Gr 10 Data Handling 3 - Maths Excellence

DATA HANDLING (3)
Learning Outcomes and Assessment Standards
Learning Outcome 4: Data Handling and Probability
The learner is able to collect, organise, analyse and interpret data to establish statistical and
probability models to solve related problems.
Assessment Standards
We know this when the learner is able to:
• Collect, organise and interpret univariate numerical data in order to determine:
◦ measures of dispersion: range, percentiles, quartiles, interquartile and semi-inter-quartile
range.
Lesson 30
Overview
In this lesson you will learn about:
●
Measures of dispersion.
Lesson
Quartiles
Quartiles are measures of dispersion around the median, which is a good
measure of central tendency. The median divides the data into two halves. The
lower and upper quartiles further subdivide the data into quarters.
There are therefore three quartiles:
The lower quartile (Q 1
):
The median (Q 2
):
The upper quartile (Q 3
):
Note:
This is the median of the lower half of the values.
The value that divides the data into halves.
This is the median of the upper half of the values.
DVD
If there is an odd number of data values in the data set, then the specific quartile
will be a value in the data set. If there is an even number of data values in the data
set then the specific quartile will not be a value in the data set. A number which will
serve as a quartile will need to be inserted into the data set (the average of the two
middle numbers).
Example 1
(a) Consider the following set of marks obtained on a class test out of 10
marks. The number of marks is odd (11 marks).
2 2 3 4 5 5 6 7 7 8 9
Since there is an odd number of data values, the median will be a
number in the data set. M = Q 2
= 5.
2 2 3 4 5 5 6 7 7 8 9
The lower quartile is the median of the lower half of the data set. The
lower half contains an odd number of data values. Therefore Q 1
= 3.
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In summary:
2 2 3 4 5
The upper quartile is the median of the upper half of the data set.
The upper half contains an odd number of data values.
Therefore Q 2
= 7.
6 7 7 8 9
2 2 3 4 5 5 6 7 7 8 9
Lower Quartile
Q 1
Median
Q 2
Upper Quartile
Q 3
(b) Consider the following set of 13 marks obtained on a class test out of
10 marks:
2 3 4 5 5 5 6 7 7 8 8 10 10
Since there is an odd number of data values, the median will a number in
the data set. M = Q 2
= 6.
2 3 4 5 5 5 6 7 7 8 8 10 10
The lower quartile is the median of the lower half of the data set. The
lower half contains an even number of data values. Therefore we need to
find the average of the two middle numbers, 4 and 5 and then insert this
number into the data set:
Q 1
= _ 4 + 5 = 4,5
2
Q 1
2 3 4 4,5 5 5 5
The upper quartile is the median of the upper half of the data set. The
upper half contains an even number of data values. Therefore we need to
find the average of the two middle numbers, 8 and 8 and then insert this
number into the data set:
Q 3
= _ 8 + 8 = 8
2
Q 3
7 7 8 8 8 10 10
In summary:
2 3 4 4,5 5 5 5 6 7 7 8 8 8 10 10
Q 1
Q 2
Q 3
Example 2
(a) Consider the following set of 12 marks obtained by a class on a class
test out of 100 marks. The number of marks is even.
20 32 43 54 55 61 73 78 89 90 91 98
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Since there is an even number of data values, the median will not be a data
value in the data set. We will need to find the average of the two middle
numbers, 61 and 73 and then insert this number into the data set:
61 + 73
M = Q 2
= _ = 67
2
20 32 43 54 55 61 67 73 78 89 90 91 98
The lower quartile is the median of the lower half of the data set. The
lower half contains an even number of data values. Therefore we need to
find the average of the two middle numbers, 43 and 54 and then insert this
number into the data set:
43 +54
Q 1
= _ = 48,5
2
Q 1
20 32 43 48,5 54 55 61
The upper quartile is the median of the upper half of the data set. The
upper half contains an even number of data values. Therefore we need to
find the average of the two middle numbers, 89 and 90 and then insert this
number into the data set:
89 + 90
Q 3
= _ = 89,5
2
Q 3
73 78 89 89,5 90 91 98
In summary:
20 32 43 48,5 54 55 61 67 73 78 89 89,5 90 91 98
(b) Consider the following set of 10 marks obtained by a class on a class
test out of 150 marks. The number of marks is even.
12 60 95 105 120 125 130 135 140 142
Since there is an even number of data values, the median will not be a data
value in the data set. We will need to find the average of the two middle
numbers, 120 and 125 and then insert this number into the data set:
120 +125
M = Q 2
= __ = 122,5
2
12 60 95 105 120 122,5 125 130 135 140 142
The lower quartile is the median of the lower half of the data set. The
lower half contains an odd number of data values. Therefore Q 1
= 95.
