Guaranteed

95.5% Pass Rate

CPA
Foundation Leval
Quantitative Analysis November 2019
Suggested solutions

Quantitative analysis
Revision Kit

QUESTION 1a

Q In the context of the above statement, describe what "business analytics" entails.
A

Solution


Description of 'Business analytics'


Business Analytics - refers to the skills, applications, and practices of continuously exploring data to gain insights that drive business decisions.

Business analysis can help an organization keep up with competition and maintain prospects for future growth in several ways, such as:

(i) Help organizations anticipate and more proactively forecast outcomes and trade-offs.

(ii) Enables businesses to create an innovation-receptive environment to design strategies

(iii)Helps organizations gain a competitive advantage by providing models for decision making purposes such as simulations, decision trees, etc..

(iv) Modern computation applications such as tax analysis e.t.c can help businesses recover taxes on overpaid transactions and prevent future overpayments.

(v) Enables businesses to protect their data with digital watermarking, malware detection measures, and more.

(vi) Prescriptive analytics can also be enhanced by developing optimization techniques for superior business performance.

More info

QUESTION 1b(i-ii)

Q (i) A binomial distribution.

(ii) A Poisson distribution.
A

Solution


A binomial distribution.


P = probability of being absent
q = probability of not being absent
r = total number of “successes” ( heads or tails etc.)
n = number of consoltant workers
c = Tip: You can use the combinations calculator to figure out the value for nCr

Pr = nCr x Prqn-r

n = 6, Q = 1 - 0.2 = 0.8

P(r ≥ 3) = 1 - [P(r = 0) + P(r = 1) + P(r = 2)]

P(r = 0) = 6C0 x (0.2)0(0.8)6-0 = 0.2621

P(r = 1) = 6C1 x (0.2)¹ (0.8)6-1

= 6C₁ x (0.2)¹(0.8)5 = 0.3932

P(r=2)= 6C2x (0.2)2(0.8)4 = 0.2458

Thus, P(r ≥ 3) = 1 - (0.2621 + 0.3932 + 0.2458)

=1 - 0.9011

= 0.0989

Therefore, there is a chance or probability of 0.989 that more than two consultants will be absent from work on a given day.

A Poisson distribution.


P(r)
=
e-λ λn
n!


r is a random variable following a Poisson distribution
n is the number of times an event occurs
P(X = n) is the probability that an event will occur k times
e is Euler’s constant (approximately 2.718)
λ (lambda) is the average number of times an event occurs
! is the factorial function

P(r ≥3) = 1 - [P(r = 0) + P(r = 1) + P(r = 2)]

e = 2.7183

λ = 20% x 6 = 1.2

P(r = 0)
=
2.7183-1.2 + 1.2o
0!


= 0.3012

P(r = 1)
=
2.7183-1.2 + 1.21
1!


= 0.3614

P(r = 2)
=
2.7183-1.2 + 1.22
2!


= 0.2169

∴P(r ≥) = 1 - (0.3013 + 0.3614 + 0.2169)

=1 - 0.8795 = 0.1205

Therefore, there is a chance or probability of 0.1205 that more than two consultants will be absent from mark on a given day.

More info

QUESTION 1c

Q The required gross output of each sector in order to meet the new demand.
A

Solution


Production sector Gross output(units)
x1 (500+800+200) 1,500
x2 (600+1,400+400) 2,400


Matrix of input-output coefficient(A)



∴
I - A =
-



(I-A)-1
=
1
|(I-A)|
[Adjoint(I-A)]


∴ |(I-A)| = (4/5 x 5/12) - (-2/5 x -1/3)

1/5

∴
(1-A)-1 = 5/1



X
=
[
X1
X2
]


D
=
[
d1
d2
]


Where;

X= Gross output production matrix

D-Final demand matrix

D
=
[
400
800
]


X
=
[
400
800
]


=
[
2,166.67
4,000
]


