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CPA
Foundation Leval
Quantitative Analysis November 2015
Suggested solutions

Quantitative analysis
Revision Kit

QUESTION 1(a)

Q Star Manufacturers Limited specialises in the production of two products, A and B. The manufacturer sells the products at a fixed selling price to its customers. The following table shows the requirements for production of products A and B:

Product

Material (Kilogrammes)
Labour (Hours)
A
5
3
B
7
4
Available resources
13,400
7,800

Product A is sold for Sh.2.080 per unit whereas product B is sold for Sh.7.939 per unit. The variable costs of production are uncertain with the following margins of error:

Product

Labour/Hour (Sh.)
Material/Kilogramme (Sh.)
A
140
236
B
265
710
Error
±10%
±5%

Required:
Using matrix algebra, determine:

(i) The total expected revenue.

(ii) The expected maximum profit.

(iii) The expected minimum profit.
A

Solution


(i) The total expected revenue.


A = units of A produced
B = units of B products

[
5
3
7
4
]
[
A
B
]
=
[
13,400
7,800
]


Using cramers rule

A
=
[
13,400
7,800
7
4
]

[
5
3
7
4
]


=
13,400 x 4 - 7 x 7,800

5 x 4 - 3 x 7


-1,000 / -1 = 1,000 units of A

B
=
[
5
3
13,400
7,800
]

[
5
3
7
4
]


=
5 x 7,800 - 13,400 x 3

5 x 4 - 3 x 7


-1200 / -1 = 1,200 Units of A

= 1200 units of A

The total Expected revenue

= 2,080 x 1,000 + 1,200 x 7,939 = Sh. 11,606,800

(ii) The expected maximum profit.


Maximum profit: = Total sales - Minimum Labour cost - Minimum material cost

Labour hours Materials
Product A 1,000 x 3 = 3,000 1,000 x 5 = 5,000
Product B 1,200 x 4 = 4,800 1,200 x 7 = 8,400
Total 7,800 13,400


Minimum labour cost:

Product A = 0.9 x 140 x 3,000 = Sh.378,000

Product B = 0.9 x 265 x 4,800 = Sh.1,144,800

Total Minimum labour costs = Sh.1,522,800

Minimum Material cost:

Product A = 0.95 x 236 × 5,000 = Sh.1,121,000

Product B = 0.95 x 710 x 8,400 = Sh.5,665,800

Total Minimum Material costs = Sh.6,786,800

Maximum profit:

Total sales - Minimum Labour cost - Minimum material cost

11,606,800 - 1,522,800 - 6,786,800 = Sh. 3,297,200

(iii) The expected minimum profit.


Minimum Expected profit: = Total Sales - maximum labour cost - maximum material cost

Maximum labour cost:

Product A = 1.1 x 140 x 3,000 = 462,000

Product B = 1.1 x 265 x 4,800 = 1,399,200.

Total maximum labour cost = 1,861,200

Maximum material cost:

Product A = 1.05 x 236 × 5,000 = 1,239,000

Product B = 1.05 x 710 x 8,400 = 6,262,200

Total maximum Material cost = 7,501,200

Minimum Expected profit:

= Total Sales - maximum labour cost- maximum material cost

= 11,606,800 - 1,861,200 - 7,501,200 = Sh. 2,244,400




QUESTION 1(b)

Q Apex Limited is planning to launch a new product in the market. It has undertaken a survey on the product's colour. brand name and packaging. The company sent questionnaires to 200 potential customers to obtain their views on the three attributes of the product. The results were as follows:

  • 24 persons liked the packaging and the brand name.
  • 77 persons liked the brand name or the colour but did not like the packaging.
  • 40 persons liked the colour only.
  • 120 persons liked the colour or the brand name.
  • 23 persons liked the colour and the packaging.
  • 43 persons liked at least two of the three attributes.
  • 5 persons did not like any of the three attributes.
  • The questionnaires of 25 persons were not received back.

The company's policy is to incorporate an attribute in the product if at least 50 per cent of the respondents liked the attribute.

Required:
(i) Present the above information in a venn diagram.

(ii) Number of persons that liked all the three attributes.

