1. REPRESENTATION
OF DATA

2. Histograms
A Histogram is a graphical representation of the distribution of data.
It consists of adjacent rectangles with an area equal to the frequency
of the observations in the interval.
The height of a rectangle is equal to the frequency density of the
interval.
Frequency
Class width
Frequency density =
The rectangles of a histogram are drawn so that they touch each other
(i.e. no gaps as a bar chart has) to indicate that the original variable is
continuous.

3. Example 1: The histogram below shows the speed in miles per hour, of
cars on a motorway. 50 60 90 70 80 1 6 5 4 3 2
Frequency
Density
Speed (m.p.h.)
Complete the frequency table. x
50-55
55-60
60-65
65-75
75-90
Frequency 12 20 30
Estimate the number of cars with a speed of between 70m.p.h.
and 85 m.p.h.
b) Find an estimate of the mean speed of the cars.

4. For 65-75: For 75-90: Frequency density = Frequency Class width50 60 90 70 80 1 6 5 4 3 2
Frequency
Density
For 65-75:
For 75-90:
Speed (m.p.h.)
Frequency density =
Frequency
Class width
f.d. × c.w = frequency
frequency = 3 × 10 = 30
frequency = 1 × 15 = 15 x
50-55
55-60
60-65
65-75
75-90
Frequency 12 20 30 30 15

5. a) Half of the cars in the 65 – 75 group have a speed of 70 m.p.h.
For the number of cars with a speed
of between 70m.p.h. and 85 m.p.h. 50 60 90 70 80 1 6 5 4 3 2
Frequency
Density
We want to find the number of cars
represented by the shaded region.
Speed (m.p.h.) x
50-55
55-60
60-65
65-75
75-90
Frequency 12 20 30 15
a) Half of the cars in the 65 – 75 group have a speed of 70 m.p.h.
or more.
Two thirds of the cars in the 75 – 90 group have a speed
of 70 m.p.h. or more.
= 25 cars have speeds between 70 m.p.h. and 85 m.p.h.

6. x
50-55
55-60
60-65
65-75
75-90
Frequency 12 20 30 15
b) For the mean, the mid-points of x are needed.
Speed x f fx
50 – 55 12
55 – 60 20
60 – 65 30
65 – 75
75 – 90 15
52.5
57.5
62.5 70
82.5
630
1150
1875
2100
1237.5
6992.5
107 =
Totals:
107
6992.5
= 65.35
The mean speed of the cars is 65.4 m.p.h. (3 sig.figs)

7. Example 2: In a fitness centre survey a random sample of 100 men
were asked how many hours, to the nearest hour, they spent
jogging in the last week. The results are summarised below.
Number of hours
Frequency
0 – 2 17
3 – 5 24
6 – 10 29
11 – 15 30
A histogram was drawn and the group (3 – 5) hours was represented
by a rectangle that was 1.5 cm wide and 12 cm high.
Calculate the width and height of the rectangle representing the
group (11 – 15) hours.
The height of each rectangle is proportional to the frequency density.
Frequency
Class width
Frequency density =

8. For the (3 – 5) group, the class width of 3 is represented by 1.5cm.
Number of hours
Boundaries
Frequency
Frequency density
0 – 2 17
3 – 5 24
6 – 10 29
11 – 15 30
2.5 – 5.5 8
10.5 – 15.5 6
2.5
5.5 8 6
15.5
10.5
12cm
1.5cm h w
For the (3 – 5) group, the class width of 3 is represented by 1.5cm.
For the (11 – 15) group, the class width of 5 is represented by
2.5cm.
For the (3 – 5) group, the f.d. of 8 is represented by 12cm.
Each unit of f.d. is represented by
1.5cm.
For the (11 – 15) group, the f.d of 6 is represented by
9cm.

9. Stem and leaf diagrams
A stem and leaf diagram is a way of displaying numerical data and
shows the shape of the data (the distribution).
A simple stem and leaf diagram contains two columns separated by a
vertical line. The left column contains the stems and the right column
contains the leaves.
To draw a stem-and-leaf diagram, the data is sorted in ascending order.
Box Plots
A box plot (or box and whisker diagram) is based on five key values
for a set of data:
The smallest value, the largest value and the three quartiles – the
upper and lower quartile and the median.
They also show outliers (extreme values).

