l3 bs 2001 s2 (m)

LCCI Examinations Board MH N T354 9 TNM >f2[EW2r@o2`8r>u3]@o2r2[8r#

Examiner’s Report and

Model Answers for

Business Statistics

THIRD LEVEL Series 2 (Code 3609) 2001 Malaysia

1

Business Statistics Third Level (Malaysia) Series 2 2001

How to use this booklet

Examiners’ Reports and Model Answers have been developed by LCCIEB to offer additional information and guidance to Centres, teachers and candidates as they prepare for LCCIEB examinations. The contents of this booklet are divided into 5 elements: (1) General Comments – assessment of overall candidate performance in this examination,

providing general guidance where it applies across the examination as a whole

(2) Questions – reproduced from the printed examination paper (3) Model Answers – summary of the main points that the Chief Examiner expected to

see in the answers to each question in the examination paper (4) Examiner’s Report – constructive analysis of candidate error, areas of weakness and

other comments that apply to each question in the examination paper

(5) Helpful Hints – where appropriate, additional guidance relating to individual

questions or to examination technique Teachers and candidates should find this booklet an invaluable teaching tool and an aid to success. The London Chamber of Commerce and Industry Examinations Board provides Model Answers to help candidates gain a general understanding of the standard required. The Board accepts that candidates may offer other answers that could be equally valid.

Note LCCIEB reserves the right not to produce an Examiner’s Report, either for an examination paper as a whole or for individual questions, if too few candidates were involved to make an Examiner’s Report meaningful.

© LCCI CET 2001 All rights reserved; no part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without prior written permission of the Publisher. The book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover, other than that in which it is published, without the prior consent of the Publisher. Typeset, printed and bound by the London Chamber of Commerce and Industry Examinations Board.

3

Business Statistics Third Level (Malaysia) Series 2 2001 GENERAL COMMENTS The paper followed the usual format and presented a fair examination of candidates’ knowledge of Third Level Business Statistics. Most candidates were well prepared and so scored high marks. As ever, there were a small number of candidates who had clearly not prepared to the Third Level standard. Candidates who rely on Second Level knowledge only will not pass this Third Level paper.

5

Business Statistics Third Level (Malaysia) Series 2 2001 QUESTION 1 A company which specialises in selling electronic products advertises its goods on a monthly basis in a series of specialist magazines. The company’s head of marketing believes that monthly sales are linked to monthly advertising expenditure. The table below contains information relating to monthly sales and monthly advertising expenditure over a twelve month period:

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Sales (RM 000)

120 110 130 120 140 160 160 150 140 180 190 200

Advertising Expenditure (RM 000)

6 6 7 5 7 8 8 8 8 9 10 10

(a) Produce a scatter graph which represents the relationship between monthly sales and monthly

advertising expenditure and comment on the relationship. (4 marks) (b) Calculate the product moment correlation coefficient between monthly sales and monthly

advertising expenditure. (10 marks) (c) Carry out a suitable test to ascertain whether the correlation coefficient is significantly different

from zero. (6 marks)

(Total 20 marks)

CONTINUED ON NEXT PAGE 6

Model Answer to Question 1 (a) Comment: Positive or linear (b) Month Sales Advertising (RM 000) Expenditure (RM 000) x y xy x

2 y

2

J 6 120 720 36 14,400 F 6 110 660 36 12,100 M 7 130 910 49 16,900 A 5 120 600 25 14,400 M 7 140 980 49 19,600 J 8 160 1,280 64 25,600 J 8 160 1,280 64 25,600 A 8 150 1,200 64 22,500 S 8 140 1,120 64 19,600 O 9 180 1,620 81 32,400 N 10 190 1,900 100 36,100 D 10 200 2,000 100 40,000 92 1,800 14,270 732 279,200

7

Model Answer to Question 1 continued Correlation Coefficient

r =

∑∑∑∑

∑∑∑

))y(–yn())x(–xn(

yx–xyn

2222

r = ( ))000,240,3(–)200,279(12()464,8(–)732(12(

)800,1()92(–)270,14(12

r = ( ))400,110()320(

640,5

r = 0.9489

(c) H0: ρ = 0, H1 ρ ≠ 0

Critical t value (10 df, 5% and 1%) = 2.23 and 3.17

t = 2r–1

2–nr

t = 9004.0–1

2–129489.0

t = 3156.0

0007.3 = 9.508

The calculated t value is greater than both critical values. There is strong evidence to reject H0. The correlation coefficient is significantly different from

zero. Examiner’s Report on Question 1 A relatively straightforward question on correlation which most candidates attempted and scored well on. Some marks were lost due to poor presentation of graphs. HELPFUL HINTS

Ensure that you identify which variable is x and which is y. This will help you to set up your graph axes correctly.

