QUARTILE AND DECILE OF GROUPED DATA

Q4-WEEK 5 & 6
WELCOME
MATHEMATICS
CLASS
FOR GRADE 10
10th grade

RECALL
It is a method that is used
to divide a distribution into
ten equal parts.
is a quantile that divides a
set of scores into 100 equal
parts.
the proportion of scores in a
distribution that a specific
score is greater than or
equal to.
DECILE
QUARTILE
PERCENTILE PERCENTILE RANK
Quartiles are the values
which divide the whole
distribution into four
equal parts.

QUARTILE FOR GROUPED DATA
The steps in computing the median are similar to that of Q1 and
Q3. In finding the median, we need first to determine the
median class. The Q1 class is the class interval where the
𝑁
4
th score is contained, while the class interval that
contains the
3𝑁
4
𝑡ℎ score is the Q3 class.
Formula :𝑄𝑘 = LB +
𝑘𝑁
4
−𝑐𝑓𝑏
𝑓𝑄𝑘
𝑖
LB = lower boundary of the of the 𝑄𝑘 class
N = total frequency
𝑐𝑓𝑏 = cumulative frequency of the class before the
𝑄𝑘 class
𝑓𝑄𝑘
= frequency of the 𝑄𝑘 class
i = size of the class interval
k = the value of quartile being asked

EXAMPLE
1: Calculate Q1, Q2 and Q3 of the Mathematics test scores of 50 students.
Class
interval
(scores)
Frequenc
y (f)
50-54 4
45-59 8
40-44 11
35-39 9
30-34 12
25-29 6
cf
6
18
27
38
46
50
𝑇𝑜𝑡𝑎𝑙 𝑛 = 50
𝑄𝑘 = LB +
𝑘𝑁
4
−𝑐𝑓𝑏
𝑓𝑄𝑘
𝑖
𝑄1 𝑐𝑙𝑎𝑠𝑠 =
50
4
= 𝟏𝟐. 𝟓
𝑄1 𝑐𝑙𝑎𝑠𝑠
𝐿𝐵 = 29.5
𝑐𝑓𝑏 = 6
𝑓𝑄𝑘
= 12
𝑖 = 5
𝑄1 = 29.5 +
12.5 − 6
12
5
= 29.5 +
6.5
12
5
= 29.5 + .54 5
= 29.5 + 2.71
𝑸𝟏 = 𝟑𝟐. 𝟐𝟏
Q1 = 32.21: 25% of the students
have a score less than or equal to

EXAMPLE
Class
interval
(scores)
Frequenc
y (f)
50-54 4
45-59 8
40-44 11
35-39 9
30-34 12
25-29 6
cf
6
18
27
38
46
50
𝑄𝑘 = LB +
𝑘𝑁
4
−𝑐𝑓𝑏
𝑓𝑄𝑘
𝑖
2(50
4
= 𝟐𝟓
𝐿𝐵 = 34.5
𝑐𝑓𝑏 = 18
𝑓𝑄𝑘
= 9
𝑖 = 5
𝑄2 = 34.5 +
25 − 18
9
5
= 34.5 +
7
9
5
= 34.5 + .78 5
= 34.5 + 3.89
𝑸𝟐 = 𝟑𝟖. 𝟑𝟗
Q2 = 38.39: 50% of the students
have a score less than or equal to

EXAMPLE
Class
interval
(scores)
Frequenc
y (f)
50-54 4
45-59 8
40-44 11
35-39 9
30-34 12
25-29 6
cf
6
18
27
38
46
50
𝑄𝑘 = LB +
𝑘𝑁
4
−𝑐𝑓𝑏
𝑓𝑄𝑘
𝑖
3(50
4
= 𝟑𝟕. 𝟓
𝐿𝐵 = 39.5
𝑐𝑓𝑏 = 27
𝑓𝑄𝑘
= 11
𝑖 = 5
𝑄3 = 39.5 +
37.5 − 27
11
5
= 39.5 +
10.5
11
5
= 39.5 + .95 5
= 39.5 + 4.77
𝑸𝟑 = 𝟒𝟒. 𝟐𝟕
Q3 = 44.27: 75% of the students have
a score less than or equal to 44.27

The interquartile range describes the middle 50% of values when
ordered from lowest to highest. To find the interquartile range (IQR),
first find the median (middle value) of the upper and the lower half of
the data. These values are Q1 and Q3. The IQR is the difference
between Q3 and Q1.
Interquartile Range (IQR) = Q3 – Q1
The quartile deviation or semi-interquartile range is one-half the
difference between the third and the first quartile.
Quartile Deviation (QD) =
𝑄3−𝑄1
2
𝑄3 = 44.27
𝑄1 = 32.21
𝐼𝑄𝑅 = 44.27 − 32.21 = 12.06
𝑄𝐷 =
12.06
2
= 6.03
𝐸𝑋𝐴𝑀𝑃𝐿𝐸:

ACTIVITY
Class Interval Frequency(f)
88 – 96 9
80 – 87 10
72 – 79 15
64 – 71 13
56 – 63 9
48 – 55 9
Consider the distribution of scores of the students in Math.
Find: a) Q1 b) Q3 c) IR, d) QD
cf
65
56
46
31
18
9
𝑄1 =
65
4
= 𝟏𝟔. 𝟐𝟓
Q1 class
𝑄3 =
3(65)
4
= 𝟒𝟖. 𝟕𝟓
Q3 class 𝑄1 = 55.5 +
16.25 − 9
9
8
= 55.5 +
7.25 8
9
= 55.5 + 6.44
𝑸𝟏 = 𝟔𝟏. 𝟗𝟒
𝑄3 = 79.5 +
48.75 − 46
10
8
= 79.5 +
2.75
10
8
= 79.5 + 2.2
𝑸𝟑 = 𝟖𝟏. 𝟕𝟎
Interquartile Range (IR) = Q3 – Q1 = 81.70 – 61.94 = 19.76
Quartile Deviation (QD) = 19.76/2 = 9.88

The formula in finding the kth decile of a distribution is
𝐷𝑘 = 𝑙𝑏𝑑𝑘 +
(
𝑘
10
)𝑁 − 𝑐𝑓
𝑓𝐷𝑘
𝑖
DECILES FOR GROUPED DATA

Let the table be scores of 45 students in a
long test in Math.
EXAMPLE
Solve for (a) D1 and (b) D6
Class f
30-34 5
25-29 11
20-24 13
15-19 6
10-14 10
LB cf
9.5
14.5
19.5
24.5
29.5
10
16
29
40
45
𝑁 = 45
i = 5
𝐷1 =
𝑘
10
𝑁 =
1
10
45 = 4.5
→ 𝐷1
𝐿𝐵 = 9.5
𝐷𝑘 = 𝐿𝐵𝑑𝑘 +
(
𝑘
10
)𝑁 − 𝑐𝑓
𝑓𝐷𝑘
𝑖
𝑐𝑓 = 0
𝑓 = 10
𝐷1 = 9.5 +
4.5 − 0
10
5
= 9.5 + (
4.5
10
)5
= 9.5 + 2.25
𝑫𝟏 = 𝟏𝟏. 𝟕𝟓

Let the table be scores of 45 students in a
long test in Math.
EXAMPLE
Solve for (a) D1 and (b) D6
Class f
30-34 5
25-29 11
20-24 13
15-19 6
10-14 10
LB cf
9.5
14.5
19.5
24.5
29.5
10
16
29
40
45
𝑁 = 45
i = 5
𝐷6 =
𝑘
10
𝑁 =
6
10
45 = 27
→ 𝐷1
𝐿𝐵 = 19.5
(
𝑘
10
)𝑁 − 𝑐𝑓
𝑓𝐷𝑘
𝑖
𝑐𝑓 = 16
𝑓 = 13
𝐷6 = 19.5 +
27 − 16
13
5
= 19.5 + (
11
13
)5
= 19.5 + 4.23
𝑫𝟑 = 𝟐𝟑. 𝟕𝟑
→ 𝐷6

ACTIVITY
100 students are given a 50-item assessment in Mathematics 10 to
determine who will be qualified to apply for club membership. The
results of the assessment are shown in the table below.
𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝐷3 𝑎𝑛𝑑 𝐷7
Scores F
46-50 19
41-45 18
36-40 12
31-35 17
26-30 14
21-25 15
16-20 3
11-15 2
LB
45.5
40.5
35.5
30.5
25.5
20.5
15.5
10.5
cf
100
81
63
51
34
20
5
2
𝑁 = 100
i= 5
𝐷3 =
𝑘
10
𝑁 =
3
10
100 = 30
→ 𝐷3
𝐿𝐵 = 25.5
𝑐𝑓 = 20
𝑓 = 14
(
𝑘
10
)𝑁 − 𝑐𝑓
𝑓𝐷𝑘
𝑖
𝐷3 = 25.5 +
30 − 20
14
5
= 25.5 +
10
14
5
= 25.5 + 3.57
= 29.07

ACTIVITY
100 students are given a 50-item assessment in Mathematics 10 to
determine who will be qualified to apply for club membership. The
results of the assessment are shown in the table below.
𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝐷3 𝑎𝑛𝑑 𝐷7
Scores F
46-50 19
41-45 18
36-40 12
31-35 17
26-30 14
21-25 15
16-20 3
11-15 2
LB
45.5
40.5
35.5
30.5
25.5
20.5
15.5
10.5
cf
100
81
63
51
34
20
5
2
𝑁 = 100
i= 5
𝐷7 =
𝑘
10
𝑁 =
7
10
100 = 70
→ 𝐷3
𝐿𝐵 = 40.5
𝑐𝑓 = 63
𝑓 = 18
(
𝑘
10
)𝑁 − 𝑐𝑓
𝑓𝐷𝑘
𝑖
𝐷3 = 40.5 +
70 − 63
18
5
= 40.5 +
7
18
5
= 40.5 + 1.94
= 42.44
→ 𝐷7

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QUARTILE AND DECILE OF GROUPED DATA

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