Quantitative Methods — July 2020

KNEC Diploma in Sales and Marketing — Module II. Paper code: 2902/204

Independent worked revision answers. These are not official KNEC marking schemes. Every printed question is covered for revision, including all subparts. Retain unrounded values until the final step.

The scan omits the examination heading. The sitting follows the original source filename recorded in the supplied archive index. A duplicated investment-row label in Question 7(b) is explained beside the answer.

Question 1(a)

Given the following matrices A, B and C:

A =

Column 1 Column 2
20 10
5 8
10 15

B =

Column 1 Column 2 Column 3
5 10 20
8 5 10
4 10 3

C =

Column 1 Column 2
10 5
20 10
8 10

Calculate:

  1. (i) B × C
  2. (ii) A + C
  3. (iii) C − A

(8 marks)

Answer

(i) B × C

B is 3 × 3 and C is 3 × 2, so BC is 3 × 2. Multiply each row of B by each column of C.

Entry Working Result
Row 1, column 1 5 × 10 + 10 × 20 + 20 × 8 410
Row 1, column 2 5 × 5 + 10 × 10 + 20 × 10 325
Row 2, column 1 8 × 10 + 5 × 20 + 10 × 8 260
Row 2, column 2 8 × 5 + 5 × 10 + 10 × 10 190
Row 3, column 1 4 × 10 + 10 × 20 + 3 × 8 264
Row 3, column 2 4 × 5 + 10 × 10 + 3 × 10 150

BC =

Column 1 Column 2
410 325
260 190
264 150
(ii) A + C

Add corresponding entries, for example 20 + 10 = 30 and 8 + 10 = 18.

Column 1 Column 2
30 15
25 18
18 25
(iii) C − A

Subtract the corresponding entry of A from C, for example 10 − 20 = −10 and 10 − 15 = −5.

Column 1 Column 2
-10 -5
15 2
-2 -5

Question 1(b)

The following information shows the advertising costs and the sales revenue of Telek Limited over a certain period.

Advertising costs (x), Ksh ‘000’ Sales revenue (y), Ksh ‘000’
21 136
28 158
22 145
30 162
26 149
25 150
  1. (i) Determine the regression equation y on x.
  2. (ii) Using the equation in (i) above, estimate the sales revenue when the advertising cost is 30,000.

(12 marks)

Answer

(i) Regression of sales revenue on advertising costs
x: advertising costs (Ksh ‘000’) y: sales revenue (Ksh ‘000’) x² xy
21 136 441 2,856
28 158 784 4,424
22 145 484 3,190
30 162 900 4,860
26 149 676 3,874
25 150 625 3,750
152 900 3,910 22,954
Step Working
Number of observations n = 6
Slope b [nΣxy − (Σx)(Σy)] / [nΣx² − (Σx)²]
Substitute [6(22,954) − 152(900)] / [6(3,910) − (152)²] = 2.595506
Intercept a ȳ − bx̄ = 150 − (2.595506)(25.333333) = 84.247191
Regression equation ŷ = 84.247191 + 2.595506x
(ii) Estimate at Ksh 30,000
Step Working
Convert advertising cost to table units x = 30,000/1,000 = 30
Substitute ŷ = 84.247191 + 2.595506(30) = 162.11236 thousand shillings
Estimated sales revenue Ksh 162,112.36

Question 2(a)

The following information relates to scores awarded to 655 sales persons based on their performance.

Scores Number of sales persons
0–5 29
5–10 195
10–15 241
15–20 117
20–25 52
25–30 10
30–35 6
35–40 3
40–45 2

Calculate the:

  1. (i) median score;
  2. (ii) lower quartile;
  3. (iii) upper quartile;
  4. (iv) 90th percentile.

