Measures of Central Tendency
A large aggregate of facts and figures are unwieldy
and it is difficult to grasp the whole data for the human mind as it becomes confusing and cannot be
analysed. Hece the statistical analysis
the measure of central tendency shall be helpful in representing the data with
a single figure whkilch is used to represent the whole series and such figures
are called measures of central tendency or
Averages. By this device it helps
the human mind in grasping the significance of large aggregad of facts and
figures. An average represents the whole
series (or population) and its values lies between the maximum and minimum
values and generally lies in the centre of the distribution.
“A measure of central tendency is a typical value
around which other figures congregate”.
- Simpson and Kafka
“One of the most widely used set of summary figures is known as measure of location which are often referred to as averages, central tendency or central location. The purpose of computing an average value for a set of observations is to obtain a single value which is representative of all the items and which the mind can grasp simply and quickly . The single value is the point of location around which the individual items clusters” -- Lawrence J. Kaplan
Objectives: -
It gives a bird’s eye view of the mass data. It helps in grasping the data dealing on aggregates. For example, it is impossible to keep mind
the individual ages or heights or weights of each students in a college. If the average age or weights or heights of
students is obtained on dividing the total ages or weights or heights by the
number of students we arrive at a single figure or value which represents the characteristic
of the series (or population) and enables one to get a bird’s eye view of the
series under study and can be remembered easily.
If there are different sets of data by computing an average the sets of data can be compared and
cncusions can be drawn about the characteristics of of the phenomena under study. For example,
e can compare the percentage results of students in different colleges
or schools and thereby determine which college or school is the best.
Similarly if there are two sets of dta on the ages of allthe inhabitants
of two countries (say) India and U.K. the
average ages of these two countries are computed t can be compered at a glance as it is impossible to remember
the ages of all inhabitants.
Another example which can be given is averages
establish precise relationship, when the income of two different groups (or data)
as expressed quantitively in terms of averages.
Measures of Central Tendency
1. Arithmetic Mean or Average
F
X1, X2, X3,………….Xn are N observations (or terms) then the
Arithmetic mean is the quotient of the sum of all the items divided by N. The summation is denoted by the capital Greek letter Ʃ (read as sigma) and the mean is denoted
by X’ (read as X- bar). The formula for
the Arithmetic Mean is X’ = (ƩX1+X2+ +Xn)/N = ƩX/N
Example{
- If the items are 14, 16, 18, 20, 27
Then ƩX = 14+16+18+20+27 = 95, N = 5
X’
= ƩX/N = 95/5 =
19
This is the direct method of calculating
AM.
Short cut method: -
The formula for the calculating AM is
given by X’ = A + Ʃd/N, where A = Assumed mean
D = deviation of each of the values from the assumed mean
N = Number of observations
In the present example by assuming 18 as
the assumed mean
i.e. A = 18, we have (14-18), (16-18),
(18-18), (20-18), (27-18)
Therefore Ʃd = -4-2+0+2+9 = +5
Therefore X’ = A + Ʃd/N = 18 +5/5 = 18 + 1 = 19
When the frequency is given then the
Arithmetic mean is given by
X’ = Ʃfx/Ʃf (Direct
Method)
Example: - Find the arithmetic mean for
the following data.
Marks No. of Students
10 3
15 5
20 7
25 5
30 10
Total 30
|
X |
f |
fX |
|||||
|
10 |
3 |
30 |
|||||
|
15 |
5 |
75 |
|||||
|
20 |
7 |
140 |
|||||
|
25 |
5 |
125 |
|||||
|
30 |
10 |
300 |
|||||
|
30 |
670 |
||||||
|
Ʃf = |
30 |
ƩfX = |
670 |
||||
|
X' = |
ƩfX/Ʃf |
22.33 |
Marks |
OR |
22 |
Marks |
|
|
|
|
|
|
|
|
|
|
By applying short-cut method
|
X |
f |
d=X-20 |
fd |
||
|
10 |
3 |
-10 |
-30 |
||
|
15 |
5 |
-5 |
-25 |
||
|
20 |
7 |
0 |
0 |
||
|
25 |
5 |
5 |
25 |
||
|
30 |
10 |
10 |
100 |
||
|
30 |
70 |
||||
|
A = |
20 |
Ʃfd = |
70 |
Ʃf = |
30 |
|
X' = |
A +Ʃfd/Ʃf |
||||
|
X' = |
22.33 |
Marks |
or |
22 |
Marks |
Find the arithmetic mean for the following data.
