look at the basic rules of this probability definition:
The three basic probability axioms can be summarised as follows:
1. P (S ) = 1
The relative frequency of an event that is certain to occur must be 1. The sample
space, S contains all possible outcomes and therefore the probability of S must
be 1.
For example, when we are rolling a dice S = {1, 2, 3, 4, 5, 6} . When A = S , ie the event
of getting any one of the numbers 1 to 6 on a dice, the probability is:
P( S ) =
6
=1
6
The probability of getting any one of the numbers 1 to 6 on a dice is certain! So this
result is saying that we give certain events a probability of 1.
It follows that for event A from sample space S that, P ( A¢) = 1  P ( A) .
Suppose the probability of an individual dying is 0.2, then the probability of the
individual not dying is 1  0.2 = 0.8 .
2. P ( A) ≥ 0 for all A Ã S
The relative frequency of occurrence of any event must not be negative, that is,
probabilities can never be negative.
If we define event A as “rolling a 7 on a dice”, then the probability of this is:
P( A) =
0
=0
6
So the probability of an impossible event is 0.
So rules 1 and 2 together are telling us that probabilities lie between 0 (impossible) and
1 (certain).
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3. P ( A » B ) = P ( A) + P (B )
if A « B = ∆
If two events cannot occur simultaneously, because they are mutually exclusive,
the probability of an event defined by their union is equal to the sum of the
probabilities of the two events. This property is known as additivity.
Mutually exclusive events are events that cannot occur simultaneously, ie A « B = ∆ .
Taking our dice rolling example again, if A = {1, 4} and B = {2, 3, 5} then A and B are
mutually exclusive.
We have P ( A) =
2
6
P( A » B) =
, P( B) =
5
6
=
2
6
3
6
and A » B = {1, 2,3, 4,5} . So:
+ 63 = P ( A) + P( B)
The reason this rule works is that there is no “overlap” between the elements in A and
B.
Suppose we now consider nonmutually exclusive events when rolling a dice.
A = {1, 2} and B = {1,3} , we have P ( A) =
P( A » B) =
3
6
π
2
6
+
2
6
2
6
, P( B) =
2
6
If
and A » B = {1, 2, 3} . So:
= P( A) + P( B)
The rule has broken down because the element 1 is in both events A and B. This
element has been counted twice on the RHS of the equation, once in P( A) and once in
P( B) .
Question 2.7
One card is picked from an ordinary pack of 52 playing cards. Consider the following
events A = {pick a 7} , B = {pick an ace} and C = {pick a club} .
Show that Axiom 3, namely P ( A » B) = P( A) + P( B) :
(i)
works for events A and B
(ii)
doesn’t work for events A and C and explain why.
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3.2
CT302: Probability
The addition rule
Axiom 3 is often known as the special addition rule. For a more general case,
where two sets are not necessarily mutually exclusive, the rule can be extended
as follows:
P ( A » B ) = P ( A) + P (B )  P ( A « B )
Axiom 3 is a special case of this rule when A and B are mutually exclusive. In which
case A « B = ∆ , so P( A « B) = 0 , and hence P ( A » B) = P( A) + P( B) .
How can we practically show that this rule is true? Recall that, for the nonmutually
exclusive events A = {1, 2} and B = {1,3} , when rolling a dice, we had P ( A) =
P( B) =
2
6
2
6
,
and A » B = {1, 2, 3} . So:
P( A » B) =
3
6
π
2
6
+ 62 = P( A) + P( B)
The rule has broken down because the element {1} is in both events A and B
(ie A « B = {1} ) . This element has been counted twice on the RHS of the equation,
once in P( A) and once in P( B) . So, to fix this error, we need to remove the {1}, ie we
need to remove A « B . Hence:
P( A » B) =
3
6
=
2
6
+ 62  16 = P( A) + P( B)  P( A « B)
We can also prove this using Venn diagrams. Core Reading now demonstrates this rule
more formally.
