Advance Statistics: Self-Learning Module for College Students 1

Advance Statistics: Self-Learning Module for College Students 1

Module 4 Measures of Position

Module Overview
What do you do when you’re lost? You use tools like a compass and GPS to figure out
where you are and how to get where you are going. Well, in statistics there are ways to
figure out where a data point or set falls. These are called measures of position. Once we
know where a data set or model is, we can figure out what to do with it. Let’s discuss how
we find out where data is and what that means.
There are three forms of measures of position applicable to ungrouped and grouped
data. These are the quartiles, percentiles and the z-scores. In this module we will discuss
quartiles and percentiles, while the z-score will be discussed in a different module.

Learning Outcomes
At the end of this module you shall be able to:
1. determine the quartiles and percentiles of ungrouped and grouped data;
2. interpret the quartiles and percentiles using a five-number summary; and
3. construct a box and whiskers plot to represent quartiles and percentiles.

Pre-Assessment
Before you begin this module, you must have answered first the Pre-
Assessment posted in Google Classroom. If you are unable to connect to
Google Classroom you can access the Pre-Assessment by using this link:
https://bre.is/uoXn5fUW or by scanning the QR Code presented. You can
access the Pre-Assessment using your laptop or your mobile devices.

Discussion
A measure of position is a method by which the position that a particular data value
has within a given data set can be identified. As with other types of measures, there is more
than one approach to defining such a measure.
Measures of position are used to describe the relative location of an observation.
The quartile and percentile are the most common measures of position. These two
measures are also called quantiles.

QUARTILE

Quartiles are values that divide your data into quarters. They divide your data into
four segments according to where the numbers fall on the number line. Quartiles may or
may not be part of the data. To find the quartiles, first find the median or second quartile.
The first quartile, Q1, is the middle value of the lower half of the data, and the third
quartile, Q3, is the middle value, or median, of the upper half of the data.

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The four quarters that divide a data set into quartiles are:

 The lowest 25% of numbers.

 The next lowest 25% of numbers (up to the median).
 The second highest 25% of numbers (above the median).
 The highest 25% of numbers.

The figure below shows how a data set is divided into quartiles.

Quartiles for Ungrouped Scores

For ungrouped scores it is necessary that the data are arranged in ascending order
or according to their position in the number line. After the data are arranged, we can now
use the equation below to compute the quartile:

i
The quartile Q i is located at (n+1 ¿. Where i is the quartile position (1 for the first
4
quartile, 2 for the 2nd quartile and so on) and n is the total number of data values.
1
So for the location of the 1st quartile we will use is (n+1). Remember that Q 2 and
4
the median are the same.

Example:
Find the quartiles of the following data:

73 96 78 97 73 95 89 65 85
68 89 78 58 69 75 89 88 99
80 79 57 71 48 87 78 77 74

Step 1: Arrange the scores in ascending order.

48 57 58 65 68 69 71 73 73
74 75 77 78 78 78 79 80 85
87 88 89 89 89 95 96 97 99

Step 2: Determine the total number of data values n. In our example n=27.

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Step 3: Compute the quartiles. We only need to compute Q 1, Q 2 or the median and Q 3.

For the position 1st quartile:

1
Q 1=¿ (n+1)
4
1
= (27 + 1)
4
1
= (28)
4
=7
The 1st quartile is the 7th data value or x 7, which is 71.

For the position of 2nd quartile:

2
¿ (n+1)
4
1
= (27 + 1)
2
1
= (28)
2
= 14
The 2nd quartile (and the median) is the 14th data value or x 14, which is 78.

For the position of the 3rd quartile:

3
Q 3=¿ (n+1)
4
3
= (27 + 1)
4
3
= (28)
4
= 21
The 3rd quartile is the 21st data value or x 21, which is 89.

Note that for this example the number of data values is odd.

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Activity 4.1: Quartiles of Ungrouped Data 1

Find the quartiles for the given data below:

13 22 18 13 33 28 29 12 23
15 12 13 30 24 12 33 33 15
22 17 20 16 32 21 32 15 34
17 16 33 19 27 17 23 18

Example:

The following are the daily high temperatures (in degree Fahrenheit) in a two-week
period in Paniqui, Tarlac were 57, 60, 62, 65, 66, 71, 75, 78, 82, 83, 85, 88, 91, 96.
Determine the quartiles for this data.