12 60 95 105 120
The upper quartile is the median of the upper half of the data set. The
upper half contains an odd number of data values. Therefore Q 2
= 135.
125 130 135 140 142
In summary:
12 60 95 105 120 122,5 125 130 135 140 142
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INDIVIDUAL
FORMATIVE
ASSESSMENT
Activity 1
1. For each set of data, determine the quartiles:
A 2 3 5 7 9 10 11 13 15 16 16 17 18 19 21 22 23 25 32
B 2 3 5 7 9 10 11 13 15 16 16 17 18 19 21 22 23
C 2 3 5 7 9 10 11 13 15 16 16 17 18 19 21 22 23 25 32 34
D 2 3 5 7 9 10 11 13 15 16 16 17 18 19 21 22 23 25
DVD
Lesson
Range
The range is the difference between the largest and the smallest value in the
data. The bigger the range, the more spread out the data is. The range cannot
be used with grouped data and it doesn’t deal with values between these
extremes. Also, a single very small or very large value would give a misleading
impression of the spread of the data.
Consider for example, the following set of marks for Class A:
Class A 1 1 2 2 3 4 4 5 5 6 7 8 8 9 10
Range = largest score – smallest score = 10 – 1 = 9
Deciles and percentiles
Deciles subdivide the data into tenths. Percentiles divide the data into
hundredths.
The interquartile range (IQR)
The difference between the lower and upper quartile is called the interquartile
range. It is a better measure of dispersion than the range because it is not
affected by extreme values. It is based on the middle half of the data. It
indicates how densely the data in the middle is spread around the median.
Example
Calculate the interquartile range for the following data set.
2 2 3 4 5 5 6 7 7 8 9
Lower
Quartile
Q 1
IQR = Q 3
– Q 1
= 7 – 3 = 4
The semi-interquartile range
Median
Q 2
Upper
Quartile
Q 3
The semi-interquartile range is half of the interquartile range.
The semi-IQR for the previous example is _
Q – Q 3 1
2
= _ 7 – 3 = 2.
2
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Activity 2
1. Class results for a test out of 30 are recorded in the table below.
10A 16 12 16 11 14 15 22 16 17 15 26 23 16 22 16 17 24 19 16 -
10B 20 19 14 10 14 9 8 13 14 30 27 23 24 28 17 29 20 16 14 18
10C 5 20 14 12 7 2 12 21 14 26 14 14 12 14 21 24 14 14 - -
INDIVIDUAL
SUMMATIVE
ASSESSMENT
(a) Calculate the mean for each class.
(b) Calculate the mode for each class.
(c) Calculate the median for each class.
(d) Calculate the range for each class.
(e) Calculate the lower quartile for each class.
(f) Calculate the upper quartile for each class.
(g) Calculate the interquartile range for each class.
(h) Calculate the semi-interquartile range for each class.
2. Over a period of 15 days, the owner of a coffee shop recorded the number
of customers who per day ordered the breakfast special. His observations
were as follows:
Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Number of orders 25 80 34 26 21 65 28 21 39 21 30 34 21 28 40
Determine:
(a) the mean
(b) the median
(c) the mode
(d) the range
(e) the inter-quartile range.
(f) Which of the mean, median or mode is the most appropriate value in
estimating how many orders will be made per day? Explain.
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ANSWERS AND ASSESSMENT
Lesson 28
Activity 1
1. (a)
600
500
400
300
200
100
0
97/98 98/99 99/00 00/01 01/02 02/03 03/04 04/05 05/06 06/07 07/08
(b) Inflation and increased population has caused the amounts to increase.
2. Marks (10) Tally Frequency (f)
1 || 2
2 ||| 3
3 ||| 4
4 |||| | 6
5 |||| 5
6 |||| |||| 9
7 ||| 3
8 ||| 3
9 || 2
10 ||| 3
Total: 40
20
10
9
8
7
6
5
4
3
2
1
0
1 2 3 4 5 6 7 8 9 10
10 LC G10 MATHS LWB.indb 20 2008/09/09 12:22:48 PM

Mass
3. Salaries R2 000 000
Grounds and maintenance R 800 000
Textbooks R 200 000
Photocopying facilities R1 000 000
Sporting facilities R 500 000
Water and lights R 300 000
Academic subject allocation R 200 000
Computer centre R 700 000
Administration R 200 000
Salaries:
Grounds and Maintenance:
Textbooks:
Photocopying:
Sport:
Water and lights
Academic:
Computer:
Administration:
2 __ 000 000
× 360° = 122°
5 900 000
__ 800 000
× 360° = 49°
5 900 000
__ 200 000
× 360° = 12°
5 900 000
1 __ 000 000
× 360° = 61°
5 900 000
__ 500 000
× 360° = 31°
5 900 000
__ 300 000
× 360° = 18°
5 900 000
__ 200 000
× 360° = 12°
5 900 000
__ 700 000
× 360° = 43°
5 900 000
__ 200 000
× 360° = 12°
5 900 000
Academic
Water and lights
Computers
Admin
Sport
Photocopying
Salaries
Textbooks
Grounds
4. (a)
3,2 .
