Using Numeracy, Data and IT for Analysis of Olympic Medals

This report discusses the use of numeracy, data and IT techniques for analyzing Olympic medals. It covers topics such as fractions, simplification, equivalent fractions, percentages, discounts, averages, calculations, rankings, graphs, charts, and replication. The report also includes a detailed analysis of Olympic medals won by different countries, including their total medals, gold medals, silver medals, and bronze medals. The report provides insights into the performance of different countries and their strengths and weaknesses in Olympic events.

USING NUMERACY,
DATA AND IT

TABLE OF CONTENTS
PART-2......................................................................................................................................3
Question 11.............................................................................................................................3
Question 12.............................................................................................................................5
Question 13.............................................................................................................................5
Question 14.............................................................................................................................7
Question 15...........................................................................................................................10
Part- 3.......................................................................................................................................12
Question 16...........................................................................................................................12
References................................................................................................................................15
Books and journal.................................................................................................................15
Online reference...................................................................................................................15

PART – 1
Question 1
Numerator and denominator can be explained as a part of fraction which consists of
two values where one of these values has been placed on the top which is known as
numerator and the other value has been placed at the bottom and is known as denominator.
The numerator and denominator are separated by a line which is also known as a fractional
bar. For example, given a fraction number is ¾, where 3 is the numerator and 4 is the
denominator and a slanting line separating these values is what the fractional bar. Also, the
numerator indicates the selected number of parts out of the available total equal number of
parts.
Question 2
Simplification process of fraction begins with the identification of HCF or Highest Common
Factor from the integer values involving numerator and denominator. The division of both
numerator and denominator will be done with the Highest Common Factor in the next step.
At last the reduced factor has been determined which is also known as simplified value of a
given fraction.
By taking the example of the given fraction, simplification process can be explained easily,
which is as follows:
HCF of 24 / 40 is 8, where after dividing both numerator and denominator with HCF, we will
get the simplified form of the fraction that is, 3 / 5.
Similarly, the second fraction given here is 18 / 42 whose HCF is 6 and by dividing both the
numerator and denominator with its HCF, we will get the simplified form of the fraction that
is, 3 / 7.
Therefore, with the help of determined value of HCF, one can easily simplified the fraction to
its simplified form.
Question 3
a) Equivalent fraction can be defined as all those fractions involving different numbers but
are indicating the same part selected out of the whole through numerator and denominator. In
other words, fraction with different values of numerator and denominator having equal
fractional value in its simplified form. For example, ½ can also be written as 5/10 and 50/100
indicating a same fractional value equivalent to half when it gets simplified.
In the given case, fractions 2/3, ¾ and 5/6 can be expresses as equivalent fractions having its
denominator equals to 12 as follows:
The equivalent fractions can be formed for the given fractions by multiplying their numerator
and denominator with the same values.
 If the fraction 2/3 gets multiplied by 4/4, the resulting fraction would be 8 / 12.
 If the fraction 3 / 4 gets multiplied by 3/3, the resulting fraction would be 9 / 12.
 If the fraction 5 /6 gets multiplied by 2 / 2, the resulting fraction would be 10 / 12.
Therefore, all these fractions have similar number in its denominator that is 12, and all these
fractions can be termed as equivalent fraction as they are indicating the similar values.

