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Ebola
Maths warm-up (2–3 hours)



Recap Stem and Leaf Diagrams

Recap types of data
Recap Histograms
Recap Cumulative Frequency and
Box Plots
2
Watch the following link. What does it
show?
http://www.breathingearth.net/
3
Ebola by numbers
4
Video clip – Ebola
https://www.youtube.com/watch?v=JNi
H18JNmqA
5
Facts and figures on Ebola
2014 West Africa outbreak:








Guinea – 2339 cases, 1454 deaths
Liberia – 7765 cases, 3222 deaths
Mali – 8 cases, 6 deaths
Nigeria – 20 cases, 8 deaths
Senegal – 1 case, 0 deaths (infection originated in Guinea)
Sierra Leone – 8014 cases, 1857 deaths
Spain – 1 case, 0 deaths
United States – 4 cases, 1 death (two infections originated in
the United States, one in Liberia and one in Guinea)
6
What is the death rate?
Calculate the percentage death rate for each
of these countries.
Country
Total cases
Total deaths
Guinea
2394
1518
Liberia
7797
3290
Sierra Leone
8273
2033
United States
4
1
United Kingdom
1
0
Total
18469
13682
7
‘If you catch Ebola, you’ll
almost certainly die’

The most widely cited figure about Ebola is that
its death rate is ‘up to 90%’. The history of Ebola,
prior to this year, is a series of short-lived and
very isolated outbreaks of different strains of the
disease, and it is true that one of these outbreaks
had a fatality rate of 90%.
 Is this is a fair representation?
8
‘If you catch Ebola, you’ll
almost certainly die’
Year
Case fatality
2014
2012
2012
2012
2011
2008
2007
2007
2005
2004
2003
2002
58%
51%
57%
71%
100%
44%
25%
71%
83%
41%
90%
75%
Does this support the
news headline?
What can we do with
this data?
What information do we
need to reach a more
accurate conclusion?
9
‘Ebola is becoming more deadly’
Guinea
Frequency of
Date
deaths
Liberia
Frequency of
Date
deaths
1/3/14 –
31/5/14
155
1/3/14 –
31/5/14
11
1/6/14 –
31/7/14
148
1/6/14 –
31/7/14
54
1/8/14 –
31/10/14
407
1/8/14 –
31/10/14
1933
1/11/14 –
30/11/14
331
1/11/14 –
30/11/14
699
1/12/14 –
31/12/14
387
1/12/14 –
31/12/14
480
10
‘Ebola is becoming more deadly’

Create a histogram to compare the
number of deaths in 2014 for both
Guinea and Liberia.

Is Ebola becoming more deadly?
11
‘40% of people that catch
Ebola are below 18 years of age’
Age
Male
Female
Total
Newborn and infant
10
14
24
1–14 years
18
22
40
15–29
31
60
91
30–49
57
52
109
50 or over
23
26
49
Unknown
2
3
5
141
177
318
Total
12
‘40% of people that catch
Ebola are below 18 years of age’


Is this statement true?
Create a histogram for males and females
to represent this data and calculate what
percentage of cases are aged under 18.
13
‘The Ebola death rate is
increasing at a rapid pace monthly’

Using the data in the frequency tables
following, create a cumulative frequency
graph to justify this statement. (Both
countries should be on the same diagram.)
14
Death rates in Guinea and Liberia
Guinea
Liberia
Date
Frequency of deaths
Date
Frequency of deaths
01/03/14 – 31/03/14
0
01/03/14 – 31/03/14
0
01/04/14 – 30/04/14
80
01/04/14 – 30/04/14
2
01/05/14 – 31/05/14
75
01/05/14 – 31/05/14
9
01/06/14 – 30/06/14
38
01/06/14 – 30/06/14
21
01/07/14 – 31/07/14
110
01/07/14 – 31/07/14
33
01/08/14 – 31/08/14
43
01/08/14 – 31/08/14
162
01/09/14 – 30/09/14
171
01/09/14 – 30/09/14
862
01/10/14 – 31/10/14
193
01/10/14 – 31/10/14
909
01/11/14 – 30/11/14
331
01/11/14 – 30/11/14
699
01/12/14 – 31/12/14
387
01/12/14 – 31/12/14
480
15
Has Guinea or Liberia been more
affected by Ebola?

Using your cumulative frequency graph, create a
boxplot to compare the death rate in Guinea to
that in Liberia.

