# 4) S1 Representation of Data - Copy

```Representation of Data
• In the last unit we have focused on the
definition of statistics and also looked at
what a population and a sample entails.
• In this unit, we shall look at Visual
representations of data, which are used often
• In short,we will be looking at ways of
representing data visually
• There will also be use of what we have learnt
so far, and when it is appropriate to use them
Representation of Data
Stem and Leaf Diagrams
You will have seen stem and leaf diagrams on your GCSE. They
are also on A-level, but you will be asked more questions on
them.
20, 9, 17, 12, 28, 31, 22, 24, 17, 25, 24, 24, 26
Stem
Leaf
Stem
0
9
1
7
2
7
2
0
8
2
3
1
4
5
4
4
The leaf will usually be the last
number, and the stem the rest.
6
Leaf
0
9
1
2
7
7
2
0
2
4
3
1
4
4
5
6
8
Make sure the data is in order!
4A
Representation of Data
Twin Stem and Leaf Diagrams
Sometimes you will have 2 sets of data on one diagram. The following
numbers represent flower widths for 2 different plants of the same
species (cm).
Plant 1
2.5
2.1
3.0
3.2
1.9
1.5
1.9
2.2
2.4
Plant 2
3.1
2.6
2.9
3.3
3.5
4.0
3.7
2.7
3.0
Plant 2
7
5
Stem
Plant 1
1
5
9
9
4
9
7
6
2
1
2
3
1
0
3
0
2
0
4
Key: 6 | 2 | 1
5
Means 2.1 for
plant 1 and 2.6
for plant 2
4A
Representation of Data
Twin Stem and Leaf Diagrams
Calculate the Median and Inter-quartile range for the following Stem
and Leaf diagram.
Stem
2
Q2 
n
2
13
2
Q1 
n
4
13
4
Q3 
3n
4
39
4
Leaf
3
6
3
1
5
7
7
4
0
3
3
4
5
2
13 Numbers
8
9
Q3 – Q1 
6.5 (7th term)
38
3.25 (4th term)
35
9.75 (10th term)
43
43 - 35 = 8
4A
Representation of Data
Twin Stem and Leaf Diagrams
Calculate the Median and Inter-quartile range for the following Stem
and Leaf diagram.
Stem
6
Q2 
n
2
14
2
Q1 
n
4
14
4
Q3 
3n
4
42
4
Leaf
1
2
7
0
1
8
1
4
9
0
14 Numbers
5
2
5
3
8
6
7
Q3 – Q1 
7 (7.5th term)
71.5
3.5 (4th term)
65
10.5 (11th term)
77
77 - 65 = 12
4A
Representation of Data
Outliers
An outlier is an extreme value that lies outside the overall pattern of data.
An outlier is any value that is;
Bigger than;
Upper Quartile + (1.5 x Inter-quartile Range)
 Q3 + 1.5(IQR)
Smaller than;
Lower Quartile – (1.5 x Inter-quartile Range)
 Q1 – 1.5(IQR)
So basically, work out ‘1.5 x IQR’. Then add it to the upper quartile, subtract it
from the lower quartile and you have the acceptable range of values.
The rules above are standard but you may be given a different rule to apply in
the exam.
4B
Representation of Data
Outliers
For the Stem and Leaf diagram below, calculate the quartiles and find any
outliers.
Stem
Leaf
2
30
2
15 (15.5th term)
n
4
30
4
7.5 (8th term)
3n
30
4
22.5 (23rd term)
3.8
Key: 3 | 1 means 3.1
2
2 2 3 3 5 7
3
1 2 6 7 7 7 8 8 8 8 9 9 9
4
0 0 0 0 4 5 6 7 8
5
1 5
30 Numbers
Q2  n
Q1 
Q3  4
3.2
4.0
Q3 – Q1  4.0 – 3.2 = 0.8
4B
Representation of Data
Outliers
For the Stem and Leaf diagram below, calculate the quartiles and find any
outliers.
