1. FRACTILES
GROUP 4

2. Fractiles are measures of location or
position which include not only central
location but also any position based on
the number of equal divisions in a given
distribution. If we divide the distribution
into four equal divisions, then we have
quartiles denoted by Q1, Q2, Q3, and Q4. The
most commonly used fractiles are the
quartiles, deciles, and percentiles.

3. Fractiles for Ungrouped Data
QUARTILES divide a distribution into four
equal parts. For example, Q1, or the first
quartile, locates the point which is greater
than 25% of the items in distribution.
Q3 is the 3rd quartile  Q3 = 3N th item
4
This means that 75% of the observations lie
below this value.

4. Q2 is the 2nd quartile  Q2 = 2N th item or
the median 4
Q1 is the 1st quartile  Q1 = N th item
4

5. DECILES are values that divide a
distribution into 10 equal parts.
D1 is the 1st decile  D1 = N th item
10
D3 is the 3rd decile  D3 = 3N th item
10
D5 is the 5th decile  D5 = 5N th item or
the median . 10

6. PERCENTILES are values that divide
the distribution into 100 equal parts. P10
or the tenth percentile means the 10th item
in the distribution which is 10% higher
than the rest of the items.
P1 is the 1st percentile  P1 = Nth item
100
P25 is the 25th percentile  P25 = 25Nth
item or Q1 100

7. P50 is the 50th percentile  P50 = 50Nth
item or the median 100
P67 is the 67th percentile  P67 = 67Nth
item 100

8. Example 1:
Calculate Q1, Q2, Q3, D1, D4, D5, D7, P10, P25,
P50 and P70 for the following IQ scores:
87 90 95 96 97 98 98 99
100 100 100 100 100 101 101 102
102 102 103 104 105 107 110
Q1 = N th = 23 th = 5.75th item, which is 98.
4 4

9. Note: for odd number of observations,
when the item number is fractional, take
the next higher item. Since the 5th item 97,
the 5.75th item is the next value, which is
98. This means that the score of 98 is
higher than 25% of the items in the
distribution. If the number of cases is
even, take the point midway between the
two items located at the middle of the
distribution.

10. Q2 = 2N th = 2(23) th = 23 th = 11.5th item,
4 4 2
which is 100. This means that the score of 100 is
higher than 50% of the items in distribution.
Q3 = 3N th = 3(23) th = 69 th = 17.25 th item,
4 4 4
which is 102.
D1 = N th = 23 th = 2.3 th item, which is 95.
10 10

11. D4 = 4N th= 4(23)th= 92 th=9.2th item, which is 100.
10 10 10
D5= 5N th= 5(23)th= 115 th= 11.5th item, which is 100.
10 10 10
D7= 7N th=7(23)th= 161 th=16.1th item, which is 102.
10 10 10
P10= 10N th=10(23)th= 23 th=2.3th item, which is 95.
100 100 10
P25= 25N th=25(23)th= 23 th=5.7th item, which is 98.
100 100 4

12. P50= 50Nth=50(23)th= 23th=11.5th item, which is 100.
100 100 2
P70= 70Nth=70(23)th= 1,610th=16.1th item,
100 100 100
which is 102.
NOTE: that median is equal to Q2, D5, and P50

13. Grouped data
• Estimate the cumulative frequencies on
the table
• Find N/4 or one-fourth of the number of
the cases in the distribution.
• Determine the class limit in which N/4
case falls
• Compute Q1 by using the formula;
Q1 = L+C (N/4-ΣCf<)
fc

14. Where ;
Q1 = is the first quartile
L = lower real limit of the first quartile
class
N = total number of cases
ΣCf< = sum of the cumulative frequencies “
lesser than” up to but below the first
quartile class
C = class interval
fc = frequency of the Q1 class

15. CLASS LIMIT FREQUENCY
CUMULATIVE FREQ.
Cf<
46-48 1 35
43-45 1 34
40-42 2 33
37-39 3 31
34-36 3 28
31-33 4 25
28-30 7 21
25-27 5 14
22-24 3 9
19-21 2 6
16-18 2 44
13-15 1 2
10-12 1 1

16. Q1 = L+C (N/4- ΣCf<)
fc
= 21.5 + 3 (8.75 – 6)
3
= 21.5 + 3 ( 2.75)
3
= 21.5 + 2.75
Q1= 24.25
N/4 = 8.75
L = 21.5
ΣCf< = 6
fc = 3
C = 3

17. Q3 = L+C (3N/4- ΣCf<)
fc
= 33.5 + 3 (26.5 – 25)
3
= 33.5 + 1.5
= 35
N/4 = 26.25
L = 33.5
ΣCf< = 25
fc = 3
C = 3

18. Decile
• Estimate the cumulative frequency “lesser
than”
• Look for 7N/10 or seven- tenth of the
number of cases in distribution
• Find the class limit in which the 7N/10 case
falls
• Compute D7 by using the formula;
D7 = L+C (7N/10-ΣCf<)
fc

19. Where ;
D7= is the seventh
L = lower real limit of the seventh decile
class
N = total number of cases
ΣCf< = sum of the cumulative frequencies “
lesser than” up to but below the first
quartile class
C = class interval
fc = frequency of the D7 class

20. D7 = L+C (7N/10-ΣCf<)
fc
= 30.5 + 3 (24.5 – 21)
4
= 30.5 + 3 (3.5)
4
= 30.5 + 10.5
4
= 30.5 + 2.625
D7= 33.125 or 33.13
7N/10 = 24.5
L = 30.5
ΣCf< = 21
fc = 4
C = 3
N= 35

21. Percentile
Formula;
P90 = L+C (90N/100-ΣCf<)
fc
Where ;
P90= is the ninetieth percentile
L = lower real limit of the ninetieth percentile class
N = total number of cases
ΣCf< = sum of the cumulative frequencies “ lesser than”
up to but below the ninetieth percentile class
C = class interval
fc = frequency of the P90 class

22. P90 = L+C (90N/100-ΣCf<) 90N/100 = 31.5
fc
solution:
= 39.5 + 3 ( 31.5 – 31)
2
= 39.5 + 3 (0.5)
2
= 39.5 + 1.5
2
= 39.5 + 0.75
P90 = 40.25
L = 39.5
ΣCf< = 31
fc = 2
C = 3
N= 35

23. On the other hand, special cases are just written as
follows
P40 = 27.5 special case
P60 = 30.5 special cases
P80 = 36.5 special cases
P40 = L+C (40N/100-ΣCf<) 40N/100 =14
fc
= 24.5 + 3 (14 – 9)
5
= 24.5 + 3 (5)
5
= 24. 5 + 3
P40 = 27.5
L = 24.5
ΣCf< = 9
fc = 5
C = 3
N= 35

24.  If it is a special case, the percentile value of an
upper real limit (2.75) can be determined by
dividing the ΣCf< by N times 100 or ΣCf< * 100.
N
For instance, the ΣCf< is 14 and N is 35. 14/35 *
100 equals to 40. hence, 27.5 is P40.

25. Another case is percentile 60 where cumulative
frequency “lesser than” is exactly 21.
P60 = L+C (60N/100-ΣCf<)
fc
= 27.5 + 3 (21 -14)
7
= 27.5 +3 (7)
7
P60= 30.5
60N/100 = 21
L = 27.5
ΣCf< = 14
fc = 7
C = 3
N= 35

26. P80 = L+C (80N/100-ΣCf<)
fc
P80 = 33.5 + 3 (28 – 25)
3
= 33.5 + 3 (3)
3
P80 = 36.5
80N/100 = 28
L = 33.5
ΣCf< = 25
fc = 3
C = 3
N= 35