MEASURES OF

POSITION
QUARTILES, DECILES AND PERCENTILES
(UNGROUPED DATA)
You are the 4th tallest student in a group of
10. If you are the fourth tallest student,
therefore 6 students are shorter than you. It
also means that 60% of the students are
shorter than you.
Q1/P25 MEDIAN / Q3/P75
Q2/D5/P50
 QUARTILES
Measures of central tendency that divide a group
of data into four subgroups

• Q1: 25% of the data set is below the first

quartile
• Q2: 50% of the data set is below the second
quartile
• Q3: 75% of the data set is below the third quartile
 Shown below are the scores of 16 students
received on a Math quiz. They are ranked in order
from least to greatest.

60 62 62 65 70 74 74 76 78 82 85 94 96 98 98 99

Q1 Q2 Q3
𝟔𝟓 + 𝟕𝟎 𝟕𝟔 + 𝟕𝟖 𝟗𝟒 + 𝟗𝟔
𝟐 𝟐 𝟐

= 𝟔𝟕. 𝟓 = 𝟕𝟕 = 𝟗𝟓
 Find Q1, median, and Q3.
◼3, 9, 12, 13, 15, 17, 19, 20, 24
MENDENHALL AND SINCICH METHOD
𝟏
Lower Quartile (L) = Position of Q1 = (𝒏 + 𝟏)
𝟒

And round to the nearest integer. If L falls halfway between two

integers, round up.
𝟑
Upper Quartile (U) = Position of Q3 = (𝒏 + 𝟏)
𝟒

And round to the nearest integer. If U falls halfway between two

integers, round down.
 Find Q1, median, and Q3.
◼3, 9, 12, 13, 15, 17, 19, 20, 24

Locate the position of Q1

Q1 =
1
(𝑛 + 1) Q1 = 2.5 𝑡ℎ 𝑖𝑡𝑒𝑚
4

1
9 + 0.5(12-9)
Q1 = (9 + 1) 9 + 0.5 (3)
4

Q1 =
1
(10) 9 + 1.5
4
10.5
Q1 = 2.5

Q1 = 2.5 𝑡ℎ 𝑖𝑡𝑒𝑚
Q2 / Median

3, 9, 12, 13, 15, 17, 19, 20, 24

2
Q2 = 4
(𝑛 + 1)
1
Q2 = 2
(9 + 1)
1
Q2 = 2
(10)

Q2 = 5𝑡ℎ 𝑖𝑡𝑒𝑚
Q3 / Upper Quartile

3, 9, 12, 13, 15, 17, 19, 20, 24

3
Q3 = 4
(𝑛 + 1) Q3 = 7. 5𝑡ℎ 𝑖𝑡𝑒𝑚
3 Q3 = 19 + .5(20 − 19)
Q3 = (9 + 1)
4 Q3 = 19 + .5(20 − 19)
3
Q3 = 4
(10) Q3 = 19 + .5(1)
Q3 = 7.5𝑡ℎ 𝑖𝑡𝑒𝑚 Q3 = 19.5
 Q1, median, and Q3.

3, 9, 12, 13, 15, 17, 19, 20, 24

Q1 Q2 Q3
10.5 or 19.5
median
 QUARTILES, continued
Q1 is equal to the 25th percentile
Q2 is located at 50th percentile and equals the median
Q3 is equal to the 75th percentile

Quartile values are not necessarily members of the data

set
DECILES
 DECILES FOR UNGROUPED DATA
The deciles are the nine score points which divide a distribution
into ten equal parts. They are denoted as D 1, D2, D3…, D9. They are
computed in the same way that the quartiles are calculated.

↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑
D1 D2 D3 D4 D5 D6 D7 D8 D9

𝑘
Dk = 10
(𝑛 + 1)
Example
The ages of the guests at a certain
party are16, 19, 17, 15, 25, 20,
19, 17, 14, 19, 20, 19, 18, 16, 17.
a. Find the upper and lower quartile
b. Find the value of D5. What does it
mean?
16, 19, 17, 15, 25, 20, 19, 17, 14, 19, 20, 19, 18, 16, 17
Arrange in ascending order

14, 15, 16, 16, 17, 17, 17, 18, 19, 19, 19, 19, 20, 20, 25

Q1 D5 Q3
or
median
PERCENTILES
Percentiles are points that divide a distribution into
100 equal parts.