b) In order to determine the percentage of computing books within the library, in the first step
the remaining number of books needed to be determined as follows:
Remaining books after subtracting books on other subject from the total number of books
= 60000 – 14000 – 22000 – 12000 = 12000 books.
It is given that computing books form 2 / 3 part of the remaining books that is, 2/3 * 12000 =
8000. So, there are 8000 books in the library on the subject computing.
To determine the percentage value of computing books, the number of computing books will
be divided by the total number of books in the library and the resulting value again gets
multiplied by 100 as follows:
8000 / 60000 * 100 = 13.33% of the total books in the library are on computing.
Question 4
Total number of shoe pairs bought = 2
Amount received by sales attendant from Liz = 3 * 50 = £150
Amount returned by sales attendant to Liz = £10.5
Therefore, the derived value of each pair of shoes charged by the sales attendant
= £150 - £10.5 / 2 = £139 .5 / 2 = £69.75.
Cost of each pair = £69.75.
Question 5
a) Computation of 240.5 * 19.54 = 4699.37 which is showing relevant results as it is having
two significant values.
b) In the form of power of 10, the number 52100 can be written as 5.21 * 104.
Question 6
a) Amount paid for 3 people = £210 (discounted charge @ 30%)
£210 = 70%
So, the 100% amount would be = 210/70 * 100 = £300
In the absence of discount, the amount paid would be = £300
Total discount availed = £300 - £210 = £90.
b) Total discount availed is equals to total savings made that is, £90 for 3 persons.
Average savings per person = £90/3 = £30 per person.
Question 7
a) 3 / 4 – 7/9 + 2/3 = 0.75 – 0.78 + 0.67 = 0.64 = 16 / 25.
b) 0.1 is the largest of all the given numbers.
Question 8
Total people question = 150 (90 men and 60 women)

3/5 of 150 said yes, so 2/5 said no.
People saying yes = 150*3/5 = 90
People said no = 150 – 90 = 60
Given that 3/10 of women said yes, so 60*3/10 = 18.
Women said no = 60 – 18 = 42.
Total number of people denied = 60 of which 42 were women. Therefore, number of men
denied = 18
So, the percentage of men said no = 18/150*100 = 12%.
Question 9
Annabelle required to leaver her home by 8.00 am to speak at 10.30 am in Birmingham.
Time required to arrive at Euston Rail Station: 8:00 a.m. + 1 hour = 9:00 a.m.
She gets the train by 9.05am
Arrived at Birmingham = 9:05 am + 1 hour 10 minutes = 10:15 a.m.
Time required to arrive at the venue of meeting = 10:15 am + 5 minutes = 10:20 am
Therefore, Annabelle reached the venue earlier by 10 minutes.
Question 10
Weight of wheat box = 0.35 kg
Weetabix box weight = 0.36 kg
Therefore, Weetabix box is heavier by 0.01 kg from that of Shredded box of wheat.
PART-2
Question 11
(a) Hungary got lowest number of medals out of 10 countries. As it has got 491 medals.
(b) China and Soviet Union has played lest number of games and both countries
performed 10 matches.
(c) Mode is to be defined as highest frequency value in the given set of data. Under total
number of games, some values are repeated like 26-2, 10-2, 28-2, 27-3. Out of this, 27
has repeated highest number of times.
(d) Range can be evaluated that by taking maximum difference between highest value
and the lowest value. 293 is the figure of the range, which has been determined by
taking difference of 440 (highest value) and 147 (lowest figure).
(e) Here, there are four countries who got grater number of silver medals than bronze.
These countries are China, Great Britain, Soviet Union and United States.

(f) Apart from united states, there are two countries who have got greater number of
silvers, bronze and gold medals than Great Britain.
(g)
Team Total games
Total number of
medals Medals per game
Australia 26 497 19
China 10 543 54
France 28 713 25
Germany 24 937 39
Great Britain 28 847 30
Hungary 26 491 19
Italy 27 577 21
Soviet Union 10 1122 112
Sweden 27 494 18
United state 27 2520 93
Medals per game is evaluated by dividing total number of medals to total games. From the
above table it can be interpreted that Australia has got 19 medals per game. Moreover,
highest number of medals per game is won by Soviet Union.
(h) Jamaica has excellent player but doesn’t get featured in Olympic (top 10 countries). It
is because Jamaica has lack of resources and large number of athletic players.
Government could not afford to provide training to all the players, this will lead to
athletics avoid to participate in Olympic. Moreover, Jamaica’s population is very low
due to this reason government can not send all the people towards games and sports in
order to run the economy.
(i) From the given information, it can be represented that Soviet Union is the nearest
competitor of united states. It is because it has played better than others. As united
states has got 1022 gold, 794 silver and 704 bronze medals. Whereas Soviet Union
has achieved 440 (gold), 357 (silver) and 179 bronze medals. So the percentage of
Gold medals 440/1022 = 43.05%
Silver medals percentage 357/794 = 44.96%
Bronze medals percentage 179/704 = 46.16%
United states has better performed in case of gold medals.
(j)
Team Gold Silver Bronze Range(maximum-minimum)
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 = 22
Italy 206 178 193 206-178 = 28
Soviet Union 440 357 325 440-325 = 115
Sweden 147 170 179 179-147 = 32
United States 1022 794 704 1022-704 = 318
Italy has the smallest range.
Question 12