Q1 = 3.2
Stem
Leaf
Key: 3 | 1 means 3.1
2
2 2 3 3 5 7
3
1 2 6 7 7 7 8 8 8 8 9 9 9
4
0 0 0 0 4 5 6 7 8
5
1 5
30 Numbers
Q2 = 3.8
Q3 = 4.0
Lowest
acceptable
value
IQR = 0.8
Highest
acceptable
value
Q1 – 1.5(IQR)
Q3 + 1.5(IQR)
3.2 – 1.5(0.8)
4 + 1.5(0.8)
2
 5.2
So 5.5 is the only outlier.
4B
Representation of Data
Box Plots and comparing data
Smallest
value
Lower
Quartile
Outlier
25%
10
20
30
Median
25% 25%
40
50
Upper
Quartile
Largest
value
25%
60
70
80
 Any outliers are plotted as crosses outside the main plot
 Each ‘section’ contains 25% of the observations in the sample
4C/4D
Representation of Data
Drawing the box plot
Stem
Leaf
Q1 = 3.2
Key: 3 | 1 means 3.1
2
2 2 3 3 5 7
3
1 2 6 7 7 7 8 8 8 8 9 9 9
4
0 0 0 0 4 5 6 7 8
5
1 5
Q2 = 3.8
Q3 = 4.0
Lowest
acceptable
value
IQR = 0.8
Highest
acceptable
value
Q1 – 1.5(IQR)
Q3 + 1.5(IQR)
3.2 – 1.5(0.8)
4 + 1.5(0.8)
2
 5.2
So 5.5 is the only outlier.
2
2.5
3
3.5
4
4.5
5
5.5
4C/4D
Representation of Data
•
Drawing the box plot
The blood glucose level of 30 males is recorded. Below is a summary of the
results.
Lower Quartile = 3.6
Lowest Value = 1.4
Upper Quartile = 4.7
Median = 4
Highest Value = 5.2
Given that there was only one outlier, draw a box plot for the data.
As we do not know
the actual lowest
value, we use the
lower boundary (1.95)
IQR = 4.7 – 3.6
= 1.1
Max value
= 4.7 + 1.5(1.1)
= 6.35
Min value
= 3.6 – 1.5(1.1)
= 1.95
So 1.4 is the outlier.
1
2
3
4
5
6
4C/4D
Representation of Data
Comparing Box Plots
Females
Males
1
2
3
4
5
6
Glucose Level
When you compare 2 box plots you should always comment on the
Median and the Inter-quartile range.
This is because Median is a measure of location (average), and the
Inter-quartile range is a measure of spread.
The median is higher for males, and they also have a larger Interquartile range. This indicates that males have a higher blood glucose
level on average, and also have a wider range of values.
4C/4D
Representation of Data
Histograms
A Histogram is similar to a bar chart but there are 2 major differences
 There are no gaps between bars (continuous data)
 The area of a bar is proportional to the frequency
When drawing a Histogram, use Frequency Density rather than frequency.
Frequency
Frequency
=
Density
Class width
You may also need to use the following formula when interpreting a Histogram.
Area of Bar = k x Frequency
Usually the Area of the bar is equal to the frequency. But it may be that all
areas have been halved (ie k = 0.5) in order to make the diagram smaller.
4E
Representation of Data
Frequency
=
Density
Histograms
The following table shows how long a
sample of 200 students took to complete
their homework. Draw a Histogram to
represent the data.
Frequency
Class width
14
12
Frequency
Frequency
Density
25-30
55
11
(55 &divide; 5)
8
30-35
39
7.8
(39 &divide; 5)
6
35-40
68
13.6
(68 &divide; 5)
40-50
32
3.2
(32 &divide; 10)
50-80
6
0.2
(6 &divide; 30)
Frequency Density
Time (mins)
10
4
2
0
20
30 40 50 60 70 80 90
Time (mins)
4E
Representation of Data
As Area represents Frequency, we need to
calculate the Area of each Rectangle we
are including.