X1 = 2,166.67 units
X2 = 4,000 units

Hence, the gross output of each sector required to meet the new demand is
X₁ = 2.166.67 units and X₂ = 4000 units


More info

QUESTION 2a(i)

Q Cumulative profit in the break-even time interval.
A

Solution


Profit = 10x - x² - 5
At breakeven point,
Profit = 0
Thus,
10x - x² - 5 = 0

-x² + 10x = 5

x² - 10x = -5

(x - 5)² = -5 + (5)²

= -5 + 25 = 20

x - 5 = √20

X = 5 + 4.4721

= (5 - 4.4721) or (5 + 4.4721)

= 0.5279 months or 9.4721 months

Therefore, the cumulative profit between 0.5279 months and 9.4721 months is,

Profit (000) = [10(9.4721) - (9.4721)² - 5] - [10(0.5279) - (0.5279)²-5]

= 0.00032159 - 0.00032159

More info

QUESTION 2a(ii - iii)

Q (ii) The best time to end the production.

(iii) The total profit based on your result in (a) (ii) above.
A

Solution


(ii) The best time to end the production.


The best time to end the production is when the profit reaches the optimum point

At maximum profit:

Marginal profit (MP) = 0

MP= dtp/dx

MP = 10 - 2x

10 - 2x = 0

-2x = -10

x = 5 months

Therefore, the production must be stopped at month 5, because at this point the profit has reached its maximum point.


iii) The total profit based on the result in (a) (ii) above


Profit(000) = 10(5) - (5)² - 5 = 20

Thus, profit Sh. 20,000

More info

QUESTION 2b

Q Test at 95% level of confidence whether the company should accept the new plant.
A

Solution


Since the sample size "n" is very small and the population standard deviation (8) is unknown, a suitable test statistic is the t-test.

Where:

t
=
X̄ - μ
S / √n


μ= 18.5, X̄ = 19.7, n = 25, S = 1.5

∴
t
=
19.7 - 19.5
1.5 / √25
= 0.2 / 0.3 = 0.667(3dp)


Formulating the hypothesis;

HO: u 19.5 That is, the mean reflectometer is less than or equal to 19.5
HI: u > 19.5 that is, the mean reflectometer is greater than 19.5 greater than 19.5

α = 1 - 0.95 = 0.05

The H1, formulated above, implies that the test is one failed.

Thus, tα,n-1 = t 0.05,25-1

t0.05,24 = 1.711



The observed to = 0.667 is less then the critical tc = 1.711, and therefore lies in the acceptance region for the null hypothesis (HO). We accept the H0 and conclude that the sample evidence suggests that the mean reflectometer is less than 19.5. the company should therefore reject the new plant.

More info

QUESTION 2c

Q Explain two advantages and two disadvantages of decision trees as used in decision theory.
A

Solution


Advantages of Decision Trees


  • Simplicity: Decision trees are easy to understand and interpret, making them a valuable tool for both experts and non-experts in decision-making processes.
  • Monetary simulation: Allows monetary values ​​to be incorporated into decision trees to help identify and isolate the best alternative course of action.
  • Visualization: Decision trees provide a visual representation of decision-making processes, which aids in the communication and comprehension of complex decision scenarios.
  • Handling Non-Linearity: Decision trees can model complex, non-linear relationships in the data, which may be challenging for other algorithms that assume linearity.
  • No Assumption of Data Distribution: Decision trees do not require assumptions about the distribution of data, making them suitable for various types of datasets.
  • Doesn't require scaling of data

Disadvantages of Decision Trees


  • Overfitting: Decision trees are prone to overfitting, especially when they are deep and capture noise in the training data. This can lead to poor generalization on new, unseen data.
  • Instability: Decision trees are sensitive to variations in the training data, and small changes can result in different tree structures. This instability can impact the reliability of the model.
  • Data Partitioning Bias: Decision trees may create biased splits in the data, particularly when certain classes are dominant, leading to imbalanced branches and potential performance issues.
  • Difficulty in Capturing XOR-like Relationships: Decision trees struggle to capture exclusive OR (XOR) relationships between features, as they rely on simple splits based on threshold values.
  • Complexity: Extensive knowledge and deeper understanding of decision theory is required.
  • Scaling: As the amount of data increases, building decision trees becomes more complex
  • Time consuming
More info

QUESTION 2d

Q (i) The average number of customers in the system.

