Using Numeracy, Data & IT in Desklib - Solved Assignments, Essays, Dissertation

This article discusses the use of numeracy, data, and IT in various scenarios. It covers topics such as numerators, denominators, equivalent fractions, data analysis, and more. The article also includes solved assignments, essays, and dissertations related to these topics. The subject, course code, course name, and college/university are also mentioned wherever applicable.

Numeracy, Data & IT

TABLE OF CONTENTS
PART 1 – USING NUMERACY....................................................................................................4
Q1................................................................................................................................................4
Q2................................................................................................................................................4
Q3................................................................................................................................................4
Q4................................................................................................................................................5
Q5................................................................................................................................................5
Q6................................................................................................................................................6
Q7................................................................................................................................................6
Q8. ..............................................................................................................................................6
Q9................................................................................................................................................7
Q10. ............................................................................................................................................7
PART 2 – USING DATA................................................................................................................8
f)..................................................................................................................................................8
g).................................................................................................................................................8
h).................................................................................................................................................9
i)..................................................................................................................................................9
j)..................................................................................................................................................9
Question 11...............................................................................................................................10
a)................................................................................................................................................10
b)...............................................................................................................................................10
c)................................................................................................................................................10
d)...............................................................................................................................................11
PART 3 – USING IT.....................................................................................................................11
Question 12...............................................................................................................................11
Question 13...............................................................................................................................12
a)................................................................................................................................................12
b)...............................................................................................................................................13
c)................................................................................................................................................14

d)...............................................................................................................................................15
e)................................................................................................................................................15
Question 14...............................................................................................................................16
a)................................................................................................................................................16
b)...............................................................................................................................................17
c)................................................................................................................................................17
d)...............................................................................................................................................18
Question 15...............................................................................................................................19
a)................................................................................................................................................19
b)...............................................................................................................................................20
c)................................................................................................................................................21
d)...............................................................................................................................................22
Question 16...............................................................................................................................22
a)................................................................................................................................................22
b)...............................................................................................................................................23
REFERENCES..............................................................................................................................25

PART 1 – USING NUMERACY
Q1.
a.) Numerator : Numerator is a Latin word that means 'number'. It is the part of a fraction that
show at above the line of fraction to show the number of parts of the whole. For example, a
fraction 3/9 in that 3 is the numerator.
b.) Denominator : Denominator is also a Latin word that means 'namer'. It is the part of a fraction
that show at below the line of fraction for work as the divisor of the numerator (Loc, Tong and
Chau, 2017). It represents the equal parts of the number in an item. For ex. A fraction 3/9 in that
9 is the denominator.
Q2.
Numerator 24 18
Denominator 40 42
The Simplest form 24/40 = 3/5 18/42 = 3/7
The Simplest forms of fraction means only 1 is the common factor of its numerator and
denominator. For this, first find the HCF ( the highest common factor) of the fraction, then
divide both numerator & denominator by HCF & then fraction is reduced at its lowest term.
Q3.
a.)
Numerator 2 3 5
Denominator 3 4 6
Equivalent fractions
with a denominator of
12
4 3 2
New fraction 2/3 * 4/4 = 8/12 3/4 * 3/3 = 9/12 5/6 * 2/2 = 10/12
Two or more fractions are equal to the same fraction when simplified is said to be equivalent. In
thus process both numerator & denominator is simplified with same value (Barahmand, 2020).

b.)
Total books in library 60,000
Business books 14,000
Healthcare books 22,000
Psychology and law books 12,000
Remaining books 12,000
Computing books are 2/3 of remaining books 12,000 * 2/3 = 8,000
Computing books % 8,000/60,000 * 100 = 13.33%
In above solution, computing books are calculated. For this first calculate the remaining total
books of library then from remaining books calculate computing books that are 2/3rd of the
remaining books.
Q4.
No. Of pairs of shoes bought by Liz 2
She gave three crisp £50 notes
Total money she gave £150
Money she returned £10.50
Total price of 2 pairs of shoes £139.50
Price of 1 pair of shoes £69.75
In this solution, Liz purchase two pairs of running shoes. She pay three crisp £50 notes and given
change of £10.50. According to that given information, first calculate the total price of two pair
of shoes that is 139.50 by total money she gave 50×3=150 less given change of 10.50. From
139.50 calculate the price of each pair of shoes that is 69.75 by dividing 139.50/2.
Q5.
a.) 240.50×19.54 = 4699.37

