Solutions Manual
This document contains solutions to all the practice questions that appear at the end of each
chapter.
Chapter 2 Solutions
1.
Year
Cost
(£000s)
Index
2006
50.1
100.0
2007
65.3
130.3
2008
68.6
136.9
2009
72
143.7
2010
76.6
152.9
2011
78.3
156.3
2012
88.7
177.0
2013
90.5
180.6
2014
99.3
198.2
2015
112.9
225.3
2.
Year
Cost (£)
Index (Year 1 = 100)
1
650
100
2
580
89.2
3
410
63.1
3.
Calculating the index for 2015 using 2013 as the base year
Resource
Index (2013 = 100)
Ink
138
Card
160
Labour
107
So card has increased the most (up 60%)
4.
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Calculating both indices for 2014 based on 2013
2013
2014
p0
q0
pn
qn
p0q0
pnq0
p0qn
pnqn
Ink
0.55
4000
0.58
5000
2200
2320
2750
2900
Card
0.05
115000
0.05
124000
5750
5750
6200
6200
Labour
6.5
25
6.9
34
162.5
172.5
221
234.6
8112.5
8242.5
9171
9334.6
8242
Laspeyres’ index = 8112.5 × 100 = 101.6
Paasch’s index =
9334.6
9171
= 101.8
For 2015 based on 2013
2013
2015
p0
q0
pn
qn
p0q0
pnq0
p0qn
pnqn
0.55
4000
0.76
3000
2200
3040
1650
2280
0.05
115000
0.08
110000
5750
9200
5500
8800
6.5
25
7
30
162.5
175
195
210
8112.5
12415
7345
11290
12415
Laspeyres’ index = 8112.5 × 100 = 153.0
Paasch’s index =
11290
7345
= 153.7
From 2013 to 2014 there has been very little change as prices have not changed much.
However during 2014 to 2015 prices for ink and card have shown a big increase.
5.
2005
2015
p0
q0
pn
qn
p0q0
pnq0
p0qn
pnqn
Bread
28
6
78
6
168
468
168
468
Milk
20
15
150
12
300
2250
240
1800
Tea
96
1
75
2
96
75
192
150
564
2793
600
2418
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Laspeyres'
495.21
Paasche's
403.00
6.
1998
Company
2011
p0
q0
pn
qn
p0q0
pnq0
p0qn
pnqn
160
200
520
500
32000
104000
8000
260000
A
Company
0
350
650
265
250
227500
172250
B
Company
66250
0
105
600
140
400
63000
84000
C
Company
8750
4200
56000
0
53
100
159
200
5300
15900
D
1060
31800
0
327800
376150
2201
414050
00
Laspeyres'
114.75
Paasche's
7. Simple index for year 2 using year 1 as the base year
Supervisor: 14/12 x 100 = 116.7
Skilled: 10/9 x 100 = 111.1
Unskilled: 7/6 x 100 = 116.7
For year 3
Supervisor: 15/12 x 100 = 125.0
Skilled: 11/9 x 100 = 122.2
Unskilled: 8/6 x 100 = 133.3
So unskilled increased wages the most.
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188.12
Supervisor
Skilled
Unskilled
Year 2 based on year 1
p0
q0
pn
12
10
9
120
6
200
Laspeyres index
Supervisor
Skilled
Unskilled
qn
14
10
7
p0q0
12
160
200
pnq0
120
1080
1200
2400
140
1200
1400
2740
113.8
114.2 Paasche index
Year 3 based on year 1
p0
q0
pn
12
10
9
120
6
200
Laspeyres index
qn
15
11
8
p0q0
15
160
180
p0qn
pnq0
120
1080
1200
2400
144
1440
1200
2784
p0qn
150
1320
1600
3070
126.9
127.9 Paasche index
pnqn
168
1600
1400
3168
pnqn
180
1440
1080
2700
225
1760
1440
3425
Both indices very similar and show a steady increase in total wages over the 3 years.
8.
186.7
£105,000 is worth 105,000 × 214.8 = £91,264
So price has dropped by £140,000 – 91,264 = £48,736
9.
Year
Turnover
RPI
Real T/O
2005
15.3
192.0
15.30
2006
10.3
198.1
9.98
2007
12.1
206.8
11.23
2008
15.2
214.8
13.6
2009
24.4
213.7
21.9
2010
34.7
223.6
29.8
10.
2003
p0
2011
q0
pn
qn
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p0q 0
pn q 0
p0 q n
pn q n
7.50
120
9.00
158
900
10.00
41
12.50
52
410
8.00
25
10.00
30
18.00
21
22.40
Laspeyres’ 122.51
1080
1185
1422
512.5
520
650
200
250
240
300
25
378
470.4
450
560
sum
1888
2312.9
2395
2932
Paasche’s 122.42
11.
2006
2011
p0
q0
pn
qn
p0q 0
pn q 0
p0 q n
pn q n
3.63
3
4.49
2
10.89
13.47
7.26
8.98
2.11
4
3.26
6
8.44
13.04
12.66
19.56
10.03
1
12.05
1
10.03
12.05
10.03
12.05
4.01
7
5.21
5
28.07
36.47
20.05
26.05
57.43
75.03
50.00
66.64
Laspeyres’ 130.65
Paasche’s 133.28
12(a)
Price
Index
Aug
155
100.00
Sept
143
92.26
Oct
120
77.42
Nov
139
89.68
Dec
165
106.45
Jan
162
104.52
Month
(b)(i)
August to January change = 104:52 - 100 = 4:52%
(ii)
September to December change =
165−143
143
= 15.38%
13.
Year
Average
wage
Average
RPI
Wages deflated
to 2008 values
2008
255.1
214.8
255.1
2009
271.3
213.7
272.7
2010
290.7
223.6
279.3
Real wages grew more between 2008 and 2009, due to the
decline in the RPI.
14.
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Year
Average RPI
House price
Real house price
2008
214.8
£162 000
£162 000
2010
223.6
£157 500
£153 030
Real drop in price = £11 179 at 2010 prices, i.e. 6.9%
15
(a)
(b) (i) (ii)
(iii)
% change
Year
Average
RPI
2008
214.8
2009
213.7
99.5
-0.5
2010
223.6
104.1
4.1
RPI
(2008 = 100)
since
2008
since
2009
100
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4.9
Chapter 3 Solutions
1.
Year
Population
Sample
Total
Female
Male
Total
Female
Male
1
320
192
128
20
12
8
2
250
150
100
16
10
6
3
230
138
92
14
8
6
Total
800
480
320
50
30
20
2.
Need to ask council tax payers. A survey could be sent out with council tax demands but that
may produce a bias sample as only those people with strong views will respond. Maybe better
to send questionnaire to random sample and follow up with letters. Could also allow residents
to use online questionnaire.
3.
The survey will probably be in the form of a simple questionnaire, possibly web based. Not
necessary to have a random sample for this type of survey so anyone who applied for tickets
could be asked to complete the questionnaire.
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4
Various issues including asking age (use ranges), Vague questions (question 5), no
option for ‘other’ in question 6. Unlikely that respondents would know what socio-economic
group they are in.
5
1, 2 and possibly 5 would have a sampling frame.
6
(a)
The sample would be Steve, Kim, Chris, Jane, Stuart, Jill. Average age is 40.3
and 3 read the Mirror, one the Sun, one The Times and one the Express.
(b)
As we have 6 males and 3 females we need a sample of 4 males and 2 females.
If we separate out the two sexes and use the same random numbers for each group then we
will have Steve, Chris, Stuart, Kim, Julie and Jill. Average age is 39.2 and two read the Sun,
two the Mirror and one each of The Times and Telegraph.
(c)
Julie
7. Answers depend on sample chosen.
8.
Target population would be all council tax payers. A good sampling frame would be a
database of council tax payers in a local council region. Alternatively the electoral register
could be used although this will contain people not paying council tax (e.g. children over 18
living with parents or at university).
9.
•
Not all supporters belong to the supporters club so their views will be ignored.
•
A simple random sample may not contain the right proportion of male/female
supporters or the different ages.
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•
A better target population may be all people who have ever watched a game (if this
data is kept) or even all people in the local town.
•
A stratified sample would be best as it would take into account male/female and ages
issue.
•
A systematic sample may also be possible as every nth person leaving the ground
could be sampled.
10.
Probably use multi-stage sampling methods. Randomly select a number of council regions
then randomly select a number of schools within each region. A random sample of school
leavers could then be selected from each chosen school.
11.
The simplest and cheapest survey would be a quota sample of shoppers in the town centre.
However, this would ignore people who don’t shop in the town. An alternative would be an
advert in a local newspaper, but this would bias towards those who read the paper and have a
strong opinion. The most expensive would use the electoral register to contact a stratified
sample of voters in the town.
12
(a)
Both stratified and quota sampling aim to produce a sample that represents the
target population in terms of the proportion contained within relevant sub-groups. However,
stratified sampling requires the use of a sampling frame and is therefore a probabilistic
sampling method.
Although it would be possible to obtain a list of all students it would probably be sufficient to
employ a quota sampling method as this will be much cheaper to administer. The proportions
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of male/female and age ranges should be easy to obtain, which would make the quota sample
reasonably accurate.
(b)
No, since all students would be on campus.
(c)
Possible answers include:

Misleading or ambiguous questions. Could be overcome by a pilot study.

Asking the wrong question.

Asking personal questions.

Asking leading questions.

Making the questionnaire too long.

Difficulty in obtaining information about the student population.

Method used to conduct survey. Face-to-face will probably be most effective but this
may be too expensive.

Poor response rate. This can be improved by face-to-face interviews.

Bias sample if only certain members of the population are included in the sample (e.g.
if only interview respondents in a particular part of the university).

Conducting survey during vacations or exam periods!

Randomly select 40 employees from the database of all 200 employees

Choose every 5th employee from the list

For the shop-floor department there are 80 employees which is 40% of the total so we
13.
(a)
need a random sample of 40% of 40 = 16 employees from shop-floor. The other
departments are calculated in a similar way
Dept
Number
Proportion of total
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Sample size
Shop floor
80
.4
16
Service engineers
15
.075
3
Quality control
20
.1
4
Marketing
25
.125
5
Accounts
15
.075
3
Personnel
10
.05
2
Administration
25
.125
5
Catering
10
.05
2
Total
200
1
40
Randomly select the required number from each department

For quote sampling a survey could be conducted in the works canteen using the above
numbers as the quota for each department; i.e. once the required number of employees
has been chosen from one department no one else from that department would be
selected
14
(a)

Target population – all first year students and all staff.

Sampling frame – database of student and staff records.

Stratified sampling – sample contains same proportion of relevant categories as
population.

Multi-stage sampling – hierarchical structure. If the students were based on different
campuses it might be necessary to pick one or two of these campuses randomly to
reduce travelling time.
(b)
Possible questions are:

‘Are you student or staff?’ (It will probably be necessary to be able to compare the
responses between staff and students.)
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
How strongly are you in favour of a 3rd semester?’ using a Likert scale. (This will be
better than asking for a ‘yes’ or ‘no’ answer.)

‘How much would you be prepared to pay/receive to do a 3rd semester?’ (Will need
to ascertain how financially viable the idea is. Students will need to pay in order for
lecturing staff to be paid.)

‘At what faculty are you based?’ (There may be differences between faculties.)

For students only – ‘What is your age (within a range)?’ (Older students might be
more interested in completing their degrees quicker.)
(c) A postal or email questionnaire will probably be the best method here. Alternatively,
could be face-to-face if quota sampling used.
15.
There are many reasons such as:

The bins are being refilled when empty so difficult to judge how empty the bin is.

Were all the bins filled to the same level?

Do people prefer different colours?

Positioning of the bins might make a difference. Do you take the nearest regardless of
your views?

Do people associate colour with flavours?

And, of course, this was not a random sample!
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Chapter 4 Solutions
1.(a) Continuous as there could be a fraction of an hour. However if the answer is obtained by
counting the number of hours of sunshine then it could be classed as discrete.
(b) Continuous
(c) Continuous
(d) Ordinal
2. (a) Time spent online
(b) Download speed
(c) Number of websites visited
(d) Rating given to a site
(e) Choice of social networking site
3.
Class interval
Frequency
0 to 2
5
3 to 5
4
6 to 8
2
9 to 11
2
12 to 14
3
15 to 19
0
20 to 30
2
A bar chart could be drawn as below
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This bar chart is not very good as it doesn’t take into account the different class intervals.
(See the solution to question 8 for a better chart).
4.
(a)
Type of coffee
Frequency
Relative frequency (%)
Instant
Filter
Ground
8
8
4
40
40
20
(b), (c) See Excel file (Chapter 4 Q4)
5.
See Excel file (Chapter 4 Q5)
Component bar chart shows that sales have increased since 2009 but this has been the result of a good
performance by the Menswear department. Furniture sales were the largest proportion of total sales in
2006 but this proportion has declined since then.
6.
(a), (b), (c), (d) See Excel file (Chapter 4 Q6)
(e)
Total sales declined during 2007 to 2009 but some recovery since then. South East gives
largest proportion of sales with Wales the smallest. Multiple bar chart shows that the decline in South
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East sales occurred later but the recovery has not started despite improvement in other regions. Wales
has shown very little change during the 5 years.
7. (a)
Weight (gm)
Frequency
20-<22
2
22-<24
8
24-<26
16
26-<28
17
28-<30
21
30-<32
16
(b)
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(c)
2 20 45
2 21
5 22 148
10 23 25588
16 24 055688
26 25 2223455688
33 26 1225588
(10) 270000124468
37 28 0000122568
27 29 00022234466
16 30 000024446688
4 31 0444
20|4 20.4 gms
(d)
Could use this chart to find the median and quartiles
(e)
Data varies from 20 to 32 gm. The mean is 27.2 gm and the median is a little higher at around
27.5 gm. The distribution is left skewed
8.
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As the class widths vary it would be easier to plot the frequency density. This is achieved by
dividing each frequency by the class interval (see table below). As this is discrete data the
ends of each group will be 2.5, 5.5, 8.5, 11.5, 14.5 and 19.5. A sketch of the histogram is
shown below (note it is not possible to use SPSS to draw this histogram).
Class interval
Boundaries
0 to 2
Frequency
Frequency density
5
2.5
3 to 5
2.5
4
2
6 to 8
5.5
2
1
9 to 11
8.5
2
1
12 to 14
11.5
3
1.5
15 to 19
14.5
0
0
20 to 30
19.5
2
0.2
9.
The histogram shows that the vast majority of people work around 40 hours but the range is from
almost zero to nearly 100 hours. The histogram is fairly symmetrical.
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10.
This clearly shows that over 80% of households have at least one car.
11.
This bar chart clearly shows that the majority of family type is still the traditional married
couple with children.
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12.
(a)
(b)
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(c)
(d)
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
The ‘No qualifications’ is the highest category.

