Paper CPA F1.1 BUSINESS MATHEMATICS & QUANTITATIVE METHODS Revision Guide
User Manual:
Open the PDF directly: View PDF 
.
Page Count: 46
LOSSARY
CPA
Certified Public Accountant Examination
Stage: Foundation F1.1
Subject Title: Business Mathematics &
Quantitative Methods
Revision Guide
LOSSARY
INSIDE COVER - BLANK
Page 2
CONTENTS
Title
Page
Study Techniques
3
Examination Techniques
4
Assessment Strategy
9
Learning Resources
10
Sample Questions and Solutions
11
Page 3
BLANK
Page 4
STUDY TECHNIQUE
What is the best way to manage my time?
• Identify all available free time between now and the examinations.
• Prepare a revision timetable with a list of “must do” activities.
• Remember to take a break (approx 10 minutes) after periods of
intense study.
What areas should I revise?
• Rank your competence from Low to Medium to High for each topic.
• Allocate the least amount of time to topics ranked as high.
• Allocate between 25% - 50% of time for medium competence.
• Allocate up to 50% of time for low competence.
How do I prevent myself veering off-track?
• Introduce variety to your revision schedule.
• Change from one subject to another during the course of the day.
• Stick to your revision timetable to avoid spending too much time on one topic.
Are study groups a good idea?
• Yes, great learning happens in groups.
• Organise a study group with 4 – 6 people.
• Invite classmates of different strengths so that you can learn from one another.
• Share your notes to identify any gaps.
Page 5
EXAMINATION TECHNIQUES
INTRODUCTION
Solving and dealing with problems is an essential part of learning, thinking and intelligence.
A career in accounting will require you to deal with many problems.
In order to prepare you for this important task, professional accounting bodies are placing
greater emphasis on problem solving as part of their examination process.
In exams, some problems we face are relatively straightforward, and you will be able to deal
with them directly and quickly. However, some issues are more complex and you will need to
work around the problem before you can either solve it or deal with it in some other way.
The purpose of this article is to help students to deal with problems in an exam setting. To
achieve this, the remaining parts of the article contain the following sections:
• Preliminary issues
• An approach to dealing with and solving problems
• Conclusion.
Preliminaries
The first problem that you must deal with is your reaction to exam questions.
When presented with an exam paper, most students will quickly read through the questions
and then many will … PANIC!
Assuming that you have done a reasonable amount of work beforehand, you shouldn’t be
overly concerned about this reaction. It is both natural and essential. It is natural to panic in
stressful situations because that is how the brain is programmed.
Archaeologists have estimated that humans have inhabited earth for over 200,000 years. For
most of this time, we have been hunters, gatherers and protectors.
In order to survive on this planet we had to be good at spotting unusual items, because any
strange occurrence in our immediate vicinity probably meant the presence of danger. The
brain’s natural reaction to sensing any extraordinary item is to prepare the body for ‘fight or
flight’. Unfortunately, neither reaction is appropriate in an exam setting.
The good news is that if you have spotted something unusual in the exam question, you have
completed the first step in dealing with the problem: its identification. Students may wish to
use various relaxation techniques in order to control the effects of the brain’s extreme
reaction to the unforeseen items that will occur in all examination questions.
Page 6
However, you should also be reassured that once you have identified the unusual item, you
can now prepare yourself for dealing with this, and other problems, contained in the exam
paper.
A Suggested Approach for Solving and Dealing with Problems in Exams.
The main stages in the suggested approach are:
1. Identify the Problem
2. Define the Problem
3. Find and Implement a Solution
4. Review
1. Identify the Problem
As discussed in the previous section, there is a natural tendency to panic when faced with
unusual items. We suggest the following approach for the preliminary stage of solving and
dealing with problems in exams:
Scan through the exam question
You should expect to find problem areas and that your body will react to these items.
PANIC!!
Remember that this is both natural and essential.
Pause
Take deep breaths or whatever it takes to help your mind and body to calm down.
Try not to exhale too loudly – you will only distract other students!
Do something practical
Look at the question requirements.
Note the items that are essential and are worth the most marks.
Start your solution by neatly putting in the question number and labelling each part of your
answer in accordance with the stated requirements.
Actively reread the question
Underline (or highlight) important items that refer to the question requirements. Tick or
otherwise indicate the issues that you are familiar with. Put a circle around unusual items that
will require further consideration.
Page 7
2. Define the Problem
Having dealt with the preliminary issues outlined above, you have already made a good start
by identifying the problem areas. Before you attempt to solve the problem, you should make
sure that the problem is properly defined. This may take only a few seconds, but will be time
well spent. In order to make sure that the problem is properly defined you should refer back
to the question requirements. This is worth repeating: Every year, Examiner Reports note that
students fail to pass exams because they do not answer the question asked. Examiners have a
marking scheme and they can only award marks for solutions that deal with the issues as
stipulated in the question requirements. Anything else is a waste of time. After you have re-
read the question requirements ask yourself these questions in relation to the problem areas
that you have identified:
Is this item essential in order to answer the question?
Remember that occasionally, examiners will put ‘red herrings’ (irrelevant issues) into the
question in order to test your knowledge of a topic.
What’s it worth?
Figure out approximately how many marks the problem item is worth. This will help you to
allocate the appropriate amount of time to this issue.
Can I break it down into smaller parts?
In many cases, significant problems can be broken down into its component parts. Some parts
of the problem might be easy to solve.
Can I ignore this item (at least temporarily)?
Obviously, you don’t want to do this very often, but it can be a useful strategy for problems
that cannot be solved immediately.
Note that if you leave something out, you should leave space in the solution to put in the
answer at a later stage. There are a number of possible advantages to be gained from this
approach:
1) It will allow you to make progress and complete other parts of the question that you are
familiar with. This means that you will gain marks rather than fretting over something
that your mind is not ready to deal with yet.
2) As you are working on the tasks that you are familiar with, your mind will relax and you
may remember how to deal with the problem area.
3) When you complete parts of the answer, it may become apparent how to fill in the
missing pieces of information. Many accounting questions are like jigsaw puzzles: when
Page 8
you put in some of the parts that fit together, it is easier to see where the missing pieces
should go and what they look like.