This can be shown as follows:
Consider the mutually exclusive events:
A « B ¢ elements of A which are not in B with probability x
A¢ « B elements of B which are not in A
with probability y
A « B elements in both A and B
with probability z
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On a Venn diagram we have:
A
B
x
z
S
y
It can be seen that P ( A) = x + z , and P (B) = y + z
P ( A » B ) can also be expressed as P ( A¢ « B ) + P ( A « B ¢) + P ( A « B)
= x +y +z
= ( x + z ) + (y + z )  z
= P ( A) + P (B )  P ( A « B )
Question 2.8
A contestant on a game show is asked two questions. The probability that she gets the
first question correct is 0.3 and the probability that she gets the second question correct
is 0.4. Given that the probability that she gets both questions correct is 0.1, calculate
the probability that:
(i)
she gets either the first, the second or both questions right
(ii)
she gets both questions wrong.
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CT302: Probability
The addition rule can be extended to three events A , B and C . Consider the Venn
diagram:
A
B
b
a
c
e
d
f
g
C
Using the letters a to g to stand for the appropriate probabilities, we can see that:
P( A » B » C ) = a + b + c + d + e + f + g
and:
P ( A) = a + b + d + e
P( B) = b + c + e + f
P(C ) = d + e + f + g
Question 2.9
Use the letters a to g to write down expressions for:
(i)
P( A « B)
(ii)
P( A « C )
(iii)
P( B « C )
(iv)
P( A « B « C )
We can then show that:
P ( A » B » C ) = P( A) + P( B) + P(C )  P( A « B)  P ( A « C )  P( B « C )
+ P( A « B « C )
Question 2.10
Use your results from Question 2.9 to prove the above result.
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Example
Students at a performing arts college can choose to study one or more classes of acting,
dance or singing. The probability that a student is studying acting is 0.5, dance 0.65,
singing 0.55, acting or dancing 0.8, acting and singing 0.25, dancing and singing 0.25.
Find the probability that a student studies all three classes.
Solution
We are given:
P ( A) = 0.5 P( D) = 0.65 P( S ) = 0.55
P ( A » D) = 0.8 P( A « S ) = 0.25 P( D « S ) = 0.25
Since students must study at least one of these classes, we also have:
P( A » D » S ) = 1
We require P ( A « D « S ) , but to use our threeevent addition rule we also require
P ( A « D) . We obtain this from:
P ( A » D) = P( A) + P( D)  P( A « D)
ﬁ 0.8 = 0.5 + 0.65  P( A « D) ﬁ P( A « D) = 0.35
Hence:
1 = 0.5 + 0.65 + 0.55  0.35  0.25  0.25 + P ( A « D « S )
ﬁ P( A « D « S ) = 0.15
Question 2.11
Customers at a restaurant may order any combination of chips, salad or onion rings.
The probability that a customers chooses onion rings is 0.3, salad 0.4, chips and salad
0.l5, chips and onion rings 0.15, salad or onion rings 0.55, all three 0.05, none 0.2.
Calculate the probability a customer chooses:
(i)
chips
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(ii)
chips only
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CT302: Probability
Conditional probabilities
Consider the two events, A and B. We might wish to know the probability that
event A occurred, given the occurrence of the event B. This is known as a
conditional probability and is denoted thus:
P ( A  B)
P( A  B) is read as “the probability of event A occurring given that event B has already
occurred” or “probability of A given B” for short. This is called a conditional
probability as the probability depends (ie is conditional) on event B.
The conditional probability of A occurring given B can be expressed as:
P ( A  B) =
P ( A « B)
P ( A « B) + P ( A¢ « B)
The above formula can be explained as representing the occasions that event A
occurs with B relative to the occasions that B occurs (with or without A).
Example
Consider picking a card from an ordinary pack of playing cards. If we have the events:
A = {pick a spade}
B = {pick an 8}
calculate the probability of picking a spade given that we have picked an 8, ie calculate
P( A  B) .
Solution
Since there are only four 8’s in the pack and only one of them is a spade, we conclude
that P( A  B) = 14 .