10. b) Show that 75 is the only outlier.
Example 3: In a study of how students use their mobile phones, the
usage of a random sample of 19 students was examined for
a particular day. The length of the calls for the 19 students are shown in
the stem and leaf diagram.
(2)
(4)
(6)
(5)
(0)
Key: 1 | 6 means a time of 16 minutes 6 7 5
(1)
a) Find the median and
quartiles for these data.
A value that is greater than Q × (Q3 – Q1) or smaller than
Q1 – 1.5 × (Q3 – Q1) is defined as an outlier.
b) Show that 75 is the only outlier.
c) Draw a box plot for these data.

11. a) For non-grouped data the median is the 10th = 37
(2)
(4)
(6)
(5)
(0)
Key: 1 | 6 means a time of 16 minutes 6 7 5
(1)
(n + 1)th 2 =
(19 + 1)th 2 =
a) For non-grouped data the median is the
10th
= 37
This leaves 9 values above and 9 values below the median.
The lower quartile and upper quartile are the middle values of these sets.
(9 + 1)th 2
i.e. Q1 =
= 5th value
= 26
Q3 = 5th value from the largest value
= 44

12. Hence 75 is the only outlier.
We now have:
Q1 = 26
1 5 6 5
6 0
7 5
The median, Q2 = 37
Q3 = 44
Q × (Q3 – Q1) =
× (44 – 26) = 71
Q1 – 1.5 × (Q3 – Q1) =
26 – 1.5 × (44 – 26) =
– 1
Hence 75 is the only outlier.
We also have:
The smallest value is 15.
The largest value is 60 (excluding the outlier).
(Note: The line on the box plot here can also be placed at 71). 20 10 40 30 60 50 70 80
Time taken (minutes)

13. Skewness
Skewness is a measure of the asymmetry of a set of data.
A distribution which is symmetrical
has zero skewness.
A distribution which has a longer tail on
the right is positively skewed.
The mean > median and Q3 – Q2 > Q2 – Q1.
A distribution which has a longer tail on
the left is negatively skewed.
The mean < median and Q3 – Q2 < Q2 – Q1.

14. The data is negatively skewed.
1 5 6 5
6 0
7 5
In Example 3, we found:
Q1 = 26
Q3 = 44
The median, Q2 = 37
The mean is: …… 19
= 36.3
So the mean < median
The data is negatively skewed.
Also, Q3 – Q2 = 44 – 37 = 7
Q2 – Q1 = 37 – 26 = 11
So, Q3 – Q2 < Q2 – Q1
The data is negatively skewed.

15. Measures of Average
There are three main measures of an average or typical value for a
set of data:
The mean – the arithmetic average
The median – the middle value
The mode – the most common value.
Measures of Spread
There are several ways to measure the spread of a set of data:
The range : The largest value minus the smallest
The interquartile range: The range of the middle half of the data
IQR = Q3 – Q1.
We shall also look at the standard deviation and variance later.

16. Again in Example 3, we found:
1 5 6 5
6 0
7 5
Again in Example 3, we found:
Q1 = 26
Q3 = 44
The median, Q2 = 37
The range =
75 – 15 = 60
The interquartile range =
44 – 26 = 18
The mode =
26 and 33
In this case there are two modes, this is known as a bimodal
distribution.

17. A histogram consists of adjacent rectangles with an area equal to the
Summary of key points:
Histograms
A histogram consists of adjacent rectangles with an area equal to the
frequency of the observations in the interval. The height of a rectangle
is equal to the frequency density of the interval.
Frequency
Class width
Frequency density =
Stem and leaf diagrams
A simple stem and leaf diagram contains two columns separated by a
vertical line. The left column contains the stems and the right column
contains the leaves.
Box Plots
A box plot is based on five key values for a set of data: The smallest value, the largest value and the three quartiles – the upper quartile, the lower quartile and the median.
This PowerPoint produced by R.Collins ; Updated Feb. 2014