3. Find and Implement a Solution
In many cases, after identifying and defining the problem, it will be easy to deal with the
issue and to move on to the next part of the question. However, for complex problems that
are worth significant marks, you will have to spend more time working on the issue in order
to deal with the problem. When this happens, you should follow these steps:
Map out the problem
Depending on your preferred learning style, you can do this in a variety of ways including
diagrams, tables, pictures, sentences, bullet points or any combination of methods. It is best
to do this in a working on a separate page (not on the exam paper) because some of this work
will earn marks. Neat and clearly referenced workings will illustrate to the examiner that you
have a systematic approach to answering the question.
Summarise what you know about the problem
Make sure that this is brief and that it relates to the question requirements. Put this
information into the working where you have mapped out the problem. Be succinct and
relevant. The information can be based on data contained in the question and your own
knowledge and experience. Don’t spend too long at this stage, but complete your workings as
neatly as possible because this will maximise the marks you will be awarded.
Consider alternative solutions
Review your workings and compare this information to the question requirements. Complete
as much of the solution as you can. Make sure it is in the format as stipulated in the question
requirements. Consider different ways of solving the problem and try to eliminate at least one
alternative.
Implement a solution
Go with your instinct and write in your solution. Leave extra space on the page for a change
of mind and/or supplementary information. Make sure the solution refers to your workings
that have been numbered.
4. Review
After dealing with each problem and question, you should spend a short while reviewing your
solution. The temptation is to rush onto the next question, but a few moments spent in
Page 9
reviewing your solution can help you to gain many marks. There are three questions to ask
yourself here:
Have I met the question requirements?
Yes, we have mentioned this already. Examiner Reports over the years advise that failure to
follow the instructions provided in the question requirements is a significant factor in causing
students to lose marks. For instance, easy marks can be gained by putting your answer in the
correct format. This could be in the form of a report or memo or whatever is asked in the
question. Likewise, look carefully at the time period requested. The standard accounting
period is 12 months, but occasionally examiners will specify a different accounting period.
Is my solution reasonable?
Look at the figures in your solution. How do they compare relative to the size of the figures
provided in the question?
For example, if Revenue were 750,000 and your Net Profit figure was more than 1 million,
then clearly this is worth checking.
If there were some extraordinary events it is possible for this to be correct, but more than
likely, you have misread a figure from your calculator. Likewise, the depreciation expense
should be a fraction of the value of the fixed assets.
What have I learned?
Very often in exams, different parts of the solution are interlinked. An answer from one of
your workings can frequently be used in another part of the solution. The method used to
figure out an answer may also be applicable to other parts of your solution.
Conclusion
In order to pass your exams you will have to solve many problems. The first problem to
overcome is your reaction to unusual items. You must expect problems to arise in exams and
be prepared to deal with them in a systematic manner. John Foster Dulles, a former US
Secretary of State noted that: The measure of success is not whether you have a tough
problem to deal with, but whether it is the same problem you had last year. We hope that, by
applying the principles outlined in this article, you will be successful in your examinations
and that you can move on to solve and deal with new problems.
Page 10
Stage: Foundation 1
Subject Title: F1.1 Business Mathematics and Quantitative
Methods
Examination Duration: 3 Hours
Assessment Strategy
Examination Approach
This subject deals with the collection and organisation of key business facts into meaningful
data and the presentation and analysis of this data into useful information. Questions are
framed in a business context, with each question having a number of sub-sections. The first
sub-section may require quantitative analysis, while others may require qualitative analysis,
that is, to provide an interpretation of the quantitative data.
Examination Format
The examination is unseen, closed
book and 3 hours’ in duration.
Students are required to answer 5
questions out of 6.
Marks Allocation
Each question carries 20 marks The
total for the paper is 100 marks.
Learning Resources
Core Texts
Curwin and Slater / Quantitative Methods for
Business Decisions, 6th Edition / Cengage
(2008) ISBN 1844805743
Clare Morris / Quantitative Approaches in
Business Studies, 7th Edition / Pearson
Education (2010) ISBN 027379417
Donald Waters / Quantitative Methods for
Business, 4th Edition / Pearson Education
(2007) ISBN 0273694588.
For greater depth on statistics, the following is
recommended:
W.M. Harper / Statistics, 6th Edition / Pearson
Education / ISBN 0273634267.
Manuals
Institute of Certified Public Accountants of Rwanda – F1.1 Business Mathematics &
Quantitative Methods
Page 11
Supplementary Texts and
Journals
Les Oakshott / Essential Quantitative Methods for
Business, Management and Finance, 4th Edition/
/Palgrave (2009)/ ISBN 9780230218185.
Lucy T. / Quantitative Techniques 6th ed. /
Continuum Publications (2002) / ISBN
0826458548.
Wisniewski M. / Quantitative Methods for
Decision Makers 5th ed / Pearson 2009 / ISBN
9780273712077 / ISBN 0273712071.
Buglear J. / Quantitative Methods for Business
Elsevier (2004) / ISBN 0750658983.
Buglear J. / Stats Means Business / Butterworth
Heinemann (2001) / ISBN 0750653647.
Soper J. / Mathematics for Economics and
Business / Blackwell (2004) / ISBN
1405111275.
Useful Websites
(as at date of publication)
http://www.icparwanda.com/services.php
http://ubalt.edu/ntsbarsh/Businessstat/
opre504.htm - Professor Hossein Arsham’s,
(FOR, FRSS, FWIF), Statistical Thinking for
Managerial Decisions.
Page 12
REVISION QUESTIONS AND
SOLUTIONS
Stage: Foundation F1.1
Subject Title: Business Mathematics
& Quantitative
Page 13
QUESTION 1
Two companies have provided tenders for the replacement of production machines for DIY Ltd.
Machine 1 will cost
RWF
25,000 and Machine 2 will cost
RWF
20,000. The financial
accountant estimates that the other cash flows for the machines over their four year lifetime
will be:
Cash inflows (
RWF
)
Year 1 Year 2 Year 3 Year 4
Machine 1 18,000 22,000 12,000 8,500
Machine 2 10,000 12,000 16,000 17,000
Cash outflows (
RWF
)
Year 1 Year 2 Year 3 Year 4
Machine 9,000 12,000 3,000 3,000
Machine 2 7,000 10,000 3,500 3,500
The loan rate charged by the bank is 14%. You may assume that the chosen machine is paid
for upon delivery on site (purchased now) and the other cash flows will occur at the end of
the year.