Male and female are approximately the same on all categories except the ‘other
qualifications where males there are more males.

Levels 2 and levels 4/5 have about the same total percentage (just under 20%).
13.
Assuming that this is discrete data then the ends of each group will be 2, 5.5, 9.5, 13.5 and
21.5. As the class intervals vary from 2 to 7 it is easiest to divide each frequency by the class
interval to get the frequency density as in the table below.
Days off work
Boundaries
Number of
Frequency density
employees
Less than 2
2
45
22.5
2 to 5
5.5
89
29.7
6 to 9
9.5
40
13.3
10 to 13
13.5
25
8.3
14 to 21
21.5
5
0.7
22 to 29
29.5
2
0.3
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The distribution is right skewed.
14.
(a) (b) See Excel file (Chapter 4 Q14) Comments should include:
General decrease in sales from 1990 to 1992.
The average sales growth has fallen fastest during 1990 to 1991.
Inner London shows least reduction.
South Eastern shows largest reduction.
South Western and Outer London are only regions to show some recovery during 1991 to 1992.
15.
(a)
See Excel file (Chapter 4 Q15) for histogram. The distribution is right skewed.
(b)
See Excel file (Chapter 4 Q15) for ogive.
(i)
About 20% of sales are in excess of £70
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(ii)
About 62% (84 - 22) of sales are between £52 and £72
(iii) The value of purchases exceeded by 10% of sales is about £75.50
(These answers can be seen on the ogive in the Excel file.)
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Chapter 5 Solutions
1.
(a) Mean is £189.53 and median is £174.34. The mean is influenced by the high value of
£274.50 so the median is probably the better average to use.
(b) Standard deviation is £48.83 (based on n−1)
(c) The mean is increased by £20 to £209.53 but the standard deviation stays the same at
£48.83
(d) The mean increases by 5% to £199.01 and the standard deviation also by 5% to £51.27
2.
(a) Mean = 24.5 0C, Median = 24.5 0C Mode = 19 0C
A holiday maker normally takes a holiday during the summer so an average annual
temperature is not that helpful.
(b) Standard deviation is 6.82 0C
(c) The mean should be 26.5 0C; the standard deviation will not change.
(d) The mean temperature in 0F is found by using the formula on the mean and is 76.10. The
standard deviation is found by multiplying by 9 and dividing by 5 only and is 12.27 0F
3.
(a) Median because it tells the students where in the class they are placed. Mean might also
be useful in looking at the distribution of scores.
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(b) Both median and mean. Median is useful as it tells you that 50% of people earn less than
this amount. The mean is generally the average quoted.
(c) Mean as it can be used in statistical analysis (quality control).
(d) Mode (the most frequent sold) is the only sensible average.
(e) Mean as it can be used in statistical analysis (quality control).
4.
Number
Frequency
of rejects
f
x
0
12
1
45
2
36
3
30
4
20
5
5
6
0
Sum
148
fx
0
45
72
90
80
25
0
312
fx2
x2
0
1
4
9
16
25
36
0
45
144
270
320
125
0
904
Mean = 2.1, standard deviation = 1.3
5.
(a) Mean and median are 1.0794 and 1.0730 respectively.
(b) Standard deviation is 0.0339
(c) This data is a time series and as the exchange rate appears to be increasing during the 10
days the mean and median represents the exchange rate half way through the period.
6.
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The mean and median are around 40 hours and the modal class 40 to 50 hours. If the
distribution is assumed normal then the range is about 80 which represents 6 standard
deviations so one standard deviation will be 13.3.
7.
Average mark = 74×.4 + 56×.6 = 63.2%
8.
Total for 12 light bulbs is 12×180.6 = 2167.2
Total for 13 bulbs = 2167.2 + 200 = 2367.2
Mean for 13 is 2367.2/13 = 182.1
9.
(a) 5.3 days
(b) 4 days
(c) 2 to 5 days
(d) 5.5 days
(e) 4.3 days
(f) 82%
(g)
2
0
4
7.5
10
20
The distribution is right skewed.
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30
10. (a)
A
B
mean
24.8
26.2
median
25
26.3
mode
24.26
26.28
(b) IQR = 2.5 (A); 2.7 (B)
SD = 1.96 (A); 2.04 (B)
(c) A: 7.9% B: 7.8%
(d)
Machine B
24.9
26.3 27.6
Machine A
23.5
25 26
(e)

Machine B produces item with slightly longer lengths than Machine A

The spread of lengths is about the same on both machines

The distribution of lengths is approximately symmetrical although for Machine A
the distribution has a slight left skew since the median is to the right of centre of the
box and is greater than the mean.
11.
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Mean = £57.67 Median about £58.80
12.
The mean wage is £134.18
Proposal 1: Increase in pay = 10% of current mean wage = £13.42
This will cost the company 98 x £13:42 = £1315:16
The median wage is about £130
Proposal 2: Increase in pay = 12.5% of current median wage
= 0:125 × £130 = £16:25
This will cost the company 98 × £16:25 = £1592:50 = approximately £1600.
13. The purchasing manager has simply calculated the mean of the four percentage figures:
10+4+0+46
4
= 15%
However, the mean increase in costs must be calculated as a weighted mean, i.e., weighting
each raw material percentage price increase by the mean expenditure on that raw material.
Thus, the mean increase in costs is:
51000
= 10.022%
5000
The purchasing manager unwittingly gave far too much emphasis to the huge percentage
increase in Additives, which in fact accounts for a very small proportion of total
expenditure.
14.
Production target is 300 widgets per week
Total production target in all 13 weeks = 13(300) widgets.
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Now let x = average to be reached in final 3 weeks in order to achieve target. Production in
first 10 weeks = 10(284.7) widgets.
Production in final 3 weeks = 3x widgets.
Total production in all 13 weeks = 10(284.7) + 3x widgets.
We want:
Total production = Total production target.
i.e.,10(284.7) + 3x =13(300)
Solving for x, we get x = 351
Thus an average of 351 widgets per week must be produced in the final 3 weeks in order to
achieve the original target of 300 widgets per week over all 13 weeks.
15.
Total time to cover all 200 miles =
100
80
+
100
60
= 2.9167 hours
200
Average speed over the 200 miles = 2.9167 = 68.57 mph
16.
(a) (i) Total sales = £6745
(ii) Mean = £26.66
(iii) Standard deviation = £19.60
(iv) coefficient of variation = 19.6/26.6 ×100 = 73.7%
(b) Mid points used
(c) (i)
Less variation in second shop
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(ii) Could be either
(d)
(ii)
(i) Median = £20
Q1 = £12, Q2 = £34, IQR = 34 - 12 = £22
(iii) Over £50 = 11%
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Chapter 6 Solutions
1.
0.25
2.
0.125
3.
0.2778
4.
There are 5 ways of getting 8 and 36 combinations altogether so probability is 5/36 =
0:1389
5.
P(first ace) = 4/52 = 0:07692;
P(second ace) = 3/51 = 0:05882;
P(third ace) = 2/50 = 0:04;
P(all 3 aces) = :07692 x 0:05882 x 0:04 ¼ 0:00181
26
25
24
6. P(3 reds) = 52 × 51 × 50 = 0.1176
7.
(a) (i) 5/10 = 0.5
(ii) 3/10
(b) (i) ¼
(ii) 3/20
(c) (i) 2/9
(ii) 1/6
(iii) 7/10
8.
(a) 4/52
(b) 26/52
(c) 4/52 + 26/52 – 28/52 = 28/52
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9.
4/50 × 3/49 = .00490
10.
(a) .05×.05×.05 = .000125
(b) If P(sale) = S= .05 and P(no sale) = S = .95
Then either SSS or SSS or SSS will give exactly 1 sale and the probabilities for each of these
is the same so.
P(1 sale) = 3×.05×.95×.95
= .1354
(c) P(at least 1 sale) = 1 – P(no sales)
= 1 - .953
= .1426
11.
Let R = respond and R don’t respond (= .8)
(a) All 8 respond = (.2)8 = 2.56×10-6
(b) No one responds = (.8)8 = 0.1678
(c) There are many combinations; for example RRRRRRRR (in fact there are 28 and in
Chapter 8 you will see that this is in fact a binomial situation).
P(exactly 2 respond) = 28 × (.2)2×(.8).6 = .2936
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12.
(a) subjective
(b) empirical
(c) ‘a priori’
(d) subjective
13.
The 3 machines operate independently, so for example the probability of Machine A
breaking down in independent of the probability of Machine B breaking down. Thus we use
the multiplication rule for independent events:
P(A and B) = P(A) × P(B)
(a)
P(all three machines will be out of action)
= P(Machine A will be out of action)
× P(Machine B will be out of action)
× P(Machine C will be out of action)
= 0:1 × 0:05 ×0:20
= 0.001 or 0.1% or 1 in 100
(b)
P(none of the machines will be out of action)
= P(Machine A will not be out of action)
× P(Machine B will not be out of action)
×P(Machine C will not be out of action)
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= (1 – 0.1) × (1 – 0.05) × (1 – 0.20)
= 0.90 × 0.95 × 0.80
= 0.684 or 68.4%
14.
We assume that bad weather and geological problems are independent events.
Thus we again use the multiplication rule for independent events:
P(A and B) = P(A) × P(B)
P(project will be delayed by both bad weather and geological problems)
= P(project will be delayed by bad weather)
× P(project will be delayed by geological problems)
= 0.30 × 0.20
= 0.06 or 6%
15.
P(both plugs will be substandard)
= P(first plug will be substandard)
× P(second plug will be substandard j first plug was substandard)
6
20
5
× 19= .0789 or 7.89%
16.
P(shares will rise in value)
= P(shares will rise in value │ FT30 index rises) × P(FT30 index rises)
+ P(shares will rise in value │ FT30 does not rise)
× P(FT30 index does not rise)
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= 0.8 × 0.6 + 0.3 × (1 – 0.6)
= 0.48 + 0.12 = 0.60 or 60%
17.
(a) .9 ×.6×.7 = .216
(b) P(at least 1) = 1 – P(none)
= 1 – (.1×.4×.3)
= .988
c) .012
18.
(a)
(i)
P(bulb defective) = 0.3
(ii)
P(bulb defective │ bulb is 60 w) = 20/60 = 0.333
(b) The probability that a box will be accepted is the sum of the probabilities on the Accept
branch ends:
P(box is accepted) = 0.4879 + 0.1494 + 0.1494
= 0.7867 or 78.67%
19.
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P(research says high) = .12 + .32 = .44
P(H│research says high) = .12/.44 = .273
20.
.9
Has Condition
.2
8
.02
Test +ve
.04
.05
.76
P(test +ve) = .18 + .04 = .22
P(test –ve) = .02 + .76 = .78
P(not have condition|Test +ve) = .04/.22 = .1818
P(have condition|Test –ve) = .18/.78 = .2308
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Chapter 7 Solutions
1.
5
2.
P(3H) = 10 ×.53 × .52 = 0.3125
C3 = 10
3.
(a) Either get a job or doesn’t so 2 states and probabilities are constant and independent.
(b) P(r>2) = 1 –[P(0) + P(1) + P(2)]
P(0) = (.52)9 = .00278
P(1) = 9×.48×(.52)8 = .02309
P(2) = 36×(.48)2(.52)7 = .08527
P(r>2) = 1-.1111
= .8889
4(a)
P(male) =105/205
= .5122
(b)
P(3 males) = 4C3 ×(.5122)3× .4878
= .2622
(c)
P(2 males) =
4
C2 x (.5122)2× (.4878)2
= .3746
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5.
Want to find the probability of two or more defective items given a defective rate of 3%
P(at least 2 defectives) = 1 – (P(0)þ + P(1))
P(0) = 20C0 × (.03)0 × (.97)20
=.544
P(1) = 20C1 × (.03)1 × (.97)19
= .336
P(≥ 2) = 1 - .544 - .336
=.12
So there is a 12% change of getting two or more defectives from a batch of 20. This is not
particularly surprising.
6.
Probability of egg undamaged = .8
(a)
P(0) = 4C0 × (.8)0 × (.2)4
= .0016
(b)
P(≥ 3) = 1 – [P(0)+ P(1) + P(2)]
P(1) = .0256
P(2) = .1536
P(≥3) = 1 – (.0016 + .0256 + .1536)
= .8192
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7.
(a)
P(< 4 calls) = P(0) + P(1) + P(2)þ + P(3)
P(0) = e-6:7 = .00123
Using the recursive properties of the formula:
P(1) = :00123 x 6:7 = .00825
P(2) = .00825 x 6.7/2 =.02763
P(3) = .02763 x 6.7/3 = .06170
P(< 4) = .00123 + .00825 + .02763 + :06170 .0988
(b)
P(> 7) = 1 – P(≤ 7)
P(4) = .1033; P(5) = .1385; P(6) = .1546; P(7) = 1480
P(> 7) = 1 – (.0988 + .1033 + .1385 + .1546 + .1480) = .3568
(c)
Mean number of calls in 1 minute = 6.7/5 = 1.34
8.
This is a Poisson problem.
(a)
Mean number of flaws 1500/1000 = 1.5 per mm
P(≥2) = 1 –[ P(0) + P(1)]
P(0) = e-1.5 = .2231
P(1) = .2231 x 1:5 = .3347
P(≥2) = 1 –[.2231 + .3347) = .4422
(b)
Need to find the probability of no flaws in 5 mm.
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Mean number of flaws in 5 mm is 5 x 1:5 = 7:5
P(0) = e-7:5 = :0006
9.
(a) P(0) = e-3.2 = .04076
(b) P(r>4) = 1-[P(0) +P(1)+P(2)+P(3)]
P(1) = .13044
P(2) = .20870
P(3) = .22262
P(r>4) = 1 - .6025 = .3975
(c) In 8 minutes m = 6.4
Using Tables P(r<4) = .2351
10.
0.1056
11.
.0735 − .0071 = .0664
12.
1 – (.2514 + .1469)
= .6017
13.
(a) 𝑍 =
6−5
1.5
= .667
P(Z>.667) = .2514 (from tables)
Area = .5-.2514 = .2486
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(b) 𝑍 =
4−5
1.5
= −.667
Same area as in part (a), that is .2486
(c) area = 2×.2486 = .4972
(d) For 6.5, 𝑍 =
6.5−5
1.5
= 1.0
P(Z>1) = .1587
For 7,5 𝑍 =
7.5−5
1.5
= 1.67
P(Z>1.67) = .0475
Area = .1587 - .0475
= .1112
14.
(a) 𝑍 =
25000−20000
7200
= .694
P(Z>.694) = .2451
So probability that demand greater than 25000 litres is .2451 or about 24.5%
(b) 𝑍 =
17000−20000
7200
= −.417
P(Z<-.417) = .3372
So P(demand>17000) = 1-.3372
= .6628 or about 66.3%
(c) P(20000<demand<25000) = .5 - .2451
= .2549 or about 25.5%
(d) For demand = 30000
𝑍=
30000−20000
7200
= 1.39
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P(Z>1.39) = .0823
For demand = 35000
𝑍=
35000 − 20000
= 2.08
7200
P(Z>2.08) = .0188
P(30000<demand<35000) = .0823 - .0188
= .0635 or about 6.4%
15.
For £80000
𝑍=
80000 − 45121
= 1.438
24246
P(Z>1.438) = .0749 which is the probability that someone will earn more than £80000
For £100,000
𝑍=
10000 − 45121
= 2.26
24246
P(Z>2.26) = .0119
So number of staff earning more than £100,000 = 500000×.0119
= 5950
16.
𝑍=
60 − 48
= 1.2
10
P(Z>1.2) = .1151
P(Call answered in less than 60 seconds) = 1 - .1151) = .8849
So there is approximately a 88.5% chance of the call being answered in less than a minute
(note in theory this probability includes time less than zero but error is negligible).
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17.
Need to find the probability less than 4.5 mm and greater than 6 mm.
𝑍=
4.5 − 5.5
= −2.5
.4
P(Z<-2.5) = .0062
𝑍=
6 − 5.5
= 1.25
.4
P(Z>1.25) = .1056
P(between 4.5 and 6 mm) = 1-(.0062 + .1056)
= .8882
So, about 88.8% of bolts will be accepted.
18.
(a)
𝑍=
(i)
50−54
2
Undersize
= -1
P(Z<-1) = P(Z>1) =.1587
So 15.87% are undersize
(ii)
Oversize
Z = .5
P(Z > :5) = .3085
So 30.85% will be oversize
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(b)
Out of 1000 parts, 53.28% will be usable and 308.5 on average will need to be
shortened. The 158.7 parts that were too short will have to be remade. Total cost will
therefore be:
1000 x 10 + 158:7 x 10 + 308:5 x 8 = £14 055
(Note: this is an underestimate as the 158.7 parts that need to be remade will also contain
some rejects.)
19.
For 5% in the upper tail, Z = 1.645
Let x be the minimum customer spend to get a free gift
1.645 =
𝑥 − 135
55
x = £225.48
If the upper tail is now 8% then Z = 1.405 and
1.405 =
225.48−𝜇
55
µ = 225.48 – 55 × 1.405 = £148.21
20.
(a)
300  320
 1.92
10.4
(i) Z 
P(Z < 1.92) = 0.0274
(ii) Z 
325  320
 0.481
10.4
P(Z > 0.481) = 0.3156
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(iii) P(318 <x< 325) = 1 – ((P(x<318) + P(x>325))
P(x<318) = P(Z<
318  320
 0.192 )
10.4
= 0.4247
P(318 <x< 325) = 1 – (.4247 + .3156)
= 0.2597
(b)
500 × 0.0274 = 13.7 (around 14 packets would be underweight)
(c) The Z value corresponding to an area of .03 is –1.88
 1.88 