𝑘
Pk = (𝑛 + 1)
100
Example
The prices of a souvenir T-shirt in Tagaytay
from 18 souvenir shops located at Sky
Ranch are as follows: 250, 100, 150, 180,
205, 220, 300, 190, 160, 180, 110, 200,250,
170, 200, 190, 210 and 300. What is the
value of the 50th percentile and what
does it mean?
 100, 110, 150, 160, 170, 180,180, 190,190,
200, 200, 205, 210, 220, 250, 250, 300, 300
𝑘
Pk = (𝑛 + 1)
100
P10 =
10
(𝑛 + 1) 2nd integer
100
1 P10 = 110
P10 = (18 + 1)
10
1
P10 = (19)
10
P10 = 1.9 𝑡ℎ 𝑖𝑡𝑒𝑚 − 𝑟𝑜𝑢𝑛𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
 100, 110, 150, 160, 170, 180,180, 190,190,
200, 200, 205, 210, 220, 250, 250, 300, 300

Pk =
𝑘
(𝑛 + 1) 10 𝑡ℎ 𝑖𝑡𝑒𝑚
100
50 P50 = 200
P50 = (𝑛 + 1)
100
1
P50 = (18 + 1)
2
1
P50 = (19)
2
P50 = 9.5𝑡ℎ 𝑖𝑡𝑒𝑚 − 10𝑡ℎ 𝑖𝑡𝑒𝑚
• 21, 22, 23, 24, 25, 25,
• 26,27,27, 28,29,29, 27,30, 26,26,28,28
• 31,32,32,33,34,35,33,31,34
• 36,36,37,37,38,38,38,40, 36,39,39
 THE QUARTILE FOR GROUPED DATA
In computing the quartiles of grouped data, the following
formula is used:
𝑘𝑁
−𝑐𝑓𝑏
Qk =LB + ( )i
4
𝑓𝑄𝑘

LB = Lower boundary of the Qk class

N = total frequency
Cfb – cumulative frequency of the class before the Qk class
fQ1 = frequency of the Qk class
i -= size of class interval
K = nth quartile
n= 1, 2, and 3
Calculate the Q1, Q2, and Q3 of the Mathematics test scores
of 50 students.
SCORES FREQUENCY
46-50 4
41-45 8
36-40 11
31-35 9
26-30 12
21-25 6
CLASS INTERVAL FREQUENCY LOWER BOUNDARIES LESS THAN CUMULATIVE
SCORES (f) FREQUENCY (< cf)

46-50 4 45.5 50
41-45 8 40.5 46
36-40 11 35.5 38
31-35 9 30.5 27
26-30 12 25.5 18
21-25 6 20.5 6
 CLASS FREQUENCY LOWER LESS THAN
INTERVAL (f) BOUNDARIES CUMULATIVE
SCORES FREQUENCY
(< cf)

46-50 4 45.5 50
41-45 8 40.5 46
36-40 11 35.5 38
31-35 9 30.5 27
26-30 12 25.5 18 (7th -18th score) Q1 class

21-25 6 20.5 6
LB = 25.5
N = 50
N = 50
Cfb = 6
𝑁 50
Q1 class = = fQ1 = 12
4 4
= 12.5 i= 5
 CLASS FREQUENCY LOWER LESS THAN
INTERVAL (f) BOUNDARIES CUMULATIVE LB = 25.5
SCORES FREQUENCY 𝑁 50
(< cf) Q1 class = = N = 50
4 4
Cfb = 6
46-50 4 45.5 50
= 12.5
fQ1 = 12
41-45 8 40.5 46 i= 5
36-40 11 35.5 38
31-35 9 30.5 27
26-30 12 25.5 18 (7th -18th score) Q1 class

21-25 6 20.5 6
N = 50 50
−6 6.5
Q1 = 25.5 + ( 4
)5 Q1 = 25.5 + ( )5
12 12

) 5 Q1 = 28.21
12.5 −6
Q1 = 25.5 + (
12
This implies that 25% of the students have a
score less than or equal to 28.21.
 CLASS FREQUENCY LOWER LESS THAN
INTERVAL (f) BOUNDARIES CUMULATIVE LB = 30.5
SCORES FREQUENCY
(< cf) Q2 class = =
2𝑁 2(50) N = 50
4 4 Cfb = 18
46-50 4 45.5 50 = 25
fQ2 = 9
41-45 8 40.5 46 i= 5
36-40 11 35.5 38
31-35 9 30.5 27 (19th -27th score) Q2 class
26-30 12 25.5 18
21-25 6 20.5 6
N = 50 2(50) 7
−18
Q2 = 30.5 + ( 4
)5 Q2 = 30.5 + ( ) 5
9 9
25 −18 Q2 = 34.39
Q2 = 30.5 + ( )5
9
This implies that 50% of the students have a
score less than or equal to 34.39.
 CLASS FREQUENCY LOWER LESS THAN
INTERVAL (f) BOUNDARIES CUMULATIVE LB = 35.5
SCORES FREQUENCY
(< cf) Q3 class = =
3𝑁 3(50) N = 50
4 4 Cfb = 27
46-50 4 45.5 50 = 37.5
fQ3 = 11
41-45 8 40.5 46 i= 5
36-40 11 35.5 38 (28th - 38th score) Q3 class
31-35 9 30.5 27
26-30 12 25.5 18
21-25 6 20.5 6
N = 50 3(50)
10.5
−27
Q3 = 35.5 + ( 4
)5 Q3 = 35.5 + ( )5
11 11
37.5 −27 Q3 = 40.27
Q3 = 35.5 + ( )5
11
This implies that 75% of the students have a
score less than or equal to 40.27.
THE DECILE FOR GROUPED DATA
𝑘𝑁
−𝑐𝑓𝑏
Dk = LB + ( 10
)𝑖
𝑓𝐷𝑘