REQUIRED:
(a) Set out the net cash flows for each machine. (4 Marks)
(b) Derive the Net Present Value for each machine. (6 Marks)
(c) Based on your analysis advise the company which machine to purchase. (4 Marks)
(d) The supplier of Machine 1 has now offered to modify its payment terms. Instead of
requiring an initial payment of
RWF
25,000, they will accept an initial payment of
RWF
15,000 with the balance of
RWF
10,000 payable in year 1. Which machine would
you now advise the company to purchase? Support your answer with relevant calculations.
(6 Marks)
[Total: 20 Marks]
Page 14
Page 15
QUESTION 2
To demonstrate that it is providing loans to small companies, the GV bank provided the
following data. This data summarises the number and level of loans to companies in both Kigali
and Butare.
Loans (
RWF
000s) Number of Companies
Kigali Butare
24 < 28 8 19
28 < 32 41 36
32 < 36 77 47
36 < 40 90 58
40 < 44 58 27
44 < 48 26 13
REQUIRED:
(a) Compare the mean loan and standard deviation for both Kigali and Butare.
(10 marks)
(b) Derive the co-efficient of variation for both regions. (6 marks)
(c) Explain the principles underpinning the calculations in (a) and (b) above.
(4 marks)
[Total: 20 Marks]
QUESTION 3
The local supermarket recently received a consignment of cereals from the national distributor.
The management accountant believed that the consignment is incorrectly packed and randomly
selected 49 bags to be sampled. The mean weight was found to be 42.4kgs with a standard
deviation of 4kgs. According to the distributor, the bags should have had a mean weight of
40kgs.
REQUIRED:
(a) Calculate the range of weights with a 95% confidence level. (10 Marks)
(b) Test if the population mean is greater than 40kg with a 95% confidence level.
(10 marks)
[Total: 20 Marks]
Page 16
QUESTION 4
The SME company wishes to establish a relationship between the costs of production and
the output of the company. The following data, on units of output and total costs, has been
collected over the last eight quarters:
Quarter Output (units) Cost
RWF
1
2
3
4
5
6
7
8
10,000
20,000
40,000
25,000
30,000
40,000
50,000
45,000
32,000
39,000
58,000
44,000
52,000
61,000
70,000
64,000
REQUIRED:
(a) Using linear regression analysis, derive the relationship between the variables. (10 Marks)
(b) Interpret the equation in terms of both fixed and variable costs of production. (6 Marks)
(c) Forecast the costs that should be incurred at an output level of 55,000 units (4 Marks)
[Total: 20 Marks]
QUESTION 5
(a) Your company is purchasing a Systems 102 shredder for
RWF
12,000. Estimate the
equal annual payments, if the machine is being purchased with a 5 year loan compounded
annually at 14%. (6 Marks)
(b) You wish to purchase a forklift for your business. Your bank quotes you an annual
percentage rate (APR) of interest of 15% on a loan of
RWF
20,000 over 4 years.
Calculate the total interest paid. Assume that the interest is paid at the end of each year.
(6 Marks)
(c) Your company’s internal auditors have examined the debtors accounts. The data is
normally distributed with a mean value of
RWF
3,000 and a standard deviation of
RWF
500. You consider that 5% of the accounts are ‘doubtful’. In this context you are
asked to determine the probability that, if an account is randomly chosen, its value will be
between
RWF
3,200 and
RWF
3,500.
Page 17
(8 Marks)
[Total: 20 Marks]
QUESTION 6
“Time series data can be used as a basis for estimating both trend and seasonal variation.
Such estimates can then be used as a basis for forecasting.”
In this context:
(a) Outline the structure of a time series including trend, seasonal, cyclical and irregular
components.
(8 Marks)
(b) Describe the moving average method for smoothing a time series. (6 Marks)
(c) Describe how the time series can be used for forecasting. (6 Marks)
[Total: 20 Marks]
QUESTION 7
CPAR Consultants Ltd. is undertaking an analysis of a proposal by Superior Products Ltd. The
company is investing in a moulding machine at a cost of
RWF
125
,
000
.
The machine will last
for 5 years and will be sold for scrap at the end of year 6 for
RWF
5
,
000
.
The machine will
require regular annual maintenance. At the end of year 1 this will amount to
RWF
1
,
500 and
will increase by 20% each year until the end of year 4. The revenue produced by the
machine is estimated at
RWF
30
,
000 at the end of year 1 and this will increase by 10% at the
end of each year over the life of the machine.
As a consultant with CPAR you are required to:
(i) Tabulate the net cash flows for the machine. (8 Marks)
(ii) Establish the Internal Rate of Return (IRR) on the investment. (8 Marks)
(iii) Interpret your result if the overdraft rate provided by the bank is 15%. (4 Marks)
[Total: 20 Ma
r
ks
]
Page 18
QUESTION 8
DIB Ltd. has two divisions with employees doing similar types of work. The data below gives
the distribution of earnings of a sample of employees from the two divisions.
Earnings/week
RWF
Number of e
m
p
l
o
y
ee
s
Division
1
Division
2
400 – 420
420 – 440
440 – 460
460 – 480
480 – 500
500 – 520
11
20
35
40
25
10
12
30
50
60
40
35
As the management accountant to the company you are asked to:
(i) Calculate the average earnings and standard deviation for employees of both divisions
(12 Marks)
(ii) Determine if there is any significant difference between the earnings in both divisions.
Test at the 1% significance level. (8 Marks)
[Total: 20 Ma
r
ks
]
QUESTION 9
Fado Ltd is a training company which pays a Government allowance to trainees during various
stages of training. A recent notice by the Office of the Comptroller General has indicated that the
company will be subject to an audit in the near future. You have compiled the following data
from the company records.
Monthly
Allowance
Number of
T
r
a
i
n
ee
s
260-280
280-300
300-320
320-340
340-360
360-380
380-400
400-420
8
14
16
15
9
7
6
5
Page 19
In order to prepare for the audit you have been requested by the financial controller to prepare
a report for the Board. As part of your report you are requested to:
(i)
Present the data in a histogram.
(8 Marks)
(ii)
Derive the modal and median allowances paid to trainees.
(8 Marks)
(iii)
Explain your results.
(4 Marks)
[Total: 20 Ma
r
ks]
QUESTION 10
A company is negotiating with Superior Products Ltd for the supply of batteries for its torches.
Superior Products needs to plan its production to meet the needs of the torch company. It uses the
companyʼs quarterly sales figures over the past three years to forecast future demand.