Checking this intuitive answer using the formula:
A « B = {8 of spades} ﬁ P( A « B) =
1
52
A¢ « B = {8 of clubs, 8 of diamonds, 8 of hearts} ﬁ P ( A¢ « B ) =
P( A « B)
ﬁ P( A  B) =
=
P ( A « B ) + P ( A¢ « B )
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52
1
52
+
3
52
=
3
52
1
4
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CT302: Probability
Page 17
( A « B) and ( A¢ « B) are mutually exclusive. Noting that ( A « B) » ( A¢ « B) = B ,
this can be rearranged thus:
P ( A  B) =
P ( A « B)
P (B)
This is the version of the formula that is commonly used in practice. We have
simplified the denominator using the addition rule for mutually exclusive events:
P[( A « B) » ( A¢ « B)] = P( A « B ) + P( A¢ « B) = P( B)
Example
Consider our previous example of picking a card from an ordinary pack of cards:
A = {pick a spade}
B = {pick an 8}
Calculate the probability of picking a spade given that we have picked an 8, ie calculate
P( A  B) .
Solution
We obtain the same answer as before – but our calculation is much simpler:
A « B = {8 of spades} ﬁ P( A « B) =
B = {pick an 8} ﬁ P( B) =
ﬁ P( A  B) =
P( A « B)
=
P( B)
1
52
4
52
1
52
4
52
=
1
4
Question 2.12
In a group of 24 actuaries, 20 have worked for Life Office A and 12 have worked for
Life Office B. Everyone in the group has worked for at least one of the two companies.
What is the probability that an actuary picked at random has worked for:
(i)
Life Office A and Life Office B
(ii)
Life Office A given that they have worked for Life Office B?
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4.1
CT302: Probability
The multiplication rule
Rearranging our formula for conditional probabilities, we obtain:
P( A « B) = P( B) P( A  B)
Recall that A « B can be interpreted as A and B. We can use this rule to calculate the
probability of events A and B both happening.
4.2
Independent events
Events A and B are said to be independent if whether or not event B has
occurred gives us no information on whether event A has occurred. This can be
expressed algebraically as follows:
A and B are independent if P ( A) = P ( A  B) = P ( A  B ¢)
Given that:
P ( A  B) =
P ( A « B)
P (B)
Then if A and B are independent:
P ( A « B) = P ( A)P (B)
This is a special case of the multiplication rule when events A and B are independent.
Example
Two dice are thrown. Find the probability of rolling a 5 on both dice.
Solution
A = {roll a 5 on the 1st dice} ﬁ P( A) =
B = {roll a 5 on the 2nd dice} ﬁ P( B) =
1
6
1
6
Since these events are independent:
P ( A « B) = P( A) P ( B) = 16 ¥ 16 =
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CT302: Probability
4.3
Page 19
Tree diagrams
A tree diagram is a convenient representation of probabilities:
Example
A card is picked from an ordinary pack of 52 playing cards, without replacement, and
then another one is picked. What is the probability of picking:
(i)
two red cards
(ii)
one of each colour?
Solution
25
51
R
RR
25
1 25
=
2 51 102
26
51
B
RB
1 26 13
=
2 51 51
26
51
R
BR
1 26 13
=
2 51 51
25
51
B
BB
1 25
25
=
2 51 102
R
1
2
1
2
B
1st card
2nd card
outcomes
probabilities
A tree diagram is used to show the possible outcomes. In this case we can either get red
or a black card. So the number of branches is the number of different possibilities, and
the numbers on the branches are the probabilities of those particular events happening.
Notice that the probabilities on any particular set of branches add up to one.
To calculate probabilities, go along the branches of the tree from left to right to get to
the end, and multiply together any probabilities that you have passed – using the
multiplication rule. If there is more than one route through the tree to give the answer
you require, then sum the answers from the different routes – using the addition rule for
mutually exclusive events.
(i)
P ( RR) =
(ii)
P (1 R and 1 B) = P ( RB) + P ( BR ) =
1
2
¥
25
51
25
= 102
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( 12 ¥ 2651 ) + ( 12 ¥ 2651 ) = 1351 + 1351 = 2651
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CT302: Probability
Question 2.13
A box of chocolates contains 8 milk chocolates and 4 plain chocolates. A chocoholic
eats three chocolates. Calculate the probability that:
(i)
all three are milk chocolates
(ii)
exactly one is a plain chocolate.
