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_
EXAMPLE 13
In a particular primary school in Pietermaritzburg, it was found that
ninety of their Foundation Phase learners (Grades 1, 2 and 3) were
accompanied to school by someone. The ages of the person
accompanying the child were recorded, as shown in the table below.
Age (in years)
(x)
Frequency
(f)
0 <_x≤ 10 12
10 < x _ 20 30
20 < x _ 30 18
30 < x _ 40 12
40 < x _ 50 9
50 < x _ 60 6
60 < x _ 70 3
Use the information given in the table to
a) Determine the modal interval.
b) Estimate the mean age of the person accompanying a learner from
the Foundation Phase.
c) Estimate the median age of the person accompanying a learner from
the Foundation Phase.
SOLUTION:
a) The modal interval is 10 < x ≤_20.
This means that more people in this age group accompanied the learners
to school
than any other age group.
b) To find the mean we have to take the midpoint of each class interval
and then
calculate frequency midpoint for each class interval.
Age (in years)
x
Midpoint
X
Frequency
f
f.X
0 <x ≤ 10 _/__
J = 5 12 12 5 = 60
10 <x ≤_20 __/J_
J = 15 30 30 15 = 450
20 <x ≤_30 J_/L_
J = 25 18 18 25 = 450
30_< x ≤ 40 L_/__
J = 35 12 12 35 = 420
40 < x ≤_50 __/K_
J = 45 9 9 45 = 405
50 < x ≤ 60 K_/,_
J = 55 6 6 55 = 330
60_< x ≤ 70 ,_/-_
J = 65 3 3 65 = 195
n = 90 _Q. M = 2 310
Mean = MN = __.O
_ _ J_L___X!_"$
__ = 25,7 years old
The mean tells us that if all the ages were added together, and then
shared out
equally amongst the 90 people, then each one would be 25,7 years old.
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EXAMPLE 13 (continued)
c) 90 people accompanied the learners to school.
90 ÷ 2 = 45. So the median lies half-way between the 45 th and 46th
person.
Age (years) Frequency
0 <_x≤ 10 12 … 12 people to here
10 < x _ 20 30 … 12 + 30 = 42 people to here
20 < x _ 30 18 … 42 + 18 = 60 people to here.
30 < x _ 40 12
40 < x _ 50 9
50 < x _ 60 6
60 < x _ 70 3
Both the 45th and 46th people lie in the interval 20 <_x ≤_30.
So the median interval is 20 years < x _ 30 years, and the approximate
value of the
median is 25 years.
The value of the median tells us that 50% of the people accompanying the
Grade 1,
2 and 3 learners to school are less than or equal to 25 years old, and 50%
are more
than or equal to 25 years old.
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_ EXERCISE 1.3
1) The table below represents the ages of the 90 people accompanying
Foundation
Phase learners to another primary school in Pietermaritzburg:
Age (in years)
(x)
Frequency
(f)
0 ≤ _ < 10 4
10 ≤ _ < 20 12
20 ≤ _ < 30 25
30 ≤ _ < 40 14
40 ≤ _ < 50 10
50 ≤ _ < 60 20
60 ≤ _ < 70 5
n = 90
a) Use the information given in the table to
i) Determine the modal interval.
ii) Estimate the mean age of the people accompanying a learner from the
Foundation Phase.
iii) Estimate the median age of the people accompanying a learner from
the
Foundation Phase.
b) What do the modal class, the mean and the median tell you about the
ages of the
people who accompany the children to school?
2) The table below has been adapted from Census 2011. It lists the
property values of
262 properties in a part of the Ikwezi municipality.
Property Value in Rand
(x)
Frequency
(number of households)
(f)
0 ≤ _ < 50_000 172
50_000 ≤ _ < 100_000 51
100_000 ≤ _ < 150_000 18
150_000 ≤ _ < 200_000 12
200_000 ≤ _ < 250_000 9
n = 262
a) Use the information given in the table to
i) Determine the modal interval.
ii) Estimate the mean property value of the properties.
iii) Estimate the median property value of the properties.
b) What does the modal class, the mean and the median tell you about
these
property values?