Y
ea
r
Q1
Q
2
Q
3
Q
4
2007
2008
2009
350
450
550
285
420
530
197
325
415
390
460
615
Using this data:
(i) Derive the trend line equation using linear regression analysis (10 Marks)
(ii) Illustrate the data on a diagram and forecast the demand for the four quarters of 2010
(10 Marks)
[Total: 20 Ma
r
ks]
QUESTION 11
As an independent financial adviser you are presented with the following problems from
clients
(i) The SPP Credit Union has launched a new saving scheme for investors. If your client
invests
RWF
10
,
000 now and
RWF
5
,
000 at the end of each year for the next 3 years,
he will receive 10% in year 1 and 15% in years 2 and 3. Advise your client of the sum
receivable at the end of year 3 and the overall return on the investment. (6 Marks)
(ii) Mr P. Orr has been offered an encashment value of
RWF
10
,
000 now by the Bank of
Kigali on his endowment mortgage. However, he estimates that it will mature to
RWF
13
,
500 in 5 years time applying an interest rate of 4%. Advise Mr P.Orr on the
best option. (6 Marks)
(iii) “Paulʼs”, the local builderʼs provider, consults you regarding the weights of concrete
provided to construction sites. He believes that the quantities supplied exceed the required
Page 20
amounts. He wants to ensure that only 5% of any orders of concrete exceed the standard
order weight of 1,000kg. The specification of the automatic weighing machine states that
orders weigh with a standard deviation of
10kg. Advise Paul on the level that the machines average should be set to, assuming a
normal distribution. Provide a diagram to support your advice. (8 Marks)
[Total: 20 Ma
r
ks
]
QUESTION 12
You are frequently asked by the financial director to make presentations to the management
team on modern business concepts. Discuss the following for inclusion in your next
presentation:
(i)
Structure of a time series.
(5 Marks)
(ii)
Investment Appraisal.
(5 Marks)
(iii)
Network Analysis.
(5 Marks)
(iv)
Measures of dispersion.
(5 Marks)
[Total: 20 Ma
r
ks]
Page 21
SUGGESTED SOLUTIONS
SOLUTION 1
(a) and (b) Cash flows are set out in the table below.
Machine 1
Year
Cash Inflows
Cash Outflows
Net Cash Flows
Discount Factor
Present Value
-1
-2
(1) - (2)
-14%
0
---------
-25,000
-25,000
1.000
-25,000
1
18,000
- 9,000
-9,000
0.877
7,893
2
22,000
-12,000
10,000
0.769
7,690
3
12,000
-3,000
9,000
0.675
6,075
4
8,500
-3,000
5,500
0.592
3,256
Net Present Value
(86)
Machine
2
Year
Cash
Inflows
(1)
Cash Outflows
(2)
Net Cash Flows
(1) - (2)
Discount Factor
(14%)
Present Value
0
---------
(20,000)
(20,000)
1.000
(20,000)
1
10,000
(7,000)
3,000
0.877
2,631
2
12,000
(10,000)
2,000
0.769
1,538
3
16,000
(3,500)
12,500
0.675
8,438
4
17,000
(3,500)
13,500
0.592
7,992
Net Present Value
2 Marks
599
3 Marks
Page 22
c) The Net Present Value for machine 1 is a net loss of
RWF
86 while machine 2
provides a positive Net Present Value of
RWF
599. On this basis machine 2 is more
financially attractive and should be selected by the company.
However, the difference between the NPV for both machines is small. Machine 1 has larger
cash inflows during the early years while machine 2 has larger cash inflows during the latter
years. The basic difference between the two proposals is the initial cost. A more appropriate
analysis is to derive the internal rate of return which will provide a % return of both proposals
which can be compared with the rate provided by the bank. (4 Marks)
d) Based on the offer of phased payments for Machine 1, the revised Net Present Value
for Machine 1 is a positive
RWF
1,144. Although Machine 2 also provides a positive Net
Present Value of
RWF
599, it is less positive than Machine 1. On this basis Machine 1 is the
more financially attractive and should be selected by the company. (4 Marks)
Year
Cash Inflows
Cash Outflows
Net Cash Flows
Discount Factor
Present Value
-1
-2
(1) - (2)
-14%
0
---------
-15,000
-15,000
1
-15,000
1
18,000
-19,000
-1,000
0.877
-877
2
22,000
-12,000
10,000
0.769
7,690
3
12,000
-3,000
9,000
0.675
6,075
4
8,500
-3,000
5,500
0.592
3,256
Net Present Value
1,144
(2 Marks)
[Total: 20 Marks]
SOLUTION 2
(a) The mean loans and standard deviations are calculated below.
Kigali Region
Lower Upper Midpoint No companies fx (x-x)
f(x-x)
2
Boundary
Boundary
(x)
(f)
24
28
26
8
208
-11.03
973.29
28
32
30
41
1230
- 7.03
2026.26
Page 23
32
36
34
77
2618
- 2.03
156.31
36
40
38
90
3420
0.97
84.68
40
44
42
58
2436
4.97
1432.60
44
48
46
26
1196
8.97
2091.96
∑
300
6765.10
Mean, ͞x = ∑fx = 11,108 =
RWF
37.03 (2 marks)
∑f 300
Standard deviation, σ = √ ∑f(x - x)2 = √6765.1 = √ 22.55 = 4.75 (3 marks)
∑f 300
Butare Region.
Lower Upper Midpoint No companies Fx (x-x) f(x-x)
2
Boundary
Boundary
(x)
(f)
24
28
26
19
494
-9.54
1729.22
28
32
30
36
1080
- 5.54
1104.89
32
36
34
47
1598
- 1.54
111.47
36
40
38
58
2204
2.46
350.99
40
44
42
27
1134
6.46
1126.75
44
48
46
13
598
10.46
1422.35
∑
200
7108
5845.67
Mean, ͞x = ∑fx = 7,108 =
RWF
35.54 (2 marks)
∑f 200
Standard deviation, σ = √ ∑f(x - ͞x)2 = √5845.7 = √ 29.22 = 5.40 (3 marks)
∑f 200
(b) Co-efficient of variation. This measures the relative dispersion between distributions and is
derived from σ .
x
Using this co-efficient gives:
Kigali: 4.75 x 100 = 12.83%; Butare:
5.40
x 100 = 15.19%
37.03
(3 marks)
35.54
(3 marks)
(c) From observations of the data from both series it is clear that the relative dispersion
between the distributions is not wide. This calculation is more appropriate where there is
substantial difference between the means of the distributions. The co-efficient is a measure that
enables us to quantify the relative dispersion. (4 marks)
[Total: 20 Marks]
Page 24
SOLUTION 3
(a) Establish the hypothesis that the population mean weight, µ, is 40kgs, that is, the Null
Hypothesis (Ho). This means that we are assuming that there is no difference between the mean
and the sample mean and that any difference can be ascribed to chance. The Alternative
Hypothesis (H1) can be stated that the population mean does not equal 40kg.