Using probability tree diagrams makes it much easier to calculate conditional
probabilities:
Question 2.14
In a restaurant, 45% of the customers are female. 74% of females choose from the à la
carte menu, whilst only 37% of males do. The rest choose from the set menu. What is
the probability that:
(i)
a customer orders from the set menu
(ii)
a customer ordering from the à la carte menu is female?
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4.4
Page 21
Law of total probability
Consider the set space, S , being divided into a partition of n mutually exclusive
events, Ei where i = 1, 2, 3, , n , then:
E i « E j = ∆ , and
E1 » E2 » E3 » » E n = S
For example, a partition of our dice sample space S = {1, 2, 3, 4, 5, 6} could be:
E1 = {1} , E2 = {2, 3} , E3 = {4, 5, 6}
There are many other possibilities, as long as the Ei ’s are mutually exclusive (ie there
is no overlap between them) and together make up the whole sample space (ie they are
exhaustive) it is a partition.
And for any A Ã S :
A = ( A « E1) » ( A « E2 ) » ( A « E3 ) » » ( A « En )
For example, if A is the event “roll an even number” we have A = {2, 4, 6} . This can be
made up of all the intersections with our partition from above:
A « E1 = ∆
A « E2 = {2}
A « E3 = {4, 6}
Hence:
A = {2, 4, 6} = ∆ » {2} » {4, 6}
Therefore:
P ( A) = P ( A « E1) + P ( A « E2 ) + P ( A « E3 ) + + P ( A « E n )
=
n
Â P (A « E j )
j =1
This result is known as the law of total probability.
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4.5
CT302: Probability
Bayes’ Theorem
This theorem is named after the Reverend T Bayes and is used extensively in
Bayesian methods of statistical inference, which will be covered in more detail in
Subject CT6, Statistical Methods.
The result is as follows:
Let E1, E2 , E3 , , En be a partition of S and define event A Ã S
Using the conditional probability:
P (Ei  A) =
P (Ei « A)
P ( A)
(1)
also, the relationship:
P (E i « A) ∫ P ( A « E i ) = P (E i )P ( A  E i )
(ie using P ( Ei « A) ∫ P ( A « Ei ) and then the multiplication rule for P( A « Ei ) )
and the law of total probability:
P ( A) =
n
Â P (A « E j )
j =1
then, by substituting for P ( Ei « A) and P( A) in equation (1),
the result is:
P (E i )P ( A  Ei )
P (E i  A) = n
,
P (E j )P ( A  E j )
Â
i = 1, 2, 3, , n
j =1
Essentially Bayes’ formula allows us to “turnaround” conditional probabilities,
ie calculate P( Ei  A) given only information about P( A  Ei ) .
The values P (E j ) are known as prior probabilities, the event A is some event
which is known to have occurred and the conditional probability P (E i  A) is
known as the posterior probability.
This formula, with slightly different notation, is given on page 5 of the Tables.
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Page 23
Example
The punctuality of trains has been investigated by considering a number of train
journeys. In the sample, 60% of trains had a destination of Manchester, 20%
Birmingham and 20% Edinburgh. The probabilities of a train arriving late in
Manchester, Edinburgh or Birmingham are 30% , 20% and 25% respectively.
If a late train is picked at random from the group under consideration, what is the
probability that it terminated in Manchester?
Solution
We want P (Manchester  Late) .
If M is the event “A train chosen at random terminated in Manchester” (and E and B
have corresponding definitions), and L is the event “A train chosen at random runs
late”, then:
P ( M  L) =
=
P( M ) P( L  M )
P( M ) P( L  M ) + P( E ) P( L  E ) + P( B) P( L  B)
0.6 ¥ 0.3
(0.6 ¥ 0.3) + (0.2 ¥ 0.2) + (0.2 ¥ 0.25)
= 66.7%
Question 2.15
A person has three routes to get to work. The probability that he arrives on time using
routes A, B and C are 60% , 62% and 70% respectively. If he is equally likely to
choose any of the routes, and arrives at work on time, what is the probability that he
chose route B?