300  310

10
 5.32
1.88
21.
Let x be the demand for the product.
(a)
We want to find the probability that x is greater than 2500
𝑧=
2500 − 2000
=1
500
P(Z > 1) = 0.1587 from the normal table
(b)
x > 2800
𝑧=
2800 − 2000
= 1.6
500
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P(Z > 1:6) = 0.0548
(c)
x < 1600
𝑧=
1600 − 2000
= −0.8
500
P(Z < -0.8) = P(Z > 0.8) by symmetry
= 0:2119
22.
(a) Z 
130  100
2
15
P(Z>2) = .0228 or 2.28% will spend over £130
Z
(b)
120  100
 1.33
15
(Z>1.33) = .0918 or 9.18% will spend over £120
(c) Z 
70  100
 2
15
P(Z<-2) = .0228 or 2.28% will spend less than £70
(d) (100<x<130) = 50 – 2.28
= 47.72%
(e) For £115
Z
115  100
1
15
P(Z>1) = .1587 or 15.87%
(115<x<130) = 15.87 – 2.28
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= 13.59
(f) For 10% in the tail Z = 1.28
1.28 
x  100
15
x = 15×1.28 + 100
= £119.20
(g) Z = 1.88
1.88 
x  100
15
x = £128.20
23.
Let x be the IQ scores.
𝑧=
115 − 100
=1
15
𝑧=
130 − 100
=2
15
P(Z > 1) = 0.1587
(b)
x > 130
P(Z > 2) = 0.0228
(c)
115 < x < 130
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P(115 < x < 130) = P(x > 115)- P(x > 130)
= 0.1587 – 0.0228
= 0.1359
24.
Let x = number of gallons.
Need to find the value of x that gives an area of 0.06 in the right hand tail. From tables, Z
=1.555
1.555 =
𝑥−2000
500
=1
x – 2000 = 500 × 1.555
x = 2338 gallons
25.
(a) Find area less than 15.75 and greater than 16.75 g and add to give required proportion
Z
15.75  16
 0.5
0.5
P(Z<-.5) = .3085
Z
16.75  16
 1.5
0.5
P(Z>1.5) = .0668
So required area is .3085+.0668
= 0.3753
So 37.53% of output will be scrapped
(b)
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Number scrapped = 10000×.3753 = 3753
Cost = 3753×7 = £26,271
(c) Need to recalculate the Z values
Z
15.75  16
 1.25
0.2
P(Z<-1.25) = .1056
Z
16.75  16
 3.75
0.2
P(Z>3.75) ≈ 0
So number scrapped = .1056×10000 = 1056
Cost of scrapping = 1056×7 = £7392
Total cost including new machine = £7392 + £10,000 = £17,392
As the combined cost is less than £26,271 it would be worth hiring the machine.
26.
Need to find the chance of profit being less than £0
For product A
Z
0  15
 1.67
9
P(Z<-1.67) = .0475
Product B
Z
0  10
 2
5
P(Z<-2) = .0228
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So should choose product B
27.
For the Poisson approximation the value of p should be small and the number of trials large.
Both are true. We need the mean which is 5. If we use Excel or tables we get the probability.
( x> 8) = 1 - .8666
=0.1334
Which agrees well with the binomial answer of 0.13
For the normal approximation n×p and n× (1-p) should be greater than 5. n×p is the mean
and is exactly 5
The standard deviation is
𝜎 = √100 × .05 × .95
= 2.179
Now need the probability that x is greater than 7.5 (because of the continuity correction)
𝑍=
7.5 − 5
2.179
= 1.147
Using tables this is a probability of .1251
So again a good approximation even though the conditions were only barely met
28.
m = 2.65
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P(r=8) =
𝑒 −2.65 ×2.658
8!
= 0.00426
So, a very low probability.
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Chapter 8 Solutions
1.
(a) 16,000
(e) SEP = √
(b) 50
(c) 20/50 = 40%
(d) 40%
40×60
50
= 6.9
95% confidence interval = 40 ±1.96×6.9
= 26.5% to 53.5%
2.
𝑥̅ =9:62 g
𝜎̂ = 0.7731
Sem =
0.7731
√6
= 0:3156
The value on t degrees of freedom = 2.571
95% confidence intervals = 9.62 ± 2:571 x 0.3156
= 9:62 ± .81
= 8.81 g to 10.43 g
3.
Sem = 0.26
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Confidence interval = 169.5±1.96×0.26
= 168:99 cm to 170.01 cm
4.
P = 75%
75×25
SEP = √
60
= 5:5902
95% confidence interval = 75 ± 1.96 x 5.5902
= 75 ± 10.96
= 64% to 86%
5.
Mean = 8.72 g standard deviation = 1.2523 g
Sem = .3960,
(a) t = 2.262
95% confidence interval: 8.72 ±2.262×0.3960
= 7.82 g to 9.62 g
(b) t = 3.250
99% confidence interval: 8.72 ±3.250×0.3960
= 7.43 g to 10.01 g
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6.
(a)
SEM =
3675
= 245:0
√225
99% confidence interval = 16 450 ± 2.575 x 245
= 16 450 ± 631
= £15 819 to £17 081
(b)
2.575 ×
3675
√𝑛
= 200
√𝑛 =
2.575 × 3675
= 47.316
200
n = 2239
So a sample of about 2240 is required.
7.
P = 30%
Sep = 7.2457
Need to apply the finite population correction factor as population is small. This is 0.8967
So Sep becomes 7.2457×0.8967 = 6.497
95% confidence interval is 30 ±1.96×6.497
= 17.3% to 42.7%
8.
320
P = 800 × 100 = 40%
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40×60
SEP =√
800
= 1.732
π = 40 ± 1.96 x 1.732
= 36.6% to 43.4%
9.
P = 80%
Sep = 2.8284
95% confidence interval is 80 ±1.96×2.8284
= 74.5% to 85.5%
10.
n = 400
𝑥̅ = £230
σ = 42
SEP =
4.2
√400
= 2.1
(a)
95% confidence interval is given by
230 ± 1.96 x 2.1 = 230 ± 4.12 or 225.1 to 234.1
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(b)
99% confidence interval is given by
230 ± 2.58 x 2.1 = 230 ± 5.42 or 224.6 to 235.4
11.
σ = 60 g
Margin of error required = 2
Margin of error = 1:96 x SEM
2 = 1.96 x
2 = 1.96 x
𝜎
√𝑛
60
√𝑛
Square both sides
4 = 1.962 x
n = 1.962 x
602
𝑛
602
4
= 3457
12.
1.96×SEP = 4%
(a) P=10%
10 × 90
4
√
=
𝑛
1.96
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Square both sides
900
= 4.1649
𝑛
n= 216
(b)
If no estimate then use P=50%
As before then n = 600
13.
SEM =
55524
√16
= 13881
(a) 95% confidence interval is 129500 ±1.96×13881
= 129 500 ± 27207
= £102 293 to £156 706
(b) Half width needs to be 13604
So 1.96 ×
55524
√𝑛
= 13604
So n = 64
(note to half the confidence interval need to multiple n by 4 so 4×16 = 64
(c)
The Finite population correction factor needs to be applied
(200−16)
Which is √
199
= 0.9616
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This is multiplied by SEM which results in a smaller half width ( £26161 compared to
£27207) so the confidence interval will be narrower
14.
P = 51%
51×49
SEP = √ 1975 = 1.1249
Confidence interval = 51 ±1.96×1.1249
= 48.8% to 53.2%
As the confidence interval straddles 50% we cannot be certain that there would be a majority
in favour of leaving the EU.
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Chapter 9 Solutions
1. 1.96
2. 718 (assuming a single sample)
3. 12.592
4.
H0: μ = 5.8
H1: μ < 5.8
(This is a one-tailed test as it is required to see if the action has reduced the time taken.)
SEM = 2.3/√10
= 0.7273
As σ is known we can use the Z test
Z=
4.9−5.8
0.7273
= -1.237
The critical value at 5% is -1.645 and as this is less than the test statistic then we cannot
reject H0 and therefore there is no evidence that the mean time has been reduced.
5.
H0: μ = 54
H1: μ ≠ 54
SEM = 5/√12
= 1.4434
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As σ is known we can use the Z test
Z=
50.5−54
1.4434
= -2.424
The critical value at 5% is -1.96 so can reject H0 and conclude that there is evidence that
the fuel consumption is not 54 mpg
6.
H0: μ = 1500
H1: μ ≠ 1500
SEM = 90/√50
= 12.728
Can use the Z test as the sample size is large
Z=
1410−1500
12.728
= -7.07
The critical value at 5% is -1.96 so can reject H0 and conclude that there is evidence that
the mean lifetime is not 1500 hours. (Note as no one would complain if the lifetime was
greater than 1500 hours it could be argued that this is a one-tailed test. However, we
would still reject the null hypothesis.)
7.
H0: μ = 500
H1: μ < 500
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(This is a one-tailed test as it is required to see if the mean weight is less than 500 g.)
From the sample the mean = 497.3 g and the standard deviation = 11.448 g
SEM = 11.448/√6
= 4.674
t=
497.3−500
4.674
(Use the t test statistic as the sample size is small and we are estimating
the standard deviation from the sample)
= -0.578
The critical value at 5% is -1.645 and as this is less than the test statistic then we cannot
reject H0 and therefore there is no evidence that the company is selling underweight
products.
8.
H0: Π = 30%
H1: Π > 30%
P = 60/150 = 40%
30×70
SEP = √
150
= 3.7417
40−30
Z = 3.7417
= 2.67
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The critical value at 5% significance level is 1.645 so as the test statistic is greater than
this can reject H0. Therefore we can conclude that the claim by the Building Society is
correct.
9.
This is a goodness of fit test.
H0: Accidents equally likely on any day H1: Accidents more likely on certain days
If H0 is true there should be 37/5 = 7.4 accidents each day on average
O
E
(O-E)2
(O-E)
(O-E)2/E
6
7.4
-1.4
1.96
0.264865
5
7.4
-2.4
5.76
0.778378
6
7.4
-1.4
1.96
0.264865
8
7.4
0.6
0.36
0.048649
12
7.4
4.6
21.16
2.859459
Sum
4.216216
Degrees of freedom = 5-1 = 4
Critical value on 4 degrees of freedom at 5% significance level = 9.488
As this is greater than the test statistic of 4.216 we cannot reject H0 and conclude that
there is no evidence that accidents are more likely on certain days.
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10. (a)
H0: There is no association between age and radio station people listened to
H1: There is an association
The expected values are
Age range
Less than 20
20 to 30
Over 30
Total
BBC
32.4
15.4
40.1
88
Local radio
12.2
5.8
15.1
33
Commercial
18.4
8.8
22.8
50
63
30
78
171
Total
O
E
O-E
(O-E)2
(O-E)2/E
22
32.4
-10.42
108.60
3.35
6
12.2
-6.16
37.92
3.12
35
18.4
16.58
274.86
14.92
16
15.4
0.56
0.32
0.02
11
5.8
5.21
27.15
4.69
3
8.8
-5.77
33.32
3.80
50
40.1
9.86
97.21
2.42
16
15.1
0.95
0.90
0.06
12
22.8
-10.81
116.79
5.12
Sum
37.50
Degrees of freedom = (3-1)×(3-1) = 4
Critical value on 4 degrees of freedom = 9.488
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As the test statistic is greater than the critical value we can reject H0 and conclude that
there is an association between age and radio station people listened to.
(b) The largest contribution to the test statistic is the number of people under 20 who
listened to a commercial station. Only 18.4 would be expected from the null hypothesis
but the observed value was nearly twice this at 35
11.
H1: µ ≠ 10
H0:µ = 10
It is a two tailed test so critical values at 5% significance level are ±1:96, at 1%
they are ±2:57 and at 0.1% they are ±3:29
n = 36, 𝑥̅ = 9.94:94, σ = 0:018
0.018
SEM =
√36
= 0.003
Test statistic Z =
9.94−10
.003
= -20.0
Since Z < -3.29 we reject H0 and conclude that there is very strong evidence at the 0.1%
level that the production process is not meeting specification.
12.
H0: Π = 36%
H1: Π ≠36%
80
n = 200, P = 200 x 100 = 40%
36×64
SEP =√
200
= 3.3941
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Test statistic Z =
40−36
3.3941
= 1.18
Since 1.18 is less than 1.96 we cannot reject H0 so there is therefore no evidence to suggest
that there has been a significant change in the percentage of people who bought the product.
13.
H1: µ ≠ £40000
H0: µ = £40000
n = 40; 𝑥̅ = £37 000; σ = £6000
SEM =
6000
√40
= 948.68
Test statistic Z =
37000−40000
948.68
= -3.16
Since -3.16 is less than -2.57 we can reject H0 at the 1% significance level. We therefore
conclude that there is strong evidence that the mean takings were significantly different from
the target figure.
14.
H0: π = 60%
H1: π ≠ 60%
250
n = 500, P = 500 x 100 = 50%
60×40
SEP = √
500
= 2.1909
50−60
Test statistic Z = 2.1909 = −4.56
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Since -4.56 is less than -3.29 we can reject H0 so there is very strong evidence to suggest that
the success rate is significantly different from 60%.
15.
Hypothesis test is inappropriate here as we are given population information not sample
information. Thus we can say with certainty that the mean size of order has increased from
last year.
16.
H0 :µA = µB
H1: µA ≠ µB
t = 2:228 on 10 degrees of freedom (two tailed so 0.025 in each tail
𝑥̅𝐴 = 5:517
SA = 1:507;
-x𝑥̅ 𝐵 B = 4:567
SB = 1:750
̂
5 × (1.507)2 + 6 × (1.750)2
𝜎=√
10
= 1:633
1
1
6
6
Standard error = 1:633 ×√ +
= 0.9428
t=
5.517−4.567
0.9426
= 1.008
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Since t is less than 2.228 H0 cannot be rejected so it is not possible to use these results to