In this solution, answer is given in two significant figures.
b.) 52100 = 5.21×1000
= 5.21×10^4
Represent the number 52100 as a power of 10
Q6.
a.)
Discount offer 30%
Patty and her 2 siblings pay total amount of
(70% payment)
£210
1% = (£210 / £70) £3
100% £300
Total savings £300 - £210 = £90
Patty and her 2 siblings means total 3 persons had a discount of 30% means that they paid 70%.
To find the total savings they made, first calculate the total cost (100%) that is total of discount
(30%) and payment (70%) they made (Dodoo, Gustavsson and Tettey, 2017). If 1% is equal to
[210/70 = £3] then 100% the gym offers will be £300 then total savings thay made is 300-210 =
90.
b.)
Average saving per person 30
Average saving of per person is calculated by total saving (90) divided by number of total person
(3).
Q7.
a.)
0.75 – 0.78 + (0.66) 0.63
b.)

The Largest number 0.1
Q8.
Total no. of men 90
Total no. of women 60
Total no. of people 150
Total people said yes 90
From total person women said yes 3/10 * 90 = 27
Total men said yes 90 – 27 = 63
Total men said no 90 – 63 = 27
In this solution, calculate the % of men said no. 90 men and 60 women are there i.e. total 150
persons are there from those 150 altogether 3/5 of people said yes that is 150*3/5 = 90. From
those 90 people calculating the 3/10 of women who said yes that is 27 that means total men said
yes is 90-27=63. Total men said no (27) are calculated by total men (90) less the number of men
said yes (63).
Q9.
Conference time in Birmingham is at 10:30 am.
For that, she has to leave her home at 08:00 am.
Described the steps below :
Time to reach Euston Rail station from her home is 1 hour, so she will be station at 09:00 am
Birmingham trains run at the following times:
5 minutes past the hour, 25 minutes past the hour and 45 minutes past the hour i.e. she will get
her train at 9.05 am.
Journey of train from Euston to Birmingham is 1 hour & 10 min. so she will reach to
Birmingham at 10:15 am.
Meeting venue in Birmingham is a 5-minute walk from the station according to that she will
reach the venue at 10:20 am.

According to that, she will reach exactly 10 minutes before the meeting.
Important note : If Annabelle wants to overcome 10 minutes for that she leaves at 8:10 am from
her home, then she will reach at station 9:10 am. According to the timings of train, she will get
her train at 9:25 am & at the end she will reach at venue 10 minutes late (Rawson, 2019).
Q10.
Weight of a Shredded wheat box is 0.35 kg
Weight of a Weetabix box is 0.36 kg
Weetabix box is heavier by 0.01 kg
In the above solution, calculate the weight of heavier box by using information given in the
question. Weight of Shredded wheat box is 0.35 kg & weight of a Weetabix box is 9/25 i.e. 0.36
kg, so the weight of Weetabix box is more than Shredded wheat box (Baidoo, 2019).
PART 2 – USING DATA
f)
Team Total Games Gold Silver Bronze Total
Australia 26 147 163 187 497
China 10 227 165 151 543
France 28 212 241 260 713
Germany 24 275 313 349 937
Great Britain 28 263 295 289 847
Hungary 26 175 147 169 491
Italy 27 206 178 193 577
Soviet Union 10 440 357 325 1122
Sweden 27 147 170 179 496
United States 27 1,022 794 704 2,520

In the given table, the country Great Britain has gold medal (263), silver medal (295) and bronze
medal (289). The country Germany & Soviet Union has gold medal (275, 440), silver medal
(313, 357) & bronze medal (349, 325). Apart from united states, the country Germany and Soviet
Union has more gold medals, more silver medals and more bronze medals than Great britain.
g)
Team Comparing no. of total games with overall no.
of medals received
Australia 497/26 = 19.11
China 543/10 = 54.3
France 713/28 = 25.46
Germany 937/24 = 39.04
Great Britain 847/28 = 30.25
Hungary 491/26 = 18.88
Italy 577/27 = 21.37
Soviet Union 1122/10 = 112.2
Sweden 496/27 = 18.37
United States 2,520/27 = 93.33
In the above table, it shows that the comparing of total no. of medals with total no. of games in
country participated & it gives the result that country United States gives the best performance
i.e. the highest no. of medals received per game (93.33).
h)
Two reasons of why a country like Jamaica does not feature in the top 10 medals:

 Lack of resources : the country has lack of resources for practicing such as trainer,
practicing ground, players, system of rules and so on. Due to lack of this resources
country is not participating in olympic games.
 Low population : the country has low population, due to this reason they do not
participate in other games other than athletes. They perform well in athletes and mostly
participate in that games (Farahani, Rabiee and Vatanpour, 2019).
i)
From the above table, it shows that country Soviet Union has close competitor of United States.
The answer is justified by table given below. In that table it show that in bronze medal category
country United States is far outperformed from its closest competitor that is country Soviet
Union.
Gold Silver Bronze
1,022 794 704
440 357 325
1,022/440 = 2.32 794/357 = 2.22 704/325 = 2.17
j)
Team Gold Silver Bronze Range (highest-
lowest)
Australia 147 163 187 187-147 = 40
China 227 165 151 227-151 = 76
France 212 241 260 260-212 = 48
Germany 275 313 349 349-275 = 74
Great Britain 263 295 289 295-263 = 32
Hungary 175 147 169 175-147 = 28
Italy 206 178 193 206-178 = 28

Soviet Union 440 357 325 440-325 = 115
Sweden 147 170 179 179-147 = 32
United States 1,022 794 704 1,022-704 = 318
Through the above working table, it is shown that country Hungary, Italy, Sweden & Great
Britain had the most evenly distributed number of gold, silver & bronze medals i.e. least smallest
range (Pak, 2020).
Question 11
a)
The lowest number of overall medals out of the ten countries is Hungary. It is because the
country has awarded only 491 medals which is minimum among all the others. The lowest
number is calculated in excel using the simple minimum function whose formula is
=Minimum(number1, number2, number3.....) (Sinaga and Simarmata, 2020).
b)
The name of countries competed in the least number of games are as follows:
 China
 Soviet Union
This is basically two countries which involve in only 10 total games thus considered as the least
number of games countries. This is also easily calculated in excel with the use of only minimum
formula.
c)
Mode is basically the value which appears most frequently and maximum time in a data set.
Whenever, a particular data arises more times than it is considered as mode. In the given case,
the mode of number of games countries participated is 27 as it arises three time. Such as Sweden,
United States and Italy are three countries which are participated in total 27 games.
d)
In order to identify the countries which got more silver medals than bronze medals, there is a
need proper comparisons. After comparison the list of countries are as follows:
Countries Silver Medal Bronze Medal

China 165 > 151
Great Britain 295 > 289
Soviet Union 357 > 325
United States 794 > 704
So, on this basis there are four countries whose silver medals are higher than bronze. This is
easily calculated in excel application with the use of IF formula. The formula of IF is involve
=IF(logical test, value if true, value if false)
Here, the logical test involve condition, value if true is the result and value if false is zero.
PART 3 – USING IT
Question 12
The spreadsheet shown below which is covered in excel file attached with report file. Excel are
as follows:
Question 13
a)
Excel rank function basically helps in identifying the returns of a numerical value as compared to
a list of other numeric value. The steps need to follow to calculate rank is as follows:

Step1: Firstly, the formula of rank in inserted in formula box that is =RANK(number, ref,
[order]).
Step2: After that select the numerical value from the rows and columns which need to be rank
and then press enter button (King and Purpura, 2021).
Step3: To repeat the same, the drop and drag method is used to get all columns and countries
result i.e., rank.
Step4: In the order section, there is need to select ascending and descending order. When there is
a need to value the largest number with 1, then decending (0) type is used and vice-versa.
For example; in the given question the rank is done in ascending order:
As per the above calculations, it is identified that United state with total medals 2520 ranked 1st
and Hungary with total medals 491 ranked 10th.
b)
The specific action which is need to be adopted in order to identify and display the list of
countries whose total medals is 800 or more is using IF excel formula. The formula of IF in excel
are as follows:
=IF (logical test, [value if true], [value if false].