(12 marks)

Answer

Scores Frequency f Cumulative frequency
0–5 29 29
5–10 195 224
10–15 241 465
15–20 117 582
20–25 52 634
25–30 10 644
30–35 6 650
35–40 3 653
40–45 2 655

N = 655 and class width h = 5. The adjoining intervals are treated as continuous: a class 5–10 means 5 ≤ score < 10. No 0.5 adjustment is applied.

Grouped quantile = L + [(target position − cumulative frequency before the class)/class frequency] × h.

Measure Target position Working Estimated score
(i) Median 327.5 10 + [(327.5 − 224)/241] × 5 12.1473
(ii) Q₁ 163.75 5 + [(163.75 − 29)/195] × 5 8.4551
(iii) Q₃ 491.25 15 + [(491.25 − 465)/117] × 5 16.1218
(iv) P₉₀ 589.5 20 + [(589.5 − 582)/52] × 5 20.7212

Question 2(b)

A market survey of 250 residents of Maputo town revealed that they read the following newspapers:

  1. 130 read the National Newspaper;
  2. 70 read the Sunlight Newspaper;
  3. 140 read the Times Newspaper;
  4. 80 read Times and National Newspapers;
  5. 30 read National and Sunlight Newspapers;
  6. 30 read Sunlight and Times Newspapers;
  7. 20 read all the three newspapers.
  1. (i) Present the information above in a Venn diagram.
  2. (ii) Calculate the probability that a resident chosen at random reads only one of the newspapers.

(8 marks)

Answer

(i) Venn diagram
Region Working Residents
All three Given 20
National and Sunlight only 30 − 20 10
National and Times only 80 − 20 60
Sunlight and Times only 30 − 20 10
National only 130 − 10 − 60 − 20 40
Sunlight only 70 − 10 − 10 − 20 30
Times only 140 − 60 − 10 − 20 50
At least one 40 + 30 + 50 + 10 + 60 + 10 + 20 220
None 250 − 220 30
Venn diagram for 250 residents: National only 40, Sunlight only 30, Times only 50; pair-only regions 10, 60 and 10; all three 20; none 30.
Venn diagram for 250 residents: National only 40, Sunlight only 30, Times only 50; pair-only regions 10, 60 and 10; all three 20; none 30.
(ii) Reads only one newspaper
Step Working
Exactly one 40 + 30 + 50 = 120 residents
Probability 120/250 = 12/25 = 0.48

Question 3(a)

The following are the prices and quantities of three commodities in the years 2010 and 2017.

Commodity 2010: Price (Ksh) 2010: Quantity (kg) 2017: Price (Ksh) 2017: Quantity (kg)
Rice 60 1 170 2
Sugar 140 2 200 2
Maize flour 240 4 450 5

(i) Calculate the:

  1. (I) Paasche’s quantity index;
  2. (II) Laspeyre’s quantity index;
  3. (III) Fisher’s ideal quantity index.

(ii) Interpret the result in (III) above.

(12 marks)

Answer

Take 2010 as the base year: p₀, q₀ are base prices and quantities; p₁, q₁ are 2017 prices and quantities.

Commodity p₀q₀ p₀q₁ p₁q₀ p₁q₁
Rice 60 120 170 340
Sugar 280 280 400 400
Maize flour 960 1200 1800 2250
Total 1300 1600 2370 2990
Index Formula and substitution Result (2010 = 100)
(i)(I) Paasche quantity index QP = (Σp₁q₁/Σp₁q₀) × 100 = (2,990/2,370) × 100 126.1603
(i)(II) Laspeyres quantity index QL = (Σp₀q₁/Σp₀q₀) × 100 = (1,600/1,300) × 100 123.0769
(i)(III) Fisher ideal quantity index QF = √(QL × QP) 124.6091

(ii) Fisher’s index is 124.6091. The weighted aggregate quantity in 2017 was approximately 24.61% higher than in 2010.


Question 3(b)

The following data shows the total annual wages paid to employees of Mulika County for a period of ten consecutive years.