decide on the best filter.
17.
H0 : Consumers equally likely to select each package
H1 : Consumers not equally likely to select each package
(𝑂 − 𝐸)2
𝐸
Package
O
E
O-E
A
106
100
9
0.36
B
92
100
-8
0.64
C
102
100
2
0.04
D
100
100
0
0
1.04
Test statistic χ2 = 1:04
5% significance level, df = 4 - 3 = 1, χ2 = 7:815 (from tables) Accept
H0
Results are consistent with the hypothesis that the consumers are equally likely to select each
package.
18.
<21
21–35
>35
Total
Liked
20
40
80
140
Disliked
30
20
10
60
Total
50
60
90
200
H0: No association between age and preference
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H1 : There is an association between age and preference
(𝑂 − 𝐸)2
𝐸
O
E
O-E
20
35
-15
6.428
40
42
2
0.095
80
63
17
4.587
30
15
15
15.000
20
18
2
0.222
10
27
-17
10.703
37.037
Test statistic χ2 = 37:037
5% critical value, df = (2 – 1)(3 – 1) = 2, from tables χ2= 5:991
Reject H0
Evidence to suggest that age and preference are associated (would also reject at 1% and 0.1%
levels)
More under-21s and less over-35s than expected appear to dislike the design.
19.
H0:Pattern of demand hasn’t changed
H1 Pattern of demand has changed
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Direct lines
Switchboard
Fax and others
O
E
120
80
100
180
75
45
(O-E) 2
-60
3600
5
25
55
3025
Sum
O-E
(O-E) 2/E
20
0.33
67.22
87.56
The critical value on 2 degrees of freedom at 5% SL= 5.991
Reject H0 and conclude that there has been a significant change in the pattern of demand.
20.
H0 : µD = 0
H1 : µD > 0
(one tailed since it is hoped that the campaign would not decrease sales)
Critical value of t = 1:895 at 5% (one tailed on 7 degrees of freedom)
Differences are:
A
B
C
D
E
F
G
H
28
-25
40
100
60
-20
0
50
𝑥̅ = 29.125
SD = 42.633
Standard error =
42.633
√8
= 15.073
29.125
𝑡 = 15.073 = 1.932
Since 1.932 is greater than 1.895 H0 can be rejected at 5%. Therefore it seems likely that the
campaign has worked.
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21.
This is a one tailed test and the null and alternative hypotheses are:
H0 :π1 – π2 = 0
H1 : π1 – π2 < 0
30
P1 = 250 × 100
= 12%
40
P2 = 220 × 100
= 18:2%
𝑃̂ =
250 × 12 + 220 ×
250 + 220
= 14:9%
The standard error of the differences is
𝜎(𝑃1 −𝑃2 ) = √14.9 × (100 − 14.9)(
1
1
+
)
250 220
= 3:2917
The test statistic is:
𝑍=
(12 − 18.2) − 0
3.2917
= - 1:88
As this is less than -1.645 we can reject H0 and therefore conclude that there is evidence to
suggest that students are working longer in 2000 than in 1995
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22.
H0:Lunch habits hasn’t changed
H1 Lunch habits have changed
Bring
lunch
Skip lunch
Local shop
Canteen
O
E
42
95
27
63
88.53
27.24
40.86
70.37
O-E
(O-E)2
(O-E)2/E
-46.53 2165.041 24.45545
67.76 4591.418
168.55
-13.86 192.0996
4.70
-7.37 54.3169
0.77
Sum
198.48
The critical value on 3 degrees of freedom is 7.815
Reject H0 and conclude that there has been a significant change in lunch habits since
moving to the new site. In particular the number of employees who skip lunch has
increased greatly and the number who bring their lunch has gone down
23.
H0:Demographic mix hasn’t changed
H1 Demographic mix has changed
Less 18
18-19
20-24
>24
O
E
O-E
(O-E)2
6
118
102
26
6.80
75.35
134.57
35.28
-0.80
42.65
-32.57
-9.28
0.65
1819.19
1060.67
86.12
Sum
(OE)2/E
0.10
24.14
7.88
2.44
34.56
Critical value on 3 degrees of freedom = 7.815
Reject H0 and conclude that there has been a significant change in demographic mix of
the users. This is caused by the number of people in the 18-19 age group using the site
more.
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Chapter 10 Solutions
1.
Earnings (£000s)
45
40
35
30
25
20
15
10
5
0
0
20
40
Age
60
80
There is a weak positive correlation between age and Earnings as shown by the scatter of
points on the chart above. This means that as someone gets older they might be paid more but
there is a lot of variation. Age range would be from 18 to 65.
2.
50
45
Consumption (units)
40
35
30
25
20
15
10
5
0
0
10
20
30
40
50
60
70
Age
Older people tend on average to consume less alcohol than younger people but this is
nowhere near a perfect correlation.
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3.
x is already ranked. y is ranked as follows
y
9
6
7
9
5
2
1
Rank
6.5
4
5
6.5
3
2
1
As two values of y are equal we rank them as 6.5
x
1
2
3
4
5
6
7
d
-5.5
-2
-2
-2.5
2
4
6
d2
30.25
4
4
6.25
4
16
36
100.5
M Acc
5
1
3
4
2
6
d
-1
2
-2
1
0
0
y
6.5
4
5
6.5
3
2
1
Sum
6×100.5
R = 1 − 7(49−1)
= 1 – 1.795
= −0.795
4.
Stats
4
3
1
5
2
6
A
B
C
D
E
F
Sum
d2
1
4
4
1
0
0
10
6×10
R = 1 − 6(36−1)
= 1 – 0.286
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= 0.714
5.
y = 3 + 5×2
= 13
6.
(a) and (b) See Excel file (Chapter 10 Q6) for scatter graph.
The scatter graph shows there is strong positive association between height and weight for
males but much weaker for females.
(c)
All data: r = 0.760; males: r = 0.906; females: r = 0.0102
7.
R = -0.3875
8.
See Excel file (Chapter 10 Q8) for scatter graph.
(a)
The scatter graph suggests that there is a strong association between sales and
advertising expenditure and that this is a positive association.
(b)
y = -0.24 + 9.292x, where y is sales and x the advertising expenditure.
The coefficient of determination is 93.3% so it appears that the regression explains the data
quite well.
(c)
£46 200
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9.
(a)
Positive (weak for league as a whole but probably strong for a particular club)
(b)
Negative, strong
(c)
Positive, weak
(d)
Positive, strong
(e)
Negative, strong
10.
(a)
(i)
See Excel file (Chapter 10 Q10)
(ii)
r = 0.8898
(b)
a = 11.129, b = 0.04037
So time to pay = 11:13 + 0:0404 × size of bill
(c)
The 11.13 is the minimum time to pay (in days) and for every £1 increase in the size
of bill the time to pay will increase by 0.0404, i.e. the marginal increase.
(d)
See Excel file (Chapter 10 Q10)
(e)
(i)
for £125 time to pay = 11.13 + 0.0404 × 125
= 16.18 or just over 16 days
(ii)
for £1000 time to pay = 11.13 + 0.0404 ×x 1000
= 51.53 or 51.5 days
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The prediction for £1000 is unreliable because it is outside the range of data (the largest
amount is only £480).
(f)
79.2% of the variation in time to pay is explained by the size of the bill. The
remaining 20.8% is not explained by this factor. (Other factors might help in explaining this
remaining variation.)
11.
(a) See Excel file (Chapter 10, Q11)
(b)
% Collected
Rank
% arrears
Rank
d
d2
1
6
13
1
5
25
12
3
10
4
-1
1
16
2
8
5
-3
9
11
4
12
2
2
4
20
5
11
3
2
4
6
5
11
3
2
4
Sum
68
6×68
R = 1 − 6(62 −1)
= 1-1.943
= -0.943
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12.
(a) See Excel file (Chapter 10 Q12)
(b) Pearson’s correlation coefficient, r = -0.5717, Spearman’s correlation coefficient = -0.515
H1: r ≠ 0
(c) H0: r = 0
Critical value on 8 df = 0.6319 for Pearson’s so cannot reject H0. There is no evidence of a
significant correlation.
(d)
(ii)
17.4 is for someone aged 0 and so has no practical meaning. The -0.2118
means that for every year older the days off work reduces by 0.2118
(iii) 17.4 – 0.2118 × 65 = 3.6 days. But 65 is outside the range of the data and the prediction
is unreliable.
(iv) r2 = 0.3268 or 32.7%. This tells us that only about 33% of the variation in days off sick
is explained by age.
(v)
Length of service, sex, marital status, health, etc.
13.(a)
Sum
𝑏=
Advertising x Sales y
50
51
60
65
45
49
55
375
xy
1
1.02
1.3
1.45
1.2
1.08
1.25
8.3
50
52.02
78
94.25
54
52.92
68.75
449.94
x^2
2500
2601
3600
4225
2025
2401
3025
20377
7 × 449.94 − 375 × 8.3
7 × 20377 − (375)2
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= 0.0184
𝑎=
8.3
375
− 0.0184 ×
7
7
= 0.199
The regression equation is therefore
Sales = 0.199 + 0.0184×Advertising
(b)
(i) Sales =0.199 + 0.0184×40
= £0.935m
(ii) Sales =0.199 + 0.0184×80
= £1.671m
The prediction for £80,000 pounds advertising is outside the range of the date and therefore
not reliable.
14.
SPSS was used for this question.
(a)
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This shows a strong positive correlation.
It looks quite a weak correlation. If the point corresponding to size of store = 12 is removed
there may be no correlation.
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Very weak correlation. Could be negative.
(b)
Correlations
Sales
Sales
Pearson Correlation
Population
1.000
Sig. (2-tailed)
N
10.000
size of store
Years experience
.937**
.699*
-.365
.000
.024
.300
10
10
10
**. Correlation is significant at the 0.01 level (2-tailed).
*. Correlation is significant at the 0.05 level (2-tailed).
Correlation for population is as expected. Other variables are affected by outliers.
(c)
For all variables
Standardized
Unstandardized Coefficients
Model
1
B
Std. Error
(Constant)
-4.152
22.351
Population
4.759
.949
Coefficients
Beta
t
1.154
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Sig.
-.186
.859
5.013
.002
size of store
-5.574
4.374
-.288
-1.274
.250
Years experience
-1.427
2.339
-.079
-.610
.564
a. Dependent Variable: Sales
R2 = .909
For population only
Coefficientsa
Standardized
Unstandardized Coefficients
Model
1
B
Std. Error
(Constant)
-12.183
16.726
Population
3.864
.509
Coefficients
Beta
t
.937
Sig.
-.728
.487
7.589
.000
a. Dependent Variable: Sales
R2 = ..878
The regression equation with the highest R2 is: Sales = -4.152 + 4.759P – 5.574S – 1.427Y
Where P = population and S= size of store and Y = years’ experience.
However neither size of store or years’ experience are significant, so the most reliable
equation is Sales = -12.183 + 3.864P
(d) Using the full equation
Sales = -4.152 + 4.759×28 – 5.574×6– 1.427×12
= 112.78 or £112,780
Using the shorter equation
Sales = -12.183 + 3.864×28
= 96.01 or £96,010
15.
H0 β0 = 0
H1 β0 ≠ 0
From question 13 we know that ∑x = 375 and ∑x2 = 20377 and the slope is .0184
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We obtain Se from the table below
Advertising
x
Sales y
50
51
60
65
45
49
55
1
1.02
1.3
1.45
1.2
1.08
1.25
Predicted
1.119
1.1374
1.303
1.395
1.027
1.1006
1.211
Residual Residual sq
-0.119
-0.1174
-0.003
0.055
0.173
-0.0206
0.039
Sum
0.014161
0.01378276
9E-06
0.003025
0.029929
0.00042436
0.001521
0.06285212
.062852
Se= √
5
= 0.1121
Sb =
.1121
√20377−(375)2
7
= .00661
t=
.0184−0
.00661
= 2.784
The critical value on 5 degrees of freedom is 2.571 so we can reject H0 and state that the
regression slope is significantly different from zero
The confidence interval for the slope is
.0184-2.571×.00661 and .0184+2.571×.00661
.00141 to .0354
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Chapter 11 Solutions
1.
Emv = .2×2 + .8×.5
= 0.8
2
(a)
0.4;
(b)
26;
(c)
14;
(d)
Decision 1;
(e)
0.509
3.
(a) (i) Manufacture
(ii) Sell
(iii) Take royalties
(b) Manufacture (emvs are 30, 28, 20)
(c) Find the value of perfect information (30 - 80×.2+40×.5 + 20×.3) = 12 (or £12,000)
4.
(a) (i) Bar
(ii) Bar
(iii) Bar
(b) Bar (emvs are 53, 68.5, 114.5)
5.
(a) (i) B
(ii) C
(iii) C
(b) C (emvs are 2.7, 3.3, 4)
(c) Calculate value of perfect information = 4 - 5×.5+7×.3 + 4×.2 = 1.4 (or £1.4m)
As this exceeds £0.5m it would be worth employing the economist.
6.
EMV = 2000×.25 + 1200×.5 + 500×.25 = £1225
So he would make a profit of £225
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If he invested the £1000 for 3 months he would get 1000 ×8.75/4 = £21.875
As this is less than £225 he should buy the car.
7.
Expected cost of taking the car = 1000×.02 + 10 = 30
As this is less than £50 his decision on financial grounds should be to take the car.
If let the probability of being caught = x then indifferent when
1000×x + 10 = 50
x = .04 which is a 2 in 50 chance.
8.
(a) Best decision is to wait
(b) Probability should decrease to 0,5
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£19m
9.
Rival store
£10m
0.7
2
Expand
Move
0.3
E
x
p
a
n
d
£20.8m
a
t
3
No rival
store branch
Rival store
£40m
£16m
0.7
t
o
w
n
c