8

QUESTION 2 A study of sickness absence amongst hospital nursing staff was carried out. A random sample of 200 nurses was taken and the number of days of sickness absence per year was recorded. The grade of each nurse was also recorded.

Number of days of sickness absence per year

Grade of nurse Less than 5 5 to 15 More than 15

Student 10 30 10

Newly qualified 10 40 50

Experienced 5 25 20

(a) Carry out an appropriate statistical test to investigate whether there is a relationship between the

grade of a nurse and the number of days of sickness absence per year. (12 marks) A previous study of sickness absence amongst nurses showed that the percentage of all nurses

taking less than 5 days absence per year was 15, the percentage taking between 5 and 15 days absence was 50 and the percentage taking more than 15 days absence was 35.

(b) Combining the grades of nurses from part (a), test whether there has been a change in the

distribution of days of sickness absence per year. (8 marks)

(Total 20 marks)

9

Model Answer to Question 2 (a)

10 30 10 50

10 40 50 100

5 25 20 50

25 95 80 200

6.25 23.75 20.00 50.00

12.50 47.50 40.00 100.00

6.25 23.75 20.00 50.00

25.00 95.00 80.00 200.00

Contribution to χ 2

2.25 1.644737 5

0.5 1.184211 2.5 0.25 0.065789 0 χ 2

= 13.3947

Null Hypothesis: There is no relationship between the grade of nurse and the number of

days of sickness absence pa Alternative Hypothesis: There is a relationship between the grade of nurse and the number of

days of sickness absence pa Significance Level 5% or 1% Degrees of Freedom (3 – 1) (3 – 1) = 4 Critical χ 2

Value 9.49 or 13.28

Calculated χ 2

Value = 13.3947

The calculated χ 2

value is greater than both critical values, there is strong evidence to reject the

null hypothesis. There does appear to be a strong relationship between the grade of nurse and the number of days of sickness absence taken each year. (b) Observed 25 95 80 Expected (% of 200) 30 100 70 Contribution to χ 2

0.8333 0.25 1.4286

Calculated χ 2 Value = 2.5119

Null Hypothesis: No change in distribution of days of sickness absence Alternative Hypothesis: There is a change Significance Level 5% or 1% Degrees of Freedom n – 1 = 3 – 1 = 2 Critical χ 2

Value 5.99 or 9.21

The calculated χ 2

value is less than both critical values, there is no evidence to reject the null

hypothesis. There does not appear to be any change in the distribution of days of sickness absence since the last study.

Observed

Expected

10

Examiner’s Report on Question 2 Most candidates scored well on part (a) of this chi squared question. Many seemed confused as to what was required in part (b). This is a common second part to a chi squared question and candidates need to be prepared for it. HELPFUL HINTS � Where the null hypothesis is rejected at 5%, candidates should go on to test at 1%. � Look back at past examination papers to see the different part (b) elements of a chi squared

question.

11

QUESTION 3 (a) Explain the difference between a Laspeyres price index and a Paasche price index. Include in

your answer, an advantage and disadvantage of using one of the methods. (4 marks) A manufacturing company regularly holds five items in stock. The company measures material price inflation by constructing an all items Paasche price index. The table below contains the latest data regarding the quantity and total cost of materials purchased. Similar information is provided in respect of 1999.

Material Quantity Purchased 1999 (Kg)

Total Cost 1999 (RM)

Quantity Purchased 2000 (Kg)

Total Cost 2000 (RM)

A 500 3,500 650 5,200

B 300 4,200 300 4,200

C 100 1,750 120 2,400

D 200 4,500 250 6,250

E 400 4,800 350 4,900

(b) Calculate an all material items Paasche price index for 2000 based upon 1999 and comment on

the result. (8 marks) A small retail outlet buys and sells a product on a daily basis. Daily demand for the product varies according to the following probability distribution.