|
Marks |
No. of students |
|
` 0 – 10 |
1 |
|
10 – 20 |
5 |
|
20 - 30 |
20 |
|
30 - 40 |
10 |
|
40 - 50 |
14 |
|
Total |
50 |
|
Mid |
No. of |
||||||
|
Marks |
Value |
students |
|||||
|
x |
f |
fx |
|||||
|
0 |
10 |
5 |
1 |
5 |
|||
|
10 |
20 |
15 |
5 |
75 |
|||
|
20 |
30 |
25 |
20 |
500 |
|||
|
30 |
40 |
35 |
10 |
350 |
|||
|
40 |
50 |
45 |
14 |
630 |
|||
|
50 |
1560 |
||||||
|
Ʃf = |
50 |
Ʃfx= |
1560 |
||||
|
AM = |
X' = |
Ʃfx/Ʃf = |
31.2 |
Marks or |
31 Marks |
||
By short-cut method
Step - deviation method
If we take 10 as the common class interval , then the
A.M> is computed by using the
formula X’ = A + (Ʃfd/Ʃf)*i
|
Marks |
Mid points x |
d = (x-25)/10 |
f |
fd |
|
0 – 10 |
5 |
-2 |
1 |
-2 |
|
10 - 20 |
15 |
-1 |
5 |
-5 |
|
20 - 30 |
25 |
0 |
20 |
0 |
|
30 – 40 |
35 |
1 |
10 |
10 |
|
40 - 50 |
45 |
2 |
14 |
28 |
|
|
|
|
50 |
31 |
A =
25 Ʃf = 50 Ʃfd = 31 i
= 10
X’
= A + (Ʃfd/Ʃf)*I = 25 + (31/50)*10
= 25 + 6.2
= 31.2
Therefore
X’ = 31.2 Marks or 31 Marks.
Median
The Median divides the data into two equal parts. In ungrouped data, the data should be arranged
in an array i.e. in ascending or descending order to find the Median.
Symbolically the Median is written as M = size of the
(n+1/2)th item.
Example: To find the Median of 2, 8, 3, 4, 5
By arranging in ascending order, the above numbers can
be written as
2, 3, 4, 5, 8
By applying the formula Median = (n+1/2)th items =
(5+1/2)th item = (6/2)th items = 3rd item. (Here n = 5, n+1 = 5+1 = 6). The third item is 4.
The middle most value is 4.
Hence the Median = 4
To find the Median
1. Make an
array of the raw data
2. Count the
items and find the middle item
3. Take the
middle item as the Median.
Hence the Median = 4.If the number of items is even,
then the average of the middle items is taken as the Median.
For grouped data:
For grouped data the formula for Median is
Md
= l1 + [ (N/2 – Ʃfc)/fmed]*i where l1= the lower limit of the
median class
N =
Total frequency
Ʃ fc = the cumulative
frequency total before the median class
fmed
= frequency in the median class interval
i = size of the class interval
Example: - Find the Median from the following data.
x
5 10 15 20 25
f 1 3 5 7 9
|
x |
f |
cf |
|
5 |
1 |
1 |
|
10 |
3 |
4 |
|
15 |
5 |
9 |
|
20 |
7 |
16 |
|
25 |
9 |
25 |
|
|
25 |
|
Median = size of the [(n + 1)/2]th item
=
size of [(25+1)/2]th item = 13th item.
Hence
the Median = 20.
Find out the Median from the following data:
|
Marks |
frequency |
|
0-5 |
5 |
|
5-10 |
10 |
|
10-15 |
20 |
|
15-20 |
30 |
|
20-25 |
45 |
Sol.
|
Marks |
frequency |
Cumulative
freqency |
|
0-5 |
5 |
5 |
|
5-10 |
10 |
15 |
|
10-15 |
20 |
35 |
|
15-20 |
30 |
65 |
|
20-25 |
45 |
110 |
n = 110
Median size of n/2th item = 110/2 = 55
Median lies in (15-20)th class interval
l1 = 15
fmed = frequency in the Median class = 30
Ʃfc
= the cumulative frequency totoal before the Median Class = 35
I = size of the class interval = 5
Median
= Md = l1 + [(N/2 – Ʃfc))/fmed]*i
= 15 + [(55 - 35)/30]*5
= 18.33
Hence
Median = 18.33 Marks or 18 Marks.
Mode
The most frequently occurring item is known as Mode.
Example:- To find the Modal value for the following
data
22, 12, 7, 8, 10, 14, 10, 16, 10, 23, 10, 19, 25
Mode
= 10
Find the Mode for the following data
10, 15, 20, 25, 30, 35, 40
Modal value is absent in this series.
Find the Mode for the following data
6, 9, 10, 12, 10, 21, 20, 25, 20, 35, 20, 7, 10
Here
the Mode I is 10
Mode II is 20
Mode for grouped data
The Mode for the grouped data is calculated by the
following formula.
Mode
= l1 + [(f1 – f0/(2f1 – f0
– f1)]* i
where l1 = lower limit of the class
in which the “mode” lies
f1
= frequency of the class in which Mode lies
f0 = frequency of the class
preceding the modal class
f2
= frequency of the class succeeding of the modal class
i =
the class interval
Find the Mode for the following data
|
Daily wage
(Rs.) |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
|
frequency |
65 |
100 |
90 |
50 |
38 |
12 |
|
Daily
wages (Rs.) |
Frequency |
|
20-30 |
65 |
|
30-40 |
100 |
|
40-50 |
90 |
|
50-60 |
50 |
|
60-70 |
38 |
|
70-80 |
12 |
|
Total |
305 |
The Mode
lies in the class interval (30 – 40)
Here l1 = 30, f1 = 100, f0
= 65, f2 = 90, i = 10
Mode = l1 + [(f1 – f0/(2f1
– f0 – f1)]* I = 30 + [(100-65)/{200-65-90)]*10
Hence
Mode = 37.78
Make a frequency table of the following scores and
find the Mean, Median and Mode.