e
n
t
r
e
0.3
No rival
store branch
£32m
To maximize expected profit,
move to the new site
M
o
v
e
t
o
n
e
w
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10
(a)
9.7
Fails
–20
0.01
Earthquake
9.97
0.99
.1
10
Strengthen
.9
12.9
10
–6
12.9
0.3
.1
1
.9
15
Fles
Dictator
11
–15
0.7
Earthquake
Don’t
Tut
Fails
–5
.2
8
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15
.
The best decision is to invest in Tutamolia but not to strengthen the building, giving an EMV
of £12.9m.
- 6p + 15(1 – p) = 11
(b)
21p = 4
p = 0:19
If probability greater than 0.19, then should choose Flesomnial. (Note: This won’t affect
decision for Tut branch as the 9.97 value will be reduced.)
11.
(a)
(i)
This would be the maximax rule and would choose Investment
(a)(ii) This would be the maximin rule. Max(-100, 0, - 10) = 0. Would
choose Investment B.
(b)
Investment
Rising
Stable
Falling
Largest
A
0
25
150
150
loss
B
60
0
50
50
160
40
0
160
C
Min(150, 60, 160) = 60 so choose Investment B
(c)
EMV A = -25, B = 28 and C = 30 so choose Investment C
(d)
The expected value with perfect information
= :2 x 150 ~+ :2 x 50 + :6 x 50 = 70, so EVPI = 70 - 30 = 40
1
(e)
(i)
u(150) = 1. u(-100) = 0; u(90) = .5; u(25) = :3
1.2
1.0
0.8
0.6
0.4
0.2
-150 –100
(ii)
–50
50
100
150
u(50) = .4. u(10) = .25; u(0) = .2; u(-10) = .15
Expected utilities are therefore:
A: .2 x 1 +.2 x .3 + .6 x 0 = .26
B: .2 x .5 + .2 x .4 + .6 x .2 = .3
C: .2 x .15 + .2 x .25 + .6 x .4 = :32
So the best investment is still C.
12.
(a)
NPV of High demand = 300.909 + 400.8264 + 400.7513 – 20 = £884
2
200
Similarly NPV for Low demand = £473
(b)
H
854
710
.6
5
success
443
699
employ
699
H
551
720
.55
Internet
699
H
884
.55
345
581
H
473
.55
750
375
Best decision is to launch internet site but not to employ consultants. EMV is £699,000
(d) Maximum amount is 699 – 678 = £21,000
(e) Find probability that makes you indifferent between two decisions. If it takes only a small
probability to change your decision then the decision is sensitive to this probability.
3
Joint probabilities
13 (a)
.009
positive
.9
allergic
.1
.001
.01
positive
.99
.297
.3
.7
.693
.1
(b)
b) P(test positive) = .009 + .297 = .306
(c)
P(allergicpositive) = .009/.306
= .0294 (ii)
P(allergicnegative) =.001/.694 :0014
(d)
positive
-11
-11
2.85
.306
9.7
A1
kit
8.96
allergic
-21
.0014
9.7
.01
1.50
-20
9
10
So should sell the A1 product, but don’t supply kit in order to maximise the EMV.
4
(e) If supply kit annual profit = 500,0002.85 =£1,425,000
If don’t supply kit profit = 50,0009.7 = £4,850,000
If sell older version profit = 50,0001.50 = £750,000
14.
We first use Bayes’ theorem as follows:
Joint probabilities
0.18
Positive
0.9
Has condition
0.1
0.002
Negative
0.02
Positive
0.98
Doesn’t
0.049
0.05
0.95
Negative
0.931
P(Positive response) = 0.018 + 0.049
= 0.067
P(Negative response) = 1 – 0.067
= 0.933
0.018
P(Condition+ve response) =
0.067
= 0.2687
P(Condition-ve response) =
0.018
0.933
= 0.0193
5
350
+ve
88.1
.067
screen
Condition
1050
69.3
-ve
.0193
.9807
Doesn’t
Don’t
screen
50
65
So cheaper not to screen.
15.
Car
A
B
C
D
E
weight
normalised
Consumption
75
0
100
58
92
100
52.6
Top
speed
64
100
0
71
79
30
15.8
Environmental
50
0
100
60
30
60
31.6
Agg.
Benefits
65.4
15.8
84.2
60.9
70.1
The car with the highest aggregate benefits is car C.
16.
Possible criteria could be floor area, parking spaces, closeness to town centre, attractiveness
of area, etc.
6
100.0
C
90.0
H
80.0
A
Efficient
frontier
70.0
60.0
B
50.0
40.0
30.0
20.0
10.0
0.0
5
7
9
11
13
15
17
19
Cost (£000s)
Should consider shops A, B, C and H
17. (a)
£1.4m
0.24
H
a
1.13
.3
.7
L
(£0.3m)
Full
c
Full
.85
L
.15
1.13
(£0.4m)
2
0.269
1
H
£1.5m
Survey
F/H
0.269 .3
b
F/L
.7
(£0.1m)
£1.4m
H
(0.1)
3
(0.13)
Full
d
.15
L
Abandon
.85
(£0.4)m
0
F/H Forecast high
L/H Forecast low
(£0.1m)
7
21
The decision should be to conduct a survey first and then if the survey indicates a favourable
response go into full production, otherwise abandon project. The EMV would be £0.269 m.
(b)
If the probability is greater than 0.35, decision changes from survey first to full production. If
the probability is less than 0.08 decision changes from survey first to abandon project.
8
Chapter 12 Solutions
1.
Year
Investment A
Investment B
Investment C
0
-20
-15
-5
1
12
0
3
2
10
10
3
3
5
5
3
4
3
5
3
5
0
5
3
Payback
2
3
2
Average income
6
5
3
ARR
30%
33.3%
60%
Investment A and C would be preferred in terms of payback period while investment C is a
clear winner in terms of ARR.
2.
Payback
first project: 3 years 2nd project: 5 years
ARR
1st project 1/2.5 ×100 = 40%
2nd project: the average income is 15/5 = £3m so ARR is 3/10 × 100 = 30%
The first project is preferred on both the payback and ARR methods.
3.
Month
Property A
Property B
Property C
0
-£200,000
-£300,000
-£400,000
1
£400
£500
£800
2
£400
£500
£800
Payback
200,000/400 = 500
600 mths =50 years
500 mths = 41.7
mths = 41.7 years
years
9
Average income pa
400×12 = 4800
6000
9600
ARR
4800×100/200,000 =
2%
2.4%
2.4%
Property A and C are best (and equal) on both measures.
4.
Using the simple interest formula
£10,000 × 6% = £600
5.
£20,000 × 5% = £1000
6.
Using the compound interest formula
P5 = 20000(1+.05)5
= £25,525.63
7.
(a) P6 = 50000(1+.055)6
= £68942.14
(b) £68942.14 – 50000 = £18942.14
8.
Using the constant repayments formula
P5 = 3000(1.08)5 + 1000[(1.08)5 – 1]/.08
= 4407.98 + 5866.60
= £10,274.58
10
9.
Using the present value formula
P0 = 50000 × discount factor
Discount factor = 1/(1.06)6
= 0.70496
P0 = 50000×0.70496
= £35248.03
10.
The answer depends on the interest rate. The table below gives the decision for different
discount rates.
r%
2
4
6
8
20
25
£20,000
£ 19,607.84
£ 19,230.77
£ 18,867.92
£ 18,518.52
£ 16,666.67
£ 16,000.00
£25,000
£ 24,029.22
£ 23,113.91
£ 22,249.91
£ 21,433.47
£ 17,361.11
£ 16,000.00
As you can see the £20,000 in 1 years’ time is the better option unless the interest rate
exceeds 25% which is highly unlikely.
11.
The sum of the present values over the 10 years is £27,026.07
12.
Year
0
1
2
3
4
5
NPV
Present value
-2
0.566
0.534
0.504
0.475
0.448
0.527
11
As the NPV is positive this investment is worthwhile.
13.
Year
0
1
2
3
4
5
Investment A
PV
-20
12
10
5
3
0
NPV
-20
11.32
8.90
4.20
2.38
0.00
6.80
Investment B
PV
-15
0
10
5
5
5
-15
0
8.90
4.20
3.96
3.74
5.79
Investment C
PV
-5
3
3
3
3
3
-5
2.83
2.67
2.52
2.38
2.24
7.64
All 3 investments show a positive NPV but investment C gives the greatest value so is
preferred on this measure.
14.
Payback period = 4 years
ARR = 25%
NPV = £0.05m
IRR = 8% (Note to find IRR you can use GoalSeek in Excel to find the interest rate that gives
the NPV of zero)
15.
See Excel file (Chapter 12 Q15). The Goal Seek function can be used to find the value
of r to give a NPV of zero. This was found to be 9.7% and as this is less than 14% the project
should not be accepted.
16.
P0 = £1000, r = 5:4%, n = 5
5.4
P5 = 1000 × (1 + 100)5
= 1000 x 1.30078
= £1300.78
12
17.
Using the sinking fund formula (assume that the customer pays the first annual instalment
now giving 4 instalments in total). Let £x be the annual instalment.
First instalment grows to £x(1.0575)3 = 1.1826x at the end of 3 years
2nd instalment grows to £x(1.0575)2 = 1.1183x at end of 3 years
3rd instalment grows to £x(1.0575) at end of 3 years
Final instalment is £x
Total = x(1.1826 + 1.1183 + 1.0575 + 1) = 20000
4.3584x = 20000
x = 4588.84
So annual instalment is £4588.84
(Note if assume that first payment is not until the end of the first year then x = £6297.63).
18.
Need to discount the 5 annual payments of £1000. So cost at today’s prices is
£8546
19.
Profit
Payback
ARR
Machine A
Machine B
6000 -4000 = 2000
4
2x
= 1:8 years*
4.5
2000
= 50%
4000
6000 -3900 = 2100
3900
2x
¼ 1.95 years*
4000
2000
= 51.3%
3900
*Note: It could be argued that the payback is 2 years for both if it is assumed that the revenue
is obtained at the end of each year.
Machine B is better on profit and ARR, but not on payback.
13
Machine A
Discount
factor
Cash flow
Machine B
Cash flow
Present
value
Present
value
1
0.9259
-4000
2000
2
0.8573
2500
2143.3
2500
2143.3
3
0.7938
1500
1190.7
2000
1587.7
0
NPV
-4000
1851.9
-3900
1500
-3900
1388.9
1185.9
1219.9
Machine B is preferred as it has a higher NPV.
Machine A
Machine B
Discount rate
NPV
Discount rate
NPV
8
1186
-32
8
1220
-76
25
25
An IRR of about 24.5%
Using formula:
1186 × 25 − (−32) × 8
1186 − (−32)
= 24.6%
For Machine B the IRR is 24.0%
Not much difference although Machine A has a slightly higher IRR and would therefore be
preferred using this measure.
20.
Investment 1
Profit
Payback
ARR
18 000 -15 000 = 3000
3 years
6000
= 40%
15 000
Investment 2
19 000 - 15 000 = 4000
3 years
6333
= 42%
15 000
Investment 2 gives a better profit and ARR.
14
Investment 1
Discount
factor
Investment 2
Cash flow
Present
value
Cash
flow
Present
value
-15 000
5 714.3
-15 000
0
-15 000
0.0
1
0.9524
-15 000
6 000
2
0.9070
6 000
5 442.2
0
0.0
3
0.8638
6 000
5 183.0
19 000
16 412.0
0
NPV
1 339.5
1 412.9
Investment 2 gives a higher NPV Using formula for investment 1
1340 × 10 − (−79.2) × 5
1340 − (−79.2)
= 9.7%
Using formula for investment 2
= 8:3%
So Investment 1 gives a higher IRR and is preferred on this measure.
21.
(a)
P3 = 800(1.04)3
= £899.89
(b)
Let instalment = x
First instalment will grow to x(1.052)3 at the end of the plan
Second instalment will grow to x(1.052)2 at the end of the plan
Third instalment will grow to x(1.052) at the end of the plan
Total amount accumulated = x{(1.052)3 + (1.052)2 + 1.952}
= 3.322296x
Machine will cost £899.89 so
3.32296x = 899.89
899.89
And x = 3.32296
= £270.81
Instalments should be £270.81
15
22.
Year
Cash flow
Present value
0
1
-3750
1310
-3750
1178
2
1310
1059
3
1310
953
4
1310
857
5
1310
770
(a)
NPV = £1067
(b)
By trial and error (using Excel) a discount rate of 25% gives a NPV of -£227.
Using formula
IRR =
1067×25−(−227)×11.2
1067−(−227)
= 22:6%
23.
The annual amount can be found from:
𝑥[(1.07)11 − 1]
= 1000000
. 07
15.78x = 1 000 000
x = £63 371
An annual investment of £63 371 would give a sinking fund of £1m in 10 years.
24.
(a)
0 = −650001.075)20 +
𝑥[(1.075)20 −1]
.075
0 = -276110 + 43.305x
x = £6376
So annual repayment would have to be £6376
(b)
This time we know the annual repayment (£7000) but need to find n, the number of
years to pay off the loan.
16
0 = −65000 × 1.075𝑛 +
7000[(1.075)𝑛 − 1
. 075
28 333 x (1:075)n = 93333
(1.075)n = 3.294
n × log(1.075) = log(3.294)
n = 16.48 years
How much money is saved: Paying £7000 for 16.48 years gives a total of
£115 374 and paying £6376 for 20 years gives a total of £127 520. So the amount saved is
£12 146.
25.
Using the Constant repayments formula
0 = -150000(1.07)25 + x[(1.07)25 – 1]/.07
0 = -814114.896 + 63.249x
x = 12871.59
So annual repayments are £12,871.59
26.
Using the depreciation formula
5000 = 50000(1 + r/100)5
(1 + r/100) = (0.1)1/5
r = -36.904
So rate of depreciation is 36.9%
27.
(a)
5000 = 12 000 x (1 + r/100)3
1 + r/100 = (.4167)1/3
= .7469
17
r = -25:3
So the rate of depreciation is 25.3%
(b)
At end of first year
P1 = 12 000 × (1 - .253)
= £8964
So car will be worth £8964 after one year.
28.
Take £1000 invested for one year.
If compounded annually at 7%
P1 = 1000 × 1.07
= £1070
If compounded continuously at 5%
P1 = 1000 x e5/1000
= £1051
So better to use the first option
18
Chapter 13 Solutions
1
(a) and (b)
See Excel file (Chapter 13 Q1)
(c)
Average seasonal differences (adjusted) are -10.05, -6.17, -5.38, -4.72,
8.66, 17.66
(d)
Forecasts using regression line through CMA
Trend forecast
2015
2
(a)
See Excel file (Chapter 13 Q2)
Forecast
1
43.5
33.4
2
43.6
37.4
3
43.7
38.3
4
43.8
39.1
5
43.9
52.5
6
The additive model is best as it gives smallest MSE. (b) and (c)
44.0
61.6
See Excel file (Chapter 13
Q2)
d) Forecasts
Period
Trend
Sales
2
26.7
22.9
3
27.5
18.6
4
28.3
32.5
3. (a) and (b) see Excel file (Chapter13 Q3)
(c) The additive is the better model as it has the lower MSE )0.04 compared to 0.55 for
multiplicative model)
19
(d) Forecasts are
