Numeracy and Data Analysis for Business - Semester 1 Assignment

This document presents a comprehensive solution to a Numeracy and Data Analysis assignment designed for a business course. The assignment covers a range of topics, including the application of BODMAS for calculations, calculations involving percentages, fractions, and ratios. The solution includes step-by-step workings for each question, ensuring clarity and understanding. Furthermore, the assignment delves into the importance of probability in business decision-making, explaining concepts like population, sample, and normal distribution, along with types of skewed data. The document also provides practical examples related to weekly sales analysis, calculating mean, median, mode, range, and standard deviation. Overall, this document offers a complete guide to the assignment, equipping students with the necessary knowledge and skills to excel in their coursework.

Numeracy and Data
Analysis for Business
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Contents
Contents...........................................................................................................................................2
Question 1........................................................................................................................................3
PART A...........................................................................................................................................3
Part B...............................................................................................................................................6
Question 4........................................................................................................................................6
Question - 5......................................................................................................................................7
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Question 1
PART A
I. Work out step-by-step the answers to the following questions. (Demonstrate
BODMAS / rule of priority to calculate answers)
a) 2220 ÷ (5 2 + 7 2) (2 × 2 ÷ 2)
 2220/ (52+72) (2X2/2)
 2220/ (25+49) (2X1)
 2220/ (74) (2)
 2220/ (148)
 15
b) (1800 ÷ 302) (375 ÷ 75)
 (1800/ 900) (5)
 (2) (5)
 10
c) 490 ÷ 7 (3 × 3 -3) – 150
 490/ 7(9-3) -150
 490/ 7(6) -150
 70 (6)-150
 420-150
 270
d) 4 × [402 ÷ (5 × 4 3)]
 4x[1600/(5X64)]
 4x [1600/ (320)]
 4x [5]
 20
e) [140 ÷ (2 × 3 + 8)] ÷ 5 + 12 – 3 + [(6 2× 3) 2]
 [140/ (6+8)]/ 5+12-3+ [(36x3) 2]
 140/ (14)]/ 5+12-3+ (108x2)
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 10/ 5+ 12-3+216
 2+12-3+216
 227
II. Calculate the answers for the following expressions.
a) (-55) × (-6) (1 marks)
 -(-330)
 330
b) 44 - (-44) (1 marks)
 1936
c) (-81) ÷ 9 – (-4)
 (-9) +4
 -5
III. Carry out the following calculations:
a) 2/ 7 + 2/ 3 (2 marks)
 6+14/21
 20/21
b) 5/ 6 – 2/ 5 (2 marks)
 25-12/ 30
 13/30
c) 6 1/ 3 + 2 3/ 4
 19/3 + 11/4
 79+33/12
 112/12
 28/ 3
iv) Evaluation
A) Total number of tea bag = 36850
Percentage of normal tea bag = 52.8%
Total number of normal tea bag = 36850*52.8/100 = 19456
Total number of green tea bag = 36850-19456 = 17394
B) Total number of students = 9855
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Percentage of students based in England = 61.6%
Number of student based in England = 9855*61.6/100 = 6070
Thus, number of students that are not based in England = 9855 – 6070 = 3785
V) Calculation
A) Total investment = £320,000
Business ratio for different partners were
A=6
B= 2
C= 5
D= 3
Actual investment for A = £320,000 * 6/16 = £120,000
Actual investment for B = £320,000 * 2/16 = £40,000
Actual investment for C = £320,000 * 5/16 = £100,000
Actual investment for D = £320,000 * 3/16 = £60,000
B) Investment of y = £49,000
Ratio of investment for three investors were = 4:7:5
Total investment will be = £49,000 *16/7 = £112000
Investment of X = £112000 * 4/16 = £28000
Investment of Z = £112000 * 5/16 = £35000
VI) Importance of understanding the concepts of probability
 Investment = Only by using probability mostly as prediction tool will a company take
these uncertainties into investors when designing financial decisions. A company may
use risk models to assess which investments or investment configurations yield the
highest predicted return after evaluating the odds of gains and losses correlated among
each financial decisions.
 Competitive strategy = Despite the fact that behavioural economics is an integral part of
deciding business policy, it does not incorporate complexity in its models. A random
variable like this can't help an organization fully refine its risk plan. Companies may use
probability models like Markov chains to construct a collection of techniques that are not
always risk-aware but also self-altering. From the light of fresh facts about rival
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businesses Furthermore, Markov chains enable businesses to test long-term approaches
numerically to determine which ones produce the best returns.
Part B
Question 4
Week Weekly sales
1 67
2 68
3 72
4 53
5 72
6 67
7 64
8 68
9 55
10 68
i) Mean = Total sum of all observations/ Total number of observations
654/10 = 65.4
ii) Mode = The value which is repeated most number of times in the given series, which
is 68
iii) Median = In case of even series the median formula is = [(n/2)th term + {(n/2)+1}th]/2
Arranging in ascending order = 53, 55, 64, 67, 67, 68, 68, 68, 72, 72
10 + (10/2) +1 / 2
5, 6
So median will be = 68 + 67 / 2
67.5
iv) Range
Difference between highest and lowest value from the values
72 – 53 = 19
v) Standard Deviation
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mean = 64.5
n = 10
X = values for each point
formula 𝝈 = √ ∑(𝑿−μ)/ n = 6.47
Question - 5
I. Using appropriate examples, explain what data and information is.
Data and information is being both different terms which are used for each other.
Data refers to facts or details from which information is being derived. It is to make
sure by the collector of data to form effective information that is to be used for
conduction of any report project or any research.
Data
They are termed are unorganized facts that needs to adjusted as per need of
investigator. It is to be considered as useless, until it is organised properly in
respective area of operations. Example of data is, scores achieved by a student in a
test paper is an example of data.
Information
It refers to stage being when a data is being organised and well-structured so as to
fulfil all the requirement of a given task in referred to as an information. Example of
information is, average score attained by students in a class.
II. Using appropriate examples, explain what population and sample is and the
importance of sampling in the business world.
Population refers to the total number of candidates present in a group or a class. On
the other hand, sampling refers to a specified group of individual that are selected for
conduction of an activity or for any other purpose respectively.
Examples of population
Advertisement for a job in UK
All countries of the world
Post graduate students in UK
Example of Sample
Top results in search for advertisement for a specific job in UK
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Countries with data available on GDP and NDP
400 PG students from Harvard university
III. Explain the characteristics of normal distribution and three types of skewed data can
be found in the business world.
Normal distribution, also known as the Gaussian distribution, is a probability distribution that is
symmetric about the mean, showing that data near the mean are more frequent in occurrence than
data far from the mean. In graph form, normal distribution will appear as a bell curve.
Characteristics of normal distribution
The mean, mode and median are all equal.
The curve is symmetric at center.
Total area under curve is 1
Exactly half of value are to the left of center and exactly half value is at right
Types of Skewed data
Positive Skewed data
If the given distribution is shifted to the left and with its tail on the right side, it is
a positively skewed distribution.
Negative Skewed data
If the given distribution is shifted to the right and with its tail on the left side, it is
a negatively skewed distribution
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