(ii) The average time spent by a customer in the system.

(iii) The average number of customers in the queue.

(iv) The utilisation factor of the service utility.
A

Solution


(i) The average number of customers in the system.


L s
=
λ
μ - λ


λ = 2 customers per hour
μ = 3 customers per hour

L s
=
2
3 - 2


2customer

(ii) The average time spent by a customer in the system.


Ws
=
1
μ - λ


Ws
=
1
3 - 2


1 hour per customer

(iii) The average number of customers in the queue.


Lq
=
λ2
μ(μ - λ)


∴
Lq
=
(2)2
3(3 - 2)


1.3333

(iv) The utilisation factor of the service utility.


e = λ / μ

2 / 3

More info

QUESTION 3a

Q In the context of time series analysis, describe three differences between "additive" and "multiplicative" models.
A

Solution


Additive model Mltiplicative model
1 Represents the components of time series i.e. seasonality(s) trends(t) cyclical(c) and residual (r) as The sum: Y = S + T + C + R Represents the components as the product of time series i.e Y = S x T x C x R
2 Changes in the time series data is usually expressed interms that are absolute The changes in data are expressed in percentages
3 Is used where seasonal component(s) which over time tends to be constant. Used where seasonality component (s) tends to change over time.
4 Considers the time series components as being independent of each other. Treats the components as being interdependent of each other.


More info

QUESTION 3b(i)

Q Illustrate, using a graph, the number of website hits against the amount spent in website promotion.

Comment on any relationship between website hits and the extent of promotion.
A

Solution


Website promotion (Ksh."000")
Website Hits(No.of visitors)


It can be seen from the graph that there is no clear relationship between click-through rate and the degree of promotion on direct mortgage websites.


More info

QUESTION 3b(ii)

Q Calculate the correlation coefficient and give an interpretation to its value.
A

Solution


r2
=
nΣxy - ΣxΣy
√(nΣx2 -(Σx)2) x (nΣy2 - (Σx)2)


X Y
"000"		 XY X2 Y2
25 1 25 625 1
24 1.2 28.8 576 1.44
56 1.6 89.6 3,136 2.56
54 1.4 75.6 2,916 1.96
55 1.2 66 3,025 1.44
58 1.8 104.4 3,364 3.24
∑X = 272 ∑Y = 8.2 ∑XY = 389.4 ∑X2 = 13,642 ∑Y2 = 11.64


n=6

r
=
6(389.4) - 272(8.2)
√[6(13,642) - 272] x [6(11.64) - 8.22]


r
=
106
√20,456.8


r = 0.7411

The value of "T" above is positive and therefore indicating a strong positive correlation between the website hits and the extent of promotion.

More info

QUESTION 3b(iii)

Q Determine the regression line.
A

Solution


y = a + bx

a
=
∑y
n
-
b∑x
n


b
=
n∑xy
n∑x2
-
∑x∑y
( ∑x )2


b
=
6(389.4) - 272(8.2)
6(13642) - (272) 2


106/7868 = 0.0135

a = 8.2/6 - (0.0135(272))/6

= 1.3667 - 0.612 = 0.7547

Therefore regression line y = 0.7547 + 0.0135x

More info

QUESTION 4a(i-ii)

Q (i) Explain the meaning of a "transition matrix'

(ii) Outline two features of a transition matrix.
A

Solution


(i) Meaning of a "transition matrix"


It is a matrix of probability that an element in a markov chain, will move from one state to another.

ii) The features of a transition matrix


➢The sum of the elements of each row of the transition matrix is ​​1
➢All transition matrices are random i.e. probability
➢ The elements of the matrix are either positive or null, that is to say not negative.

