IBA, JU WMBA Program Course Instructor: Dr Swapan Kumar Dhar
2 (continued)
Quartiles, Deciles and Percentiles Median divides the set of observations into two equal parts so that the number of observations less than median is equal to the number of observations greater than median. Similarly we can divide the set of observations into a fixed number of equal parts. These are called partitions. The values that divide the set of observations into different partitions are called partition values or Quantiles. Quantiles. Some of the important partition values are Quartiles, Deciles and Percentiles.
Q1 , Q 2 , Q3 , which divide the set of observations into four equal parts. They are named as first quartile (Q1 ) , second quartile (Q 2 ) and third quartile (Q3 ). Of them Q2 is the median and Q1 and Q3 are also called lower a nd upper quartiles respectively.
Quartiles: Quartiles: There are three Quartiles,
(a) From From Ungrou Ungrouped ped Data: Data:
Qi
=
Value of the
= Value of
iN + 1 4
1 iN
2 4
th term if N is odd, i = 1, 2, 3.
iN + 1 th observation, if N is even, i= 1, 2, 3. 4
th +
Example 1: Find the quartiles of the following numbers. 10
72
18
45
32
56
64
27
60
Solution: Arranging the numbers in ascending order of magnitude, we get 10
18
Here, n = 9,
( 2n + 1) 4 ∴1
st
=
∴
18 18 + 1 4
quartile
27
n +1 4
=
= 4.75,
(Q1 )
32
9 +1
56
60
64
72
= 2.5
4 (3n + 1) 4
=
27 + 1 4
= value of 2.5 th term
= Value of 4.75 th term = 45.
Q2
45
Q3
=
=7 =
2 nd
ter term + 3rd ter term 2
=
18 + 27
=
2
45 2
= 22.5 .
Value of 7th term = 60.
Example 2: Consider the temperature (in Celsius) of several days during a summer season. 33
33
32
32.5
32
32
31.5
31.5
30.8
31
30.8
30.5
30.8
29.5
29
30
Find (i) maximum temperature of the first 25% lowest temperature days (ii) minimum temperature of the last 25% high temperature days (iii) median temperature. Solution: Solution : Arranging the numbers in ascending order of magnitude, we get 29
29.5
30
30.5
30.8
30.8
30.8
31
31.5
31.5
32
32
32
32.5
33
33
(i) Maximum temperature of the first 25% lowest temperature days is given by first quartile Q1 , where
1 N
N
Q1 = Value of th + + 1 th 24 4 = Value of
1 2
[ 4th + 5th ]
observation as n = 16 and it is even.
observation =
1 2
30.5 + 30.8 30.8] = 30.65 0 C . [ 30.5
(ii) Minimum temperature of the last 25% highest temperature days is given b y third quartile Q3 , where
1
1 3 N
Q3 = Value of = Value of
2
1 2
4
3 N + 1 th 4
th +
[ 12th + 13th ]
observation as n = 16 and it is even.
observation =
1
[ 32 + 32 ] = 320 C .
2
(iii) Median temperature is Median = Q2 = Value of = Value of
1 2
1 2 N
2
4
[ 8th + 9th ]
2 N + 1 th 4
th +
observation =
1 2
observation as n = 16 and it is even.
31.5] = 31.250 C . [ 31+ 31.5
(b) From Grouped Data or from Frequency Table : Example 3: Find the quartiles of the following distribution . Height (in inches) 58 59 Number of Students 15 20
60 32
61 35
62 33
63 22
64 20
65 10
Also find the median median Solution: Height (in inches) 58 59 60 61 62 63 64 65 66
Frequency(f) 15 20 32 35 33 22 20 10 8
Cumulative frequency 15 35 67 102 135 157 177 187 195
N + 1 th item = Size of 195 + 1 th item = 49 th item = 60. 4 4 N + 1 th item = Size of 2 195 + 1 th item = 98 th item = 61. Q2 = Size of 2 4 4 N + 1 th item = Size of 3 195 + 1 th item = 147 th item = 63. Q3 = Size of 3 4 4
Q1 = Size of
Since Q2 = Median, we can write median = 61. From Frequency distribution with class interval : The formula to calculate quartiles is
iN Qi
= L+
4
− F 1
f m
× C .Where
L = Lower limit of the quartile class N = Total frequency F 1 = Cumulative frequency for the class just above the quartile class f m
= Frequency of the quartile class.