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5
CT302: Probability
Exam questions
Past Exam Question (Subject C1, April 1994, Q1)
In a certain constituency, 30% of the voters are “blue collar” workers, of whom 46%
voted Conservative at the last United Kingdom election. Of the remaining voters, 36%
voted Conservative.
Consider a voter selected at random from those who voted Conservative in this
constituency. What is the probability that this voter is a “blue collar” worker?
Past Exam Question (Subject C2, September 1997, Q5)
A coin is selected at random from a pair of coins and tossed. Coin 1 is a doubleheaded
coin (ie a head on both sides). Coin 2 is a standard unbiased coin.
The result of the toss is a head. What is the probability that it was coin 1 which was
tossed?
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Page 25
Chapter 2 Summary
A set is a collection of objects, called elements. A is a subset of B, written A Ã B , if all
the elements in A are contained in B. The complement of A, written A¢ , is the set of all
the elements not in A. The empty set is denoted ∆ .
The union of A and B, written A » B , is the set of all elements in A or B or both. The
intersection of A and B, written A « B , is the set of all elements in A and B.
Venn diagrams are used to represent sets and the relationships between them.
A sample space, S, is the set of all the possible outcomes from an experiment. An event
is anything we might wish to observe from our experiment.
Probabilities are a numerical way of describing how likely an event is to happen. A
formula for equally likely elements is given overleaf. Probabilities lie between 0
(impossible) and 1 (certain).
We can use the addition rule and the multiplication rule (see overleaf) to calculate
probabilities. Tree diagrams are a helpful way of working out probabilities.
The conditional probabilities of A occurring given that B has already occurred is written
P( A  B) . The formula is given overleaf.
Events A and B are mutually exclusive if A « B = ∆ . Events A and B independent if
P ( A  B) = P( A) .
E1, , En is a partition of S if the Ei ’s are mutually exclusive and together make up
the whole set S.
Bayes’ Theorem (see overleaf) allows us to “turnaround” conditional probabilities, ie
calculate P( Ei  A) given only information about P( A  Ei ) .
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CT302: Probability
Chapter 2 Formulae
Probabilities
For equally likely elements:
P ( A) =
number of elements in A
number of elements in S
Addition rule
P ( A » B) = P( A) + P( B)  P( A « B)
For mutually exclusive events A « B = ∆
P ( A » B) = P( A) + P( B)
Multiplication rule
P ( A « B) = P( A) P( B  A) or P( B) P( A  B)
For independent events P ( A  B) = P( A) and P( B  A) = P( B) , so:
P ( A « B) = P( A) P( B)
Conditional probabilities
P( A  B) =
P( A « B)
P( B)
Bayes’ Theorem
For a partition Ei , i = 1, 2, , n
P ( Ei  A) =
P( Ei ) P( A  Ei )
n
Â P( E j ) P( A  E j )
,
i = 1, 2, 3, , n
j =1
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CT302: Probability
Page 27
Chapter 2 Solutions
Solution 2.1
When throwing two dice, we can get a total from 2 to 12. Hence:
S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
Note that (unlike the sample space for a single dice) each element in this sample space
is not equally likely. For example, a total of 7 occurs more than, say, a total of 8.
Solution 2.2
We have:
A = {2, 8, 11, 16, 121}
B = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121}
Hence:
A « B = {16, 121}
A » B = {1, 2, 4, 8, 9, 11, 16, 25, 36, 49, 64, 81, 100, 121}
Solution 2.3
(i)
Since all the elements in A are contained in B, we have A Ã B . This is shown
on a Venn diagram as follows:
S
B
2
6
A
1
4
3
5
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(ii)
CT302: Probability
We have three sets. Taking care to ensure each number is in the correct overlaps
between sets, we obtain:
S
B
A
1
3
4
7
9 8
5
C
2
6
Solution 2.4
(i)
A¢ « B is everything not in set A and in set B:
A
(ii)
S
B
A¢ « B ¢ is everything not in set A and not in set B:
A
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B
S
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CT302: Probability
(iii)
Page 29
( A « B) ¢ is everything not in set A and set B (ie everything outside of the
intersection of A and B).
A
(iv)
B
S
( A » B) ¢ is everything not in set A or set B (ie everything outside of the union
of A and B).
