Online Exam - Desklib

Get solved assignments, essays, dissertations and more for your online exam preparation at Desklib. Find study material for various subjects and courses.

ONLINE EXAM

TABLE OF CONTENTS
PART A...........................................................................................................................................3
QUESTION 1..................................................................................................................................3
I....................................................................................................................................................3
II...................................................................................................................................................4
III.................................................................................................................................................4
IV.................................................................................................................................................5
V..................................................................................................................................................5
VI.................................................................................................................................................6
PART B...........................................................................................................................................7
QUESTION 3..................................................................................................................................7
I....................................................................................................................................................7
II...................................................................................................................................................7
III.................................................................................................................................................8
IV.................................................................................................................................................8
QUESTION 4..................................................................................................................................8
Mean............................................................................................................................................9
Mode............................................................................................................................................9
Median.........................................................................................................................................9
Range.........................................................................................................................................10
Standard deviation.....................................................................................................................10

PART A
QUESTION 1
I.
a.
300/ (82 + 62) (3*2 + 2)
300/ (8*8 + 6*6) (3*2 + 2)
300/ (64 + 36) (6 + 2)
300/ (100) (8)
300/ (800)
0.375
b.
(400 / 202) (625 / 125)
(400/ 20 * 20) (5)
(20/ 20) (5)
(1) (5)
5
c.
810 / 9 (5*5-5) – 200
90 (5*5-5) – 200
90 (25 – 5) – 200
90 (20) – 200
1800 -200
1600
d.
2* [502 / (2*54)]
2*[50*50/ (2*5*5*5*5)]
2* [10*10/ 2*5*5]
2* [1*10/ 5]
2* [2]
4
e.
3

[144/ (10+8)] / 2+10-2+ [(52*2)2]
[144/ 18]/ 12-2+ [(5*5*2)2]
[8]/ 10 + [(50)2]
0.8+ 100
100.8
II.
a.
(-44) * (-4)
176
b.
33 – (-3)
33 + 3
36
c.
(-72) / 12 – (-5)
-6 / 12 + 5
-0.5 + 5
4.5
III.
a.
3/8 + 1/3
(9+8) /24
17/24
0.708
b.
5/8 – 1/5
(25 – 8)/40
17/ 40
0.425
c.
5(1/2) + 3(1/4)
4

11/2 + 13/4
(44 + 26)/ 8
70/ 8
8.75
IV.
a.
Normal tea bags= 47.8% of (52628)
47.8% * 52628
25156 (approx)
Green tea bags = total tea bags – normal tea bags
52628 – 25156
27472 (approx) are green tea bags
b.
Students based in England = 63.8% of 15635
63.8% * 15635
9975 (approx)
Students not based in England = Total students- Students based in England
15635 – 9975
5660 students (approx) are not based in England
V.
a.
A:B:C:D
8:3:5:4
Investment of A = 8/ (8+3+5+4) * 100000
= 8/20 * 100000
= 8* 5000
= £40000
Investment of B = 3/ (8+3+5+4) * 100000
= 3/20 * 100000
= 3* 5000
= £15000
5

Investment of C = 5/ (8+3+5+4) * 100000
= 5/20 * 100000
= 5* 5000
= £25000
Investment of D = 4/ (8+3+5+4) * 100000
= 4/20 * 100000
= 4* 5000
= £20000
b.
Let x be the total investment amount
X:Y:Z
5:6:7
Investment by Y = £36000
6/ (5+6+7)*x = 36000
6/ 18 *x = 36000
x/3 = 36000
x = £108000
Therefore, total investment = 108000
Investment of X
= 5/ 18 * 108000
= £30000
Investment of Z
= 7/18 * 108000
= £42000
VI.
Risk basically indicates the different nature or categories of potential crisis situations that can
arise before an organisation and uncertainty is the lack of surety or guarantee regarding the
happening of an event. When combines together they formulate the concept of probability in raw
terms where there is lack of any certainty regarding arousal of any risk or event for a business. It
is very important for businesses to understand the concept of probability because it can assist
immensely in the improvement of business policies and also improving their decision making
6