Which country has been more affected by the
outbreak of Ebola?
16
The Ebola exponential
17
Exponential growth


What’s the same? What’s different?
What gives these graphs their shape?
18
Exponential growth

Exponential growth occurs when the growth
rate of the value of a mathematical function
is proportion to the function’s current value.

Can you think of examples where exponential
growth would happen in real life?
19
Exponential growth
Video on the ‘burial boys’ http://nyti.ms/1pqVE2T
20
Exponential growth
How does a rumour spread among a population?
On day 1, a single person tells someone else a rumour; suppose
that on every subsequent day, each person who knows the
rumour tells exactly one other person the rumour.
How many days until:
 50 people have heard the rumour?
 The whole school?
 The whole country?
What is the percentage transmission rate?
If we can half the transmission rate (one person every two days),
how much longer will it take to infect the whole country?
21
Exponential growth
Graph showing the number of confirmed
cases and the number of deaths
22
Exponential growth
Confirmed cases each month in 2014
23
‘Ebola is the most deadly
virus of our generation’
24
Visual spread of diseases
Http://www.washingtonpost.com/wp-srv/special/health/how-ebola-spreads/
25
Comparing how deadly
Ebola is to other diseases
26
‘Ebola is the most deadly
virus of our generation’
Is this headline correct?
You must use all the data given to write a report
about your findings.
You must compare data from other diseases and
finish with a conclusion (minimum 500 words).
27
Sampling
28
Sampling – what is it?
There are five types of sampling:
1.
2.
3.
4.
5.
Random
Systematic
Stratified
Quota
Cluster
29
Starter
Methods of sampling
Random
Systematic
What is it?
Advantages
Disadvantages
30
Stratified
Which is most effective?

… if you are asking people what shoe size
they are in the class.

… if you are looking at the fatality rate when
catching a disease.
31
Stratified sampling

When the population is composed of
different groups of people (e.g. different
genders, different ages, different social
classes etc.), we may wish to choose our
sample so that it contains the same
proportion of each group as the entire
population.

For example, if 60% of our population is
female, a stratified sample would be 60%
female too.
32
Stratified sampling – Example 1
The number of students in each year group at a
school is shown in the table:
Year group
Number of students
7
180
8
200
9
170
10
190
11
160
Total
900
Suppose that a stratified sample of size 10%
(90 students) needs to be chosen.
33
Stratified sampling – Example 1
As our sample size is 10% of the entire population of the
school, we would choose 10% of each year group. Our sample
would therefore be composed as follows:
Year group
Number of students
7
18
8
20
9
17
10
19
11
16
Total
90
Having decided on the number of students from each year
group, the actual students would then be picked at random
from those in the year group (e.g. using a random number
generator).
34
Stratified sampling – Example 2
The table shows the number of boys and the number of
girls in each year group at Springfield Secondary School.
There are 500 boys and 500 girls in the school.
Year group
7
8
9
10
11
Total
Number of
boys
100
150
100
50
100
500
Number of
girls
100
50
100
150
100
500
10% of girls
10
50
10
15
10
50
Aaron took a stratified sample of 10% of the girls, by year group.
Work out the number of Year 8 girls in his sample.
Have a go at questions 1, 2, and 3…
35
Stratified sampling – Example 3
The table shows the number of people in each
age group who watched the school sports.
Martin did a survey of these people.
Age group
0–16
17–29
30–44
45–59
60+
Number of
people
177
111
86
82
21
He used a stratified sample of exactly 50 people
according to age group.
Work out the number of people from each age group
that should have been in his sample of 50.
36
Stratified sampling – Example 3
We first find the total number of students who
watched the school sports:
177 + 111 + 86 + 82 + 21 = 477
The number of people aged 0–16 in the sample
needs to be of size:
177
´ 50 = 18.55....
477
As the number of people chosen for the sample
must be a whole number, we would round this to 19.
37
Stratified sampling – Example 3
We can repeat this approach for all other age groups:
Age group
Number of
people in
sample
0–16
19
17– 29
30–44
45–59
60+
Total
111
 50
477
86
 50
477
82
 50
477
21
 50
477
= 11.64…
= 9.01…
5
1
= 8.60...
= 2.20…
= 12
=9
=9
=2
Notice that our total is incorrect. This is due to rounding. To correct this we
must reduce the number of students from one of the age groups by 1. The
calculation for the number of students in the 0–16 age group resulted in
the answer 18.55… (only just big enough to round up). We could reduce
the number of students in this age group to 18.
Have a go at questions 4, 5, and 6…
38
Case study
The table shows the number of Ebola cases
and deaths in the main countries affected.
Country
Total cases
Total deaths
Guinea
2394
1518
Liberia
7797
3290
Sierra Leone
8273
2033
United States
4
1
United Kingdom
1
0
18469
13682
Total
39
Case study

Make another column stating the total
deaths as a percentage.