3,1
3,0
2,9
2,8
2,7
2,6
2,5
2,4
2,3
2,2
2,1
0
. .
.
.
.
1 2 3 4 5 6 7 8 9 10
Week
.
.
.
.
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Frequency
Frequency
(b) 6 th week
(c) Average mass =
Activity 2
__ 2,3 + 2,6
= 2,5 kg
2
1. (a)
200
180
160
140
120
100
80
60
40
20
(b)
2. (a)
200
180
160
140
120
100
80
60
40
20
0-1,9 2-3,9 4-5,9 6-7,9 8-9,9 10-11,9 12-13,9 14-15,9
Duration
0,95
2,95
.
.
4,95
.
6,95
0-1,9 2-3,9 4-5,9 6-7,9 8-9,9 10-11,9 12-13,9 14-15,9
Duration
.
8,95
.
10,95
.
12,95
14,95
2 4
3 6 6 8
4 2 2 3 7 8 8 8
5 2 4 4 4 4 5 5 5 5 6 6 7 7 8 8 8 9 9
6 0 1 1 1 1 2 2 3 3 3 4 4 4 4 4 4 5 5 5 6 6 6 8 8 8 8 8 9 9 9 9
7 0 0 2 2 2 3 3 3 3 4 4 4 4 4 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 6 7 8 8 8 8 9 9
8 0 0 1 2 2 2 2 2 2 2 4 4 4 4 4 4 5 5 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 8 8 9 9 9 9 9 9 9
9 0 0 0 1 1 4 4 4 5 5 5 5 6
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Frequency
Frequency
(b) Class interval Frequency (number of clients per mass group)
10 – 19 kg 0
20 – 29 kg 1
30 – 39 kg 3
40 – 49 kg 7
50 – 59 kg 18
60 – 69 kg 31
70 – 79 kg 38
80 – 89 kg 49
90 – 99 kg 13
100 – 109kg 0
(c)
50
40
35
30
25
20
15
10
5
10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89
Masses
90-99 100-109
(d)
50
40
35
30
25
20
15
10
5
44,5
34,5
24,5
14,5
10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89
Masses
. .
.
.
54,5
.
64,5
.
74,5
.
84,5
.
94,5
104,5
90-99 100-109
(e) The histogram shows that the majority of people were in the 80-89
kg group. This group would probably either want to lose body fat
and keep or gain muscle mass. It would be unlikely that this group
would want to gain too much weight or lose too much. It seems that
the health product company should market both products in Gauteng.
The 50-69 kg group will benefit from the Muscle Grow product. The
90-99 kg group will benefit from the Fat Go product.
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Lesson 29
Activity 1
1. First arrange in ascending order.
10A 11 12 14 15 15 16 16 16 16 16 16 17 17 19 22 22 23 24 26 -
10B 8 9 10 13 14 14 14 14 16 17 18 19 20 20 23 24 27 28 29 30
10C 2 5 7 12 12 12 14 14 14 14 14 14 14 20 21 21 24 26 - -
333
(a) Mean for 10A: _
19 = 17,5
367
Mean for 10B: _
20 = 18,4
260
Mean for 10C: _
18 = 14,4
(b) Mode for 10A: 16
Mode for 10B: 14
Mode for 10C: 14
(c) Median for 10A: 16
Median for 10B: 17,5
Median for 10C: 14
2 4
3 6 6
4 2 2
5 2 4
6 0 1
7 0 0
8 0 0
9 0 0
2. (a) Marks (10) Tally Frequency (f) Mark × f
Activity 2
Discussion
Activity 3
1 || 2 2
2 |||| 4 8
3 |||| 5 15
4 |||| 4 16
5 | 1 5
6 |||| 4 24
7 |||| |||| 9 63
8 |||| 4 32
9 |||| 4 36
10 ||| 3 30
Total: 40 Total: 231
Mean = _ 231 = 5,8 Mode = 7 Median = 5,5
40
1. (a) 6 – 7,9 interval.
(b) 6 – 7,9 interval.