Ho: µ = 40kg
H1: µ ≠ 40kg (2 marks)
If Ho is found to be true H1 is rejected while if Ho is found to be false, H1 is accepted. With a
95% confidence level the sample mean must lie within ± 1.96 standard errors (the acceptance
zone), so that, if Ho is true, 95% of the means of all samples must be within the range µ ±
1.96sx where sx is the standard error, that is, sx = σ/√n. (2 marks)
Therefore, sx = 4/√49 = 0.57.
The range is 40 ± 1.96 x 0.57, that is, 38.88kgs to 41.12kgs (2 marks)
or the z-score [(x – µ)/sx = (42.4 -40)/0.57 = 4.21] must fall outside ± 1.96
As the sample mean of 42.4kg falls outside the acceptance zone the null hypothesis is rejected
and H1 is accepted, that is, µ ≠ 40kg. The difference between the population and sample
means is significant at the 5% level. (4 marks)
(b) In this test it is necessary to test if the population mean is greater than the assumed
value of 40kgs, that is,
Ho: µ = 40kgs; H1: µ > 40kgs. (4 marks)
This is a one tailed test where the number of standard errors at the 5% level is 1.65. Since the
calculated z score is 4.21 and is outside the 5% value the Null Hypothesis is rejected.
(2 marks)
It can therefore be stated that the population mean is greater than 40kgs. (2 marks)
The analysis is shown in the diagram for a two tailed test.
Page 25
(2 marks)
[Total: 20 Marks]
Page 26
SOLUTION 4
To determine the relationship between the variables, and the regression line Y = a + bX, we
solve the following equations
∑Y = na + b∑X
∑XY = a∑X + b∑X2 or substitute the above values into the following (2 Marks)
b = ∑XY - ∑X x ∑Y
n
∑X2 - (∑X)
2
n
a = ∑Y - b∑X
n n
X (000’s) Y (000’s) XY X
10 32 320 100
20 39 750 400
40 58 2320 1600
25 44 1100 625
30 52 1560 900
40 61 2440 1600
50 70 3500 2500
45 64 2880 2025
260 420 14900 9750
Quarter
Output (units)
Total Cost RWF
1
10,000
32,000
2
20,000
39,000
3
40,000
58,000
4
25,000
44,000
5
30,000
52,000
6
40,000
61,000
7
50,000
70,000
8
45,000
64,000
Page 27
Therefore, b = 14900 - 260 x 420/8 = 14900 - 13650
9500 - (260)/8 9750 - 8450
= 1250/1300
= 0.9615 (2 Marks)
a = 420/8 - 0.9615 x 260/8
= 52.5 - 31.25
= 21.25 (2 Marks)
Y = 21,250 + 0.962X (4 Marks)
Page 28
(b) Total Cost = Fixed cost + Variable cost per unit of production. (2 marks)
In the above equation the fixed cost is
RWF
21,250; this cost is independent of the
level of production. X represents the variable cost of production i.e. for each unit increase in
production the costs increases by a factor of 0.962. At some stage during the production
process the variable cost profile will intersect the fixed cost to give the breakeven value. The
sum of the fixed and variable costs gives the total cost profile. (4 marks)
(c) The projected cost at 55,000 units: At output levels of 55,000 units, the likely incurred
costs are:
Y = RWF21,250 + 0.962X 1,250
(2 marks)
= RWF21,250 + 0.962 (55,000) = RWF52,910
This projection is outside the boundaries of the relationship and the accuracy is predicated on
the principle that the linear relationship will continue. On the assumption that the linearity of
the relationship continues, the extrapolated figure can be accepted as being relatively accurate.
(2 marks)
[Total: 20 Marks]
SOLUTION 5
(a) In this case we are talking about an amortisation annuity over the 5 years, that is, the
amount borrowed (
RWF
12,000) is equal to the present value of the annual payments (say
A) over the 5 years at the interest rate of 14%.
RWF
12,000 = A + A + A + A + A
1.14 (1.14)2 (1.14)3 (1.14)4 (1.14)5
(3 Marks)
= A (0.8772 + 0.7695 + 0.6749 + 0.5921 + 0.5194)
= A (3.4331)
Therefore, A =
RWF
12,000 =
RWF
3,495. (3 Marks)
3.4331
Page 29
(b) The interest will be compounded over the four years. The compound interest is
derived as follows (the total cost at the end of four years less the initial cost equals the interest
paid)
CI = P(1 + i)t - P, where I = 15%, P =
RWF
20,000, t = 4. (3 Marks)
Therefore, CI = 20,000(1 + 0.15)4 - 20,000
= 20,000 x 1.749 - 20,000
= 34,980 - 20,000 = RWF14,980
(3 Marks)
(c) Probability of more than
RWF
3,200 is given by
Z = (3200 - 3000)/500
= 0.4
The probability is 0.3446 from the Normal distribution tables. (3 Marks)
Probability of more than
RWF
3500 is given by
Z = (3500 - 3000)/500
= 1.00
The probability is 0.1587 from the Normal distribution tables. (3 Marks)
Therefore, probability is within the range 3200 to 3500, that is, 0.3446 – 0.1587, i.e. 0.1859.
(2 Marks)
[Total: 20 Marks]
Page 30
SOLUTION 6
(a) Structure of a time series.
A time is a set of results for a particular variable of interest taken over a period of time. There are
four separate elements to consider
- Trend component: in a set of time series data the measurements are taken at regular
intervals. However, there may be random fluctuations in the data but in some cases the
data will show a shift to lower or higher values over the time period in question. This
movement is called the trend and is usually the result of some long term factors such as
changes in expenditure, output, etc. There may be various trend patterns such as increasing
linear trends, decreasing linear trend, non-linear trend or no trend.
(2 Marks)
- Cyclical component: a time series may show a trend of some sort but may also show a
cyclical pattern of alternative sequences of observation above and below the trend line.
This moves from the medium to longer term around the trend line. Any pattern of this
type which lasts longer than a year is called a cyclical element for the time series. This is
generally a very common occurrence. (2 Marks)
- Seasonal component: there may be a pattern of variability in regular cycles within a year.