A
B
S
Solution 2.5
Since 25% have passed both CT1 and CT3, there must be 60%  25% = 35% who have
only passed CT1. There must be 45%  25% = 20% who have only passed CT3. This
leaves 100%  35%  25%  20% = 20% who have passed neither exam:
S
CT1
CT3
35%
25%
20%
20%
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CT302: Probability
Solution 2.6
(i)
13
52
=
(ii)
4
52
1
= 13
(iii)
1
52
(iv)
12
52
1
4
3
= 13
Solution 2.7
(i)
We have P( A) =
4
52
, P( B) =
4
52
and:
A » B = {7 hearts, 7 clubs, 7 diamonds, 7 spades,
Ace hearts, Ace clubs, Ace diamonds, Ace spades}
So:
P( A » B) =
8
52
=
4
52
4 = P ( A) + P ( B )
+ 52
This rule works if that there is no “overlap” between the elements in A and B.
(ii)
We have P( A) =
4
52
, P(C ) =
13
52
and:
A » C = {7 hearts, 7 diamonds, 7 spades, Ace clubs, 2 clubs, 3 clubs,
4 clubs, 5 clubs, 6 clubs, 7 clubs, 8 clubs, 9 clubs, 10 clubs,
Jack clubs, Queen clubs, King clubs}
So:
P ( A » C ) = 16
=
52 /
4
52
+ 13
= P( A) + P( B )
52
This is because the 7 of clubs is in both A and C and so is counted twice on the
RHS of the equation.
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CT302: Probability
Page 31
Solution 2.8
If A = {get 1st question correct} and B = {get 2nd question correct} then we are told:
P( A) = 0.3 , P ( B) = 0.4 and P( A « B) = 0.1
(i)
We want P( A » B) , so using the addition rule:
P ( A » B) = 0.3 + 0.4  0.1 = 0.6
(ii)
We want the probability that both questions are wrong, that is the complement of
getting at least one question right. Hence the probability is:
= 1  0.6 = 0.4
Alternatively, we could have used a Venn diagram as follows:
B
A
0.2
0.1
S
0.3
0.4
Since, P( A) = 0.3 and P( A « B) = 0.1 there must be 0.3  0.1 = 0.2 just in set A alone.
Similarly we get 0.4  0.1 = 0.3 in set B alone. It is then easy to see the total
probability in both sets is 0.2 + 0.1 + 0.3 = 0.6 and the probability that neither question
is right, ie the area outside both sets, is 0.4.
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CT302: Probability
Solution 2.9
(i)
P( A « B) = b + e
(ii)
P( A « C ) = d + e
(iii)
P( B « C ) = e + f
(iv)
P( A « B « C ) = e
Solution 2.10
Replacing the terms on the RHS of the equation we obtain:
P ( A) + P( B) + P(C )  P( A « B)  P( A « C )  P( B « C ) + P( A « B « C )
= (a + b + d + e) + (b + c + e + f ) + (d + e + f + g )  (b + e)  (d + e)
 (e + f ) + e
= a+b+c+d +e+ f + g
= P( A » B » C )
Solution 2.11
We are given:
P (O) = 0.3 P( S ) = 0.4
P (C « S ) = 0.15 P(C « O) = 0.15 P( S » O) = 0.55 P(C « S « O) = 0.05
Since we also have P (none) = 0.2 , this gives us:
P (C » O » S ) = 1  0.2 = 0.8
(i)
We require P (C ) , but to use our threeevent addition rule we also require
P( S « O) . We get this from:
P ( S » O) = P( S ) + P(O)  P( S « O)
ﬁ 0.55 = 0.4 + 0.3  P( S « O) ﬁ P( S « O) = 0.15
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CT302: Probability
Page 33
Hence:
0.8 = P(C ) + 0.4 + 0.3  0.15  0.15  0.15 + 0.05
ﬁ P(C ) = 0.5
(ii)
We require P (C « S ¢ « O ¢) . This is not so easy to get by the threeevent
addition rule (as too many new events need to be calculated). However, it easy
to get if we place all the known probabilities into a Venn diagram:
C
S
0.1
0.25
0.1
0.05
0.15
0.1
0.05
O
0.2
We can see that P (C only) = 0.25 .