A newspaper article stated: ‘25% of people
who are diagnosed with Ebola die from it.’
Is this true? What is this data based on?
What would be a more accurate statement?
40
Case study

The government would like to look further
into the relationship between the number of
people diagnosed with Ebola and the death
rate. They use two different methods.
41
Method 1
The government would like to look further into the
death rate of Ebola patients. They cannot look at all
13,000 so they take 300 people from each country.
1.
What is the problem with this?
2.
What would a solution be?
3.
If 300 people are asked from Guinea, what
would the death rate for these people be?
42
Method 2
The government would like to look further into the
death rate of Ebola patients. They cannot look at all
13,000 so they take a stratified sample of 1,500.
1.
How many people are asked from each
country?
2. From this sample, what was the death rate for
each country?
43
Case study

Which method is more accurate, and why?
44
Standard deviation
45
Maths warm-up

Recap calculating the mean of grouped
data.

Complete the worksheet on calculating
mean from a table.
46
Two classes took a recent quiz. There were
10 students in each class, and each class
had an average score of 81.5
47
Since the averages are the same, can we
assume that the students in both classes all
did pretty much the same in the exam?
48
The answer is… No.
The average (mean) does not tell us anything
about the distribution or variation in the
grades.
49
Following are Dot-Plots of the grades
in each class…
50
Mean
51
So, we need to come up with some way of
measuring not just the average, but also the
spread of the distribution of our data.
52
Why not just give an average and the range
of data (the highest and lowest values) to
describe the distribution of the data?
53
Well, for example, let’s say from a set of data,
the average is 17.95 and the range is 23.
But what if the data looked like this:
54
But really, most
of the numbers
are in this area,
and are not
evenly distributed
throughout the
range.
Here is the average
And here is the range
55
The standard deviation is a number that
measures how far away each number in
a set of data is from their mean.
56
If the standard deviation is large,
large, it means
the numbers are spread out from their mean.
small, it means the
If the standard deviation is small,
numbers are close to their mean.
57
Here are the
scores in the
maths quiz
for Team A:
72
76
80
80
81
83
84
85
85
89
Average:
81.5
58
The standard deviation measures how far away
each number in a set of data is from their mean.
For example, start with the lowest score, 72.
How far away is 72 from the mean of 81.5?
72 – 81.5 = - 9.5
Team A Quiz Grades
- 9.5
59
Or, start with the highest score, 89.
How far away is 89 from the mean of 81.5?
89 – 81.5 = 7.5
Team A Quiz Grades
- 9.5
60
7.5
Distance
from mean
So, the first
step to finding
the standard
deviation is
to find all the
distances from
the mean.
72
- 9.5
76
80
80
81
83
84
85
85
7.5
89
61
Distance
from mean
So, the first
step to finding
the standard
deviation is
to find all the
distances from
the mean.
72
- 9.5
76
- 5.5
80
- 1.5
80
- 1.5
81
- 0.5
83
1.5
84
2.5
85
3.5
85
3.5