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QUARTILES AND THE FIVE NUMBER
SUMMARY
a) Quartiles
_ Quartiles are the three values Q1, Q2 and Q3 that divide a data set into
four
approximately equal parts. Each part consists of approximately 25% of
the
elements of the data set.
_ Q1 is the lower quartile; Q2 is the middle quartile or median and Q3 is
the upper
quartile.
_ The median divides an ordered data set into two halves.
_ The quartiles divide an ordered data set into four quarters.
_ The median is also the 2nd quartile (M or Q2).
25% of the data 25% of the data 25% of the data 25% of the data
_ From the above diagram, one can see that:
Approximately one quarter or 25% of the data is less than Q 1.
Approximately three quarters or 75% of the data is more than Q 1.
Approximately one half or 50% of the data is less than Q 2 and one half or
50% is more than Q2.
Approximately three quarters or 75% of the data is less than Q 3.
About one quarter or 25% of the data is more than Q 3.
Approximately one half or 50% of the data lies between Q 1 and Q3.
HOW TO FIND THE QUARTILES:
i) Put the data items in order and find the median.
ii) Find the midpoint of the data items to the left of the median. This is the
lower quartile (Q1).
iii) Find the midpoint of the data items to the right of the median. This is
the
upper quartile (Q3).
Lower
quartile
Q1
Median
M or Q2
Upper
quartile
Q3
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_
EXAMPLE 14
For each of the following sets of data
a) 23 65 33 101 23 21 102 18 26 9
b) 65 33 101 23 21 102 18 26 9
i) Find the median (M)
ii) Find the lower quartile (Q1) and the upper quartile (Q3)
SOLUTION:
a) First arrange the data in ascending or decending order.
9 18 21 23 23 26 33 65 101 102
i) There are ten data items (an even number of data items). To find the
median, we
have to find the mean of the middle two numbers.
10 ÷ 2 = 5, so the median lies between the 5 th and the 6th terms.
9 18 21 23 23 26 33 65 101 102
So the median = M = JL/J,
J = __
J = 24,5
ii) To find the lower quartile (Q1), take the data before the median and find
the
median of that.
There is an odd number of data items below the median. Take the middle
one.
9 18 21 23 23 |26 33 65 101 102
So, Q1 = 21.
To find the upper quartile (Q3), take the data after the median and find the
median of that:
There is an odd number of data items above the median; take the middle
one.
9 18 21 23 23 | 26 33 65 101 102
So, Q3 = 65.
So, the three quartiles are 21; 24,5 and 65.
b) First arrange the data in ascending or decending order.
9 18 21 23 26 33 65 101 102
i) There are nine data items (an odd number of data items).
9 ÷ 2 = 4,5 so the median is the 5th term.
9 18 21 23 26 33 65 101 102
So the median = 26
ii) To find the lower quartile (Q1), take the data before the median (26)
and find the
median of that.
There is an even number of data items below the median.
4 ÷ 2 = 2, so the lower quartile, Q1, lies between the 2nd and 3rd terms.
9 18 | 21 23 26 33 65 101 102
So the Q1= _*/J_
J = L_
J = 19,5
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EXAMPLE 14 (continued)
To find the upper quartile (Q3), take the data after the median and find the
median of that.
There is an even number of data items above the median.
4 ÷ 2 = 2, and 5 + 2 = 7, so Q3 lies between the 7th and the 8th terms.
9 18 | 21 23 26 33 65 |
101 102
So the Q3 = ,K/___
J = _,,
J = 83
So, the three quartiles are 19,5; 26 and 83.
b) The Five Number Summary
_ The five number summary consists of 5 items
1) The minimum value in the data set;
2) Q1, the lower quartile;
3) M, the median;
4) Q3, the upper quartile;
5) The maximum value in the data set.
_ Use the following method to find a five number summary:
STEPS EXAMPLE
1) Put your numbers in
the data set in order.
Find the five-number summary of the following data
set: 1, 3, 5, 6, 12, 15, 23, 28, 31
2) Find the minimum and
maximum values.
Minimum value = 1
Maximum value = 31
3) Find the median There are 9 data items.