(iii) Proportion of persons that liked the colour.

(iv) Proportion of persons that liked the brand name.

(v) Proportion of persons that liked the packaging.

(vi) Attribute(s) to be incorporated in the product.
A

Solution


(i) Present the above information in a venn diagram.


Let A be the customers who like the products colour
Let B be the customers who like the products brand name
Let C be the customers who like the products packaging

E + F = 24

A + D + B + G + E + F = 120

A + D + B = 77

77 + 24 + G = 120

G = 19

A = 40

Venn diagram

A(Color)
B(Brand)
A=40
E=4
D=0
B=37
F=20
G=19
C=50
C(Packaging)


G + E = 23

19 + E = 23

E = 4
F = 24 - 4 = 20
D + E + F + G = 43
D + 4 + 19 + 20 = 43
D + 43 = 43

D = 0

A + B + D = 77

40 + B + 0 = 77

B = 77 - 40 = 37

C = 170 - 19 - 4 - 40 - 20 - 37 = 50

ii) Number of people who liked three attributes


E = 4

iii)Proportion of persons that liked the colour.


= (40 + 4 + 19) ÷ 175 = 0.36 or 36%

iv) Proportion of persons that liked the brand name.


=(37 + 20 + 4) ÷ 175 = 0.35 or 35%

v) Proportion of persons that liked the packaging.


(50 + 19 + 4 + 20) ÷ 175 = 0.5314 or 53.14%

vi) Attribute(s) to be incorporated in the product.


only packaging exceeds 50%.

∴Packaging.




QUESTION 2(a)

Q Explain how differential calculus could be used in solving optimisation problems.
A

Solution


Optimization with Differential Calculus


In mathematics, optimization involves finding the maximum or minimum value of a function. Differential calculus plays a crucial role in solving optimization problems.

Key Steps in Optimization Using Differential Calculus:


  1. Define the Objective Function: Formulate the problem by defining a mathematical function that represents the quantity to be optimized. Let's call it f(x) .
  2. Identify Critical Points: Find the critical points of the function by setting its derivative f'(x) equal to zero. Critical points occur where the slope of the function is horizontal (slope = 0).
  3. Apply the Second Derivative Test: Determine whether each critical point is a minimum, maximum, or neither. Use the second derivative f''(x) to analyze concavity. If f''(x) > 0 , the critical point is a local minimum. If f''(x) < 0 , it is a local maximum.
  4. Check Endpoints (if applicable): If the optimization problem is constrained to a specific interval, check the function's values at the endpoints of the interval to ensure the solution is valid within the given constraints.
  5. Interpret the Results: The values of (x) obtained from the critical points and endpoints represent the locations where the function is optimized. Substitute these values back into the original function to find the corresponding optimized quantity f(x).

By following these steps, you can use differential calculus to find the optimal solutions for various problems, such as maximizing profit or minimizing cost in business applications.





QUESTION 2(b)

Q The marginal cost and demand functions for Ujenzi Limited are given as follows:

MC = 2𝑥 + 16 (in Sh.million)

and

P = 𝑥² - 24 x 117 (in Sh.million)

Where:

MC is the marginal cost function

P is the price of a building constructed

𝑥 is the number of buildings constructed in a year.

The total annual fixed costs of the company amount to Sh.39 million.

Required:
(i) The profit function.

(ii) The selling price per building constructed that will maximise profit
A

Solution


(i) The profit function.


Total cost = 𝑥² + 16𝑥 + 39

Total revenue = PQ

𝑥(𝑥² - 24𝑥 + 117)

=𝑥³ - 24𝑥² + 117𝑥

Profit funation

Profit (n) = TR - TC

(𝑥³ - 24𝑥² + 117𝑥) - (𝑥² + 16𝑥 + 39)

= (𝑥³ - 24𝑥² - 𝑥² + 117𝑥 - 16𝑥 - 39

=𝑥³ - 25𝑥² + 101𝑥 - 39.