On the basis of calculation in excel, it is identified that there are basically four countries whose
total number of medals is 800 or more. The list and name of such countries is as follows:
 Germany
 Great Britain
 Soviet Union
 United States
c)
In order to represent only gold medal information, the column chart is used to because it is the
best suitable to represent single information at a specific point if time. It is basically a vertical
bar chart which represent how many gold medals awarded to all the ten countries with proper
chart labelling. For example, the column chart representing the gold information are as follows:

d)
Replication is basically used for writing and copying the same data to different columns and
rows. In case, if one formula is put in one column to than it can easily be replicated to other
column to identify their results. But the only action need to acquired is using drop and drag
method.
e)
The overall total medals awarded is calculated with the use of Sum excel formula. This formula
is as follows:
=Sum(number1, number2, number3 …......)
Australia
China
France
Germany
Great Britain
Hungary
Italy
Soviet Union
Sweden
United States
0
200
400
600
800
1000
1200
147 227 212 275 263
175 206
440
147
1022
Gold

This formula is helpful in calculating the sum of large data very easily at a given point of time
and is the best function to add values. The calculation was done in excel which are as follows:
After calculation, the overall total medals which are awarded to all the ten countries are 8743
which indicate the sum or addition of ten countries particular medals.
Question 14
a)
In order to calculate the total number of medals for Germany and Great Britain, the Sum formula
of excel is being used. It is because it helps in adding the numerical value shown in all the cells.
With the help of this function, the addition of individual values, cell references, range and the
mixture of three are possible. In the given question, the total medals of Germany is 937 while of
Great Britain is 847 using the sum formula. The formula is =Sum(number1, number2 …....)

b)
The average formula is used to calculate the arithmetic mean of a group of numbers where the
average formula is available in auto sum section in excel. Here, in the given question in order to
calculate the average of European countries silver medals first need to identify the countries
which are member of Europe. The countries are France, Germany, Hungary, Italy, Sweden. So,
the average of these countries silver medal is 209.8.
c)
In order to determine the sum of medals for gold of those countries with less than 20 games
involvement, first we need to identify the name of those countries. For the identification of these
countries, it is required to use the IF formula. The IF formula is as follows: =IF (logical test,
[value if true], [value if false]. After identification of these countries the Sum formula of excel is
used to calculate the total number of gold medals for the countries with the less than 20 total
games. The calculation are as follows:

Step1:
Step2:
d)
In order to search the database related to Italy and its corresponding total medals the VLOOKUP
formula of excel is used. The formula is as follows:
=VLOOKUP(“Italy”; A109:A119; 6; False)

Here, in the false position 0 is used as it is helpful for calculating the sum of the totals. After
calculation, it is identified that Italy has total 577 medals.
Question 15
a)
Median is basically a middle number of the whole database in a sorted, ascending and
descending order. In case if the database has odd number than the middle value is the number
that exist between the middle with the same value of number above and below.
The formula of Median is {(n + 1)/ 2}th
Median of Gold Medals is 219.5 using the formula.
The median of gold, silver and bronze medal using excel formula is 219.5, 282.3 and 280.6
respectively.

b)
Mean is also known as average which is generally a central value of a group of values and data
sets. The formula for calculating mean is as follows:
Arithmetic Mean = X1 + X2 + X3 …..../ n, where n is total counts.
Mean of Bronze Medals: 2806/10 = 280.6
On the basis of excel calculations, the mean of gold, silver and bronze is 311.4, 282.3 and 280.6
respectively. It is basically calculated using the same formula i.e., average as the result will
remain same in both the functions.

c)
Standard deviation formula is √ ∑ (x – u)2 / N which defines the distance of each value with its
mean.
Team Total medals (x) (x – u) (x - u)2
Australia 497 497 – 874.1 = -377.1 142204.41
China 543 543 – 874.1 = -331.1 109627.21
France 713 713 – 874.1 = -161.1 25953.21
Germany 937 937 – 874.1 = 62.9 3956.41
Great Britain 847 847 – 874.1 = -27.1 734.41
Hungary 491 491 – 874.1 = -383.1 146765.61
Italy 577 577 – 874.1 = -297.1 88268.41
Soviet Union 1122 1122 – 874.1 = 247.9 61454.41
Sweden 494 494 – 874.1 = -380.1 144476.01
United States 2520 2520 – 874.1 =
1645.9
2708986.8
Total / ∑ 8741
N = 10
3432426.9