Quarter
Trend
Sales
2
123.5
112.1
3
124.5
136.0
4
125.5
138.1
4. See Excel file (Chapter 13 Q4)
Forecast on day 12 is 112.4.
5.
See Excel file (Chapter 13 Q5) for detailed calculations. Multiplicative was used as it had the
lowest MSE. Forecasts are:
Period
Trend
Sales
1 196.517
70.2
6
2 199.220
151.1
4
3 201.923
405.9
2
4 204.626
178.9
If demand is expected to rise by 5% from 2011 then demand in quarter 3 and 4 of 2015
should be no more than 331 and 148 respectively. As the forecasts exceed this it looks as if
they won’t have enough titanium to meet demand.
20
6.
(a)
There is a downward trend and if a straight line was fitted the sales would at some point
become negative! It is more likely that the trend would flatten off.
(b) Multiplicative as the MSE is lower. The chart shows that the seasonal swings are getting
less and not constant.
(c)
Year
2013
2014
Quarter
Turnove
Forecas
error in
New
r
t
forecast
forecast
1
138
2
134.6
138
-3.4
137.32
3
129.5
137.32
-7.82
135.76
4
124.6
135.76
-11.16
133.52
1
127.6
133.52
-5.92
132.34
2
125.8
132.34
-6.54
131.03
3
125.3
131.03
-5.73
129.89
4
117.7
129.89
-12.19
127.45
The forecast for quarter 1 of 2015 is 127.45
(d)
The squared errors using the figures above are
11.56
61.15
21
124.46
35.10
42.77
32.85
148.49
And the average of these is 65.2. This is the MSE
7.
(a) Simple exponential smoothing with a low alpha tends to dampen the forecasts and puts
less weight on more recent observations. This means that for the oil consumption graph the
forecast is very slowly adapting to the increase in oil consumption from from1984 to 1996
Simple exponential smoothing is not very good at forecasting a series with trends which is
what is apparent here.
(b) If the smoothing constant was increased the forecasts would try and follow the actual
series. However, there would always be a lag between the actual and forecast. A small value
of the smoothing constant is useful if it is required to predict the long term value of
consumption. However a larger value would give more weight to more recent observations
which might be what is required here.
(c) New forecast = last forecast + α×error in last forecast
= 1777 + 0.5×(1696-1777)
= 1873
8.
(a) See Excel file (Chapter 13 Q8).
22
(b) See Excel file (Chapter 13 Q8). The trend line suggests that the sales are rising slowly.
The Additive model might be best as the fluctuations are fairly constant from year to year.
(c) See Excel file (Chapter 13 Q8) The MSE for the additive and multiplicative models are
.52 and .63 which shows that both models would be equally good although the additive
model is slightly better which is consistent with (b)
(d) See Excel file (Chapter 13 Q8). The deseasonlised figures for quarters 1 and 2 of 2015
are 124.8 and 123.4 respectively. The first figure is consistent with the trend but the figure
for quarter 2 is less showing that the trend may be changing.
(e) See Excel file (Chapter 13 Q8). The forecasts are shown below. (Note the trend forecast
were used by fitting a regression line to the trend values. The regression line was trend =
0.7491 + 113.82)
Period
Trend
Sales
1 126.554
116.6
7
2 127.303
136.2
8
3 128.052
143.2
9
4 128.802
114.6
23
Chapter 14 Solutions
1.
50
A
45
40
Optimal solution is at B which
is x= 20 and y = 25 giving a Z
of 137,500
35
30
B
25
20
15
10
5
C
0
0
10
20
30
40
2.
Let P = number of units of product P and
Q = number of units of product Q
Max 600P + 400Q
Subject to
2P + Q ≤ 10
P+Q≤7
P, Q ≥ 0
24
50
60
70
Solution is P = 3 and Q = 4 giving a profit of £3400.
3.
Let D = number of diesel engines and
P = number of petrol engines to be produced each day
Max 60D + 45P
Subject to
4D + 2P ≤ 16
2D + 2P ≤ 12
P≥3
D≥0
Solution is D = 3 and P = 2 giving a profit of £270
4.
(a)
Let
L = number of low wattage bulbs to produce each day
Let H = number of high wattage bulbs to produce each day
20L + 30H ≤ 36 000 (10 x 60 x 60)
10L + 30H ≤ 36 000
L +H ≤ 1500
L, H ≥ 0
25
(b)
See Excel file (Chapter 14 Q4) for Solver solution.
Solution is L = 900 and H = 600 giving a profit of 8700p (£87)
(c)
There are 9000 seconds (2.5 hours) left in testing but both fitting and supply of
shells are scarce resources. The shadow price of the fitting resource is .2p per second (£7.20
per hour) and 1p per extra shell. Increasing either the fitting time or the number of shells
would increase total profits.
5.
19
Deluxe
18
17
16
15
14
13
12
11
10
A
9
A possible integer
solution
8
7
6
5
4
B
Feasible region
3
Iso profit line
2
1
Basic
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
As the isoprofit line is parallel to AB (part of the resin constraint) there will be multiple
optimal solutions as an point on AB is optimal. A sensible one would be to make 12 Basic
and 5 Deluxe dinghies a day.
(b)
(i) Shadow price of the resin resource is £5.
26
(ii) Can store an extra 34kg which will give an extra profit of 5 ×34 = £170 per day.
6.
(a) Max 60P +85S
ST
4P +8S ≤ 400
2P + 3S ≤ 250
10S ≥ 150
5P + 7S ≤ 480
P – 2S ≥ 0
(b) P = 70, S =15 giving a profit of £5475 per day
(c) Components A and c are binding (tight) constraints. Shadow prices of A and C are £15
and £3.50 respectively. (Note increasing C would reduce profits). It would be worthwhile
increasing A since shadow price greater than cost.
(d) No upper limit but lower limit of £42.50
7.
(a) 50 radio and 30 TV ads giving a total rating of 28000
(b) Radio and budget constraints binding. Shadow price of radio ads constraint is 50 so If
radio ads increased by 50 then extra rating would be 50 x 50 = 2500
(c) Can’t reduce TV ads rating.
27
8.
(a)
Let x1 = Amount invested in government bonds
Let x2 = Amount invested in corporate bonds
Let x3 = Amount invested in FTSE 100 stocks
Let x4 = Amount invested in Aim stocks
Max .04x1 + .05x2 + .06x3 + .08x4 + .03(550000 - x1 - x2 - x3 - x4)
i.e. Max 16500 + .01x1 + .02x2 + .03x3 + .05x4
s.t
x1 + x2 + x3 + x4 ≤ 550 000
x2 ≥ 50 000
x2 + x3 + x4 ≤ 300 000
x1 + x2 - x3 - x4 ≤ 0
x1 + x2 + x3 - 3x4 ≥ 0
x1 , x2 , x3 , x4 ≥ 0
(b)
(i)
£200 000 in government bonds, £50,000 in Corporate bonds, and £125,000 in
each of FTSE 100 and Aim stocks. This means that £50,000 is left in the bank. Total return
is
200 ×.04 + 50 × .05 + 125 × .06 + 125 × .08 + 50 × .03
= £29 500
28
As a percentage on capital invested = 5.36%
(ii)
The shadow price for constraint 2 is -.04 which means that reducing the RHS
increases the return. Can reduce RHS by 25 000 so increase in return would be 25000 × .04 =
£1000.
The shadow price for constraint 3 is 0.05 so increase the RHS by the maximum which is
25000. Increase in return would be 5000 × .05 = £1250. So it is better to increase the RHS of
constraint 3.
(iii) The allowable decreases for x1 , x2 , x3 , x4 are .015, infinite, .08 and .02 respectively.
The implication of this is that x1 would have to fall to 2.5% which is below the bank rate. x2
and x3 would mean a negative rate so is not realistic. x4 could fall to 6% and still remain in
solution. The allowable decrease for corporate bonds is infinite because constraint 2 says that
must have at least £50,000.
9.
(a)
L = No. of LXT models to make each week
H = No. of HXT models to make each week
Max 580L +440H
ST
5L + 10H ≤ 40
(1)
9L + 2H ≤ 40
(2)
7L + 5H ≤ 40
(3)
29
L ≥ .6(L + H)
Or
0.4L - .6H ≥ 0
(4)
L, H ≥ 0
(b)
20
19
18
17
16
15
14
9 L + 2 H = 40
13
Optimal solution is L = 4 units per week and
H = 2 units per week
12
H
11
The value of the optimal solution is 580 x 4 +
440 x 2 = £3200
10
9
8
7 L + 5 H = 40
7
6
5
0.4L -.6H = 0
4
3
A
2
Optimum
B
5 L + 10 H = 40
1
C
0
0
1
2
3
4
5
6
7
L
30
8
9
(c)
Substituting optimal values into the constraints
5 × 4 + 10 × 2 = 40
so binding
9 × 4 + 2 × 2 = 40
so binding
7 × 4 + 5 × 2 = 38
Slack of 2 hours
.4 × 4 - .6 × 2 = 0.4
Surplus variable = 0.4
(d)
If profit becomes less than £220 then optimal moves to point A (3.4, 2.3). If greater than
£1980 then moves to C (4.4,0)
(e)
Need to consider the loss in profits of making one unit of ZXT
For each hour lost in the Body shop the company loses £35 so if the new model requires 4
hours in this shop the loss of profits will be 4 × 35 = £140
Simly for the Assembly shop the loss of profits is 2 × 45 = £90
As there are 2 spare hours in the Painting shop there will be no loss of profits
Total loss of profits is therefore 140 + 90 = £230
So profit on the ZXT will need to be at least £230 to make it worthwhile to produce.
(f)
The new constraint is
31
5L + 4H ≤ 30
Substituting L = 4 and H = 2 gives
5 × 4 + 4 × 2 = 28
As this constraint is not binding it does not affect the optimal solution found in (a)
10. (a)
Max Z = 1.5S + L
ST
0.5S + 0.4L ≤ 6
0.25S +0.7L ≤ 14
S + L ≤ 30
L≥5
S≥0
(b)
32
30
25
No. of Lentil Puff (L)
20
Z = £15
15
10
Optimal solution, Z = £17
Feasible
region
5
0
0
5
10
15
20
25
30
No. of Shepherds Ecstasy (S)
So make 8 Shepherd’s Ecstasy and 5 Lentil Puff giving a profit of £17
(c) The isoprofit line is now parallel to the ingredient A constraint so there are multiple
optimal solutions all along this line. The profit would now become £15
(d) (i) Yes, it could increase to 7 pies today as this is within the allowable limits, but profit
would fall by £0.4 as the shadow price is −£0.2
(ii) Shadow price of A is £3 so net profit would be £2 per kg. We could increase A by 8.5 kg
so profits would increase by 2 × 8.5 = £17
(iii) The shadow price of A is £3 per kg so using 0.6 kg on Corn Delight the ‘loss’ in profits
would be £1.8 per pie. As B is not scarce there is no cost in using this ingredient. However
as the profit is less that £1.8 she shouldn’t make this new pie.
11.
33
Please note an error in the formulation on page 382. The demand constraints should be
greater than or equal to otherwise the optimal solution is to deliver nothing!
Xap
Supply A
Supply B
Supply C
Demand P
Demand Q
Demand R
Demand S
Solution
Xbp
5
1
Xcp
13
Xaq
7
Xbq
6
1
1
1
1
Xcq
4
Xar
8
Xbr
5
1
1
1
1
1
0
0
35
Xbs
3
1
Xcs
9
Total
1
1
1
65
0
35
1025
1
1
1
1
0
50
1
0
1
0
Avail/Req'd
4
175
35
35
60
35
50
100
1
1
0
Xas
20
1
1
60
Xcr
22
<=
<=
<=
>=
>=
>=
>=
175
80
150
60
35
50
100
12.
(a) Using the transport algorithm the optimal solution is:
Watford to London 330 boxes
Watford to NE 170 boxes
Birmingham to London 200 boxes
Birmingham to Wales 250 boxes
Birmingham to NW 450 boxes
Bristol to Wales 100 boxes
Bristol to SW 400 boxes
This allocation means that the NE were short by 330 boxes
The total cost of this allocation is £6316
(b) There is an alternative solution (as the shadow price of the cell Bristol to NW is zero. The
alternative solution is:
Watford to London 430 boxes
34
Watford to NE 70 boxes
Birmingham to London 100 boxes
Birmingham to Wales 350 boxes
Birmingham to NW 450 boxes
Bristol to NE 100 boxes
Bristol to SW 400 boxes
(c) The shadow price of using the Bristol to London route is 0.70 per box and this will be the
additional cost of using this route.
(d) The shadow cost of this route is 0.2 therefore if the cost could be reduced by this much
(that is £9.80 or less) it would be become economic to use this route.
13.
(a) Cost = £106,000. It is not optimal because there of the negative shadow price (-1) in
route Colchester to Warehouse IV
(b) The optimal solution is
Aberdeen to
I (60)
Bristol to
II (35)
III (50)
IV (65)
Colchester to IV (35)
Cost is £102,500
35
14.
(a)
Solution is to make 3.75 convection heaters and 1.5 thermal heaters giving a profit of
£35.25
(b)
Solution is same as part (a). Can meet all goals except Goal 4 (short by 3.25
convection heaters)
15.
Goal 1: L + H + d1 - d1+ = 1500
Minimise P1d1-
Goal 2: 10L + 30H + d2- - d2+ = 28800
Minimise P2d2+
Goal 3: 5L + 7H + d3- - d3+ = 10000
Minimise P3d3-
Using solver (see Excel file Chapter 14 Q15) solving each goal sequentially we get
Low
High
d-
d+
Total
Limit
Goal 1
1
1
0
0
1500
=
1500
Goal 2
10
30
0
0
28800
=
28800
Goal 3
5
7
1120
0
10000
=
10000
#Bulbs
810
690
So we produce 810 Low wattage bulbs and 690 High wattage bulbs. Both goals 1 and 2 are
met but we fall short of goal 3 by 1120p or £11.20