Δπ/Δ𝑥 = 3𝑥²- 50𝑥 + 101 = 0

Using quadratic funation


𝑥
=
-b+b² - 4ac

2a


Where;

b = - 50
a = 3
c= 101


𝑥
=
50+2,500 - (4 x 3 x 101)

6


𝑥
=
50 + 35.89

6


𝑥 = 14.32 or 2.35

ii) The selling Price that will maximise profit


P = 𝑥² - 24𝑥 + 117

= 14.32² - (24 x 14.32) + 117

= Sh. (21.62)

P = 2.35² - (24 x 2.35) + 117 =sh. 66.1225

Profit maximizing price Sh. 66.1225 million




QUESTION 2(c)

Q The data below show the number of cars imported by a certain car dealer over a four-year period:

Year
2011
2012
2013
2014
Quarter 1
20
21
23
27
Quarter 2
32
42
39
39
Quarter 3
62
75
77
92
Quarter 4
29
31
48
53

Required:
(i) The trend equation, using the least squares method.

(ii) Average seasonal index for each quarter using the multiplicative model.

(iii) Year 2015 seasonally adjusted import forecasts for each quarter.
A

Solution


(i) The trend equation, using the least squares method.


X Y XY X2 T = a + bx Y/T
1 20 20 1 30.575 0.654
2 32 64 4 32.415 0.987
3 62 186 9 34.255 1.810
4 29 116 16 36.095 0.803
5 21 105 25 37.935 0.554
6 42 252 36 39.775 1.056
7 75 525 49 41.615 1.802
8 31 248 64 43.455 0.713
9 23 207 81 45.295 0.508
10 39 390 100 47.135 0.827
11 77 847 121 48.975 1.572
12 48 576 144 50.815 0.945
13 27 351 169 52.655 0.513
14 39 546 196 54.495 0.716
15 92 1380 225 56.335 1.633
16 53 848 256 58.175 0.911
∑X =136 ∑Y=710 ∑XY=6,661 ∑X2=1,496


b
=
nΣxy - ΣxΣy

nΣx2 - (Σx)2


b
=
16 x 6,661 - 136 x 710

16 x 1,496 - 18,496


10,016 / 5,440 = 1.84

a
=
Σy

n
-
bΣx

n


a
=
710

16
-
1.84 x
136

16


44.375 - (1.84 + 1.84) = 28.735

trend equation y = 28.735 + 1.84x

(ii) Average seasonal index for each quarter using the multiplicative model.


Year 1 2 3 4
2011 0.654 0.987 1.810 0.803
2012 0.554 1.056 1.802 0.713
2013 0.508 0.827 1.572 0.945
2014 0.513 0.716 1.633 0.911
Total 2.229 3.586 6.817 3.372
Average 0.557 0.897 1.704 0.843


(iii) Year 2015 seasonally adjusted import forecasts for each quarter,


Seasonal factors

Quarter 1 = 0.557
Quarter 2 = 0.897
Quarter 3 = 1.704
Quarter 4 = 0.843

Quarter 1 = (28.735 + 1.84 x 17) x 0.557 = 33.43
Quarter 2 = (28.735 + 1.84 x 18) x 0.897 = 55.48
Quarter 3 = (28.735 + 1.84 x 19) x 1.704 = 108.54
Quarter 4 = (28.735 + 1.84 x 20) x 0.843 = 55.25



QUESTION 3(a)

Q Outline four applications of the programme evaluation and review technique (PERT) in the planning and management of projects.
A

Solution


Applications of PERT in Project Planning and Management


1. Time Estimation:


PERT is used to estimate the time required for each activity in a project. By considering optimistic, pessimistic, and most likely time estimates, PERT helps in calculating the expected time for each task.

2. Critical Path Analysis:


PERT identifies the critical path in a project, which is the sequence of activities that determines the project's minimum duration. This information is crucial for project managers to focus on tasks that directly impact project timelines.


3. Resource Allocation:


By understanding the time requirements for each activity, PERT aids in efficient resource allocation. Project managers can allocate resources optimally, avoiding bottlenecks and ensuring a smooth workflow.


4. Risk Management:


PERT incorporates uncertainty by considering a range of time estimates. This allows project managers to identify high-risk activities and plan contingency measures accordingly, reducing the impact of unforeseen events on project timelines.