9 ÷ 2 = 4,5 so the median is the 5th data item
1 3 5 6 12 15 23 28 31
So the median = 12
4) Find Q1 and Q3. There are 4 items below the median.
4 ÷ 2 = 2, so Q1 lies between the 2nd and 3rd terms.
1 3 | 5 6 12 15 23 28 31
So Q1 =_L/K
J=4
There are 4 items above the median.
5 + 2 = 7, so Q3 lies between the 7th and 8th terms.
1 3 | 5 6 12 15 23 | 28 31
So Q2 = JL/J*
J = 25,5
5) Write down the five
number summary.
Minimum = 1
Q1 = 4
Median = 12
Q3 = 25,5
Maximum = 31
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_ EXERCISE 1.4

1) Find the five number summary for each of the following data sets:
a) 1 6 6 9 15 17 23 24 33 33 38 38 38 45 46 51
b) 9 14 19 21 24 29 29 32 33 35 36 40 46 49
c) 45 15 43 19 26 25 15 36 27 32 41 25 48
d) 4 46 6 44 10 17 34 35 31 22 10 16
2) Johan is asked to find the five number summary for the following set of
numbers:
2 23 24 12 11 23 34 12 34 12 33 19 48 25 37 38 59
Johan’s answer is as follows:
a) Give the correct solution to the question.
b) Identify the mistake that Johan has made. Describe the misconception
and
explain to John why he is incorrect.
2 11 12 12 12 19 23 23 24 25 33 34 34 37 38 48 59
Min = 2
Q1 = 12
Median = 24
Q3 = 34 or 37
Max = 59
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PERCENTILES
_ When you write a test and get a mark of 75%, it tells you how many
questions
you got right. But, it doesn’t tell you how well you did compared to the
other
people who wrote the test. Percentiles are values from 0 to 99 that tell
you the
percentage of the marks that are less than a particular mark.
If the percentile of your test mark is 75, it tells that
75% of the marks are LESS than yours.
100% – 75% = 25% of the marks are MORE than yours
_ Percentiles can be used to compare values in any set of ordered data.
You can
calculate percentiles for income, mass, etc. Percentiles are often used in
education and health-related fields to indicate how one person compares
with
others in a group.
NOTE:
There is a difference between a mark of 60% (a percentage telling you
that
you got 60 out of 100) and a mark at the 60th percentile (which tells us
that
approximately 60% of the marks are less than yours).
_ The following are some special percentiles:
The median is at the 50th percentile
The lower quartile is at the 25th percentile
The upper quartile is at the 75th percentile.
_ Scores that are in the 95th percentile and above are unusually high
while those in
the 5th percentile and below are unusually low.
Example:
a) The 89th percentile is a number that 89% of the data items are below.
We
then also know that 100% – 89% = 11% of the data items are above that
number.
b) If the 20th percentile is 12, then 20% of the data items are less than
12, and
80% of the data items are more than 12.
c) If the 90th percentile is 17, then 90% of the data items are less than
17, and
10% of the data items are more than 17.
_ We usually find percentiles of a large number of items.
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_
EXAMPLE 15
For the following set of 19 data items
72 71 65 60 62 58 67 57 70 73
50 61 51 55 64 68 69 59 63
a) At what percentile is 70?
b) Find the data item that is at the 20th percentile.
c) What is the median score (the score at the 50 th percentile)?
SOLUTION:
A stem and leaf diagram can be used to get the data into ascending order:
Stem Leaf
5015789
6012345789
70123
KEY: 6/2 = 62
a) To calculate at which percentile a data item lies:
Percentile = _________] _ _^______ __ _______ _______ ____-_
___ __________] _ _^___ × 100%
= _,
__ 100%
= 84,210…%
_ 84%
So 70 is at the 84th percentile.
a) To calculate the data item which is at a given percentile:
i) Find out which term corresponds to the 20th percentile:
Data item corresponding to the 20th percentile
= a_b__^
___ × cdefgh_iQ_jklk_mlgen
= J_
___ × 19 data items
= 3,8
_4
ii) Count along until you get to the 4th data item – It is 57
So 57 is the 20th percentile
b) To calculate the data item which is at a given percentile:
The data item corresponding to the 50th percentile
= a_b__^
___ × cdefgh_iQ_jklk_mlgen
= K_
___ × 19
= 9,5
_ 10
Count along to find the 10th data item – It is 63.