Year Annual wages (Ksh millions)
2008 148
2009 156
2010 160
2011 169
2012 176
2013 190
2014 202
2015 208
2016 216
2017 228
  1. (i) Using the least squares method, determine the trend line equation.
  2. (ii) Using the equation in (i) above, estimate the annual wages in the year 2020.

(8 marks)

Answer

(i) Least-squares trend

Code time as t = year − 2012.5 so that Σt = 0. In the regression working, x represents this time code t.

t = year − 2012.5 Y: wages (Ksh millions) x² xy
-4.5 148 20.25 -666
-3.5 156 12.25 -546
-2.5 160 6.25 -400
-1.5 169 2.25 -253.5
-0.5 176 0.25 -88
0.5 190 0.25 95
1.5 202 2.25 303
2.5 208 6.25 520
3.5 216 12.25 756
4.5 228 20.25 1,026
0 1,853 82.5 746.5
Step Working
Number of observations n = 10
Slope b [nΣxy − (Σx)(Σy)] / [nΣx² − (Σx)²]
Substitute [10(746.5) − 0(1,853)] / [10(82.5) − (0)²] = 9.048485
Intercept a ȳ − bx̄ = 185.3 − (9.048485)(0) = 185.3
Regression equation ŷ = 185.3 + 9.048485x

Trend equation: Ŷ = 185.3 + 9.048485t, where t = year − 2012.5 and Ŷ is in Ksh millions.

(ii) Forecast for 2020
Step Working
Time code t = 2020 − 2012.5 = 7.5
Forecast Ŷ = 185.3 + 9.048485(7.5) = 253.163636 million
Annual wages Approximately Ksh 253,163,636.36

Question 4(a)

The following information relates to a project that is to be undertaken by Malip Limited.

Activity Preceeding activity Duration (days)
A — 6
B A 9
C A 15
D B 24
E B 18
F C 3
G C 6
H C, D 9
I E, F 21
J G, H 12
K I, J 15
  1. (i) Draw a network diagram of the project.
  2. (ii) Determine the: (I) critical path; (II) project duration.

(12 marks)

Answer

(i) Network diagram
Activity-on-node network. Gold activities A, B, D, H, J and K form the critical path; activity durations are in days.
Activity-on-node network. Gold activities A, B, D, H, J and K form the critical path; activity durations are in days.

Every arrow represents a stated precedence requirement. H waits for both C and D; I waits for E and F; J waits for G and H; K waits for I and J.

(ii) Critical path and duration

Forward pass: earliest start is the latest finish of the predecessors; earliest finish = earliest start + duration. Backward pass: latest finish is the earliest latest start of the successors; latest start = latest finish − duration. Total float = latest start − earliest start.

Activity Duration Earliest start Earliest finish Latest start Latest finish Total float
A 6 0 6 0 6 0
B 9 6 15 6 15 0
C 15 6 21 21 36 15
D 24 15 39 15 39 0
E 18 15 33 21 39 6
F 3 21 24 36 39 15
G 6 21 27 42 48 21
H 9 39 48 39 48 0
I 21 33 54 39 60 6
J 12 48 60 48 60 0
K 15 60 75 60 75 0
Result Answer
(I) Critical path A → B → D → H → J → K
(II) Project duration 6 + 9 + 24 + 9 + 12 + 15 = 75 days

Question 4(b)

Ms Jane plans to purchase a photocopying machine in 3 years time. The machine is expected to cost Ksh 250,000. In order to purchase the machine, Jane intends to deposit Ksh 100,000 in a savings account at the beginning of the first year and Ksh 60,000 at the beginning of the second year. The account pays compound at the rate of 10% per annum.

Calculate the amount that Jane should deposit into the savings account at the beginning of the third year in order to purchase the machine. (8 marks)

Answer

Use the purchase date, the end of year 3, as the common valuation date. The three deposits earn interest for 3, 2 and 1 years respectively.