5. Communication and Coordination:


PERT provides a visual representation of project activities and their dependencies through network diagrams. This helps in effective communication and coordination among team members, stakeholders, and other project participants.


6. Performance Monitoring:


During project execution, PERT allows for real-time monitoring of activities. Any deviations from the planned schedule can be identified, and corrective actions can be taken promptly to keep the project on track.


7. Optimization of Project Schedule:


PERT facilitates the optimization of project schedules by identifying non-critical activities with float or slack. Project managers can make informed decisions on how to allocate time and resources to non-critical activities without affecting the overall project timeline.





QUESTION 3(b)

Q The table below relates to the number of units packaged by nine casual employees of Bidii Limited and the packaging time taken by each of the employees:

Number of units packaged
Time (seconds)
14
230
8
110
9
130
12
190
6
109
11
181
10
154
5
79
10
144

Required:
(i) The regression line of packaging time against the number of units packaged.

(ii) The product moment correlation coefficient.

(iii) The standard error of estimate.

(iv) A 95 per cent interval estimate of the regression line

(v) The packaging time interval for 7 units.
A

Solution


(i) The regression line of packaging time against the number of units packaged.


X Y X2 XY Y2
14 230 196 3,220 52,900
8 110 64 880 12,100
9 130 81 1,170 16,900
12 190 144 2,280 36,100
6 109 36 654 11,881
11 181 121 1,991 32,761
10 154 100 1,540 23,716
5 79 25 395 6,241
10 144 100 1440 20,736
∑x=85 ∑Y=1,327 ∑x2=867 ∑xy=13,570 ∑Y2=213,335


b
=
nΣxy - ΣxΣy

nΣx2 - (Σx)2


b
=
(9 x 13,570) - (85 x 1,327)

(9 x 867) - 852


= 16.15

a
=
Σy

n
-
bΣx

n


a
=
1,327

9
-
16.15 x
85

9


147.44 - 152.53 = -5.09

Regression equation = Ye = -5.09 + 16.15x

(ii) The product moment correlation coefficient.


r
=
nΣxy - ΣxΣy

(nΣx2 - (Σx)2) x (nΣy2 - (Σy)2)


r
=
(9 x 13,570) - (85 x 1,327)

(9 x 867 - 852) x (9 x 213,335) - 1,3272)


9,335 / 9,589.15

0.9735

(iii) The standard error of estimate.


Ye = -5.09 + 16.15x

X Y Ye (y - Ye)2
14 230 221.01 80.82
8 110 124.11 199.37
9 130 140.26 105.27
12 190 188.71 1.66
6 109 91.81 295.50
11 181 172.56 71.23
10 154 156.41 5.81
5 79 75.66 11.16
10 144 156.41 154.01
∑x = 85 ∑y = 1,327 ∑(y - ye)2 = 924.55


Standard error of estimate Se

Se
=
∑(y - ye)2

n - 2


Se
=
924.55

9 - 2


= 11.492

(iv) A 95 per cent interval estimate of the regression line.


Interval estimate = y + ts.

df = n - 2

9 - 2 = 7

tcritical = t(0.05.7)= 1.895

Interval estimate = ye + 1.895 x 11.492

ye + 21.77734

(v) The packaging time interval for 7 units.


ye seven units = -5.08 + 16.15 x 7 = -5.08 + 113.05

= 107.97 seconds

Interval = 107.97 ± 1.895 x 11.492

=107.97 ± 21.77734

=86.19266 seconds to 129.74734 seconds



QUESTION 4(a)

Q Explain the following terms as used in game theory:

(i) Pure strategy.

(ii) Saddle point.
A

Solution


(i) Pure Strategy:


In game theory, a pure strategy refers to a specific and deterministic course of action that a player chooses to follow in a game. It is a strategy where the player makes a definite decision, without any randomization or uncertainty. The term "pure" distinguishes it from a mixed strategy, where a player may choose among several strategies with certain probabilities. Pure strategies are straightforward and easy to understand, representing a player's clear and unambiguous choice in a given situation.