36
Chapter 15 Solutions
1.
(a) 16 weeks
(b) 0
(c) 2 weeks
(e) 5
37
(d) Yes by 1 week
2.
31 weeks (see below)
8
10
D
2
10
0
8
8
12
12
8
12
12
8
0
B
4
15
15
E
3
C
4
A
8
0
12
4
4
17
17
15
20
I
3
H
2
15
20
17
17
24
24
20
27
K
3
J
4
20
27
24
24
31
L
4
27
27
31
6
F
2
End
31
9
13
0
1
G
1
16
17
38
3. (a)
(b)
(i) 6 workers required during weeks 1 to 3
(ii) One solution is to start C at day 4 so ends at 11. Activity D will then start at 11 and finish at 15
4.
(a)
A
20
Start
0
0 20
F
0 20
30
B
02
C
2 12
D
12 13
10
3 13
1
13 14
2
13
I
0 16
16 34 50
K
2
02
51 53
G
12 18
6
40 46
E
13 19
6
14 20
H
18 22
4
46 50
20 50
20 50
J
3
50 53
50 53
Finish
53
Critical activities are A, F and J. Time = 53 weeks. Slacks are B, C, D, E = 1 and G, H = 28, I = 34 and K = 51 weeks. Would miss trade fair by
5 weeks
(b)
(i)
I is not on critical path so not worthwhile
(ii)
Need to consider both AFJ and BCDEFJ as both exceed 48 weeks
Find crash cost per week
Activity
Weeks saved
A
2
E
1
F
2
I
8
J
1
Extra cost
6
2
12
8
10
Unit cost
3
2
6
1
10
Crash A by 2 weeks at a total cost of 6. Then F by 2. Then J by 1. Can’t crash anymore
Path
A2
F2
J1
AFJ
53
51
49
48
BCDEFJ
49
49
47
46
Crash cost
6
12
10
Cum. Crash
6
18
28
cost
Lost profits 40
24
8
0
Total
30
26
28
Crash A by 2 and F by 2 weeks to give duration of 49 weeks. Total cost including lost profits would be £26,000
5. (a)
0
.5
.5
2.5
B
C
.5
1
0
1
3
3
3
5
A
0
9
E
3
0
5
3
D
5
6.5
11
11
H
9
9
11.5
I
11
11
11.5
11.512
8
J
G
6.5
9
F
3
1
5.5
5
11.5
11.512
(b)
Job will take 12 days and critical activities are A, E, F, H, I and J (c)
(c)
Job delayed by 0.5 days
(d)
5 people required between days 6 and 8. Not possible to have F and G at the same
time. If F starts as soon as G has finished project will be delayed for 3 days.
A0
E2
F3
H4
I1
J4
B
1
1
1
C1
11
D
2
11
G2
0
2
4
6
Days
8
10
12
6.
(a)
(a)
A
E
G
6
4
4
Start
C
D
4
2
H
5
B
F
2
3
Activity
A
B
C
D
E
F
G
H
EST
0
0
4
8
10
10
16
20
EFT
4
2
8
10
16
13
20
25
LST
0
2
4
8
10
17
16
20
(b) Project takes 25 days. Critical activities are ACDEGH
LFT
4
4
8
10
16
20
20
25
End
(c) Average duration of B = 20/6 = 3.3. As this is less than the EFT of A the project isn’t
affected
(d) Cost (in £000s) is 25 + 5x6.5 = £57.5
(e) The critical path is ACDEGH but path BCDEGH takes 23 days so need to consider this
path too. Crash A by 2 days as cheapest and common to both paths and costgs £10,000 Now
both paths critical. Next crash E by 1 day which costs £7000. Now crash A and B by 1 day
each which costs £8000. Finally crash C by 1 day costing £15000
Path
ACDEGH
BCDEGH
Crash cost
Cum. Crash
Overheads
Penalty
Total
Duration
25
23
25
32.5
57.5
A-2
23
23
10
10
23
19.5
52.5
E-1
22
22
7
17
22
13
52
A-1, B-1
21
21
8
25
21
6.5
52.5
C-1
20
20
15
40
20
0
60
Cheapest option is crash by 3 days. Crash A by 2 days, and E by 1 day.
7.
( a)
Start
0
A
05
C
58
5
16
3
69
B
06
6
06
D
69
3
69
E
07
7
29
F
9 13
4
9 13
Finish
13
Critical activities are B, D and F, project will take 13 weeks and slacks are A:1, C: 1 and E:2
(b)
Project delayed by 1 week. Activity C will now become critical instead of B and D
(c)
There are 3 paths through the network, BDF (13), ACF(12) and EF(11). Need to reduce BDF
first so crash activity B by 1 week costs £12k
Now need to crash both BDF and ACF. Choice here is between activity F (£20k) and
activity B and A (£17k) so chose B and A
This reduces time to 11 weeks at a total cost of £29k
F (£20k) is common to all 3 paths and is cheaper than crashing B,C and E. Crash F
by 2 weeks
Now have to crash B, C and E by one week at a cost of £26k
Minimum time is 8 weeks at a cost of 12 + 17 + 40 + 26 = £95k
To summarize:
Path
Time
B (1)
B(1), A(1)
F(2)
BDF
ACF
EF
Crashing
cost
Cumulative
cost
13
12
11
12
12
11
12
11
11
11
17
9
9
9
40
B(1), C(1),
E(1)
8
8
8
26
12
29
69
95
8.
(a)
C
4
A
B
D
F
3
3
4
5
E
6
Activity
A
B
C
D
E
F
(b)
EST
0
3
6
6
6
12
EFT
3
6
10
10
12
17
LST
0
3
8
8
6
12
LFT
3
6
12
12
12
17
Critical activities are A, B, E and F
Time = 17 hours
Slacks are C = 2 hours and D = 2 hours
(c)
Normal cost ¼ £1650, penalty cost ¼ £500 and overheads ¼ £1700
Profit ¼ 10000 - 1650 - 500 - 1700 ¼ £6150
(d)
D is not on critical path
Path
B1
E2
C 1, D 1, E 1
ABEF
17
16
14
13
ABDF
15
14
14
13
ABCF
15
14
14
13
Crash cost
–
£100
£300
£500
Penalty cost
£500
£400
£200
£100
Overheads
£1700
£1600
£1400
£1300
Total
£2200
£2100
£1900
£1900
So crash by 4 hours (same cost as only crashing by 3 hours but better to finish earlier)
Crash B by 1 hour C by 1 hour D by 1 hour E by 3 hours
Total cost = £1650 + £1900 = £3550
Profit = £10 000 - £3550 = £6450
9.
5
10
F
5
5
5
2
10
5
D
1
B
3
2
0
6
5
6
7
6
2
2
9
E
3
C
2
A
2
7
5
7
3
3
2
16
16
10
16
Critical path is A, B, F, I, J and K
7
7
10
Total time = 24 weeks
10
21
21
16
24
24
End
K
3
J
5
I
6
6
H
3
G
1
6
10
4
21
21
24
24
(b)
A (3)
F (0)
B (2)
I (2)
J (2)
K (2)
C(2)
G (1)
H (5)
D (1)
E (5)
0
0
1
2
3
3
5
4
9
5
7
6
6
7
8
5
9
10
11
12
0
13
14
15
16
17
2
(c)
Overload during the 4th week .
Could delay H by 1 week and E by 1 week.
Then max no. of persons required is only 7 during 5th week.
18
19
20
21
22
23
24
10
(a)
B
3
C
3.5
Start
E
1
A
2
D
3.3
G3
F
10
H
6
I
2
J
4.8
End
G
3
Activity
to
tp
tm
jJ
(J2
C
2
7
3
3.5
0.6944
D
2
6
3
3.5
0.4444
J
3
6
5
4.8
0.25
(b) Details of the EST and LST and floats are given below.
Activity
EST
EFT
LFT
LST
Float
A
0
2
2
0
0
Y
B
2
5
16.5
13.5
11.5
N
C
2
5.5
5.5
2
0
Y
D
2
5.3
5.5
2.2
0.2
N
E
5.5
6.5
6.5
5.5
0
Y
F
6.5
16.5
16.5
6.5
0
Y
G
0
3
16.5
13.5
13.5
N
H
16.5
22.5
22.5
16.5
0
Y
I
22.5
24.5
24.5
22.5
0
Y
J
24.5
29.3
29.3
24.5
0
Y
Activities A, C, E, F, H, I are critical. Expected time before the national campaign can be
launched is 29.3 days.
(c)
The variance of the critical path is 0.6944 + 0.25 = 0.9444
(Note: Activity D is not on the critical path), so the standard deviation is 0.9718
To find the probability that expected time is greater than 32 days we need to calculate the Z
value which is
𝑍=
32 − 29.3
0.9718
= 2.77
From tables this give a probability of 0.0028 So there is a 0.28% chance that the expected
time will exceed 32 days, which means that there is a probability of 100 – 0.28 = 99.72% that
the launch can take place within 32 days.
55
Critical?
Chapter 16 Solutions
1. EOQ = 63.25 Time between orders = 18.9 days
2.
For less than 600 units the stock holding cost, h = .2×£70 = £14.0
2 × 9 × 4800
𝑄=√
14
= 79
Cost of this is product cost + order cost + holding cost
=4800×70 + 9×4800/79 + 14×79/2
= £337,100
If order 600
h= .2×68 = £13.60
Cost = =4800×68 + 9×4800/600 + 13.60×600/2
= £330,552
So should order 600 units which gives a total annual cost of £330,552
3.
37 units (to nearest whole number)
4.
2 × 570 × 3000
𝑄=√
. 01
= 18,493 So order 20,000 gallons (1 tanker load) every 6.7 days on average (20,000/3000)
when stock is down to 9000 gallons (3 ×3000)
Cost = 570×3000/2000 + .01×20000/2
56
= £185.50 per day
5. (a)
There is effectively a buffer stock of 300 tins as in lead time of 1 day the paint shop will only
use 200 tins. This buffer stock costs 300 ×0.25 = £75 per day to hold in store.
The normal stock holding cost is 0.25 ×500 = £125
An order is made every 5 days so the average daily order cost is 100÷5 = £20
Total cost is therefore £75 + £125 + £20 = £220 per day
2×100×200
The EOQ amount is √
0.25
= 400 (so ordering every 2 days)
Cost is now £100 a day, a saving of £120
If only order when down to 200 will not incur a buffer stock cost.
(b) At EOQ the product cost is £2000 per day giving a total cost of £2100
If order 2000 the product cost is £1960. The order and stock cost will now cost £260 per day
so total cost is £2220 which is more expensive.
(c) If x is the product cost this must be less than £1840 since x + 260 = 2100, which means
the cost per item should be less than £9.20. This represents a discount of 8%
6.
h = .2×10 = £2
2 × 20 × 80
𝑄=√
2
= 40
Cost = 80 ×10 + 20×80/40 + 2×40/2
= £880
If order 300 h = .2×9.80 = £1.96
57
Cost = 80 ×9.8+ 20×80/300 + 1.96×300/2
= £1083
If order 500
h = .2×9.70 = £1.94
Cost = 80 ×9.7+ 20×80/500 + 1.94×500/2
= £1264
7.
(a) Demand = 50×40 = 2000 p.a.
Order cost = 64×2000/200 = £640
h = .25×£2 = £0.50
Holding cost = .5 ×200/2 = £50
Total cost = £690
(b)
2 × 64 × 200
𝑄=√
0.5
= 715.5
So order 2000/715.5 = 2.795 times a year
= 50/2.795 = 17.9 days on average
Holding cost = .5×715.5/2 = £178.88
Order cost = £178.88
Total = £357.76 which is a saving of £332.24 p.a.
(c) If order 5000 then order cost = 64 ×2000/5000 = £25.6
Product unit cost = £1.80 so h = .25×1.80 = £0.45
58
And holding cost = .45×5000/2 = £1125
Total cost including cost of buying 2000 units= 2000×1.80 + 25.6 + 1125 = £4751
Total cost at EOQ = 2000×2 +357.76 = £4357.76
(d) Let discount = d%
Unit price = 2×(1-d/100)
Cost of 2000 = 4000× (1-d/100)
h= .25×2×(1-d/100)
= .5×(1-d/100)
Holding cost = .5×(1-d/100)×5000/2
= 1250×(1-d/100)
Order cost = 25.6
Total cost = 4000× (1-d/100)+25.6 + 1250×(1-d/100)
When total cost is the same as the EOQ cost then we are indifferent between ordering the
EOQ amount or 5000
The EOQ cost = £4357.76
So 4000× (1-d/100)+25.6 + 1250×(1-d/100) = 4357.76
Solving gives d = 17.48%
So a discount of at least 17.5% would be necessary to make it worth ordering in batches of
5000
8.
2 × 100 × 50000
𝑄=√
0.25
= 6324.6 or 6325 metres
59
Holding cost = .25×6325/2 = £790.625
Order cost = 100×50000/6325 = £790.51
Total = £1581
Orders per year = 50000/6325 = 7.905 on average.
Time between orders = 50/7.905 = 6.3 weeks on average.
9.
Average usage during lead time of 7 days is 7000 m.
Standard deviation during this same period = √7 × 2502
= 661.4
(a) At 5% Z = 1.645
𝑋−7000
661.4
= 1.645
The buffer level = X – 7000 = 1088
(b) At 1% Z = 2.327
𝑋−7000
661.4
= 2.327
X- 7000 = 1539
10.
h= .1 ×10 = £1
2 × 50 × 10000
𝑄=√
1
= 1000 m
Holding cost = 1 ×1000/2
60
= £500
Order cost = 50 ×10000/1000 = £50
Product cost = 10×10000 = £100,000
Total cost = £100,550
If order 5000 then order cost = 50 ×10000/5000 = £100
Product unit cost = £9.50 so h = .10×9.50 = £0.95
And holding cost = .95×5000/2 = £2375
Product cost = 9.50×10000 = £95,000
Total cost = £97,475
So should order 5000 as this saves £3075
11.
Let discount = d%
Unit price = 10×(1-d/100)
Cost of 100,000 = 100000× (1-d/100)
h= .1×10×(1-d/100)
= (1-d/100)
Holding cost = (1-d/100)×5000/2
= 2500×(1-d/100)
Order cost = £100
Total cost = 100000× (1-d/100)+2500 (1-d/100)+100
When total cost is the same as the EOQ cost then we are indifferent between ordering the
EOQ amount or 5000
The EOQ cost = £100,550
61
So 100000× (1-d/100)+2500 (1-d/100)+100 = 100550
Solving gives d = 2%
So the discount would need to fall below 2% before it makes more sense to order the EOQ
amount.
62
Chapter 17 Solutions
1.
Random numbers are allocated as follows:
Inter-arrival time (IAT)
Mid point
Random numbers
20-<50
35
00-04
50-<100
75
05-24
100-<150
125
24-54
150-<200
175
55-99
Random number 05 is equivalent to an arrival time of 75 seconds and ends at 75+45=120
This could be set out in a table as follows
RNo
IAT
Clock
Service
Ends
starts
Waiting
time
05
75
75
75
120
0
20
75
150
150
195
0
30
125
275
275
320
0
85
175
450
450
495
0
22
75
525
525
570
0
21
75
600
600
645
0
04
35
635
645
690
10
67
175
810
810
855
0
00
35
845
855
900
10
03
35
880
900
945
20
Average waiting time = 40/10 = 4 seconds
Cash dispenser has been available for 945 seconds but has been used for only 450 (10×45)
seconds so utilisation = 450/945 = 47.6%
63
2.
Inter-arrival time (IAT)
Mid point
Random numbers
0<4
2
00-29
4<6
5
30-69
6<8
7
70-89
8<10
9
90-99
Service time
Mid point
Random numbers
1<3
2
00-49
3<5
4
50-89
5<7
6
90-99
RNo
IAT
Clock
RNo
Service
Service
time
starts
Ends
Waiting
time
04
2
2
10
2
2
4
0
59
5
7
07
2
7
9
0
38
5
12
98
6
12
18
0
01
2
14
75
4
18
22
4
48
5
19
91
6
22
28
3
04
2
21
12
2
28
30
7
Mean time queuing = 14/6 = 2.3 minutes
3.
(b)
Unloading times
Mid point
Random numbers
0-<30
15
00-19
30-<40
35
20-54
40-<50
45
55-76
64
50-<60
55
77-91
60-<70
65
92-99
Random number
Unloading time
42
35
17
15
38
35
61
45
4.
(a)