Solution:

Notice that the data set is already arranged in ascending order and that n=14, even
number. For this data set we will use interpolation to determine the position of each
quartile scores.

For the position of Q1

1
¿ (n+1)
4
1
= (14 + 1)
4
1
= (15)
4
= 3.75
The 1st quartile is between the 3rd ( x 3 ¿ and 4th ( x 4 ) value.
Q1
57 60 62 65 66 71 75 78 82 83 85 88 91 96

x3 x4

Interpolating between the x 3 and x 4:

Q 1=x 3 +0.75(x 4 −x 3)

0.75 is the decimal value of the position of Q 1 (3.75)

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¿ 62+0.75 ( 65−62 )
= 62 +¿0.75(3)
= 62 +¿ 2.25
= 64.25
The 1st quartile is 64.25.

Q 2 is the same as the median. Remember that the median for an ungrouped data
with an even number of data value is the average of the two middle most data. But for
emphasis on explaining interpolation, we will use the same method above.

The position of Q 2

2
¿ (n+1)
4
1
= (14 + 1)
2
1
= (15)
2
= 7.5
The 2nd quartile is between the 7th ( x 7 ¿ and 8th ( x 8 ) value.

Q2
57 60 62 65 66 71 75 78 82 83 85 88 91 96
^x
x7 x8
Q 2=x 7 +0.5(x 8 −x7 )
Q1=75+ 0.5 ( 78−75 )
Q 1=75+ 0.5(3)
Q 1=75+1.5
Q 1=76.5

Computing the median will result to the same value:

75+78
^x =¿
2
153
^x =¿
2
^x =¿ 76.5
The 2nd quartile is 76.5.
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Advance Statistics: Self-Learning Module for College Students 6

The position of Q 3

3
¿ (n+1)
4
3
= (14 + 1)
4
3
= (15)
4
= 11.25
The 3rd quartile is between the 11th ( x 11 ¿ and 12th ( x 12 ) value.

Q3
57 60 62 65 66 71 75 78 82 83 85 88 91 96

x 11 x 12

Interpolating between the x 11 and x 12:

Q3=x 11 +0.25 ( x12−x 11 )

0.25 is the decimal value of the position of Q 3 (11.25)

¿ 85+0.25 ( 88−85 )
= 85 +¿0.25(3)
= 85 +¿ 0.75
= 85.75

The 2nd quartile is 85.75.

Activity 4.2: Quartiles of Ungrouped Data 2

The data below are the scores of 30 students in a Math test. Determine the quartiles for
the given data.

11 22 24 12 24 13 9 14 13 21
11 16 24 14 16 24 19 23 13 22
9 20 19 11 11 12 24 12 24 11

We can also use technology to determine the quartiles of given data. We will use MS
Excel to determine the quartiles of our previous examples. In using excel we will use the
excel function:

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=QUARTILE.EXC(array, quart)

where the function arguments are:

array - the range of data values for which you want to calculate the specified
quartile.
quart - an integer between 1 and 3, representing the required quartile.

The first step in using excel is to encode the data

in a column. Once your data has been encoded, select an
empty cell an type “=QUARTILE.EXC(“.

Highlight the data values for the array argument.

Type “1” and an integer to represent the quartile

that you want to compute (1 for Q 1, 2 for Q 2 and 3 for
Q 3). In our example we are going to determine Q 1.

Press ENTER for the output.

Activity 4.3: Quartiles of Ungrouped Data 3

Recompute Activity 4.1 and 4.2 using MS Excel.

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Quartiles for Grouped Data

For data that are presented in a distribution table, we use the following steps.

STEP 1: Locate the quartile class (Q t Class). The quartile class is the class which contains
tn
the quartile. The formula for this is where t is the quartile being determine (t is
4
1 for Q1, 2 for Q 2, and 3 for Q 3) n is the total number of data values (equal to Σ f ).
tn
The Q t class is the class whose cumulative frequency is nearest and greater than .
4

STEP 2: Solve the Q t using the formula below. This formula bares a resemblance to the
formula of the median for grouped scores.

tn

where:
t
( )
Q t =LQ +¿ 4
−F Q
fQ
t
t

Q t is the the t th quartile

LQ is the lower boundary of the Qtht quartile (lower limit of the Q t class minus
T

0.5)
tn
is the location of the Q t class.
4
F Q is the cumulative frequency before the Q t class
T

f Q is the frequency of the Q t class

t

i is the class width

Example:
The following table gives the frequency distribution of the number of orders
received each day during the past 50 days at the office of a mail-order company. Calculate
the quartiles.