(ii) Saddle Point:


A saddle point is a concept in game theory associated with a payoff matrix. In a two-player zero-sum game, a saddle point occurs when the maximum value in a row is the minimum value in its corresponding column. At this point, neither player has an incentive to unilaterally change their strategy because doing so would result in a worse outcome for them. The saddle point represents a stable solution where the players have reached an equilibrium. In such cases, the game is said to have a solution in pure strategies, and the saddle point is a key element in determining optimal strategies for the players.





QUESTION 4(b)

Q Highlight four applications of linear programming in business.
A

Solution


Applications of Linear Programming in Business


1. Production Planning:


Linear programming is widely used in business for optimizing production processes. It helps in determining the optimal mix of production quantities to maximize output while considering constraints such as resource availability, labor, and machine capacities.

2. Inventory Management:


Businesses use linear programming to optimize inventory levels. It aids in deciding the optimal quantity to order and when to reorder, considering factors like carrying costs, demand variability, and order processing costs.


3. Financial Portfolio Optimization:


In finance, linear programming is applied to optimize investment portfolios. It helps in allocating assets to maximize returns while adhering to constraints such as risk tolerance, budgetary limitations, and investment guidelines.


4. Marketing Resource Allocation:


Linear programming assists businesses in allocating marketing resources effectively. It helps in determining the optimal distribution of advertising budgets across various channels to maximize reach and impact.


5. Transportation and Logistics:


Linear programming is used in optimizing transportation and logistics networks. It helps in determining the most cost-effective way to transport goods from multiple sources to multiple destinations, considering factors like transportation costs and capacity constraints.


6. Personnel Scheduling:


Businesses utilize linear programming for optimizing employee schedules. It assists in determining the most efficient assignment of staff to shifts while considering factors like labor costs, employee preferences, and legal constraints.


7. Supply Chain Management:


Linear programming is applied to optimize supply chain operations. It helps in coordinating and optimizing various activities within the supply chain, including production, distribution, and inventory management.





QUESTION 4(c)

Q Quick Works Limited deals in the provision of typing services. On average, a typist at the company receives 22 letters per day for typing. The typist works for 8 hours a day and it takes an average of 20 minutes to type a letter. The company has determined that the cost of a letter waiting to be typed is Sh.8 per hour and the typing equipment operating cost plus the salary of the typist amount to Sh.400 per day. In an attempt to improve on the letter typing service, the company is planning to lease one of the two models of automated typewriters to be used together with the existing typing equipment. The additional cost per day and the increase in typist's efficiency of the two models is as given below:

Model
I
II
Additional cost per day (Sh.)
370
390
Increase in typist's efficiency (%)
50
75

Required:
Advise the company on the action that it should take in order to minimise the total daily cost.
A

Solution


Arrival rate(λ)
=
22 letters

8 hrs


λ = 2.75 letters per hour

Service rate(typing rate);(μ)
=
60 minutes x 1 letter

20 minutes


μ = 3 letters per hr

Total daily cost under the existing model:

Total service cost + total waiting cost

∴ Total service cost = Sh. 400 per day

Total waiting cost if based on the average waiting time in the system:

(Total time spent waiting by all arrivals) x Average waiting cost (Cw )

Total time spent waiting = No.of arrivals (λ) x average wait per arrival (Ws )

Ws
=
1

μ - λ


Ws
=
1

3 - 2.75


= 4 hours per arrival (letters)

Hence, Total time waiting by all arrivals (letters);

8 x 2.75 x 4 = 88 hours per day

∴ Total waiting cost = 88 hours per day x sh.8 = Sh.704 per day

Total daily cost = 400 + 704 = Sh. 1,104 per day

Total daily cost under model I of typing.

