Post on 15-Apr-2017
Unit 18Measures of Variation
Cumulative Frequency
The table shows the age distribution (in complete years) of the population of Nigeria in 1991.
Age 0 ≤ x < 15 15 ≤ x < 30 30 ≤ x < 45 45 ≤ x < 60 60 ≤ x < 75 75 ≤ x < 100
% of Pop. 32 29 20 12 6 1
Cumulative % 32 61 81 93 99 100
0 10 20 30 40 50 60 70 80 90 100
Age in Years
100
90
80
70
60
50
40
30
20
10
Lower Quartile ≈ 11
Median ≈ 24
Upper Quartile ≈ 40∙5
Estimate:a)The lower quartileb)The medianc)The upper quartile
Unit 18Measures of Variation
Cumulative Frequency 2
The table shows the distribution of marks on a test for 70 students.
Mark Frequency Cumulative Frequency
1 – 10 2 2
11 – 20 5 7
21 – 30 9 16
31 – 40 14
41 – 50 16
51 – 60 12
61 – 70 8
71 – 80 4
30
46
58
66
70
0 10 20 30 40 50 60 70 80Marks
80
70
60
50
40
30
20
10
Mark Frequency Cumulative Frequency
1 – 10 2 2
11 – 20 5 7
21 – 30 9 16
31 – 40 14 30
41 – 50 16 46
51 – 60 12 58
61 – 70 8 66
71 – 80 4 70
a) Draw a cumulative frequency curve
b) The pass mark for the test is 47. Use your graph to determine the number of students who passed the test
c) What is the probability that a student chosen at random, had a mark of less than or equal to 30?
Students Pass: 70 – 42 = 28
Unit 18Measures of Variation
Box and Whisker Plots
The goals scored in the first 11 football matches played by a National Premier League team were:
This data can be represented using a box and whisker plot.Record the data starting with the smallest
Identify:
Construct a box and whisker plot
1 0 4 2 2 3 1 2 5 0 1
0 0 1 1 1 2 2 2 3 4 5
Smallest Value Largest Value
Median
Lower quartile Upper quartile
0
2
1
5
3
0 1 2 3 4 5 6 7
Construct an additional box and whisker plot for a team with data
3 5 2 1 3 3 2 5 6 2 1
Compare the two sets of data
Unit 18Measures of Variation
Standard Deviation
The STANDARD DEVIATION (s.d.) of a set of data is a measure of the spread of the data about the mean and is defined by
a)What is the mean (m) of each set?S1 = {6, 7, 8, 9, 10}
S2 = {4, 5, 8, 11, 12}
S3 = {1, 2, 8, 14, 15}
b) The standard deviation for S1 is calculated as:
6 - 2 4
7 - 1 1
8 0 0
9 1 1
10 2 4
TOTAL 10
m = 8m = 8m = 8
The STANDARD DEVIATION (s.d.) of a set of data is a measure of the spread of the data about the mean and is defined by
S1 = {6, 7, 8, 9, 10} m = 8
S2 = {4, 5, 8, 11, 12} m = 8
S3 = {1, 2, 8, 14, 15} m = 8
c)Compare the standard deviations for S1, S2 and S3
S1 = {6, 7, 8, 9, 10} s.d. ≈ 1∙414
S2 = {4, 5, 8, 11, 12} s.d. ≈ 3∙162
S3 = {1, 2, 8, 14, 15} s.d. ≈ 5∙831