Where: LB = lower boundary of the Dk class

N = total frequency
cfb = cumulative frequency before the Dk class
fDk = frequency of the Dk class
i = size of the class interval
k = nth decile where n = 1, 2, 3, 4, 5, 6, 7, 8, 9
 CLASS FREQUENCY LOWER LESS THAN
INTERVAL (f) BOUNDARIES CUMULATIVE LB = 40.5
SCORES FREQUENCY
(< cf) D8 class =
8𝑁
=
8(50) N = 50
10 10 Cfb = 38
46-50 4 45.5 50 = 40
fD8 = 8
41-45 8 40.5 46 (39th - 46th score)
i= 5
D8 class
36-40 11 35.5 38
8(50)
31-35 9 30.5 27 −38
26-30 12 25.5 18 D8 = 40.5 + ( 10
)5
8
40 −38
21-25 6 20.5 6 D8 = 40.5 + ( )5
8
N = 50 1
𝑘𝑁
−𝑐𝑓𝑏 D8 = 40.5 + ( ) 5
4
Dk = LB + ( 10
)𝑖 D8 = 41.75
𝑓𝐷𝑘

Therefore, the 8th decile is equivalent to 80th percentile.

This implies that 80% of the students got a score less than
or equal to 41.75.
 D8 class =
8𝑁 LB = 40.5
10
N = 50
8(50) Cfb = 38
=
10 fD8 = 8
= 40
i= 5
𝑘𝑁
−𝑐𝑓𝑏 40 −38
Dk = LB + ( 10
)𝑖 D8 = 40.5 + ( )5
𝑓𝐷𝑘 8
1
8(50)
−38 D8 = 40.5 + ( ) 5
4
D8 = 40.5 + ( 10
)5 D8 = 41.75
8

Therefore, the 8th decile is equivalent to 80th percentile.

This implies that 80% of the students got a score less than
or equal to 41.75.
THE PERCENTILE FOR GROUPED
DATA
𝑘𝑁
−𝑐𝑓𝑏
Pk = LB + ( 100
)𝑖
𝑓𝑃𝑘

Where: LB = lower boundary of the Pk class

N = total frequency
cfb = cumulative frequency before the Pk class
fPk = frequency of the Pk class
i = size of the class interval
k = nth percentile where n = 1, 2, 3,…, 97, 98, 99
 CLASS FREQUENCY LOWER LESS THAN 45𝑁
INTERVAL (f) BOUNDARIES CUMULATIVE P45 class = LB = 30.5
SCORES FREQUENCY 100
(< cf) N = 50
45(50) Cfb = 18
= fP45 = 9
100
46-50 4 45.5 50 i= 5
41-45 8 40.5 46 = 22.5
36-40 11 35.5 38
(19th - 27th score)
31-35 9 30.5 27
P45 class
26-30 12 25.5 18

21-25 6 20.5 6

N = 50
 45𝑁 45(50) LB = 30.5
P45 class = = N = 50
100 100
= 22.5 Cfb = 18
fP45 = 9
𝑘𝑁 i= 5
−𝑐𝑓𝑏
Pk = LB + ( 100
)𝑖
𝑓𝑃𝑘
45(50) 4.5
−18
P45 = 30.5 + ( 100
)5 P45 = 30.5 + ( )5
9 9
P45 = 33
22.5 −18
P45 = 30.5 + ( )5
9

Therefore, the 45th percentile is equivalent to 45% of the

students got a score less than or equal to 33.
STEPS IN CONSTRUCTING A FREQUENCY
DISTRIBUTION TABLE
1. Determine the range. It is the difference between the
highest and the lowest values in the list of data.
2. Determine the number of classes or class intervals
desired. The number of classes(usually between 6 -
10) is arbitrarily selected depending upon the size of
the data and spread of the values over which
frequencies are found.
3. Determine the size of the intervals by dividing the
range by the desired number of class intervals and
then rounding the result up. The usual class sizes are 2,
3, 5 or 10.

4. Determine the lower and upper limits of the first class

interval. It should include the smallest value in the list of
data.
5. Determine the lower and upper class limits of the
succeeding class intervals by adding the size of the
class interval to the lower and upper limits of the
preceding class interval until the highest class interval
is obtained.
6. Determine the number of observations by tallying
each value into each class interval, thus finding the
class frequencies.
Consider the following number of years of teaching
experience of 48 teachers.
28 26 21 15 20 16
32 15 18 19 16 14
25 14 22 21 13 9
12 9 18 15 12 10
9 11 12 9 10 11
6 6 7 8 6 8
7 6 8 8 3 4
3 5 5 2 2 1
Range: 32 – 1 = 31
Number of classes: 6
Size of the intervals: 31/6 = 5.16
=6
 Experience in frequency LB <cf
Years
31 - 36
25 - 30
19 - 24
13 - 18
7 -12
1-6