Total service cost + total waiting cost

Total service cost = 400 + 370 = Sh 770 per day

Arrival rate (λ) = 2.75 letters per hour

New service / typing rate (μ) = 150% x 3 = 4.5 letters per hour

Average waiting cost Ws
=
1

4.5 - 2.75


= 0.5714 hours per arrival

Average waiting cost Cw = Sh. 8 per day

Total waiting cost = 22 x 0.5714 x 8 = Sh. 101 per day

Therefore, Total daily cost = 770 + 101 = Sh.871 per day

Total daily cost under model II of typing:

Total service cost + total waiting cost

Therefore, Total service cost = 400 + 390 = Sh. 790 per day

Total waiting cost

λ = 275 letters per hour

New service / typing rate: μ = 3 x 175%

5.25 letters per hour

Ws
=
1

μ - λ


Ws
=
1

5.25 - 2.75


= 0.4 hours per letter

Therefore, Total waiting cost = 22 x 0.4 x 8 = Sh.70.4 per day:

Therefore, Total daily cost = 790 + 70.4 = Sh. 860.40 per day

Advice:

The company should implement the typing model II since it has the minimum daily operating cost.



QUESTION 4(d)

Q Jane Cherop was employed by Golden Houses Limited as a sales agent last year. During the year, she was able to sell up to a maximum of 6 houses in a month. Due to good performance in the past year, the company has offered Jane Cherop one of the following three salary plans for the next year:

Plan A: A 25 per cent salary increament to Sh.50,000 per month.

Plan B: A fixed monthly salary of Sh.20,000 per month plus a commission of Sh.12.000 per house sold.

Plan C: No monthly salary but a commission of Sh.20.000 per house sold.

Required:
(i) The optimal salary plan for Jane Cherop based on the maximin criterion.

(ii) The optimal salary plan for Jane Cherop based on the minimax regret criterion.

(iii) Assume that during the past year, the distribution of the houses sold by Jane Cherop for the twelve months was as follows:

Number of house sold
Number of months
0
1
1
2
2
1
3
2
4
1
5
3
6
2

Advise Jane Cherop on the optimal salary plan based on the expected value criterion.
A

Solution


(i) The optimal salary plan for Jane Cherop based on the maximin criterion.


House sold

Salary plan
0 1 2 3 4 5 6
A 50,000 50,000 50,000 50,000 50,000 50,000 50,000
B 20,000 32,000 44,000 56,000 68,000 80,000 92,000
C 0 20,000 40,000 60,000 80,000 100,000 120,000


Salary plan Minimum pay off
A 50,000
B 20,000
C 0


Best salary is plan A

(ii) The optimal salary plan for Jane Cherop based on the minimax regret criterion.


House sold

Salary plan
0 1 2 3 4 5 6 Maximum
A 0 0 0 10,000 30,000 50,000 70,000 70,000
B 30,000 18,000 6,000 40,000 12,000 20,000 28,000 30,000
C 50,000 30,000 10,000 0 0 0 0 50,000


Plan that minimizes regret Plan B

(iii) The optimal salary plan based on the expected value criterion.


Salary plan A.

(50,000 x 12) / 12= Sh.50,000

Salary plan B

(20,000 x 1) + (80,000 x 3) + 92,000 x 2

732,000 / 12 = Sh.61,000

Salary plan C

(20,000 x 2) + 40,000 + (60,000 x 2) + 80,000 +(100,000 x 3) + (120,000 x 2)

820,000 / 12 = Sh.68,333

Best plan under expected value is plan C



QUESTION 5(a)

Q A simulation model attempts to describe a business system using a number of equations. These equations are characterised by four types of variables.

Required:
With reference to the above statement, explain the four types of variables in a simulation equation.
A

Solution


Types of Variables in Simulation Equations


1. Endogenous Variables (Output Variables):
These are the results or outcomes of a simulation. They represent the system's response to the input variables and the decisions made during the simulation.

Example: In a financial simulation, the endogenous variable could be the total revenue ( R ) generated based on pricing decisions and market demand.


2. Parameters:

Parameters are input variables with constant values for a given simulation. They are fixed values that do not change during the simulation run.

Example: In a manufacturing simulation, a parameter could be the cost per unit ( C ) of raw materials, which remains constant throughout the simulation.


3. Input Variables (Exogenous Variables):

Input variables drive the simulation. They can be either controlled (under management control) or uncontrolled (not under management control).

Example: In a project management simulation, a controlled input variable could be the project team size ( T ), while an uncontrolled variable could be external market conditions affecting project timelines.