More info

QUESTION 4b

Q Calculate the output matrix K.
A

Solution


K = T²P-1

where;

T - is an input matrix

P-1 - is an inverse of matrix P.

K - is an output matrix.

Given that T
=
[
24
20
]
&
P =
[
24
20
]


T²
=
[
24
20
]
x
[
24
20
]


[
4+88+0
4+08+0
]


T²
=
[
128
48
]


P
=
[
17
04
]


P-1
=
1
|P|
(Adjoint P)


|P| = (1 x 4) - (0 x 7) = 4 - 0 = 4

P-1
=
1
4
[
4-7
01
]


K
=
1
4
[(
128
48
)
(
+4-7
0+1
)
]


=
1
4
[
48+0-84+8
16+0-28+8
]


=
1
4
[
48-76
16-20
]


K
=
[
1219
4-5
]

More info

QUESTION 4c(i-ii))

Q (i) Construct a probability transition matrix of the switching probabilities.

(ii) Construct a vector to represent the initial market share in percentages.
A

Solution


(i) Construct a probability transition matrix of the switching probabilities.


0
M
N
S
[
M
0.6
0.4
0.1
N
0.2
0.5
0.1
S
0.2
0.1
0.8
]


Where;

M = Meta, N = Nzuri, S = Safi


(ii) Construct a vector to represent the initial market share in percentages.


M N S
(40 40 20)

More info

QUESTION 4c(iii)

Q Calculate a new market share a week after the current market share.
A

Solution


(40 40 20)
[
0.6
0.4
0.1
0.2
0.5
0.1
0.2
0.1
0.8
]


(24+16+2 8+20+2 8+4+16) = (42 30 28)

Therefore, the market share will be 42%, 30% and 28% for meta, Nzuri and safi respectively, a week after the current market share.

More info

QUESTION 4d(i)

Q Predicting business outcome.
A

Solution


Predicting business outcome.


They help managers in evaluating scenarios for forecasting purposes.

Helps organizations implement strategies to improve and enhance employee performance to achieve business goals

Enables organizations to prepare employees for complex business environments through business process modeling.

More info

QUESTION 4d(ii)

Q Managing business risks.
A

Solution


Managing business risks


The simulation itself is used as a tool to assess and analyze the uncertainty and risk surrounding the business

The simulation model helps managers generate random numbers for effective scenario analysis of risks in the business.

They are used to identify contingencies and facilitate effective cost management.

More info

QUESTION 5a(i-iv)

Q Advise the businessman on the best alternative under:

(i) Maximin criterion.

(ii) Maximax criterion.

(iii) The Hurwitz criterion, assuming a degree of optimism of 0.6.

(iv) Laplace criterion.
A

Solution


(i) Maximin criterion.


Alternative

Minimum payoff
Sh."million"
A -12
B -6
C 0 (maximin)


Therefore, the business man should select alternative c resulting in a payoff of sh. 0 in scenario.

(ii) Maximax criterion.


Alternative

Maximum payoff
Sh."million"
A 6
B 16 (maximum)
C 12


Therefore, the businessman should choose alternative B resulting in the maximum pay off Sh. 16 million, under this criterion.

(iii) The Hurwitz criterion, assuming a degree of optimism of 0.6.


It is based on weighted payoff where:
Weighted payoff = α (maximum payoff) + 1-d (minimum payoff)

1 - α = 1 - 0.6 = 0.4

Alternative Minimum payoff
Sh."million"
A 0.6(6) + 0.4(-12) - 1.2
B 0.6(16) + 0.4(-6) 7.2 (maximum)
C 0.6(12) + 0.4(0) 7.2 (maximum)


Therefore, the business man should select either alternative B or C, resultin in the maximum weighted payoff of shs. 7.2 million each.