The element of the time series which represents variability due to seasonal influences
is called the seasonal element. It normally refers to movements over a one year period
but it can also refer to any repeating pattern of less than one-year’s duration (2 Marks)
- Irregular component: this is the element which cannot be explained by the trend,
cyclical and/or seasonal elements and is entirely unpredictable. It represents the random
variability in the time series, caused by unanticipated and non-recurring factors which
are unpredictable. It gives a dramatic and unexpected departure from trend. (2 Marks)
(b) Use of the moving average method for smoothing.
Smoothing methods are used to smooth out the irregular element of time series where there are
no significant trends, cyclical or seasonal pattern.
Moving average: this method involves using the average of the most recent data values to
forecast the next period. The number of data values used can be selected in order to minimise
the forecasting error. To find the moving average, the simple average for a specified number
of items of data is found. For example, with quarterly data a simple average for four items of
data can be calculated and then move that average along. This is done by recalculating the
average having dropped the initial item of data and adding a subsequent item of data. This
gives the four quarter moving total. The moving totals fall in between the actual quarterly
data, that is, the first moving total falls between quarters 2 and 3. The respective four quarter
Page 31
moving totals are divided by 4 to find the four quarter moving average. The data is centred by
summing respective pairs of data and dividing by two. This smooths the data so that the trend
line can be derived. In this way moving averages are used to ‘smooth out’ fluctuations in the
data and obtain a trend line. This is the moving average trend which can be plotted to show
the direction. Since the trend does not vary significantly in short time periods the trend can
be projected to forecast future events.
However, there are limitations to the use of moving averages. Equal weighting is given to each
of the values used in the moving average calculation whereas the most recent data is more
relevant to current conditions. The moving average takes no account of data outside the period
of average so full use is not made of all the data. The use of the unadjusted moving average as a
forecast can cause misleading results when there is an underlying seasonal variation.
(6 Marks)
(c) Use of a time series for forecasting.
Forecasting from the time series lies in our knowledge of the behaviour of trend. Since the trend
does not vary significantly in short time periods, the trend can be projected to forecast future
events. A time series can be used to forecast where there are no significant trends, cyclical
or seasonal elements using the moving averages method as set out above. Using moving
averages, to get an accurate projection of trend without plotting the values, it is necessary to
derive the average rate of change of the trend. This can be done by deducting the first value
of the trend from the last trend value, taking the average (less one) and adding to the last
trend value. This can be done for a number of values to forecast for a number of periods into
the future. This is the moving average trend which can be plotted to show the direction. The trend
can be projected forward. However, the further the trend is projected the greater is the
possibility of inaccuracy.
However, if there are both trend and seasonal elements in the series there are other methods
which may be used. The first stage in such a forecasting technique is to calculate the seasonal
factors by smoothing out the time series using the moving averages method. If the seasonal
factors are quarterly then the moving average is calculate on groups of four data points. When
seasonal elements are involved it is necessary to work out the centred moving averages which
are averages of the moving averages. Dividing the original observation by the equivalent
centred moving average gives a seasonal factor for each observation. This is done for all
quarters.
If the time series shows a long term linear trend, linear regression may be used. A linear
regression equation linking dependent and independent variables can be derived but this
assumes that we have a linear trend and that the future will be like the past. (6 Marks)
[Total: 20 Marks]
Page 32
SOLUTION 7
1. (i) Cash Flows for the proposal.
Year
E
nd
Ma
c
h
i
n
e
Cost
RWF
Ma
i
n
t
e
n
a
n
c
e
Cost
RWF
Re
v
e
nu
e
Net Ca
s
h
Flows
RWF
0 125,000
(125,000)
1
(1,500) 30,000 28,500
2
(1,800) 33,000 31,200
3
(2,160) 36,300 34,140
4
(2,592) 39,930 37,338
5
43,923 43,923
6 5
,
000
5
,
000
(2 Marks) (2 Marks) (2 Marks) (2 Marks)
(ii) Calculation of the Internal Rate of Return. To calculate the IRR it is necessary to derive present
values. Taking discount values at 14% and 20% the following table is developed.
Y
ea
r
Net Cash
F
l
o
w
s
RWF
Discount
F
a
c
t
o
r
@ 12%
PV
RWF
Discount
F
a
c
t
o
r
@ 18%
PV
RWF
0 (125,000) 1
.
000 (125,000) 1
.
000 (125,000)
1 28,500 0
.
893 25,450.5 0
.
847 24,139.5
2 31,200 0
.
797 24,866.4 0
.
718 22,401.6
3 34,140 0
.
712 24,307.7 0
.
609 20,791.3
4 37,338 0
.
636 23,747.0 0
.
516 19,266.4
5 43,923 0.567 24,904.3 0.437 19,194.4
6 5
,
000 0.507 2,535.0 0
.
370 1,850.0
NPV
810.9
(17,356)
(2 Marks) (2
Marks)
The IRR can be derived by calculation or graphically. Using the formula
N1I2 - N2I1 , where discount rate I1 gives NPV N1 and discount rate I2 gives NPV N2
N1 - N2
N1 =
RWF
810
.
9
,
I1 = 12%; N2 = (
RWF
17
,
356)
,
I2 = 18%
Page 33
IRR = 810.9 x 0.18 - (17,356) x 0.12 = 2,229 = 12.3%
810.9 - (17,356) 18,167
(4 Marks)
Page 34
(iii) Interpretation of the result. The value of 12.2% can be interpreted as the rate of
return that the project will earn. On face value the NPV is positive at 12% and this
represents a positive result. However, if the cost of capital to the company, represented by
the rate of interest charged by the bank of 15%, is greater than this value, the project is not
viable. However, there are mechanisms for deriving the real cost of capital to the company
based on its equity and funds which may not be representative of the bank rates. The
decision of the company in this case is that the project, as a stand alone investment, will not
contribute the required return. (4 Marks)
[Total: 20 Ma
r
ks
]
SOLUTION 8
(i) The average earnings and standard deviations for employees in both divisions.
Division 1
Lower
Bound
a
r
y
Upper
Bound
a
r
y
F
r
e
q
f
Mid
point
(
x
)
F
x
f
x
2
400
420
440
460
480
500
420
440
460
480
500
520
11
20
35
40
25
10
410
430
450
470
490
510
4
,
510
8
,
600
15
,
750
18
,
800
12
,
250
5
,
100
1,849,100
3,698,000
7,087,500
8,836,000
6,002,500
2,601,000
∑
141
65,010
30,074,100
Mean = x = ∑fx = 65,010 =
RWF
461
.
