Quartiles, Deciles and Percentiles

Quartiles, Deciles and Percentiles

1. Quartiles, Deciles and Percentiles
Let us now extend the concept of
partitioning of the frequency
distribution by taking up the concept
of quantiles (i.e. quartiles, deciles and
percentiles)
 We have already seen that the
median divides the area under the
frequency polygon into two equal
halves:

2. 50% 50%
X
f
Median
Afurthersplittoproducequarters,tenthsor
hundredthsofthetotalareaunderthefrequencypolygon
isequallypossible,andmaybeextremelyusefulfor
analysis.(Weareofteninterestedinthehighest10%of
somegroupofvaluesorthemiddle50%another.)

3. QUARTILES
Thequartiles,togetherwiththemedian,achievethe
divisionofthetotalareaintofourequalparts.
Thefirst,secondandthirdquartilesaregivenby
theformulae:





 c
n
f
h
lQ
4
1 





 c
n
f
h
lQ
4
3
3
First quartile
Second quartile (i.e. median)
Third quartile mediancn
f
h
lc
n
f
h
lQ 





 2
4
2
2

4. 25%
X
f
Q1 Q2 =X
~ Q3
25% 25% 25% Itisclearfromtheformulaofthesecond
quartilethatthesecondquartileisthesameasthe
median.

5. DECILES & PERCENTILES:
Thedecilesandthepercentilesgivethedivisionofthe
totalareainto10and100equalpartsrespectively.





 c
n
f
h
lD
10
1
The formula for the first decile is
Theformulaeforthesubsequentdecilesare





 c
n
f
h
lD
10
3
3
and so on. It is easily seen that the 5th decile is the same quantity as
the median.





 c
n
f
h
lD
10
2
2

6. 





 c
n
f
h
lP
100
1 The formula for the first percentile is
Theformulaeforthesubsequentpercentilesare





 c
n
f
h
lP
100
2
2 





 c
n
f
h
lP
100
3
3
and so on.

7. Again,itiseasilyseenthatthe50thpercentileisthesameas
themedian,the25thpercentileisthesameasthe1stquartile,the75th
percentileisthesameasthe3rdquartile,the40thpercentileisthe
sameasthe4thdecile,andsoon.
Allthesemeasuresi.e.themedian,quartiles,decilesandpercentiles
arecollectivelycalledquantilesorfractiles.
Thequestionis,“Whatisthesignificanceofthisconceptof
partitioning?Whyisitthatwewishtodivideourfrequency
distributionintotwo,four,tenorhundredparts?”
Theanswertotheabovequestionsis:
In certain situations, we may be interested in describing the relative
quantitative locationof a particular measurement within a data set.
Quantiles provide us with an easy way of achieving this. Out
of these various quantiles, one of the mostfrequently used is
percentileranking.

8. FREQUENCY DISTRIBUTION OF
CHILD-CARE MANAGERS AGE
Class Interval Frequency
20 –29 6
30 –39 18
40 –49 11
50 –59 11
60 –69 3
70 –79 1
Total 50

9. Suppose we wish to determine:
The 1st quartile
The 6th decile
The 17th percentile

10. Solution
We begin with the 1st quartile (also known as lower
quartile).
The 1st quartile is given by:
Where, l, hand fpertain to the class that contains
the first quartile. 





 c
n
f
h
lQ
4
1
In this example,
n = 50, and hence
n/4 = 50/4 = 12.5

11. Class Boundaries Frequency
f
Cumulative
Frequency
cf
19.5 –29.5 6 6
29.5 –39.5 18 24
39.5 –49.5 11 35
49.5 –59.5 11 46
59.5 –69.5 3 49
69.5 –79.5 1 50
Total 50
Class
containing
Q
1

12. Hence,
l= 29.5
h= 10
f= 18
and
C = 6

13. Hence, the 1st quartile is given by: 
1
=
4
10
= 29.5 12.5 6
18
= 29.5 3.6
= 33.1
hn
Q l c
f







14. Interpretation
One-fourthofthemanagersareyoungerthanage
33.1years,andthree-fourthareolderthanthis
age.

15. The 6th Decile is given by6
6
10
hn
D l c
f

  


In this example,
n = 50, and hence
6n/10 = 6(50)/10 = 30

16. Class Boundaries Frequency
f
Cumulative
Frequency
cf
19.5 –29.5 6 6
29.5 –39.5 18 24
39.5 –49.5 11 35
49.5 –59.5 11 46
59.5 –69.5 3 49
69.5 –79.5 1 50
Total 50
Class
containing
D
6

17. Hence,
l= 39.5
h = 10
f = 11
and
C = 24
Hence, 6th decile is given by 
6
6
=
10
10
= 39.5 30 24
11
= 29.5 5.45
= 44.95
hn
D l c
f







18. Interpretation
Six-tenth i.e. 60% of the managers are younger than
age 44.95 years, and four-tenth are older than this age.