Daily demand 100 200 250 300

Probability 0.1 0.6 0.2 0.1

(c) Calculate the expected daily demand for the product. (3 marks) The product costs RM 2 per unit and sells for RM 4.50 per unit. Any items left unsold at the end of the day are resold back to the supplier for RM 1.25. (d) Assuming that the retailer regularly buys in 250 units per day, calculate the expected daily profit. (5 marks)

(Total 20 marks)


Model Answer to Question 3 (a) Difference due to origin of weights Laspeyres base period weights Paasche current period weights Advantages Laspeyres – less time consuming and less costly – comparisons can be made between periods Paasche – more up to date Disadvantages Laspeyres – may be out of date – overstates price inflation Paasche – more time consuming and costly – understates price inflation (b) p1q1 values are given in question (Total cost in 2000) p0 values need to be calculated using the total cost and quantities given for 1999

Material Unit Price Quantity

p0 q1 p1q1 p0q1

A

B

C

D

E

7

14

17.5

22.5

12

650

300

120

250

350

5,200

4,200

2,400

6,250

4,900

22,950

4,550

4,200

2,100

5,625

4,200

20,675

(8,205,500) (6,550,000)

Paasche Price Index = 10

11

qp

qp

∑

∑ x 100

10

11

qp

qp

∑

∑ x 100 =

675,20

950,22 x 100 = 111.0 (125.3)

Comments: There has been an 11% increase in overall material costs. (c) Expected value = ∑ (x.p(x))

Expected daily demand = (100 x 0.1) + (200 x 0.6) + (250 x 0.2) + (300 x 0.1) = 10 + 120 + 50 + 30 = 210 units

13

Model Answer to Question 3 continued (d)

Daily Demand Prob Conditional Profit (RM)

Expected Profit (RM)

100

200

250

300

0.1

0.6

0.2

0.1

137.50

462.50

625.00

625.00

13.75

277.50

125.00

62.50

478.75

Examiner’s Report on Question 3 The explanation in part (a) was done well by many but very poorly by some. A lack of knowledge was responsible for the poor explanation in most cases. However, in some cases answers were jumbled up and it became difficult to interpret what was actually being said. Candidates need to express themselves more clearly. A table headed Laspeyres and Paasche might help some candidates to focus their thoughts. The usual mistake was made of not separating out the price and quantities from the total cost figures given. This resulted in a substantial loss of marks in part (b). Parts (c) and (d) involved the use of expectation. Whilst most candidates scored well on part (c), almost all failed to answer part (d) correctly. The main problem appeared to be that candidates ignored the uncertain demand when calculating the expected daily profit. HELPFUL HINTS The calculation of a Laspeyres or Paasche Index is likely to involve initial separation of prices and quantities (see past examinations).


QUESTION 4 The table below shows the amount spent by a company on energy over the past three and a half years:

Energy Costs Per Quarter (RM 00s)

Year Quarter 1 Jan - Mar

Quarter 2 Apr - June

Quarter 3 July - Sept

Quarter 4 Oct - Dec

1997 56 70

1998 60 30 50 68

1999 58 32 48 66

2000 56 30 48 64

(a) Plot a graph of the time series data and comment upon the movement within the data. (4 marks) (b) Using a suitable model, calculate a four-quarter moving average trend and the quarterly seasonal

variations. (12 marks) (c) Using the results from your analysis in part (b), forecast energy costs for the first two quarters of

2001. (4 marks)

(Total 20 marks) Model Answer to Question 4 (a) Comment: Gentle downward trend/seasonal pattern present


Model Answer to Question 4 continued (b) Additive model should be used. Year/Quarter Profit Add in fours Add in twos Trend S. V S. V Y – T Y/T 1997 1 2 3 56 4 70 216 1998 1 60 426 53.250 6.750 1.127 210 2 30 418 52.250 –22.250 0.574 208 3 50 414 51.750 –1.750 0.966 206 4 68 414 51.750 16.250 1.314 208 1999 1 58 414 51.750 6.250 1.121 206 2 32 410 51.250 –19.250 0.624 204 3 48 406 50.750 –2.750 0.946 202 4 66 402 50.250 15.750 1.313 200 2000 1 56 400 50.000 6.000 1.120 200 2 30 398 49.750 –19.750 0.603 198 3 48 4 64 Additive Quarter Year 1 2 3 4 1998 6.750 –22.250 –1.750 16.250 1999 6.250 –19.250 –2.750 15.750 2000 6.000 –19.750 total 19.000 –61.250 –4.500 32.000 ASV 6.333 –20.417 –2.250 16.000 –0.333

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Model Answer to Question 4 continued Multiplicative Quarter Year 1 2 3 4 1998 1.127 0.574 0.966 1.314 1999 1.121 0.624 0.946 1.313 2000 1.120 0.603 total 3.368 1.801 1.912 2.627 ASV 1.123 0.600 0.956 1.314 3.993 (c) Predicted Trend. Use average change in trend.