1 (3 Marks)
∑f 141
Standard Deviation
σ = √ ∑ fx 2 – {∑ fx}2 = √ 3,074,100 – 212,578
∑ f {∑ f } 141
= √ 731 = RWF26.71 (3 Marks)
Division 2
Lower
Bound
a
r
y
Upper
Bound
a
r
y
F
r
e
q
f
Mid point (
x
)
F
x
f
x
2
400
420
440
460
480
500
420
440
460
480
500
520
12
30
50
60
40
35
410
430
450
470
490
510
4
,
920
12
,
900
22
,
500
28
,
200
19
,
600
17
,
850
2
,
017
,
200
5
,
547
,
000
10
,
125
,
000
13
,
254
,
000
9
,
604
,
000
9103
,
500
Page 35
∑
227
105,970
49,650,700
Mean = x = ∑fx = 105,970 =
RWF
466
.
8 (3 Marks)
∑f 227
Standard Deviation.
σ = √ ∑ fx2 – {∑ fx}2 = √ 49,650,700 – 217,928
∑ f {∑ f } 227
= √ 797 =
RWF
28
.
23 (3 Marks)
Summary
Division
1
Division
2
Mean
RWF
461
.
1
RWF
466
.
8
Std Deviation
RWF
26
.
71
RWF
28
.
23
Number 141 227
(ii) In order to test if there is a difference in the earnings between both divisions a hypothesis
can be used. In this case we are testing for the difference of means in a similar way as confidence
intervals for the difference of means. The difference between the sample estimates is first
calculated and the values given by the null hypothesis, transform the difference to a number of
standard errors, then compare with a critical value consistent with the alternative hypothesis. In
this way we are comparing if the two sample means have come from the same population,
i.e. Null Hypothesis, Ho: µ1 = µ2 and the alternative hypothesis
H1: µ1 ≠ µ2 (4 Marks)
This is a two sided test. In this particular test we are assuming H0 to be correct.
The Z statistic is X1 - X2 , that is, the difference in means divided by the standard error
√S
12 +
S
22
N1 N2
X1 - X2 = 461.1 - 466.8 = - 5.7
2
S
1
2
= 26
.
71
2
= 5.06;
S
2
2
= 28
.
23 = 3.51;
N1 141 N2 227
Z = -5.7 = - 1.94 (4 Marks)
√8.57
Z, at 1% significance level, is 2.58.
Since the calculated value is less than this level we accept the null hypothesis – the samples do
not provide evidence that there is a difference in the long-run earnings of employees.
Page 36
[Total: 20 Ma
r
ks
]
Page 37
SOLUTION 9
(i) Histogram of the data.
Monthly Allowance rwf
Number of Trainees (f)
Cumulative Frequency
260-280
280-300
300-320
320-340
340-360
360-380
380-400
400-420
8
14
16
15
9
7
6
5
8
22
38
53
62
69
75
80
Frequency
16
14
12
10
8
6
4
2
0 260 280 300 320 340 360 380 400 420 Weekly allowance
RWF
(8 Marks)
The modal value of the allowance is approximately
RWF310,
found from the histogram.
(ii) The Median.
The median is the value of the 40.5 trainee in group 320 – 340.
From the formula the median value = 320 + (40.5 – 38)/15 x 20 =
RWF
323
.
33
.
(6 Marks)
(iii) Comparison of the mode and median.
The mode is the greatest frequency of the distribution and is in the range
RWF320
-
RWF340.
The accurate value is derived from the histogram above.
Page 38
The median represents 50% of the data above and below the mid point. It means that 50% of
the trainees are receiving more than
RWF
323
.
23 and 50% less than
RWF
323
.
23
(8 Marks)
Page 39
SOLUTION 10
(i) The trend for this data can be developed using either a moving average technique or linear
regression analysis. A four quarter centred moving average could be derived allowing us to
decompose the data to isolate the trend by means of either additive or multiplicative models. In the
present case it may be substantially easier to develop the trend by means of linear regression without
smoothing the data, that is,
y = a + bx where
a = Σy - bΣx
n n
b = Σxy - (ΣxΣy)/n,
Σx2 - (Σx)2/n
Y
ea
r
Quarter
x
Sales
y
x
2
xy
1
2
3
∑
1
2
3
4
5
6
7
8
9
10
11
12
78
350
295
197
390
450
420
325
460
550
530
415
615
4
,
997
1
4
9
16
25
36
49
64
81
100
121
144
650
350
590
591
1,560
2,250
2,520
2,275
3,680
4,950
5,300
4,565
7,380
36
,
011
(5 Marks)
Inserting values gives
b = 36,011 - (78 x 4997)/12
650 - 55
2
/
12
= 3530.5 = 24.69
143
a = 4997 - 24.69 x 78 = 416.4 - 160.5 = 255.9
12 12
Therefore, y = 255.9 + 24.69x. (5 Marks)
Page 40
(ii) Plotting a graph of the points and projecting the data for 2010. (5 Marks)
Y Sales
600
300
Time
Series
Regression
Line
0 4 8 12 16
Q
uar
t
ers
The projected points for the four quarters of 2010 are calculated from the regression line for
quar
t
ers
13,14,15,16.
For x = 13, y (Sales) = 576,870
For x = 14, y (Sales) = 601,560
For x = 15, y (Sales) = 626, 250
For x = 16, y (Sales) = 650,940
(5 Marks)
(Total: 20 Ma
r
ks)
Page 41
SOLUTION 11
(i) Initial Sum invested:
RWF
10
,
000
At end year 1
+10%
RWF
1,000
RWF
5,000
RWF
16
,
000
At end year 2
+15%
RWF
2,400
RWF
5,000
RWF
23
,
400
At end year 3
+15%
RWF
3,510
RWF
5,000
RWF
31
,
910 ---- Sum Receivable (3 Marks)
Return on investment: (
RWF
31
,
910 -
RWF
25
,
000) x 100 = 27.64%
RWF
25
,
000
(3 Marks)
(ii) Discounting
RWF
13
,
500 to present values at 4% gives 13,500 = 13,500
1.045 1.2166
= R
WF
11
,
096
Since this value is greater than
RWF
10
,
000
,
the option to receive
RWF
13
,
500 in 5 years is
the best option. (6 Marks)
(iii) Standard deviation = 10kg; Mean = ?; x = 1000kg.