C = Class interval of the quartile class. i = 1, 2, 3. The quartile class is that class for which cumulative frequency ≥
2
iN 4
.
66 08
Example 4: 4: The following distribution represents the monthly salary of a group of industrial workers: Salary (in Taka) < 1500 1500 – 1700 1700 – 1900 1900 – 2100 2100 – 2300 2300 – 2500 2500 >
Number of Workers (f) 18 42 65 150 70 45 20
Cumulative Frequency 18 60 125 275 345 390 410
(i) Find the maximum salary of the first 25% low paid workers. (ii) Find maximum salary of the last 25% high paid workers. (iii) Represent the above two values by an appropriate diagram. (iv) How many workers workers have income less than 2100 Taka? Solution: Solution : (i) The maximum salary of the first 25% low paid workers is calculated by 1st quartile it is necessary to obtain the quartile class. Since (Cumulati (Cumulative ve
frequency frequency
N Q1
= L+
4
− F 1
f m
≥ 102.5),
the
410
− 60
4
× C = 1700 +
65
quartile
class ass
N 4
is
Q1 . For calculating
=102.5 lies in the class 1700 – 1900
1700
–
1900. 900. So,
it
is
given
by
× 200 = 1830.77 Taka.
(ii) The minimum salary of the last 25% high paid workers is calculated by Q3 and it is given by
3 N Q3
= L+
4
− F 1
f m
× C = 2100 +
307. 307.5 5 − 275 275 70
× 200 = 2192.86
Taka. Here Q3 lies in the class 2100 –
2300, because cumulative frequency of the class is greater than or equal to
3 N 4
= 307.5.
(iii) The two values of Q1 and Q3 can be well represented by Box – and – Whisker plot as given below:
2 N The value of Q2 is
Q2
= L+
4
− F 1
f m
× C = 1900 +
205 205 − 125 125
Q1 Q2
<1500
1700
1900
150
× 200 = 2006.67
Taka.
Q3
2100
2300
2500>
Figure: Box – and – Whisker plot to represent the values of quartiles . (iv) From the cumulative frequency it is observed that 275 workers’ salary is less than 2100 Taka. Example 5: 5: Marks obtained by 25 students are given below: Marks obtained Number of students
0-10 3
3
10-20 4
20-30 8
30-40 6
40-50 4
Find the quartiles of the above distribution.
Solution: Calculations of Quartiles Marks 0-10
Frequency (f) 3
Cumulative frequency 3
10-20
4
7
20-30
8
15
30-40
6
21
40-50
4
Total
N = 25
Calculation of Q1
N
= 6.25
4
= L+
4
− F 1
f m
× C = 10 +
2 N
= L+
4
− F 1
3 N
f m
= 18.75
4
× 10 = 18.125.
= L+
4
12.5 2.50 − 7 8
× 10 = 26.875.
:
lies between C.F. 15 and C.F. 21 and the corresponding upper classes are 30 - 40.
3 N Q3
−3
:
× C = 20 +
Calculation of Q3
4
4
= 12.5 lies between C.F. 7 and C.F. 15 and their corresponding upper class is 20 - 30.
2 Q2
:
25
Calculation of Q 2
N
-
lies between C.F. 3 and C.F. 7 and the corresponding class is 10 - 20.
N Q1
25
− F 1
f m
× C = 30 +
18.7 18.75 5 − 15 6
× 10 = 36.25.
Deciles: Deciles: Deciles divide the set of observations into ten equal parts and there are nine deciles, denoted by
D1 , D 2 , ..., D9 . (a) For Ungrouped Data: Data : First arrange the data in ascending or descending order of magnitude. i( N + 1) Di = Value of the th term if N is odd, i = 1, 2,…,9. 10 1 iN iN + 1 th observation, if N is even, i= 1, 2, 3,…,9. th + = Value of 2 10 10 Example 6: 6: Repeat the Example 2. (i) Find maximum temperature of the first 10 per cent lowest temperature days (ii) Find the minimum temperature of the last 20% high temperature days. Solution: Solution : (i) Maximum temperature of the first 10% lowest temperature days is given by first decile D1 , where
D1 = Value of
1 N
2 10
N + 1 th 10
th +
observation as N = 16 and it is even.