9

25.53–75.49 = –0.389

Forecast = predicted trend adjusted by average seasonal variation Additive 2001 1 49.75 + 3(–0.389) + 6.333 = 54.916 (RM5491.6) 2 49.75 + 4(–0.389) + (–20.417) = 27.777 (RM2777.7) Multiplicative 2001 1 49.75 + 3(–0.389) x 1.123 = 54.559 (RM5455.9) 2 49.75 + 4(–0.389) x 0.6 = 28.916 (RM2891.6) Examiner’s Report on Question 4 A time series question on which most candidates scored well. The omission of data for the first two quarters caused some confusion with a number of candidates assuming that these figures were both zero. This is clearly wrong. Many failed to offer any comment regarding the movement of data. HELPFUL HINTS � Remember to fully label graphs and diagrams � When stating answers, include appropriate units

17

QUESTION 5 A company has recently introduced a revised remuneration scheme for its manual workers. The management believe that the revised scheme will increase productivity. A random sample of nine employees was taken just prior to the introduction of the revised scheme and their productivity was measured. A second random sample of nine different employees was taken after the introduction of the new scheme and again their productivity was measured. Both sets of results are shown in the following table:

Productivity (Output per week in units)

Old Scheme Revised Scheme

48

48

52

46

40

50

54

58

52

49

50

54

44

40

52

52

56

54

(a) Briefly explain how you would decide whether to use a paired comparison t test or an

independent t test. (4 marks) (b) Carry out an appropriate test to ascertain whether the revised scheme has indeed improved

productivity. (12 marks) (c) Calculate a 99% confidence interval for the mean output per week per employee under the

revised remuneration schemes. (4 marks)

(Total 20 marks)


Model Answer to Question 5 (a) Paired comparison t test indication of pairing (before and after) eg Before and after training Independent t test no evidence of pairing samples independent eg Production from two different factories (b) Workings (b) and (c) (independent t test – no pairing) Old New

x y x2 y

2 2)x–x( 2)y–y(

48 49 2,304 2,401 3.160494 1.234568 48 50 2,304 2,500 3.160494 0.012346 52 54 2,704 2,916 4.938272 15.12346 46 44 2,116 1,936 14.2716 37.34568 40 40 1,600 1,600 95.60494 102.2346 50 52 2,500 2,704 0.049383 3.567901 54 52 2,916 2,704 17.82716 3.567901 58 56 3,364 3,136 67.60494 34.67901 52 54 2,704 2,916 4.938272 15.12346 448 451 22,512 22,813 211.5556 212.8889 old

mean 9

448x = = 49.78

new

mean 9

451y = = 50.11

Pooled st dev s = 2–mn

)y–y()x–x( 22

+

∑+∑

s = 2–99

8889.2125556.211

+

+

s = 5278.26

s = 5278.26 s = 5.15

19

Model Answer to Question 5 continued

Null yx µ=µ

Alt yx µ<µ

Sig level 5% 0r 1% degrees of freedom (n + m – 2) = 9 + 9 – 2 = 16 Critical value t = –2.12 or –2.92

Calculated value t =

m

1

n

1s

y–x

+

t =

+

9

1

9

115.5

11.50–78.49

t = 4277.2

33.0– = –0.1359 alt (-0.1373)

The calculated t value is greater than both critical values, there is no evidence to reject the null

hypothesis. The revised scheme does not appear to have improved productivity.

(c) mean 9

451y = = 50.11

St dev sd = 1–n

)y–y( 2∑ sd =

8

8889.212 sd = 5.16

Confidence Interval n

sty ±

50.11 ± 3.36 9

16.5

50.11 +/ –5.7776 44.33 units to 55.89 units Examiner’s Report on Question 5 This question was not well done. Most candidates were unable to offer a valid explanation in part (a) and many chose the wrong test to apply in part (b). At this level candidates are expected to be able to identify the appropriate test to use for the information given. This type of question is very likely to appear on future papers. HELPFUL HINTS If two samples of individuals are used it is most likely that an independent t test is appropriate.