Converting these values to the standard Normal Distribution z = x - µ
σ
For 5% (0.05), z = 1.645
Therefore, 1.645 = 1000 - µ
10
µ = 1000 - 1.645 x 10
= 983
.
55kg
To ensure that only 5% of the orders are overweight the machine should be set to 983.55kg.
(4 Marks)
Page 42
Diagram (4 Marks)
100 kg
Area
= 0.05
gives z
= 1.645
µ
[Total: 20 Ma
r
ks
]
Page 43
SOLUTION 12
An explanation of the terms is provided below.
(i)
Structure
of a time series. A time series is a set of results for a particular variable
taken over a period of time. There are four separate elements in the structure
- Trend: In many cases the data exhibits a shift either to lower or higher values over
the time period in question. This movement is called the trend and is usually the result
of some long-term factors such as changes in expenditure, sales, demographic factors, etc.
There may be many trend patterns – increasing linear trend, decreasing linear trend, non-
linear trend or no trend.
-
Cyclical:
A time series may display a trend of some form but may also show a cyclical
pattern of alternative sequences of observation above and below the trend line. Any regular
pattern which lasts longer than a year is the cyclical element of the time series. This is a
common occurrence represented by the retail sector of business.
- Seasonal: although the cyclical patter may be displayed over a number of years, there
may be a pattern of variability within one-year periods. The element of the time series which
represents variability due to seasonal influences is the seasonal element. While it normally
refers to patterns over a one-year period, it can also refer to any repeating pattern of less
than one-yearʼs duration.
- Irregular element: this is the element which cannot be explained by the trend, the
cyclical or seasonal elements. It represents the random variability in the time-series, caused
by unanticipated and non-recurring factors which are unpredictable. There are particular
methods used to smooth out the irregular element of time series where there are no
significant trends, cyclical or seasonal patterns such as moving averages or exponential
smoothing. (5 Marks)
(ii)
Investment Appraisal
.
Many use many quantitative techniques as an aid to
decision making. One of the most important applications is concerned with investment
decisions. Managers often have to make choices between alternative decisions, they need to
consider future costs and revenues and the importance of incremental changes in costs and
revenues. It is also critical to consider the time value of money because of time scales
involved. The managerʼs decision to invest is based on three important factors – the
estimate of the future based on forecasts of costs, revenues, inflation, and interest rates; the
alternatives available for investment: techniques to support the decision must be used; the
business attitude to risk: because of the uncertainty of the future and project uncertainty,
additional techniques must be used. A number of these techniques used in investment
appraisal are, the accounting rate of return; payback; discounted cash flow, such as net
present value and internal rate of return.
The accounting rate of return is the ratio of the average annual profits, after depreciation, to
the capital involved. Variations of this exist but it has some drawbacks – it does not allow
for the timing of cash flows and profit has subjective elements. Payback is a popular
technique which is defined as the period which it takes for the projects net cash flows to
recover the original investment. Payback favours quick return projects
Page 44
– the project with the shortest payback period is accepted. However, it does not measure
overall investment worth since it does not consider cash flows after the payback period or
the timing of cash flows. Net Present Value uses discounting principles and involves
calculating the present value of expected cash inflows and outflows and establishing
whether the present value of the net cash flows is positive. The internal rate of return is the
discount rate which gives zero NPV. If the calculated IRR is greater than the companyʼs
cost of capital the investment is accepted. However these two techniques do not necessarily
rank investment proposals in the same order of attractiveness. But the IRR technique is used
extensively in investment appraisal decisions. (5 Marks)
(iii) Network
Analysis
.
This deals with a range of techniques used to aid managers in the
planning and control of projects. These techniques show the inter-relationship of the
various jobs to be done to complete the project and clearly identify the critical parts of the
project. They provide planning and control information on the time, cost and resources
required. These techniques are of particular value where projects are large and contain many
related and interdependent activities, where many types of facilities, high capital
investments and a large number of staff are involved, where projects have to be completed
within target time and costs.
Path analysis. To develop the technique it is necessary to know the activities involved,
their logical relationship, an estimate of the time the activity is expected to take. It may be
necessary to have a range of estimates of times, costs, resources and probabilities. Once the
logic of the activities has been agreed and an outline network drawn, it can be completed by
inserting the activity duration times. These times may be estimated by using probabilities
(developing the expected time) or using basic time estimates. A basic feature of the analysis
is the calculation of the project duration which is the duration of the critical path. This is the
path through the network which gives the shortest time in which the whole project can be
completed. It is the chain of activities with the longest duration times. In a network there
may be more than one critical path. The activities along the critical path are vital activities
which must be completed on time otherwise the project will be delayed. If it is required to
reduce the overall project duration then the time of one or more of the activities on the
critical path must be reduced.
A further important feature of the analysis deals with the process of cost scheduling. This
implies that by using more resources the duration could be reduced but at the expense of
higher costs. The costs associated with delivery of the project, as designed, is the ʻnormalʼ
cost. It is possible to reduce the activities by means of extra wages, overtime, additional
facility costs which are higher than normal and are called crash costs. This is an element of
least cost scheduling which finds the least cost method of reducing the overall project
duration, time period by time period. (5 Marks)
(iv) Measures of
dispersion.
Measures of dispersion represent a measure of the extent of
spread of data around the mean or average of the data. The simplest measure of
dispersion is to take the absolute difference between the highest and lowest value of the
raw data – ʻthe rangeʼ. The interquartile range is the absolute difference between the upper
and lower quartiles of a distribution. These measures of dispersion involve comparing two
different points on the frequency distribution such as the maximum and minimum points
(the range) or the upper and lower quartiles. However, other measures of dispersion which
compare all the points on the frequency distributions are more important. The ʻmean
Page 45
deviationʼ, that is, the average of the absolute deviations from the arithmetic mean, may be
used. However, if is the basis for the most common measures of dispersion – the variance
and the standard deviation. The variance is derived by squaring all the deviations from the
arithmetic mean. Rather than use squared units, it is more realistic to express solutions to
problems in terms of single units. The standard deviation takes the square root of the
variance. The standard deviation is then the square root of the average of the squared
deviation from the mean and is the most commonly used measure to explain the dispersion
of data. These various methods can be used for both grouped and ungrouped data. (5 Marks)