80, 72,, 75, 52, 80, 90, 90, 50, 40, 80, 85, 82, 80
Here n = 13
By arranging the data in ascending order, we have
40, 50, 52, 72, 75, 80, 80, 80, 80, 82, 85, 90, 95
By Direct Method, x’ = Ʃx/n
Ʃx =
80+72+75+80+80+80+80+82+85+90+95 = 956
Arithmetic Mean = Ʃx/n = 956/13 = 73.54
By making the frequency table
|
Scores (x) |
40 |
50 |
52 |
72 |
75 |
80 |
82 |
85 |
90 |
|
Frequency(f) |
1 |
1 |
1 |
1 |
1 |
4 |
1 |
1 |
2 |
|
fx |
40 |
50 |
52 |
72 |
75 |
320 |
82 |
85 |
180 |
Then Ʃf = 13 and Ʃfx = 956
Arithmetic Mean = x’ = Ʃfx/Ʃf = 956/13 = 73.54
Median = (n+1/2)th size of item =[
(13+1)/2)]th size of the item
in the array
Therefore, Median is 7th size of the item =
80
By observation, Mode = 80 as it has occurred 4 times
in the frequency distribution.
Prepare a frequency distribution table of marks scored
for the following and find the three measures of central tendency.
55, 38, 44, 80, 82, 55, 55, 32, 20, 38, 45, 55, 61,
44, 44, 44, 44, 24, 28
By arranging the data in ascending order we have,
20, 24, 28, 32, 38, 38, 44, 44, 44, 44, 44, 45, 55,
55, 55, 55, 61, 80, 82
Here n = 19
Frequency Table
Marks: 20
24 28 32 38 44 45 55 61
80 82
Frequency
1 1 1
1 2 5 1 4 1 1 1
|
X |
20 |
24 |
28 |
32 |
38 |
44 |
45 |
55 |
61 |
80 |
82 |
Total |
|
f |
1 |
1 |
1 |
1 |
2 |
5 |
1 |
4 |
1 |
1 |
1 |
19 |
|
fX |
20 |
24 |
28 |
32 |
76 |
210 |
45 |
210 |
61 |
80 |
82 |
888 |
Mean
= 46.74
Median
= 44
By observation Mode = 44 as it occurs 5 times in the series.
Find the Mean and Median in the set of following
numbers.
4, 4, 5, 6. 6. 6, 7, 8, 9
Here the total number of items n = 9
Median = [(n+1)/2]th item = 10/2th item = 5th
item = 6
Also, Mode = 6 as it occurs maximum number of times.
Find the Median marks obtained by 12 students out of
50 which are given below.
16, 46, 10, 19, 25, 32, 28, 30, 22, 24, 27, 40
By arranging the marks scored by students in ascending
order of magnitude , we have the scores as
10, 16, 19, 22, 24, 25, 27, 28, 30, 32, 40, 46
The total number of items = 12
Therefore, Median = (n+1)/2th size = 13/2th
size = 6.5th size
= (6+7)/2th items
= (25+27)/2 = 26
Hence
Median = 26
Calculate the
Modal size of short collars from the following data.
|
Size of Collar(in) |
12.0 |
12.5 |
13.0 |
13.5 |
14.0 |
14.5 |
15.0 |
15.5 |
|
No. of persons |
8 |
26 |
36 |
40 |
42 |
12 |
4 |
2 |
|
|
|
|
|
|
|
|
|
|
|
size of |
No. of |
|||||
|
color |
persons |
|||||
|
I |
II |
III |
IV |
V |
VI |
|
|
12 |
8 |
|||||
|
34 |
||||||
|
12.5 |
26 |
70 |
||||
|
62 |
||||||
|
13 |
36 |
|||||
|
76 |
||||||
|
13.5 |
40 |
|||||
|
82 |
102 |
118 |
||||
|
14 |
42 |
94 |
||||
|
54 |
||||||
|
14.5 |
12 |
|||||
|
16 |
58 |
|||||
|
15 |
4 |
18 |
||||
|
6 |
||||||
|
15.5 |
2 |
Columns Size of the items
containing
Maximum frequencies
I 14.0
II 13.0 13.5
III 13.5 14.0
IV 13.5 14.0 14.5
V 12.5 13.0 13.5
VI 13.0 13.5 14.0
Total 1 3 5 4 1
The item
13.5 occurs maximum number of times. (5 times)
Hence Mode =
13.5 inches.
II Column Add
(1, 2), (3, 4), (5, 6), (7, 8) Add two numbers and enter
III Column Add
(2, 3), (4, 5), (6, 7) Leave the first
item and add two items
IV Column Add
(1, 2, 3), (4, 5, 6) Add three
items
V Column Add
(2, 3, 4), (5, 6, 7) Leave the
first item and add 3 items
VI Column Add
(3, 4, 5), (6, 7, 8) Leave the Ist
and 2nd items and add 3 items