Solution 2.12
If A is the event “has worked for company A” and B is the event “has worked for
company B”, then we are told that:
P( A) =
(i)
20
24
, P( B) =
12
24
and P( A » B) = 1
Using the addition rule, we obtain:
P( A « B) =
20
24
+ 12
1 =
24
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8
24
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Page 34
(ii)
CT302: Probability
Using the conditional probability formula, we obtain:
P( A  B) =
P( A « B)
=
P( B)
8
24
12
24
=
2
3
Solution 2.13
If M is the event “eats a milk chocolate” and P is the event “eats a plain chocolate”,
then the tree diagram is as follows:
6
10
7
11
4
11
4
10
7
10
8
11
P
MMP
8
7
4
28
=
12 11 10 165
M
MPM
8
4
7
28
=
12 11 10 165
P
MPP
8
4
4
3
=
12 11 10 55
M
PMM
28
8
7
4
=
12 11 10 165
P
PMP
4
4
8
3
=
12 11 10 55
M
PPM
4
4
3
8
=
12 11 10 55
P
PPP
4
1
3
2
=
12 11 10 55
M
3
10
8
10
P
3
11
7
6
14
8
=
12 11 10 55
P
3
10
7
10
4
12
MMM
M
M
8
12
M
P
2
10
(i)
P (3 milk chocs) = P ( MMM ) = 14
55
(ii)
P (1 plain choc) = P ( MMP ) + P ( MPM ) + P ( PMM )
28
28
28
= 165
+ 165
+ 165
=
28
55
We could also have solved this more quickly using combinations:
C1 ¥
3
(128 ¥ 117 ¥ 104 ) = 5528
where 3C1 is the number of ways of obtaining one “success” (in this case plain
chocolate) and two “failures” (in this case milk chocolate) out of three trials,
Ê nˆ
n!
and nCr or Á ˜ is calculated as
.
r !(n  r )!
Ë r¯
© IFE: 2013 Examinations
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CT302: Probability
Page 35
Please note that permutations and combinations are assumed knowledge for
CT3 and so can be (and have been) asked in the examination.
Tree diagrams and combinations are covered in further detail in the Stats Pack.
Solution 2.14
If F is the event “is female”, M is the event “is male”, A is the event “chooses from “à
la carte”, and S is the event “chooses from the set menu”, then the tree diagram is as
follows:
0.74
A
FA
0.45 0.74 = 0.333
S
FS
0.45 0.26 = 0.117
A
MA
0.55 0.37 = 0.2035
S
MS
0.55 0.63 = 0.3465
F
5
0.4
0.5
5
0.26
0.37
M
0.63
(i)
P (S) = P( FS ) + P( MS )
= 0.117 + 0.3465
= 0.4635
(ii)
This is a conditional probability as we are told that the customer has chosen
from the à la carte menu – so this is already given.
P ( F  A) =
P( F « A)
0.333
=
= 0.621
P( A)
0.333 + 0.2035
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CT302: Probability
Solution 2.15
If A is the event “chooses route A”, B is the event “chooses route B”, C is the event
“chooses route C”, and T is the event “arrives on time to work”, then:
P( B  T ) =
=
P( B) P(T  B)
P( A) P(T  A) + P( B) P(T  B) + P(C ) P (T  C )
1
3
1
3
1
3
¥ 0.62
¥ 0.6 + ¥ 0.62 + 13 ¥ 0.7
= 32.3%
Past Exam Question (Subject C1, April 1994, Q1)
Using B for “blue collar”, N for “not blue collar”, C for “vote Conservative” and V for
“vote something else” we obtain:
0.46
C
BC
0.3 0.46 = 0.138
V
BV
0.3 0.54 = 0.162
C
NC
0.7 0.36 = 0.252
V
NV
0.7 0.64 = 0.448
B
0.54
0.3
0 .7
0.36
N
0.64
We want:
P( B  C )
P( B C )
P(C )
So we need to calculate:
P (C ) P( BC ) P( NC )
0.138 0.252
0.39
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CT302: Probability
Page 37
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