Population mean (u)
= 8741 / 10 = 874.1
After application of data in formula, the standard deviation is √3432426.9 / 10 = √ 343242.69 =
585.869
d)
Jamaica is a country which perform best in the athlete but the reason they do not feature and
participate in the event are as follows:
 First, they are not large enough as compared to other ten countries to compete with them
besides having the best performance in athlete.
 Another reason for this is that the total population of Jamaica is lower than the other 10
countries, and they are not able to provide training to small population regarding all
events (Buljan and et.al., 2019).
Question 16
a)
The fully labeled data chart of comparison between the gold, silver and bronze medals of all the
ten countries are as follows:

b)
In order to calculate the contribution of each country medals against the overall total medals, the
weights is need to be calculated. The formula to calculate weights are each country total medals/
overall total medals of all ten countries (Raymer and et.al., 2018).
Now, the labeled chart describing the contribution are as follows:
Australia
China
France
Germany
Great Britain
Hungary
Italy
Soviet Union
Sweden
United States
0 200 400 600 800 1000 1200
147
227
212
275
263
175
206
440
147
1022
163
165
241
313
295
147
178
357
170
794
187
151
260
349
289
169
193
325
179
704
Comparison of gold, silver and bronze medal
Bronze Silver Gold

0.057
0.062
0.082
0.107
0.097
0.056
0.066
0.128
0.057
0.288
Contribution
Australia China France Germany Great Britain Hungary
Italy Soviet Union Sweden United States

REFERENCES
Books and Journals
Sinaga, R. and Simarmata, E. J., 2020. Constructive Play on Numeracy Skills in Elementary
School. Budapest International Research and Critics in Linguistics and Education (BirLE)
Journal. 3(4). pp.1644-1650.
King, Y. A. and Purpura, D. J., 2021. Direct numeracy activities and early math skills: Math
language as a mediator. Early Childhood Research Quarterly. 54. pp.252-259.
Buljan, I. and et.al., 2019. Health numeracy skills of medical students: cross-sectional and
controlled before-and-after study. BMC medical education. 19(1). pp.1-10.
Sinclair, N., Hilton, G. and Hilton, A., 2017. Staging a Numeracy Performance. Australian
Journal of Middle Schooling. 17(2). pp.20-31.
Raymer, D. S. and et.al., 2018. Health Literacy and Numeracy in Patients Considering a Left
Ventricular Assist Device: Findings From the DECIDE-LVAD Trial. Circulation:
Cardiovascular Quality and Outcomes. 11(suppl_1), pp.A219-A219.
Loc, N. P., Tong, D. H. and Chau, P. T., 2017. Identifying the concept fraction of primary school
students: The investigation in Vietnam. Educational Research and Reviews. 12(8).
pp.531-539.
Barahmand, A., 2020. Equivalent Fractions and Natural Numbers Bias. For the Learning of
Mathematics. 40(3). pp.20-24.
Dodoo, A., Gustavsson, L. and Tettey, U. Y., 2017. Final energy savings and cost-effectiveness
of deep energy renovation of a multi-storey residential building. Energy. 135. pp.563-
576.
Rawson, N. S., 2019. Comparison of numbers and timing of new medication regulatory
approvals in Canada and New Zealand. Regulatory Toxicology and Pharmacology. 101.
pp.24-28.
Baidoo, J., 2019. Dealing with grade 10 learners’ misconceptions and errors when simplifying
algebraic fractions. Journal of Emerging Trends in Educational Research and Policy
Studies. 10(1). pp.47-55.

Farahani, M. H. D. A., Rabiee, H. and Vatanpour, V., 2019. Comparing the effect of
incorporation of various nanoparticulate on the performance and antifouling properties
of polyethersulfone nanocomposite membranes. Journal of water process
engineering. 27. pp.47-57.
Pak, R. J., 2020. Application of the Gamma distribution approximating the asymptotic
distribution for ranges. The Korean Data & Information Science Society. 31(4). pp.513-
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