20

QUESTION 6 Company X employs three salespersons A, B and C. These salespersons visit any new customer contacts that the company receives. The number of visits made to new customer contacts and the number of new orders received in the last year, are shown in the table below. (The new customer contacts are allocated to the salespersons at random and salespersons do not visit other salesperson’s contacts).

Salesperson Visits Orders

A

B

C

400

200

400

35

25

60

(a) Find the probability that: (i) a new customer contact is visited by salesperson B (ii) an order is received following an initial visit by salesperson A (iii) salesperson C made the initial visit given that no new order was received by any

salesperson. (9 marks) The best selling product of Company X is a general high protein animal feed (HPAF) which is sold in bags with a stated weight of 25 kg. Because of the variability that exists in packaging the product, the packaging process is set to fill each bag with on average 26.5 kg of the animal feed. (You may assume that the packaging process follows a normal distribution with a standard deviation of 1.25 kg). Industry quality regulations require that no more than one in 15 bags contain less than the stated weight. (b) Advise the company as to whether they are conforming to industry quality regulations in respect

of the packaging of this product. (6 marks) The company has decided to buy a new packaging machine which will reduce the variability relating to the packaging process. The new packaging process continues to follow a normal distribution but now with a standard deviation of 400 g. (c) How much feed on average should each bag now be filled with in order to satisfy the industry

quality regulations? (5 marks)

(Total 20 marks)


Model Answer to Question 6 (a) Tabulate data

A B C Total

35 25 60 120

365 175 340 880

Order

No Order

400 200 400 1,000

(i) p(B) = 000,1

200 = 0.2

(ii) p(order | A) = 400

35 = 0.0875

(iii) p(C | no order) = 880

340 = 0.3864

Tree diagram order or no order 0.4 x 0.0875 = 0.0350 0.4 x 0.9125 = 0.3650 0.2 x 0.125 = 0.0250 0.2 x 0.875 = 0.1750 0.4 x 0.15 = 0.0600 0.4 x 0.85 = 0.3400 1.0000

(i) p(B) = 000,1

200 = 0.2

(ii) p(order | A) = 400

35 = 0.0875

4.

0875.x4.

(iii) p(Cno order) = 88.0

34.0 = 0.3864

++ 3400.1750.3650.

85.x4.

(0.0875)

(0.9125)

(0.125)

(0.875)

(0.15)

(0.85)

A(0.4)

B(0.2)

C(0.4)

o

no

o

no

o

no

22

Model Answer to Question 6 continued (b) Find probability that a bag contains less than 25 kg.

Z = 25.1

5.26–25–x=

σ

µ = –1.2

Z = –1.2 area is 0.885 area of interest = 1 – 0.885 = 0.115 ie p(x < 25 kg) = 0.115 Since this is greater than 1 in 15 (0.067) so it appears that the packaging process does not

conform to industry quality regulations. (c) Need to find the new mean value. Industry quality regulations - p(x < 25 kg) = 0.067 area of interest = 0.067 area of tables = 0.933 Using tables Z = –1.5

Z = σ

µ–x –1.5 =

4.0

–25 µ

µ = 25 + (1.5 x 0.4)

µ = 25.6 kg

Examiner’s Report on Question 6 Very few candidates attempted this question. Those who did, scored badly. Part (a) involved the application of some basic probability rules, and part (b) required the use of the normal distribution. These areas are fundamental to this paper and candidates must work to ensure that they understand and can apply them. By ignoring these areas candidates are reducing their chances of success as they reduce the choice to 5 from 7 questions. HELPFUL HINTS Work through the question!

23

QUESTION 7 (a) State four uses of the normal distribution (4 marks)

A building contractor has been working for 25 days and is part way through a building project. In order to complete the project he has to start and complete three independent tasks A, B and C which have to be done in the order ABC. The times taken to complete these tasks follow normal distributions as detailed below:

Task Mean (days) Standard Deviation (days)

A 14 2

B 7 3

C 10 1

(b) Find the probability that the time taken to complete task A is:

(i) less than 12 days

(ii) greater than 17 days

(iii) between 12 and 17 days. (6 marks)

(c) Find the probability of completing tasks A and B in less than a total of 16 days. (5 marks)

If the project takes longer than 60 days the contractor will incur a penalty payment.