Question 13
Rank can be calculated with the help of “rank formula” in excel. For getting the rank of total
medals won by 10 countries, rank formula can be applied by =rank (choose a number whose
rank need to be evaluated, range or series out of which rank is decided, select 0 for
descending arrangement). From the above analysis, it has been interpreted that United States
got 1 rank in medals out of other countries.
(b)

To find the list of all the countries who have got medals which is greater than and equal to
800. Whenever it is required to put condition then if formula will be used (Spithourakis and
Riedel, 2018). Likewise, application of if formula is done with the help of putting =if
( condition like here to get the name of countries who has achieved greater than and equal to
800 medals, resultant figure-name of the country, if false then zero).
(c )
Australia
China
France
Germany
Great Britain
Hungary
Italy
Soviet Union
Sweden
United states
0
200
400
600
800
1000
1200
Gold Medals
the data of gold can be shown with the help of graphs, charts etc. These graph and chart help
to analyse in different forms and interpret in order to generate useful information. In addition
to this, under the bar chart, data can be represented through separate pillar which make easy
to understand the data. Moreover, through pie chart, user can analyse given set of data in 360-
degree format. Further, set of data is divided in circle to analyse clear and concise manner.
(d) Replication
Replication is to be used to reduce the manual work and get quick answer of complex
problem. Replication can be implemented through copy paste method and drag and drop
method (Netemeyer and et.al., 2020). Whereas copy paste method is stated that any formula
or data which is required to wright frequently then apply copy paste method so that answer
can be get without putting manual figure. Second is drag and drop method which is used to in
place of copy paste method. It is applied by taking pointer to the right corner of the cell and
then drag so than all the column will fill automatically.
(e )

Sum formula can be applied through =sum(select a range of which addition is to be
calculated). Sum can be known as addition. Overall total medals is to be calculated with the
help of sum (addition) formula. Likewise, 8743 is the value of overall medals won by all the
ten countries.
Question 14
(a)
Total medals which have been won by Great Britain and Germany can be evaluated with the
help of Sum formula. By applying sum formula, total medals of Germany and Great Britain
can be calculated which is 961 and 847 respectively.
(b)

Average is also known as mean which has been computed with the help of average formula
(Litkowski and et.al., 2020). It is applied by =average (select the range of which average need
to be calculated) and then press enter. So as here the value of average is 209.8.
(c )
For computing gold medals of 10 countries who played less than 20 games so that “if”
formula will be applied because the resultant figure is connected with the condition. It can be
implemented like =if( condition, if true than the resultant answer, if false answer will be
zero).
(d)

VLOOKUP function is implemented where it is necessary to find certain data out of big data
set. Likewise, here the data of Italy is needed to find. This can be implimeted by
=vlookup(reference whose data is looking for, that data is presented in which column, choose
exact match or approx. match). Like here total medals of Italy is 577 can be easily computed.
Question 15
(a)
Median is represented to the mid value of the data. In order to evaluate the middle value of
gold medals which has been won by 10 countries need to follow certain process (Lin and
et.al.,2021).