So 63 is the 50th percentile or the median.
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_ EXERCISE 1.5

1) Given the following set of data:

24 34 35 37 39 40 41 45 46 48
50 52 54 55 56 56 59 59 60 61
63 66 67 68 69 70 73 75 77 77
78 79 79 80 84 84 86 86 86 89
94 95 96 97 98 98 100 101 102 103
a) At which percentile is:
i) 102?
ii) 34?
iii) 96?
iv) 70?
b) Find the item that corresponds to
i) The 30th percentile
ii) The 50th percentile
iii) The 10th percentile
iv) The 65th percentile
2) For the following stem and leaf diagram
STEM LEAF
10 0 0 2 3 3 3 4 5 7 7 7 7 8 8 9
11 1 1 2 3 5 5 5 6 6 6 8 9
12 0 0 0 1 3 4 6 7 7 9 9 9 9 9
13 1 2 2 2 5 6 7 7 7 8 8 8 9
14 0 1 1 1 1 2 3 5 5 5 5 6 8 8 8 9
KEY: 10/4 = 104
a) What is the score that corresponds to the 11 th percentile?
b) Find the data item that corresponds to the 44th percentile
c) At which percentile is a score of 135?
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MEASURES OF DISPERSION
_ A measure of central tendency such as the mean, median and mode
gives you a
single measurement to stand for a set of data. A measure of dispersion
or
measure of spread tells you how spread out the data is.
_ The data can either:
Be grouped closely together around the measure of central tendency, or
Be spread widely apart around the measure of central tendency.
a) Range
_ The range is the simplest measure of spread. It is the difference
between the
largest and smallest items of data.
Range = Largest Value – Smallest Value
_ There are some limitations to using range:
It does not take into account anything about the distribution of any
other
piece of data except the smallest and largest value.
When data is given in a grouped frequency table, the range cannot be
used.
b) Interquartile Range
_ The interquartile range (or IQR) is the difference between the upper
quartile and
the lower quartile.
Interquartile Range = Q3 – Q1
_ The interquartile range is a better measure of dispersion than the range.
It is not
affected by any extreme values (very small or very large values). It is
based on
the middle half of the data. It is the range between the upper and lower
quartiles.
_ The semi-interquartile range is sometimes used. It is half of the
interquartile
range.
Semi-interquartile range = p3q_p2
1
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_
EXAMPLE 16
a) For the following set of data: 22 17 28 19 23 18 25 29 19 29
Find
i) the range
ii) the interquartile range
iii) the semi-interquartile range
b) Approximately what percentage of the data items lie within the
interquartile range?
SOLUTION:
a)
i) Arrange the data in order: 17 18 19 19 22 23 25 28 29 29
Range = Largest Value – Smallest Value= 29 – 17 = 15
ii) First find the median:
There are 10 terms.
10 ÷ 2 = 5 which means that the median lies between the 5 th and 6th
terms
Median = JJ/JL
J = _K
J = 22,5
The find the two quartiles
17 18 19 19 22 | 23 25 28 29 29
There are five terms to the left of the median and five terms to the right of
the
median.
So Q1 = 19 and Q3 = 29
Interquartile range = Q3 – Q1 = 28 – 19 = 9
iii) Semi-interquartile range = p3q_p2
1
= 1rq2s
1 =_s1
= 4,5
b) Approximately 50% of the data items lie within the interquartile range.
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_ EXERCISE 1.6

1) Find the median, lower quartile, upper quartile and the interquartile
range for each
of the following sets of data:
a) 9 11 12 12 13 13 14 16 21 22 24
b) 4 5 5 6 7 9 10 11 12 12 13 14 14 14
c) 12 7 1 3 2 12 2 9 14 5 6 5 4 8 11 14
2) A group of 21 learners attending extra mathematics classes were
required to write a
test which was out of 50.