(iv) Laplace criterion.


It is based on the average payoff

Where;

Average payoff
=
∑(payoffs) for each alternative
Number of states of nature


Alternative

Maximum payoff
Sh."million"
A (6 + 2 - 2 - 12 + 4) ÷ 5 - 0.4
B (-6 + (-3) + 10 + 16 + 0) ÷ 5 3.4
C (12 + 8 + 4 + 0 + 6) ÷ 5 6 (maximum)


Therefore, the businessman should select alternative C, resulting in the maximum average payoff Sh. 6 million

More info

QUESTION 5b(i-ii)

Q (i). Formulate and solve the above Linear programming model graphically.

(ii). Determine the maximum selling price of Product Y at which the solution in (b) (i) above would still remain optimal.
A

Solution


(i). Formulate and solve the above Linear programming model graphically.


Let x and y represent the number of units of products x and y produced and sold respectively.

Objective:

To determine the number of units of each product x and y to be produced and sold in order to maximize profit

Objective function (z)
Zmax = 24x + 32y

Subject to constraints:

1. Material B:

(80/20)x +(80/20)y ≤ 240

x + y ≤ 60

2. Direct labour hours

(64/16)x + (32/16)y ≤ 100

2x + y ≤ 100

3Packaging time

(24/24)x + (48/24)y ≤ 100

x + 2y ≤ 100

4. Non-negativy constraints

x ≥ 0
y ≥ 0

Hence the optimal output is

x = 20 units of product x
y = 40 units of product y

∴ Zmax = 24(20) + 32(40)

1,760

Product Y


ii) Determining the maximum selling price of product y at which the solution in b, i) above would still remain optimal


Let maximum selling price of product y be represented by p"

:. Total sales = total cost + total profit.

Optimal output production of "y" = 40 units as obtained under (b) (i)above

Total cost = 40 (300 + 28)

= 40 x 328 = Sh 13,120

Total profit 40 x 32 = Sh 1,280

∴ Total sales of y = 13,120 + 1,280 = Sh. 14,400

More info

QUESTION 5c(i-ii)

Q (i) The total cost function using the least squares method.

(ii) If each bread is sold at Sh.50, predict the break-even number of units of bread.
A

Solution


(i) The total cost function using the least squares method.


y = a + bx

X Y XY X2
45 46 2,070 2,025
42 43.2 1,814.4 1,764
55 46.6 2,563 3,025
43 48 2,064 1,849
60 56.4 3,384 3,600
40 44.8 1,792 1,600
48 46.2 2,217.6 2,304
53 50.6 2,681.8 2,809
36.6 40.2 1,471.32 1,339.56
34 33 1,122 1,156
∑x=456.6 ∑y=455 ∑xy=21,180 ∑x2=21,471.56


b
=
n∑xy
n∑x2
-
∑x∑y
( ∑x )2


b
=
10(21,180.12) - (456.6 x 455)
10(21,471.56) - (456.6) 2


4,048.2 / 6,232.04

∴ b = 0.6496

a
=
∑y
n
-
b∑x
n


a
=
455
10
-
0.6496(456.6)
10


45.5 - 29.6607 = 15.8393

∴ y = 15.8393 + 0.6496x


(ii) If each bread is sold at Sh.50, predict the break-even number of units of bread.


Break even units = Total fixed costs/Contribution per unit

∴X(in hundreds) = (15,839.3 x 1,000) / (50 - 0.6496)

15,839.3 / 49.3504 = 320.96

Hence the break even number of units of bread

320.96 x 100 = 32,096 units

More info

Comments on CPA past papers with answers:

New Unlock your potential with focused revision and soar towards success
Pass Kasneb Certification Exams Easily

Comments on:

CPA past papers with answers