Representation of data
----histogram
Alpha
Part One : Class boundaries
The heights of 100 students are recorded to the nearest centimeter.
Height ,h(cm) 160-164 165-169 170-174 ….
Frequency 7 9 13 ….
Class boundaries are as below:
Height ,h(cm) 159.5-164.5 164.5-169.5 169.5-174.5 ….
Frequency 7 9 13 ….
Example :
The mass of 40 students are recorded to the nearest kilogram.
Mass ,m(kg) 45-55 55-65 65-75 ….
Frequency 7 9 13 ….
Class boundaries are as below:45 55 65 75 so the table are rewrote as below
Height ,h(cm) 45≤m<55 55≤m<65 65≤m<75 ….
Frequency 7 9 13 ….
If there is no gap between each class, then what are the boundaries?
Special case1--Age
A group of 40 motorists was asked to state the ages at which they passed their driving tests.
Age ,a (year) 17- 20- 23- 26
Frequency 6 11 7 8
Class boundaries are as below: 17 ， 20 ， 23 ， 26
Age ,a (year) 17≤a<20 20 ≤ a<23 23 ≤ a<26 a> 26
Frequency 6 11 7 8
Special case 2- Score
Score 0-9 10-19 20-39 40-49 50-59Frequency 14 9 9 3 5
Score -0.5≤x<9.5 9.5≤x<19.5 19.5≤x<39.5 39.5≤ x <49.5 49.5≤ x<59.5
Frequency
14 9 9 3 5
Class boundaries are as below: -0.5 ,9.5,19.5,29.5,39.5,59.5 so we rewrite the table as below when we draw the diagram.
Part two : frequency density
Two important properties of histogram
• The bars have no spaces between them( though there may be some bars of height zero, which look like spaces)
• The area of each bar is proportional to the frequency
Note: the first one is the main difference between bar chart and histogram
Height, h(cm)
frequency
0 ≤ x<5 3
5 ≤ x<10 5
10 ≤ x<15 11
15 ≤ x<20 6
20 ≤ x<25 3
25 ≤ x<30 2
0 10
5
20
10
30
Height, h(cm)
If each bar has the same width
Height, h(cm)
frequency
0 ≤ x<5 3
5 ≤ x<10 5
10 ≤ x<15 11
15 ≤ x<20 6
20 ≤ x<30 5
10
0 10
5
20 30
Height, h(cm)
When the widths of bar are unequal and we still take the frequencies as the height of bars. A diagram is drawn as below:
Then why this diagram is misleading?
frequency
Height, h(cm)
frequency
0≤ x<5 3
5 ≤x<10 5
10 ≤ x<15 11
15 ≤x<20 6
20 ≤x<30 5
0 10
5
20
10
30Height, h(cm)
Frequency density
So we get:
frequency height of bars= ,
width of clasWhich is known as frequency dens
sity
Now let us draw a histogram
Mass (kg) Frequency
47—54 455—62 763—66 867—74 775—82 883—90 4
Example: The grouped frequency distribution in Table 1.17 represents the masses in kilograms of a sample of 38 of the people from the datafile ”Brain size”. Represent these data in a histogram
Mass, m(kg)
Class boundaries
Class width
frequency
Frequency density
47—54 46.5 ≤x<54.5 8 4 0.555—62 54.5 ≤x<62.5 8 7 0.87563—66 62.5 ≤x<66.5 4 8 267—74 66.5 ≤x<74.5 8 7 0.87575—82 74.5 ≤x<82.5 8 8 183—90 82.5 ≤x<90.5 8 4 0.5
First, draw a table and calculate the boundaries and frequency densities.
50 60 70 9080 Mass, m(kg)
1
2
Frequency density
• Label the two axes with given labels• The height of each bars should be correct• The boundaries should be right• The scale should be 40----95 and 0---2• The frequency density should be given.
• What if the last class is open-ended?
Example: the grouped frequency distribution in the table below summarizes the mass in grams (g), measured to the nearest gram, of sample of 20 pebbles. Represent the data in a histogram
Mass (g) Frequency
101—110 1111—120 4121—130 3131—140 7140—150 2Over 150 4
Mass, m(kg)
Class boundaries Class width
frequency Frequency density
101—110 100.5 ≤x<110.5 10 1 0.1
111—120 110.5 ≤x<120.5 10 4 0.4
121—130 120.5≤ x<130.5 10 2 0.2
131—140 130.5 ≤x<140.5 10 7 0.7
141—150 140.5 ≤x<150.5 10 2 0.2
Over 150 150.5≤ x<170.5 20 4 0.2
First, draw a table and calculate the boundaries and frequency densities.
100 180120 160140Mass, m(kg)
0.2
0.4
Frequency density
0.7
Open-ended interval
Take the width of the last interval be the
twice that of the previous one
• What if the first class is open?• What are the advantages and disadvantages
of the histogram?• What are the advantages and disadvantages
of the stem and leaf diagram?
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