Simulation does not effect the current system so no disruption to operations

Cheaper, particularly if the planned changes do not work

Simulation will take less time than experimentation

Develop conceptual model

Build computer model

Verify that computer model is correct

Validate computer model against real data. The run length of the simulation should be
(b)
the same as the time of operation of the depot. The number of runs should be
sufficient to give reliable results (small confidence interval)

Repeat the simulation with two bays and compare the average waiting time of lorries
of both systems
(c)
Frequency distribution for the loading/unloading times
Loading
Time
Mid
(mins)
point
%
Cum %
Unloading
Random
No.
65
%
Cum %
Random
Nos.
0 to <30
15
20
20
00-19
30
30
00-29
30 to <40 35
35
55
20-54
40
70
30-69
40 to <50 45
22
77
55-76
25
95
70-94
50 to <60 55
15
92
77-91
4
99
95-98
60 to <70 65
8
100
92-99
1
100
99
Inter-arrival time distribution
Time
Mid point
%
Cum %
Random Nos.
0 to <10
5
15
15
00-14
10 to <20
15
40
55
15-54
20 to <30
25
30
85
55-84
30 to <40
35
5
90
85-89
40 to <50
45
5
95
90-94
50 to <60
55
3
98
95-97
60 to <70
65
2
100
98,99
Type of lorry
Random Nos.
Loading (L)
70%
00-69
Unloading (U)
30%
70-99
Arrivals
RN
IAT
Clock
Queue
Service
RN
Type
RN
Time
Starts
Ends
Wait
time
20
15
15
0
17
L
42
35
15
50
0
96
55
70
0
23
L
17
15
70
85
0
28
15
85
0
66
L
38
35
85
120
0
59
25
110
1
38
L
61
45
120
165
10
73
25
135
1
76
U
80
45
165
210
30
66
00
5
140
2
20
L
56
45
210
255
70
10
5
145
3
5
L
87
55
255
310
110
88
35
180
2
78
U
15
15
310
325
130
Average waiting time is 43.75 minutes. Maximum queue size is 3
67