Deposit Years earning interest Value at purchase date (Ksh)
Beginning of year 1 3 100,000(1.10)³ = 133,100
Beginning of year 2 2 60,000(1.10)² = 72,600
Beginning of year 3 1 X(1.10)
Step Working
Future-value equation 133,100 + 72,600 + 1.10X = 250,000
Amount still required at purchase 1.10X = 250,000 − 205,700 = 44,300
Third deposit X = 44,300/1.10 = Ksh 40,272.73

Question 5(a)

The following is a distribution of marks awarded to 210 employees of Moffat Limited based on their performance.

Marks Number of Employees
5–10 10
10–15 20
15–20 27
20–25 36
25–30 52
30–35 42
35–40 18
40–45 5

Calculate the:

  1. (i) standard deviation;
  2. (ii) interquartile range;
  3. (iii) mean deviation.

(12 marks)

Answer

Use class midpoints x and N = 210. These are grouped-data estimates. Use the population denominator N for the described distribution.

Class / value f x fx fx² f|x − mean|
5–10 10 7.5 75 562.5 176.9048
10–15 20 12.5 250 3,125 253.8095
15–20 27 17.5 472.5 8,268.75 207.6429
20–25 36 22.5 810 18,225 96.8571
25–30 52 27.5 1,430 39,325 120.0952
30–35 42 32.5 1,365 44,362.5 307
35–40 18 37.5 675 25,312.5 221.5714
40–45 5 42.5 212.5 9,031.25 86.5476
Total 210 — 5,290 148,212.5 1,470.4286
(i) Standard deviation
Step Working
Mean x̄ = Σfx/N = 5,290/210 = 25.190476
Variance σ² = Σfx²/N − x̄² = 148,212.5/210 − (5,290/210)² = 71.213719
Standard deviation σ = √71.213719 = 8.4388 marks
(ii) Interquartile range
Measure Working Result
Q₁ 15 + [(52.5 − 30)/27] × 5 19.1667
Q₃ 30 + [(157.5 − 145)/42] × 5 31.4881
IQR Q₃ − Q₁ 12.3214 marks

For Q₁ the cumulative frequency before 15–20 is 30; for Q₃ it is 145 before 30–35. The class width is 5.

(iii) Mean deviation about the arithmetic mean
Formula Substitution Answer
MD = Σf|x − x̄|/N 1,470.428571/210 7.002 marks

Question 5(b)

Determine the derivatives of each of the following functions:

  1. (i) y = 5x² − 2x + 4
  2. (ii) y = 12 − 10x + 6x² − 2x³
  3. (iii) y = 5x⁴ − 2x³ + 5x² + 10x − 12
  4. (iv) y = 3x² − 2x + 4

(8 marks)

Answer

Apply the power rule d(axⁿ)/dx = anxⁿ⁻¹. A constant differentiates to zero.

Part Differentiate term by term dy/dx
(i) 10x − 2 + 0 10x − 2
(ii) 0 − 10 + 12x − 6x² −10 + 12x − 6x²
(iii) 20x³ − 6x² + 10x + 10 − 0 20x³ − 6x² + 10x + 10
(iv) 6x − 2 + 0 6x − 2

Question 6(a)

Explain five roles of quantitative techniques in business decision making. (10 marks)

Answer

  1. Forecasting: use historical sales, demand or cost data to estimate future outcomes and support planning.
  2. Resource allocation: determine how limited labour, funds, materials or machine time can be assigned to competing activities.
  3. Comparing investments and alternatives: calculate expected returns, present values or decision payoffs using consistent criteria.
  4. Operational control: use inventory, scheduling, quality-control and queueing models to reduce delays, shortages and waste.
  5. Managing uncertainty: quantify probabilities and risks, test scenarios and assess how decisions respond to changing assumptions.

Question 6(b)

The following data shows the output (in tonnes) of Pandia Limited for a period of 30 working days.