process. This ultimately increases the profitability and revenue position for every organisation.
Areas like investment and their potential returns, customer services, competitive strategies based
on the rival firm’s decisions etc. can be greatly achieved and influenced with the help of these
tactics as these help the companies in predicting the expected outcomes and favourable or
unfavourable situations.
PART B
QUESTION 3
I.
a.
Probability (green apple) = number of green apples/ total apples
= 350/ 800
= 0.44
b.
Probability (yellow apple) = number of green apples/ total apples
= 250/ 800
= 0.31
c.
Probability (red apple) = number of green apples/ total apples
= (800-250-350)/ 800
= 200/ 800
= 0.25
II.
Here complementary events are buying at least one kind of bread and not buying any bread
which can be denoted as:
P (at least one bread) = 1- P (no bread)
P (at least one bread) = 1- [(6.5/10) * (5.5/10)]
P (at least one bread) = 1- [0.65 * 0.55]
P (at least one bread) = 1- [0.36]
P (at least one bread) = 0.64
Therefore the probability that person selects at least one type of bread is 64%
7

III.
Probability (damaged goods of machine A) = 278/8650
Probability (damaged goods of machine B) = 215/6750
Probability (both are damaged) = Probability (damaged goods of machine A)*Probability
(damaged goods of machine B)
Probability (both are damaged) = 278/ 8650 * 215/6750
Probability (both are damaged) = 0.032 * 0.031
Probability (both are damaged) = 0.0009
Therefore the probability that both the goods are damaged is 0.09%
IV.
The use of probability events in case of business and its decision making is extensive where the
companies often develop a worst- case, best- case and likely scenario for themselves where they
predict that what kind of situation is most likely to arise for them collectively. This arises in case
of scenario forecasting strategy.
Another probability event is forecasting of sales where the expected future level of sales are
identified and analysed using the worst case or best case strategy because the precise level can
never be predicted.
Lastly, probability is used in risk evaluation as well where the risk is predicted using probability
that normally uses some existent data such as sales level or revenue level. This helps companies
in developing their breakeven points and hence take better decisions.
It can be concluded that overall the decision making gets drastically improved by taking use of
probability and its techniques of estimation.
QUESTION 4
Week Weekly Sales ($)
1 65
2 72
3 68
4 78
5 86
6 72
7 65
8 80
9 77
8

10 81
Mean
Mean indicates the average of all the values that have been recorded in a set of observations. In
context of the weekly sales that have been recorded, the median can be calculated in following
manner:
Week
Weekly Sales
($)
1 65
2 72
3 68
4 78
5 86
6 72
7 65
8 80
9 77
10 81
∑X 744
μ = ∑𝑿 / N
μ = 744/ 10
μ = 74.4
Therefore, the calculation indicates that 74.4 is an average weekly sales figure that has been
achieved in the sales made during all the 10 weeks.
Mode
Mode is indicated by the most frequent or repeated observation amongst all the individual
observations that have been recorded throughout.
Since 65 and 72 are the twice repeated value and all other observations are unique numbers, \ in
the present case there is bimodal value i.e. 65 and 72.
Median
Median indicates the middle value in a set of observation and for its calculation there are two
steps:
1: Arranging data in ascending order:
9

Week Weekly Sales ($)
1 65
2 65
3 68
4 72
5 72
6 77
7 78
8 80
9 81
10 86
2: Formula = (n+1)/ 2
Median Value = (10+1)/ 2
= 5.5
i.e. (5th value + 6th value)/ 2
= (72+ 77)/2
74.5
This indicates that the middle or centre value in this set of observations in 74.5 weekly sales
value amongst all the recorded weekly sales.
Range
Range is indicated by the difference of the largest observation value and the smallest observation
value indicating the oscillation range of observations.
In the present case, they can be identified as:
Largest Observation 86
Smallest Observation 65
Range = Largest - Smallest 21
Hence, here, range of the weekly sales recorded is 21.
Standard deviation
Standard deviation mainly indicates the scale at which the observed values deviate ie. move up
and down from the median i.e. average value.
The formula for this calculation is:
𝝈 = √ ∑(𝑿−μ)2 / N
Week Weekly Sales ($) (X) (X- μ) (X- μ)2
10

1 65 -9.4 88.36
2 72 -2.4 5.76
3 68 -6.4 40.96
4 78 3.6 12.96
5 86 11.6 134.56
6 72 -2.4 5.76
7 65 -9.4 88.36
8 80 5.6 31.36
9 77 2.6 6.76
10 81 6.6 43.56
Total ∑(X- μ)2 = 458.4
𝝈 = √ 458.4/ 10
𝝈 = √45.84
𝝈 = 6.77
Therefore, it can be clearly identified that observations deviate by the scale of 6.77 as compared
with the average value of 74.4.
11

12