89
7.5
62
Next, you need
to square each
of the distances
to turn them all
into positive
numbers
Distance
from mean
Distances
squared
72
- 9.5
90.25
76
- 5.5
30.25
80
- 1.5
80
- 1.5
81
- 0.5
83
1.5
84
2.5
85
3.5
85
3.5
89
7.5
63
Next, you need
to square each
of the distances
to turn them all
into positive
numbers
Distance
from mean
Distances
squared
72
- 9.5
90.25
76
- 5.5
30.25
80
- 1.5
2.25
80
- 1.5
2.25
81
- 0.5
0.25
83
1.5
2.25
84
2.5
6.25
85
3.5
12.25
85
3.5
12.25
89
7.5
56.25
64
Add up all of
the distances
Distance
from mean
Distances
squared
72
- 9.5
90.25
76
- 5.5
30.25
80
- 1.5
2.25
80
- 1.5
2.25
81
- 0.5
0.25
83
1.5
2.25
84
2.5
6.25
85
3.5
12.25
85
3.5
12.25
89
7.5
56.25
65
Sum:
214.5
Divide by n
where n
represents the
amount of
numbers you
have.
Distance
from mean
Distances
squared
72
- 9.5
90.25
76
- 5.5
30.25
80
- 1.5
2.25
80
- 1.5
2.25
10
81
- 0.5
0.25
= 21.45
83
1.5
2.25
84
2.5
6.25
85
3.5
12.25
85
3.5
12.25
89
7.5
56.25
66
Sum:
214.5
Finally, take the
square root of
the average
distance
Distance
from mean
Distances
squared
72
- 9.5
90.25
76
- 5.5
30.25
80
- 1.5
2.25
80
- 1.5
2.25
(10 – 1)
81
- 0.5
0.25
= 23.8
83
1.5
2.25
= 4.88
84
2.5
6.25
85
3.5
12.25
85
3.5
12.25
89
7.5
56.25
67
Sum:
214.5
This is the
standard
deviation
Distance
from mean
Distances
squared
72
- 9.5
90.25
76
- 5.5
30.25
80
- 1.5
2.25
80
- 1.5
2.25
10
81
- 0.5
0.25
= 21.45
83
1.5
2.25
= 4.63
84
2.5
6.25
85
3.5
12.25
85
3.5
12.25
89
7.5
56.25
68
Sum:
214.5
Now find the
standard
deviation for
the other class
grades
Distance
from mean
Distances
squared
57
- 24.5
600.25
65
- 16.5
272.25
83
1.5
2.25
94
12.5
156.25
(10)
95
13.5
182.25
= 228.05
96
14.5
210.25
= 15.1
98
16.5
272.25
93
11.5
132.25
71
- 10.5
110.25
63
-18.5
342.25
69
Sum:
2280.5
Now, let’s compare the two
classes again:
Team A
Team B
Average in
the Quiz
81.5
81.5
Standard
deviation
4.63
15.1
70
Hint:
1. Find the mean of the data.
2. Subtract the mean from each value
– called the deviation from the mean.
3. Square each deviation from the mean.
4. Find the sum of the squares.
5. Divide the total by the number of items
– result is the variance.
6. Take the square root of the variance
– result is the standard deviation.
71
Solve:
 A maths class took a test with these five test
scores: 92, 92, 92, 52, 52.
 Find the standard deviation for this class.
ANSWER NOW
72
The maths test scores of five
students are: 92, 92, 92, 52 and 52.
1. Find the mean: (92+92+92+52+52)/5 = 76
2. Find the deviation from the mean:
92–76=16
92–76=16
92–76=16
52–76= –24 52–76= –24
3. Square the deviation from the mean:
4. Find the sum of the squares:
256+256+256+576+576 = 1920
73
The maths test scores of five
students are: 92, 92, 92, 52 and 52.
5. Divide the sum of the squares by the number of
items: 1920/5 = 384 variance
6. Find the square root of the variance:
384  19.6
Thus the standard deviation of the second
set of test scores is 19.6.
74
Standard deviation practice
Complete questions…
75
Example 2
Find the mean and standard deviation for the following distribution:
x
f
fx
fx2
0
1
0
0
1
3
3
3
2
7
14
28
We must now complete the table, and
calculate the totals of f, fx and fx2:
3
5
15
45
Note: f x 2 means f × x2 and not (f × x)2.
4
4
16
64
20