(d) Find the probability that the contractor will have to pay the penalty payment. (5 marks)

(Total 20 marks)

24

Model Answer to Question 7 (a) Finding probabilities associated with normally distributed variables Estimation - confidence intervals Hypothesis testing Statistical process control - quality control charts

(b) (i) Z = σ

µ–x Z =

2

14–12 = –1

area = 0.841 area of interest = 1 – 0.841 = 0.159 p(x < 12 days) = 0.159

(ii) Z = 2

14–17 = 1.5

area = 0.933 area of interest = 1 – 0.933 = 0.067 p(x > 17 days) = 0.067 (iii) p(12 < x < 17) = 1 – (0.159 + 0.067) = 0.774 (c) mean = 14 + 7 = 21

st dev = 22 32 + = 3.61

61.3

21–16 = –1.39 (say –1.4)

area = 0.919 area of interest = 1 – 0.919 = 0.081 p(y < 16 days) = 0.081 (d) mean = 14 + 7 + 10 = 31 days

st dev = 222 132 ++ = 3.74

If remaining 3 tasks take longer than 35 days then the penalty will need to be paid.

74.3

31–35 = 1.07 (say 1.1)

area = 0.864 area of interest = 1 – 0.864 = 0.136 p(paying penalty payment) = 0.136

25

Examiner’s Report on Question 7 Another question on the normal distribution. This time many more candidates attempted the question and most scored well. There was some confusion shown in dealing with part (d) with many candidates forgetting that 25 days had already elapsed. A number of candidates confused the term ‘uses’ with the term ‘features’ in answering part (a).

26

QUESTION 8 A random sample of 50 retail outlets reported their weekly sales figures for the year 2000 for a particular product. The data has been set out in the following table:

Weekly Demand (000s units)

Number of outlets

0 and less than 5

5 and less than 10

10 and less than 15

15 and less than 20

20 and less than 25

8

11

14

10

7

(a) Using the data above, calculate the mean, the variance, the median and a measure of skewness. Comment upon what the measure of skewness indicates.

(12 marks)

(b) In year 1999 the proportion of all retail outlets reporting weekly sales of 10,000 units or more was 0.57. Test whether there has been a significant increase in the proportion of outlets reporting weekly sales in year 2000, of 10,000 units or more.

(8 marks)

(Total 20 marks)

27

Model Answer to Question 8 (a) x f fx fx

2

2.5 8 20.0 50.00 7.5 11 82.5 618.75 12.5 14 175.0 2,187.50 17.5 10 175.0 3,062.50 22.5 7 157.5 3,543.75 50 610.0 9,462.50

mean = 50

610

f

fx

∑

∑ = 12.2 (12,200 units)

variance var = 22

f

fx–

f

fx

∑

∑

∑

∑

var = 2)122(–50

625,94

var = 189.25 – 148.84 var = 40.41 (40,410)

median = lm +

1–m

m

m F–2

n

f

c

= 10 +

19–

2

50

14

5

= 12.143 (12,143 units)

Coefficient of = 357.6

)143.12–2.12(3

sd

)Med–x(3= = 0.027

comment: positive skewness position of mean and median (b) Proportion of weeks where demand was 10,000 or more = 31/50 = 0.62 null hypothesis Π = 0.57 alt hypothesis Π > 0.57 sig level 5% and 1% critical values 1.64 and 2.33

calc value Z =

n

)–1(

–p

ππ

π Z =

50

)57.0–1(57.0

57.0–62.0

Z = 0.7141 The calculated Z value is less than both critical values. There is no evidence to reject the null hypothesis. There does not appear to be any difference in

the proportion of outlets when weekly sales were 10,000 units or more.

Skewness

28

Examiner’s Report on Question 8 A relatively simple question which was generally well done. Many candidates stated the standard deviation as a measure rather than the variance, and most failed to include units in their answers. In answering part (b), some candidates attempted to test the mean value or included a combination of mean and proportion in their test. HELPFUL HINTS � Do not forget to include comments where asked for � Set out your answer in a structured way

l3 bs 2001 s2 (m)

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