1 2 3 4 5 6
04 28 32 38 05 10
05 10 16 22 20 18
07 12 20 31 30 20
40 27 03 45 41 42
45 08 12 18 48 09
  1. (i) Construct a frequency distribution table starting with the classes 0–10, 10–20, …
  2. (ii) Using the table in (i) above, plot a cumulative frequency curve.

(10 marks)

Answer

(i) Frequency distribution

Use non-overlapping classes: 0–10 means 0 ≤ output < 10; 10–20 means 10 ≤ output < 20, and so on. A value exactly 10 belongs in 10–20.

Output (tonnes) Frequency Less-than cumulative frequency
0–10 7 7
10–20 7 14
20–30 6 20
30–40 4 24
40–50 6 30

The frequencies total 30 working days.

(ii) Less-than cumulative frequency curve

Plot the upper class limits against cumulative frequencies, including the starting point (0, 0): (10, 7), (20, 14), (30, 20), (40, 24), (50, 30). Join consecutive points to form the ogive.

Less-than ogive of output for 30 working days, plotted at upper class limits 10, 20, 30, 40 and 50 tonnes with cumulative frequencies 7, 14, 20, 24 and 30.
Less-than ogive of output for 30 working days, plotted at upper class limits 10, 20, 30, 40 and 50 tonnes with cumulative frequencies 7, 14, 20, 24 and 30.

Question 7(a)

Explain the meaning of each of the following terms as used in sampling theory:

  1. (i) sampling design;
  2. (ii) sampling frame;
  3. (iii) sampling bias;
  4. (iv) statistic.

(8 marks)

Answer

Term Meaning
(i) Sampling design The plan specifying the target population, sampling units and method used to select a sample. It includes how selection will be carried out and the intended sample size.
(ii) Sampling frame The list or operational representation of population units from which a sample is actually selected, such as a current register of all eligible employees.
(iii) Sampling bias A systematic distortion caused by the sampling or selection process, so that some population members are inadequately represented. Increasing the sample size alone does not remove it.
(iv) Statistic A numerical summary calculated from sample observations, such as a sample mean or proportion. It may be used to estimate a population parameter.

Question 7(b)

A company has three alternative investments, X, Y and Z. The payoffs from each of those investments will be determined by four possible events; A, B, C and D.

The following table shows the payoffs from each of the three alternative investments.

Alternative A (Ksh) B (Ksh) C (Ksh) D (Ksh)
X 10,000 2,000 -8,000 8,000
Y -2,000 16,000 20,000 0
X 16,000 8,000 3,000 10,000

Advise the management on the investment to choose based on:

  1. (i) maximin criterion;
  2. (ii) maximax criterion;
  3. (iii) minimax regret criterion.

(12 marks)

Answer

Source note: the question names X, Y and Z, but the printed table repeats X on its third row. In this answer, Z denotes that third row, following the alternatives named in the question. No payoff is changed.

(i) Maximin criterion

Find the minimum payoff for each investment, then choose the greatest minimum.

Investment Minimum payoff (Ksh)
X -8,000
Y -2,000
Z 3,000

Choose Z: its minimum payoff is Ksh 3,000, the highest of the three minimums.

(ii) Maximax criterion

Find the maximum payoff for each investment, then choose the greatest maximum.

Investment Maximum payoff (Ksh)
X 10,000
Y 20,000
Z 16,000

Choose Y: it offers the highest possible payoff, Ksh 20,000.

(iii) Minimax regret criterion

For each event, regret = the best payoff available under that event minus the investment’s payoff. The best event payoffs are A: 16,000; B: 16,000; C: 20,000; D: 10,000.

Investment Regret A Regret B Regret C Regret D Maximum regret (Ksh)
X 6,000 14,000 28,000 2,000 28,000
Y 18,000 0 0 10,000 18,000
Z 0 8,000 17,000 0 17,000

Choose Z: its maximum regret of Ksh 17,000 is smaller than Y’s Ksh 18,000 and X’s Ksh 28,000. These rules use different attitudes to uncertainty; no event probabilities are supplied.