Rectangle 1:
 4 x 13.6
 54.4 students
Rectangle 2:
 5 x 3.2
 16 students
Overall our
estimate would be
70.4 (70) students
between 36 and 45
minutes.
36 to 45
13.6
14
12
Frequency Density
Histograms
Use the Histogram to estimate the
number of students whose times were
between 36 and 45 minutes.
3.2
10
8
6
1
4
2
0
20
2
30 40 50 60 70 80 90
Time (mins)
4E
Representation of Data
Histograms
The Histogram to the right shows the
time taken (s) for a group of children to
complete a puzzle.
Why has a Histogram been used?
 Time is Continuous Data
What is the underlying feature of each
bar?
 It is proportional to the group
Frequency
14
16
18
20
22
24
26
28 30 32
Time (s)
4E
Representation of Data
Histograms
The Histogram to the right shows the
time taken (s) for a group of children to
complete a puzzle.
Bar A represents 78 children. What Area
represents 1 child?
27.3
Area represents Frequency
 2 x 27.3
 54.6cm2
A
78 Children = 54.6cm2
1 Child =
0.7cm2
&divide; 78
14
16
18
20
2
22
24
26
28 30 32
Time (s)
4E
Representation of Data
Histograms
The Histogram to the right shows the
time taken (s) for a group of children to
complete a puzzle.
1 Child = 0.7cm2
If the Area is 210cm2 in total, how many
children were surveyed?
x 0.7
1 Child = 0.7cm2
? Children = 210cm2
&divide; 0.7
210cm2 &divide; 0.7 =
14
16
18
20
22
24
26
28 30 32
Time (s)
300 Children
4E
Representation of Data
Skewness and Comparisons
The Skewness of data can be described using diagrams, measures of location and
Symmetrical
Positive Skew
Negative Skew
Data which is spread evenly  Symmetrical
Data which is mostly at the lower values  Positive Skew
Data which is mostly at the higher values  Negative Skew
4F
Representation of Data
Skewness and Comparisons
There are several ways of comparing Skewness. Sometimes you will be told which
to use, and sometimes you will have to choose one depending on what data you
have available.
You can see shape of the data from a box plot.
Q1 Q2 Q3
You can also look at
the quartiles
Symmetrical
Q 2 – Q1 = Q3 – Q2
Positive Skew
Q 2 – Q1 &lt; Q3 – Q2
Negative Skew
Q 2 – Q1 &gt; Q3 – Q2
Q1 Q2 Q3
Q1 Q2Q3
4F
Representation of Data
Skewness and Comparisons
There are several ways of comparing Skewness. Sometimes you will be told which
to use, and sometimes you will have to choose one depending on what data you
have available.
Another test uses the measures of location:
Symmetrical 
mean = median = mode
Positive Skew 
mean &gt; median &gt; mode
Negative Skew 
mean &lt; median &lt; mode
Low mode = lots of low values ie)
Positive Skew
High mode = lots of high values
ie) Negative Skew
4F
Representation of Data
Skewness and Comparisons
There are several ways of comparing Skewness. Sometimes you will be told which
to use, and sometimes you will have to choose one depending on what data you
have available.
The final test is a formula:
Negative Skew
3(Mean – Median)
Standard Deviation
Symmetrical
0
Positive Skew
A value of 0 implies that mean = median  Symmetrical Data
A positive value implies that median &lt; mean  Positive Skew
A negative value implies that median &gt; mean  Negative Skew
The further from 0 a positive or negative value is, the more skewed the data is.
4F
Representation of Data
Skewness and Comparisons
Find the 3 Quartiles for this data on
test marks for 50 students.