1st phase: arrange the given set of data in ascending order
147,147,175,206,212,227,263,275,440,1022
2nd phase: number of data is even so take out two middle values that is 5th and 6th term which
is 212, 227 respectively.
3rd phase: both the term is to be divided by 2
So (212+227)/2=219.5
(b)
Mean(average) can be evaluated by adding all the figures and then divide by n (which is
count of numbers). Further, mean of data of bronze medals can be computed with the help of
its formula which is ∑x/n.
∑x = 187+151+260+349+289+169+193+325+179+704 = 2807
N = 10
So ∑x/n
=∑x/n = 2807/10
So 280.7 represent the value of mean.
(c.)
Team Gold Silver Bronze Total x-mean (x-mean)2
Australia 147 163 187 497 -377.3 142355.29
China 227 165 151 543 -331.3 109759.69
France 212 241 260 713 -161.3 26017.69

Germany 275 313 349 937 62.7 3931.29
Great
Britain 263 295 289 847 -27.3 745.29
Hungary 175 147 169 491 -383.3 146918.89
Italy 206 178 193 577 -297.3 88387.29
Soviet
Union 440 357 325 1122 247.7 61355.29
Sweden 147 170 179 496 -378.3 143110.89
United
States 1022 794 704 2520 1645.7 2708328.49
average 874.3 Total 3430910.10
√ (x-
mean)2/n 343091.01
standard
deviation
585.739711
8
(d)
With the help of spread sheet user can convert complex problem in the simpler form.
Further, it also provides facility to analyse and interpret all the data in tabular, chart and
graph form (Heilmann, 2020). Moreover, standard deviation is necessary to evaluate the
amount of variation present in observation. Lower or higher value of standard deviation
represent that resultant figure is close or far to mean. =STDEV.P formula is to be used to
calculated standard deviation in spread sheet.
Part- 3
Question 16
(a)

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
Olympic Game Medals (Top 10)
Bronze Silver Gold
Bar chart is used to analyse all the data in clear and precise manner. From the above analyse,
data can be interpreted in order to reach to the conclusion. Moreover, bronze medals are
shown in silver colour (Craig, 2018). Gold medals are represented in blue colour. In addition
to this, silver medals are reflected in orange colour.
(b)

0.06
0.06
0.08
0.11
0.10
0.06
0.07
0.13
0.06
0.29
contribution of each country to overall medals
Australia China France Germany Great Britain
Hungary Italy Soviet Union Sweden United states
Contribution of each and every country has been calculated through sum of overall medals
got by 10 countries (Aunio and et.al., 2021). As it has been computed by dividing total
medals won by each and every country(separately) to overall medals.

References
Books and journal
Aunio, P. and et.al., 2021. An early numeracy intervention for first-graders at risk for
mathematical learning difficulties. Early Childhood Research Quarterly. 55. pp.252-
262.
Craig, J., 2018. The promises of numeracy. Educational Studies in Mathematics. 99(1).
pp.57-71.
Heilmann, L., 2020. Health and numeracy: The role of numeracy skills in health satisfaction
and health-related behaviour. ZDM. 52(3). pp.407-418.
Lin, J. and et.al.,2021. The relation between parent ratings and direct assessments of
preschoolers’ numeracy skills. Learning and Instruction. 71. p.101375.
Litkowski, E.C. and et.al., 2020. When do preschoolers learn specific mathematics skills?
Mapping the development of early numeracy knowledge. Journal of experimental child
psychology. 195. p.104846.
Netemeyer, R.G. and et.al., 2020. Health literacy, health numeracy, and Trust in Doctor:
effects on key patient health outcomes. Journal of Consumer Affairs. 54(1). pp.3-42.
Spithourakis, G.P. and Riedel, S., 2018. Numeracy for language models: Evaluating and
improving their ability to predict numbers. arXiv preprint arXiv:1805.08154.
Online reference
Mean, median and mode formula. 2021. [online] Available through < http://byjus.com/mean-
median-mode-formula/>