4. Status Variables:

Status variables reflect the general state of the system at various times during the simulation, capturing overall system behavior.

Example: In a healthcare simulation, a status variable might be the average patient wait time ( W ), indicating the system's efficiency in managing patient flow.





QUESTION 5(b)

Q The table below shows the probability distribution of the number of digital boxes sold by an electronics store on a daily basis:

Digital boxes sold (units)
Probability
0
0.05
1
0.05
2
0.10
3
0.15
4
0.20
5
0.15
6
0.15
7
0.10
8
0.05

Required:
(i) The probability that the number of digital boxes sold in a given day is at least 3 but less than 7.

(ii) The mean daily sales of digital boxes.

(iii) The standard deviation of digital boxes daily sales.
A

Solution


(i) The probability that the number of digital boxes sold in a given day is at least 3 but less than 7.


P(3 ≤ x ≤ 7)

= 0.15 + 0.20 + 0.15 + 0.15 = 0.65

ii) The Mean daily sales of digital boxes


(0.05 x 0) + (0.05 x 1) + (0.1 x 2) + (0.15 x 3) + (4 x 0.2) + (5 x 0.15) + (6 x 0.15) + (7 x 0.10) + (8 x 0.05)

= 0 + 0.0 5 + 0.2 + 0.45 + 0.8 + 0.75 + 0.9 + 0.7 + 0.4 = 4.25

iii) The standard deviation of digital boxes


σ = √i (xi − μ)² ⋅P(xi)



where:
xi is the number of digital boxes sold,

μ is the mean, and

P(xi) is the probability of selling

σ = √ (0 - 4.25)^2 x 0.05 + (1 - 4.25)^2 x 0.05 + (2 - 4.25)^2 x 0.1 + (3 - 4.25)^2 x 0.15 + (4 - 4.25)^2 x 0.2 + (5 - 4.25)^2 x 0.15 + (6 - 4.25)^2 x 0.15 + (7 - 4.25)^2 x 0.1 + (8 - 4.25)^2 x 0.05

=√4.1875 = 2.05




QUESTION 5(c)

Q The sales manager of Uza Limited has obtained the following data on the values of a random sample of 100 outstanding sales invoices of the company:

Value
Sh."000"

0 < 100
100 < 200
200 < 300
300 < 400
400 < 500
500 < 600
600 < 700
700 < 800
800 < 900

Number of outstanding
sales invoices

20
18
22
15
9
8
4
2
2
100

Required:
(i) The standard deviation of the random sample.

(ii) A 95 per cent confidence level of the mean value of outstanding sales invoices.
A

Solution


(i) The standard deviation of the random sample.


Midpoint x f fx fx²
50 20 1,000 2,500 50,000
150 18 2,700 22,500 405,000
250 22 5,500 62,500 1,375,000
350 15 5250 122,500 1,837,500
450 9 4,050 202,500 1,822,500
550 8 4,400 302,500 2,420,000
650 4 2,600 422,500 1,690,000
750 2 1,500 562,500 1,125,000
850 2 1,700 722,500 1,445,000
∑x = 4,050 ∑f = 100 ∑fx = 28,700 ∑x² = 2,422,500 ∑fx² = 12,170,000


∑fx²

∑f
-
[
∑fx

∑f
]
²


12,170,000

100
-
[
28,700

100
]
²


121,700 - 287²

198.32 x 1000 = 198,320

ii) a 95% confidence level of the mean value of outstanding sales invoices


Mean = Σfx / Σf = 28,700 / 100 = 287 x 1,000 = 287,000

Confidence interval ( Xe )
=
±
Z
(
s

√n
)

where:

  • X̄ is the sample mean,
  • s is the sample standard deviation,
  • n is the sample size,
  • Z is the Z-score corresponding to the desired confidence level.

For a 95% confidence level, the Z-score is typically 1.96.

Xe
=
287,000
±
1.96
(
198,320

√100
)

287,000 + 1.96 x 19,832

287,000 + 38,870.72

287,000 + 38,870.72 = 325,870.72

287,000 - 38,870.72 = 248,129.28

We are 95% confident that the true population mean lies between 248,129.28 and 325,870.72.




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