Their results were:
17 8 19 9 12 28 11 16 20 14 29
23 37 23 26 4 35 26 18 45 7
a) Find the range.
b) Find the lower and upper quartiles.
c) Calculate the interquartile range.
d) What do the lower and upper quartiles indicate about the results of the
test?
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BOX AND WHISKER DIAGRAMS
_ A graphical representation of the five number is known as a box and
whisker
diagram, also sometimes called a box plot.
Vertical lines mark the two quartiles and the median. These are joined to
make a box containing the middle half of the data. The box illustrates the
interquartile range.
From the quartiles, horizontal lines are drawn to the minimum and
maximum
values. These lines are the whiskers.
0 4 6 8 10 12 14 16 18 20 22 24 26 28 30
Minimum value
Lower quartile
Median
Upper quartile
Maximum value
HOW TO DRAW A BOX AND WHISKER DIAGRAM
STEP 1: Make sure that the data is arranged in ascending order
STEP 2: Find the five number summary
STEP 3: Draw a number line long enough to fit the minimum and
maximum values.
Make sure that the units are plotted correctly on the number line.
STEP 3: Draw vertical lines at Q1, M and Q3 and then draw two horizontal
lines to
make the box.
STEP 4: From the middle of the box, first draw a horizontal line to the
minimum
value and then draw a horizontal line to the maximum value.
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_
EXAMPLE 17
Draw a box-and-whisker diagram of the following set of data:
506 503 507 504 510 511 526 513
517 508 515 513 508 509 516
SOLUTION:
STEP 1: Arrange the data in ascending order
503 504 506 507 508 508 509 510 511 513 513 515 516 517 526
STEP 2: Find the five number summary.
503 504 506 507 508 508 509 510 511 513 513 515 516 517 526
Minimum value = 503
Q1 = 507
M = 510
Q3 = 515
Maximum value = 526
STEP 3: Draw a number line long enough to go from 503 to 526.
STEP 4: Draw vertical lines at Q1, M and Q3 and form the box
STEP 5: Join the box to the minimum and maximum values to form the
whiskers.
502 504 506 508 510 512 514 516 518 520 522 524 526
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_
EXAMPLE 18
Two box and whisker diagrams are drawn on the same number line
a) List the five number summary of
i) Set A
ii) Set B
b) Find the range and interquartile range of
i) Set A
ii) Set B
c) Explain why the dot is on the edge of the box in Set B
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140
SOLUTION:
a)
i) SET A:
Minimum value = 25
Q1 = 30
M = 50
Q3 = 70
Maximum value = 140
ii) SET B
Minimum value = 5
Q1 = 15
M = 20
Q3 = 30
Maximum value = 30
b)
iii) Range of Set A = maximum value – minimum value = 140 – 25 = 115
Interquartile range of Set A = Q3 – Q1 = 70 – 30 = 40
iv) Range of Set B = maximum value – minimum value = 30 – 5 = 25
Interquartile range of Set B = Q3 – Q1 = 30 – 15 = 15
c) The dot is on the edge of the box in Set B because the upper quartile
and the
maximum value in Set B are identical (they are both 30).
Set A
Set B
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_
EXERCISE 1.7
1) The percentages achieved by a learner for a series of mathematics
tests that he wrote
throughout his Grade 9 year are as follows:
35 45 50 28 39 49 55 35 56 49
43 37 28 53 55 38 47 51 30 58
a) Calculate the five number summary
b) Draw a box-and-whisker diagram to illustrate the five number
summary.
c) Find
i) the range of the set of data
ii) the interquartile range.
2) As part of the 2009 Census@School, the 26 Grade 10A learners
measured their
heights.
The girls’ heights (in centimetres) were:
150 150 153 155 156 158 160 161 164 164 166 170 170
The boys’ heights in centimetres were:
140 142 151 157 158 159 160 162 165 180 180 180 180
a) Find the five number summary and the interquartile range for the girls
and for
the boys
b) On the same number line draw two box and whisker diagrams to
illustrate the
girls’ heights and the boys’ heights.
c) Use the five number summaries, the interquartile ranges and the box
and
whisker diagrams to write down two conclusions you can make about the
heights of the girls and the boys.