19. The 17th Percentile is given by17
17
100
hn
P l c
f

  


In this example,
n = 50, and hence
17n/100 = 17(50)/100 = 8.5

20. Class Boundaries Frequency
f
Cumulative
Frequency
cf
19.5 –29.5 6 6
29.5 –39.5 18 24
39.5 –49.5 11 35
49.5 –59.5 11 46
59.5 –69.5 3 49
69.5 –79.5 1 50
Total 50
Class
containing
P
17

21. Hence,
l= 29.5
h = 10
f = 18
and
C = 6
Hence, 6th decile is given by 
17
17
=
100
10
= 29.5 8.5 6
18
= 29.5 1.4
= 30.9
hn
P l c
f







22. Interpretation
17% of the managers are younger than age 30.9 years,
and 83% are older than this age.

23. EXAMPLE:
If oil company ‘A’ reports that its yearly sales are at the
90th percentile of all companies in the industry, the
implication is that 90% of all oil companies have yearly
sales lessthan company A’s, and only 10% have yearly
sales exceeding company A’s:
0.1
0
Relative Frequency
Company A’s sales
(90th percentile)
Yearly Sales
0.9
0

24. EXAMPLE
SupposethattheEnvironmentalProtectionAgencyof
adevelopedcountryperformsextensivetestsonallnewcar
modelsinordertodeterminetheirmileagerating.
Supposethatthefollowing30measurementsare
obtainedbyconductingsuchtestsonaparticularnewcar
model.EPA MILEAGE RATINGS ON 30 CARS
(MILES PER GALLON)
36.3 42.1 44.9
30.1 37.5 32.9
40.5 40.0 40.2
36.2 35.6 35.9
38.5 38.8 38.6
36.3 38.4 40.5
41.0 39.0 37.0
37.0 36.7 37.1
37.1 34.8 33.9
39.9 38.1 39.8
EPA: Environmental Protection Agency

25. Whentheabovedatawasconvertedtoafrequency
distribution,weobtained:Class LimitFrequency
30.0 – 32.92
33.0 – 35.94
36.0 – 38.914
39.0 – 41.98
42.0 – 44.92
30

26. 0
5
10
15
20
25
30
35
29.95 32.95 35.95 38.95 41.95 44.95 Cumulative Frequency Polygon
or OGIVE

27. ThisOgiveenablesustofindthemedianandanyother
quantilethatwemaybeinterestedinveryconveniently.
Andthisprocessisknownasthegraphiclocationof
quantiles.Letusbeginwiththegraphicallocationof
themedian:
Because of the fact that the median is that value
before which half of the data lies, the firststep is to divide
the total number of observations n by 2.
In this example:15
2
30
2
n

Thenextstepistolocatethisnumber15onthey-axis
ofthecumulativefrequencypolygon.

28. 0
5
10
15
20
25
30
35
29.95 32.95 35.95 38.95 41.95 44.95 Cumulative Frequency Polygon
or OGIVE2
n

29. 0
5
10
15
20
25
30
35
29.95 32.95 35.95 38.95 41.95 44.95 Cumulative Frequency Polygon or OGIVE2
n
Next,wedrawahorizontallineperpendiculartothey-
axisstartingfromthepoint15,andextendthislineuptothe
cumulativefrequencypolygon.

30. 0
5
10
15
20
25
30
35
29.95 32.95 35.95 38.95 41.95 44.95 Cumulative Frequency Polygon
or OGIVE2
n

31. 0
5
10
15
20
25
30
35
29.95 32.95 35.95 38.95 41.95 44.95 Cumulative Frequency Polygon
or OGIVE2
n 9.37X
~


32. Inasimilarway,wecan
locatethequartiles,decilesand
percentiles.
Toobtainthefirstquartile,
thehorizontallinewillbe
drawnagainstthevaluen/4,
andforthethirdquartile,the
horizontallinewillbedrawn
againstthevalue3n/4.

33. 0
5
10
15
20
25
30
35
29.95 32.95 35.95 38.95 41.95 44.95 Cumulative Frequency Polygon
or OGIVE4
n
Q
1
Q
34
n3