4
1
= Value of
2
[ 1.6th + 2.6th ]
observation =
1 2
29.5 + 30.0 30.0] = [ 29.5
0
29.75 C .
(ii) Minimum temperature of the last 20% high temperature days is given by third decile D8 , where
D8 = Value of
th + 8 N + 1 th 10 10 2
1 8 N
1
= Value of
2
[ 12.8th + 13.8th ]
observation as N = 16 and it is even.
observation =
1 2
32.0 + 32.5 32.5] [ 32.0
= 32.250 C .
(b) Grouped frequency distribution: Example 7: 7: The following data represent the distribution of Jack fruit trees by number of Jack fruits in the tree: No. of Jack fruit No of trees Cumulative Frequency
10 5 5
15 18 23
18 22 45
19 36 81
20 15 96
25 42 138
30 12 150
32 18 168
40 27 195
45 08 203
50 10 213
(i) Find the maximum number of jack fruits in first 30% lower producing trees. (ii) Find the minimum number of jack fruits in last 30% high producing trees. Solution: Solution : (i) Maximum number of jack fruits in first 30% lower producing trees is obtained from D3 , where
D3 = Value of
3( N + 1)
th observation = Value of 64.2 th observation = 19 [C.F. > 64.2 or 65 is 81]
10
(ii) Minimum number of jack fruits in last 30% high producing trees is calculated by D7 , where
D7 = Value of
7( N + 1)
th observation = Value of 149.8 th observation = 30 [C.F. > 149.8 is 150]. 10 From Frequency distribution with class interval : The formula to calculate deciles is
iN Di
= L+
− F 1
10 f m
× C .Where
L = Lower limit of the decile class N = Total frequency F 1 = Cumulative frequency for the class just above the decile class f m
= Frequency of the decile class.
C = Class interval of the decile class. i = 1, 2, …,9. The decile class is that class for which cumulative frequency ≥
iN 10
.
Percentiles: Percentiles: Percentiles divide the set of observations into 100 equal parts and there are 99 percentiles,
. .., P99 . denoted by P1, P2, .. Data : (a) Ungrouped Data: i( N + 1) P i = Value of the th term if N is odd, i = 1, 2 , …,99. 100 = Value of
iN th + iN + 1 th observation, if N is even, i= 1, 2, …,99. 2 100 100
1
(b) Grouped Data without Class Intervals: Intervals : N+1 P = value of the th term 100 2(N + 1) P = value of the th term 100 . . . . . .. . . 1
2
5
= value of the
P
k
k(N + 1) 100
th term
(k
= 1, 2, ... .., 99 )
(b) Grouped frequency distribution distribution with class interval: interval : The formula to calculate percentiles is
iN Pi
= L+
− F 1
100 f m
× C .Where
L = Lower limit of the percentile class N = Total frequency F 1 = Cumulative frequency for the class just above the percentile class f m
= Frequency of the percentile class.
C = Class interval of the percentile class. i = 1, 2, …,99. The decile class is that class for which cumulative frequency ≥
iN
. 100 Example 8: 8: The following are the marks obtained by 50 students in Statistics: Marks 10 marks and less 20 marks and less 30 marks and less 40 marks and less 50 marks and less 60 marks and less 70 marks and less
Number of students 2 7 17 29 38 45 50
If 54% of the students pass the test, find the minimum pass marks. Solution: 54% students pass the test, i.e. 46% students fail in the test. In other words 46% of the student get less than pass marks, which is the 46 th percentile. So, we have to calculate 46th percentile P 46 .
46 N
P 46 =
L +
− F 1
100 f m
× C
Calculation of Percentile Marks 00 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
Frequency 2 5 10 12 09 07 05
To find the percentile class, we have to calculate
46 N
, which is 23. So blocked row is the percentile
100
class, because 23 lies lies in this class. According to the formula formula given, 46 N − F 1 23 − 17 P 46 = + 100 × 10 = 35. = 30 + × C L 12 f m Hence, 35 is the pass marks.
6
Cumulative frequency 2 7 17 29 38 45 50
