Numeracy and Data Analysis - Solved Assignments and Study Material

This page provides solved assignments and study material for Numeracy and Data Analysis. It covers step-by-step calculations, ratios, probability, and statistical analysis. The content includes solved questions on BODMAS, index numbers, mean, mode, median, range, and standard deviation. It also includes case studies and examples for better understanding. The study material is relevant for courses in Numeracy and Data Analysis offered by various colleges and universities.

NUMERACY AND
DATA ANALYSIS

Contents
PART A......................................................................................................................................3
Question 1..............................................................................................................................3
PART B......................................................................................................................................5
Question 2..............................................................................................................................5
Question 4..............................................................................................................................5

PART A
Question 1
1. Work out step-by-step the answers to the following questions. (Demonstrate
BODMAS / rule of priority to calculate answers)
a) 875 ÷ (32 + 42) (2 × 2 + 3)
Answer: 875 / (9+16) (4+3)
875/25*7
35*7
245
b) (4410 ÷ 212) (52 ÷ 5) (2 marks)
Answer 4410/441*5
50
c) 960 ÷ 12 (1 + 2 × 2) – 302 (3 marks)
Answer 80*5-900
400-900
-500
d) 33 × [(3× 5 ×2 + 62) ÷ (6 + 24)] (3 marks)
Answer 9*(30+36)/(6+16)
9*66/22
9*3
27
d) [640 ÷ (8 + 2)] ÷ 82+ 9 – 2 + [(42× 2) 2] (3 marks)
Answer (640/10)/64+9-2+(32*2)
64/64+7+64
1+7+64
72
I. Calculate the answers for the following expressions.
a) (-33) × (-4) (1 marks)
Answer 132
b) 46 + (-16) (1 marks)
Answer 30
c) (-56) ÷ 8 – (-9) (1 marks)
Answer -16

III. Carry out the following calculations:
a) + (2 marks)
Answer 33/21
b) − (2 marks)
1 2
c) 5 + 2 (3 marks)
3 5
IV. Evaluate the following:
a) A shirt manufacturing company produces blue and black coloured shirts only. If
32.8% of 65000 total production is blue shirts, how many blue shirts are there?
(2 marks)
Answer 21320
b) If 37.6% of students out of 28000 total students are based in the UK, how many
students are not based in the UK? (3 marks)
Answer 11872
Contd.
V. Calculate the following.
a) A, B, C & D investors have invested in a business in a ratio of 7:5:3:4. Total
investment is
£180,500. What is the actual investment contributed by each investor? (2 marks × 4
= 8 marks)
Answer A 66500
B 47500
C 28500
D 38000
b)X, Y & Z investors have invested in a ratio of 4: 7: 3. ‘Y’ has invested £35,000.
Calculate the individual investment made by ‘X’ & ‘Z’ and the amount of the total
investment?
(2 marks × 3 = 6
marks)
Answer X 20000
Y 15000

VI. The business world involves risk and uncertainty, explain the importance to
businesses of understanding of the concept of probability. (8 marks)
Answer It helps in making decisions that can help the company to have an upper hand
in the market so that it can sustain and survive the dynamic competition of the
industry in an effective and efficient manner.
PART B
Question 2
Year Annual sales
(£000)
2014 6.7
2015 6.9
2016 7.4
2017 7.1
2018 7.6
a) Index Number = (2018 sales / base year 2017) * 100
= (7.6 / 7.1) * 100 = 107.04
b) Index = 92.56
Sales of Base year = sales of 2017 / Index
= (7.1 / 92.56) * 100 = 7.67
c) Index = 118.76
Sales of Base year = sales of 2015 / Index
= (6.9 / 118.76) * 100 = 5.81
d) Significance of Consumer price index: CPI portrays what the progressions in cost
levels mean for heterogeneous clients. Likewise, it helps the public authority in
forming monetary approaches, cost arrangement, charge strategy, and so on.
Consequently, it helps the public authority in working out the dangers taken for
improvement without changing the cost for many everyday items.
Question 4
Days Sales in £
1 58

2 42
3 44
4 52
5 70
6 46
7 42
8 42
9 52
10 65
(i) Mean: It is the average of all the sales figure of 10 consecutive days.
Mean = sum of all observations / total no of observations
= 513 / 10 = 51.3
(ii) Mode = It is the highest number of times the sae value has occurred
Mode = 42
(iii) Median: It is the mid – value of data collected of the sales.
Median value = (N +1) / 2
(10 + 1) / 2 = 11 / 2 = 5.5
Median = (5th value + 6th value) / 2
= (70 + 46) / 2 = 116 / 2 = 58
(iv) Range = Maximum – Minimum
= 70 – 42 = 28
(v) Standard Deviation = √ (∑(𝑿−μ)2) / N
Day Sales (xi – μ)
(xi –
μ)2
1 58 6.7 44.89
2 42 -9.3 86.49
3 44 -7.3 53.29
4 52 0.7 0.49
5 70 18.7 349.69
6 46 -5.3 28.09
7 42 -9.3 86.49
8 42 -9.3 86.49

9 52 0.7 0.49
10 65 13.7 187.69
Standard Deviation = 9.613