Q2 
Q1 
Q3 
n
2
n
4
3n
4
50
2
50
4
150
4
25
(25.5th
term)
60
12.5 (13th term)
46
37.5 (38th term)
69
Stem
Leaf
Key: 6 | 1 means 61
2
1 2 8
3
3 4 7 8 9
4
1 2 3 5 6 7 9
5
0 2 3 3 5 5 6 8 9 9
6
1 2 2 3 4 4 5 6 6 8 8 8 9 9
7
0 2 3 4 5 7 8 9
8
0 1 4
4F
Representation of Data
Skewness and Comparisons
Given the two values below, calculate
the Mean and Standard Deviation of
the data.
 x  2873  x
Mean
x
n
2873
50
x  57.46
2
Stem
 177353 n  50
Standard Deviation
2
2


x
x
 2      
n
 n 
177353  2873 
2 


50
 50 
2
 2  245.4084
  15.67
(2dp)
Leaf
Key: 6 | 1 means 61
2
1 2 8
3
3 4 7 8 9
4
1 2 3 5 6 7 9
5
0 2 3 3 5 5 6 8 9 9
6
1 2 2 3 4 4 5 6 6 8 8 8 9 9
7
0 2 3 4 5 7 8 9
8
0 1 4
Q1 = 46
Q2 = 60
Q3 = 69
4F
Representation of Data
Skewness and Comparisons
Use the formula below to calculate the
Skewness of the data.
Stem
Key: 6 | 1 means 61
Leaf
2
1 2 8
3
3 4 7 8 9
4
1 2 3 5 6 7 9
3(57.46 - 60)
5
0 2 3 3 5 5 6 8 9 9
15.67
6
1 2 2 3 4 4 5 6 6 8 8 8 9 9
7
0 2 3 4 5 7 8 9
8
0 1 4
3(Mean – Median)
Standard Deviation
-7.62
15.67
= -0.486
So the data is Negatively Skewed!
Q1 = 46
Mean = 57.46
Q2 = 60
Standard Deviation = 15.67
Q3 = 69
Mode = 68
4F
Representation of Data
Skewness and Comparisons
Use another two methods to show the
data is Negatively Skewed.
1) Q2 – Q1 = 14
Q 3 – Q2 = 9
Q 2 – Q1 &gt; Q 3 – Q2
 Negative Skew
2) Mean &lt; Median &lt; Mode
57.46 &lt; 60 &lt; 68
High mode implies many higher values…
 Negative Skew
Stem
Key: 6 | 1 means 61
Leaf
2
1 2 8
3
3 4 7 8 9
4
1 2 3 5 6 7 9
5
0 2 3 3 5 5 6 8 9 9
6
1 2 2 3 4 4 5 6 6 8 8 8 9 9
7
0 2 3 4 5 7 8 9
8
0 1 4
Q1 = 46
Mean = 57.46
Q2 = 60
Standard Deviation = 15.67
Q3 = 69
Mode = 68
4F
Representation of Data
Skewness and Comparisons
A company runs two manufacturing lines, A and B. They both make 2cm
rods in different ways. Samples are taken from both lines and data
summarised in the following table. Which manufacturing line is best in
this situation?
Mean
Standard
Deviation
A
2
0.015
B
2
0.05
The rods need to be accurate…
 Standard Deviation measures
 The rods from line A have a lower
Standard Deviation
 Line A is therefore more reliable
4F
Representation of Data
Skewness and Comparisons
This table shows data on pupils taking a Statistics and Mechanics Paper.
Which will be easier to set fair grade boundaries for?
Mean
Standard
Deviation
Statistics
55
16
Mechanics
55
4
 A higher standard deviation
means the marks are more spread
out
will be more spread out for
Statistics
 And will therefore be fairier!
4F
Summary
• We have looked at using Stem and Leaf
diagrams and Histograms to represent data
• We have looked at comparing data using
these, as well as box plots
• We have learnt what outliers are
• We have learnt what Skewness is and used
several measures to test it
```