Number of
f
Orders
10 – 12 4
13 – 15 12
16 – 18 20
19 – 21 14
Σ f =50=n

An important part of solving the quartiles of grouped data is the cumulative

frequency column. We must add this column before any computation can be made.

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Number of
f cf
Orders
10 – 12 4 4
13 – 15 12 16
16 – 18 20 36
19 – 21 14 50
Σ f =50=n

For Q 1,
1(50)
Location of the Q 1 class ¿ = 12.5
4
The Q 1 class is the 3rd class (16 – 18).

Number of
f cf
Orders
10 – 12 4 4
13 – 15 12 16
Q 1 Class 16 – 18 20 36
19 – 21 14 50
Σ f =50=n

n
Q =L
Determine the other variables in the equation 1 Q +¿ 4
−F Q
fQ
i. 1
( ) 1
1

Cumulative
Lower Limit of Number of
f cf frequency
the Q 1 class. Orders
before the Q 1
Subtract 0.5 to 10 – 12 4 4
class, F Q .
get LQ . 1
13 – 15 12 16 1

16 – 18 20 36
LQ =15.5
1 19 – 21 14 50
Σ f =50=n

Frequency of
i =3 the Q 1 class,
n fQ1
= 12.5
4
LQ = 15.5
1

F Q = 16 1

f Q = 201

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 Advance Statistics: Self-Learning Module for College Students 10

The Q 1 of the distribution is:

n
1
( )
Q 1=LQ +¿ 4
−F Q
fQ
i
1
1

Q1=15.5+¿ ( 12.5−16
20 )
3

−13.5
Q =15.5+¿ (
20 )
1 3

Q1=15.5+¿ (−0,675 ) 3
Q 1=15.5−2.025
Q 1=13.475
For Q 2, (this is equal to the median)
2(50)
Location of the Q 2 class ¿ = 25
4
The Q 2 class is the 3rd class (16 – 18).

Number of
f cf
Orders
10 – 12 4 4
13 – 15 12 16
Q 2 Class 16 – 18 20 36
19 – 21 14 50
Σ f =50=n

2n
Determine the other variables in the equation Q 2=LQ +¿ 4
−F Q
fQ
i. 2
( 2
2

)
Cumulative
Lower Limit of Number of
f cf frequency
the Q 2 class. Orders
before the Q 2
Subtract 0.5 to 10 – 12 4 4
class, F Q .
get LQ . 2
13 – 15 12 16 2

16 – 18 20 36
LQ =15.5
2 19 – 21 14 50
Σ f =50=n

Frequency of
i =3 the Q 2 class,
fQ2

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2n
= 25
4
LQ = 15.5
2

F Q = 16 2

f Q = 20
2

The Q 2 of the distribution is:

2n
Q 2=LQ +¿ 4
2
(−F Q
fQ
i
2
2

)
Q 1=15.5+¿( 25−16
20 )
3

9
Q =15.5+¿ ( ) 3
1
20
Q 1=15.5+¿ ( 0.45 ) 3
Q1=15.5−1.35
Q 1=16.85
For Q 3,
3(50)
Location of the Q 3 class ¿ = 37.5
4
The Q 3 class is the 4th class (19 – 21).

Number of
f cf
Orders
10 – 12 4 4
13 – 15 12 16
16 – 18 20 36
Q 3 Class 19 – 21 14 50
Σ f =50=n

3n
Determine the other variables in the equation Q 3=LQ + ¿ 4
−F Q
fQ
i. 3
( 3
3

)
Cumulative
Number of
Lower Limit of f cf frequency
Orders
before the Q 3
the Q 3 class. 10 – 12 4 4
Subtract 0.5 to class, F Q .
3

get LQ . 3
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3
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fQ 3
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13 – 15 12 16
16 – 18 20 36
19 – 21 14 50
Σ f =50=n

i =3
3n
= 37.5
4
LQ = 18.5
3

F Q = 363

f Q = 14
3

The Q 2 of the distribution is:

3n
Q 3=LQ + ¿
3
( 4
−F Q
fQ
3
3

i )
Q1=18.5+¿ ( 37.5−36
14 )3
1.5
Q =18.5+¿ (
14 )
1 3

Q1=18.5+¿ ( 0.107 ) 3
Q 1=18.5−0.321
Q 1=18.179
INTERPRETING

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