48 140
 fx
f
x
Frequency
x=
 fx
f
0
1

1
3
2
7
3
5
x = 48 = 2.4
20
2
 x2

140
 2.4 2 =
20
σ=
76
1.114
4
4
Example 2
Find the mean and standard deviation for the following distribution:
fx
fx2
x
f
0
2
1
6
2
4
We must now complete the table, and
calculate the totals of f, fx and fx2:
3
7
Note: f x 2 means f × x2 and not (f × x)2.
4
8
x
Frequency
x=

 fx
f
 fx
f
2
 x2

σ=
77
0
2

1
6
x=
2
4
3
7
4
8
Standard deviation practice

Find the standard deviation for the averages
from your initial worksheet.
78
Compare the mean and standard
deviation for males and females
Age
Male
Female
Total
0–1
10
14
24
1–15
18
22
40
15–30
31
60
91
30–50
57
52
109
50–70
23
26
49
Total
141
177
318
79
The incubation period for Ebola
80
Incubation period for Ebola

The incubation period is the time that elapses
between exposure to a pathogenic organism, a
chemical or radiation, and when symptoms and
signs are first apparent.

Depending on the disease, the person may or may
not be contagious during the incubation period.
 The incubation period, or the time interval from
infection to onset of symptoms, is from 2 to 25
days. The patients become contagious once they
begin to show symptoms. They are not contagious
during the incubation period.
81
Incubation period
Distribution of Ebola virus incubation period, by days of
incubation
82
Days of incubation
Days of Incubation
Number of cases
1
0
2
95
3
420
4
760
5
790
6
690
7
590
8
410
9
310
10
270
11
175
12
130
13
80
14
70
15
40
16
30
17
15
18
20
19
15
20
10
21
15
22
10
23
10
24
5
25
5
What does this tell us about
the incubation period of
Ebola?
1. What is the mean?
2. What is the standard
deviation?
Extension:
How can you represent this
graphically?
83
Starter
Mode of –
567, 600, 356, 600, 400, 500
Mean of –
150, 160, 290
1st quartile of –
2, 5, 9, 11, 14, 23, 30
2nd quartile of –
2, 5, 9, 11, 14, 23, 30
3rd quartile of –
2, 5, 9, 11, 14, 23, 30
Interquartile range of – 2, 5, 9, 11, 14, 34, 36
The range of –
12, 36, 37, 41, 46, 47
Where have you seen the above before?
84
Starter – Answers
Mode of –
600
Mean of –
200
1st quartile of –
5
2nd quartile of –
11
3rd quartile of –
23
Interquartile range of – 34 – 5 = 29
The range of –
47 – 12 = 35
Where have you seen the above before?
– Averages, Cumulative frequency, Box plot…
85
Looking at data
Look at the data below. Write a 50–100 word comparison
on males to females. You may want to represent the data
on a graph and use: Averages, Range and IQR
Age
Male
Female
Total
Newborn and infant
10
14
24
1–14 years
18
22
40
15–29
31
60
91
30–49
57
52
109
50 or over
23
26
49
Unknown
2
3
5
141
177
318
Total
86
1. Quentin’s height is between the median and the
third quartile for all heights in his school. His
height’s percentile rank could be which of the
following?
(1) 45th
(3) 64th
(2) 79th
(4) 23rd
(3) 64th
2. In a certain test, a score of 90 was the 25th
percentile. If 20 students took the test, how many
received scores of 90 or below?
5
87
3. In a mathematics test, Sal scored at the 90th
percentile. Which one of the following statements
is true?
(1) Ninety per cent of the students who took the
test had the same score as Sal had.
(2) Ninety per cent of the students who took the
test had a score equal to or less than Sal’s score.
(3) Sal scored 90% in this test.
(4) Sal answer 90 questions correctly.
(2) Ninety per cent of the students who took the
test had a score equal to or less than Sal’s score.
88
Percentiles
Here are some scores from a mock test:
17, 12, 5, 36, 24, 13, 33, 27, 21
Find the 16% percentile
 Firstly we need to order them:
5, 12, 13, 17, 21, 24, 27, 33, 36
 Then we use the formula
i = (P/100) x n
Percentage to find
Amount of numbers
RULES: Always round up. If it becomes a whole number, add 0.5
 i = (16/100) x 9
so i = 1.44; i = 2 so use the 2nd number
16% percentile from the data = 12
89
Your turn

Find the… 10th and 34th percentile:
13, 85, 99, 69, 56, 42, 42, 99, 101, 157, 76, 58, 57,
140, 142, 70, 132, 77

Find the interpercentile range from the 10th to the
90th percentiles:
16, 63, 55, 32, 89, 70, 49, 89, 19, 92, 86, 17, 52,
27, 39, 30, 87, 21, 15
90
Create a cumulative frequency
Which class does
this lie in?
91
‘21% of females that catch Ebola are
aged 14 and under’
Age
Newborn and infant
1–14 years
15–29
30–49
50 or over
Unknown
Total
Female
Cumulative frequency
14
22
60
52
26
3
177
14
36
96
148
174
177
92
‘45% of males that catch Ebola are aged
over 20’
Age
Male
Cumulative frequency
Newborn and infant
1–14 years
15–29
30–49
50 or over
Unknown
Total
10
18
31
57
23
2
141
10
28
59
116
139
141
93
‘The IQR of people that catch Ebola is
between 18 and 44’
Age
Newborn and infant
1–14 years
15–29
30–49
50 or over
Unknown
Total
Total
Cumulative frequency
24
40
91
109
49
5
318
24
64
155
264
313
318
94
The interpercentile range from the 20th to the 80th
percentile of people that catch Ebola is 38 years.
Age
Male
Female
Total
Newborn and infant
1–14 years
15–29
30–49
50 or over
Unknown
Total
10
18
31
57
23
2
141
14
22
60
52
26
3
177
24
40
91
109
49
5
318
95
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