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2. (a)
4
6 6 8
2 2 3 7 8 8 8
2 4 4 4 4 5 5 5 5 6 6 7 7 8 8 8 9 9
0 1 1 1 1 2 2 3 3 3 4 4 4 4 4 4 5 5 5 6 6 6 8 8 8 8 8 9 9 9 9
0 0 2 2 2 3 3 3 3 4 4 4 4 4 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 6 7 8 8 8 8 9 9
0 0 1 2 2 2 2 2 2 2 4 4 4 4 4 4 5 5 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 7 8 8 9 9 9 9 9 9 9
0 0 0 1 1 4 4 4 5 5 5 5 6
(b)
Class interval Frequency Midpoint class interval Frequency × Midpoint
10 – 19 kg – – –
20 – 29 kg 1 24,5 24,5
30 – 39 kg 3 34,5 103,5
40 – 49 kg 7 44,5 311,5
50 – 59 kg 18 54,5 981
60 – 69 kg 31 64,5 1999,5
70 – 79 kg 38 74,5 2831
80 – 89 kg 49 84,5 4140,5
90 – 99 kg 13 94,5 1228,5
100 – 109kg - - -
Totals 160 - 11620
11 620
(c) Estimated mean = _
160 = 72,6
Lesson 30
Activity 1
A 2 3 5 7 9 10 11 13 15 16 16 17 18 19 21 22 23 25 32
B 2 3 5 7 8 9 10 11 13 15 16 16 17 18 18,5 19 21 22 23
C 2 3 5 7 9 9,5 10 11 13 15 16 16 16 17 18 19 21 21,5 22 23 25 32 34
D 2 3 5 7 9 10 11 13 15 15,5 16 16 17 18 19 21 22 23 25
Activity 2
333
1. (a) Mean for 10A: _
19 = 17,5
367
Mean for 10B: _
20 = 18,4
260
Mean for 10C: _
18 = 14,4
(b) Mode for 10A: 16
Mode for 10B: 14
Mode for 10C: 14
(c) Median for 10A: 16
Median for 10B: 17,5
Median for 10C: 14
25
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(d) Range for 10A: 26 – 11 = 15
Range for 10B: 30 – 8 = 22
Range for 10C: 26 – 2 = 24
(e) Lower quartile for 10A: Q 1
= 15
Lower quartile for 10B: Q 1
= 14
Lower quartile for 10C: Q 1
= 12
(f) Upper quartile for 10A: Q 3
= 22
Upper quartile for 10B: Q 3
= 23,5
Upper quartile for 10C: Q 3
= 20,5
(g) Interquartile range for 10A: Q 3
– Q 1
= 22 – 15 = 7
Interquartile range for 10B: Q 3
– Q 1
= 23,5 – 14 = 9,5
Interquartile range for 10C: Q 3
– Q 1
= 20,5 – 12 = 8,5
(h) Semi-interquartile range for 10A: _
Q – Q 3 1
= _ 22 – 15
= 3,5
2 2
Semi-interquartile range for 10B: _
Q – Q 3 1
= __ 23,5 – 14
= 4,75
2 2
Semi-interquartile range for 10C: _
Q – Q 3 1
= __ 20,5 – 12
= 4,25
2 2
2. (a) Mean = 34,2 orders
(b) Median = 28 orders
(c) Mode = 21 orders
(d) Range = 80 – 21 = 59 orders
(e) Interquartile range: Q 3
– Q 1
= 39 – 21 = 18
(f) The mean would probably be the most accurate estimate. The mode
and median are too small a number of orders per day. However, the
owner would need to be prepared to make more than 34 orders on a
few days, even up to 80!
26
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TIPS FOR TEACHERS
Lesson 28
●
●
Ensure that learners know the difference between bar graphs and histograms.
The points represented on a broken line graph should not be joined by means
of a continuous solid line. This is because data values between these values
are not known. The points should be joined by a dotted line.
Reference
“Mathematics textbook and workbook grade 10”
Allcopy publishers
M.D.Phillips
Lesson 29
●
Ensure that learners know that the median will be a number in the data if
the number of data values is odd. If the number of data values is even, the
median will be an insertion into the data set. In this case, the median will be
the average between the two middle values.
Reference
“Mathematics textbook and workbook grade 10”
Allcopy publishers
M.D.Phillips
Lesson 30
●
●
Ensure that learners know that the median will be a number in the data if
the number of data values is odd. If the number of data values is even, the
median will be an insertion into the data set. In this case, the median will be
the average between the two middle values.
The median is excluded from the lower and upper halves when determining
the lower and upper quartiles.
Reference
“Mathematics textbook and workbook grade 10”
Allcopy publishers
M.D.Phillips
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