For example, it is impossible to remember the each
match runs scored by Virat Kohali in One Day International cricket, but if the
average runs, scored, is obtained; we get one single value that represents the
entire One Day International cricket career score.
| Central Tendency | Definition | Formula for Ungrouped Data | Formula for Discrete Data | Formula for Continuous Data |
|---|---|---|---|---|
| Arithmetic Mean or Mean (x̄) | The arithmetic mean is defined as sum of the numerical values of each and every observation divided by the total number of observations |  n = Number of observations |
 |
 Xi = middle value of class interval N = Total frequency |
| Median (M) | Middle most or central value of observation is called as median. Median devides the given data into two equal parts. |
I] When n is odd number: M{(n+1)/2}th term II] When n is even number: M = [ (n/2) + (n+1)/2 ] / 2 |
M = (N/2)th term N = Total Frequency |
M = l + [(N/2 – C) / f] × h Where: l = Lower limit of class interval of median f = Frequency of class where median lies C = Cumulative frequency before median class N = Total frequency h = Width of class interval |
| Median (M) | Middle most or central value of observation is called as median. Median devides the given data into two equal parts. |
I] When n is odd number: M={(n+1)/2}th term II] When n is even number: M = [ (n/2) + (n+1)/2 ] / 2 |
M =(N+1)/2th term N = Total Frequency |
M = l + [(N/2 – C) / f] × h Where: l = Lower limit of median class CI f = Frequency of median class CI C = Preceding cumulative frequency of median class CI N = Total Frequency h = Width of CI |
| Mode (Mo) | Most occurring value or most repeating value of observation is called as mode. | Most repeating value of observation | The observation which corresponds to maximum frequency. |
Mo = l + [(fm – f1) / (2fm – f1 – f2)] × h Where: l = Lower limit of mode class CI fm = Frequency of mode class CI f1 = Preceding Frequency of mode class CI f2 = Succeeding Frequency of mode class CI h = Width of mode class CI |
Case II : When
data is Grouped data
Case I : When
data is Ungrouped data/Raw data
Example 1: Calculate mean from
following data
Age of eight patients in years: 14, 17, 18,
20, 21, 22, 25, 29.
Solution:
The Arithmetic mean Patient age is 20.75 years.
Case II: When data is Grouped data
Example 2: Calculate mean from following frequency distribution data
| Hemoglobin level (g/dL) | 10 | 11 | 12 | 13 | 14 | 15 |
|---|---|---|---|---|---|---|
| Number of Cases | 06 | 09 | 21 | 11 | 07 | 04 |
| Hemoglobin level in g/dL (Xi) | Number of Cases (fi) | Xifi |
|---|---|---|
| 10 | 06 | 60 |
| 11 | 09 | 99 |
| 12 | 21 | 252 |
| 13 | 11 | 143 |
| 14 | 07 | 98 |
| 15 | 04 | 60 |
| Total | N=58 | ∑〖Xifi 〗=712 |
The mean Arithematic Hemoglobin level in case is
12.27 g/dL.
Merits of Arithmetic Mean:
• It is easy to understand and calculate.
• It is rigidly defined.
• It is based on all observations.
• It is least affected by sampling fluctuation.
• It is useful for further algebraic treatment.
· Demerits of Arithmetic Mean:
· • It is very much affected by extreme values.
• It is not calculated in case of open-end interval.II] Median: [M]
Middle
most or central value of observation is known as median. Median devides the given data into two equal
parts. Fifty per cent of the observations in the data are above the value of
median and other fifty per cent of the observations are below this value of median.
The median is the value of the middle observation when the series is arranged
in order of size or magnitude (Ascending order). Data
should be arranged in increasing / decreasing order.
It
is denoted by letter ‘M.
Case I : When data is Ungrouped data/Raw data
I] If n
is an odd number:
M={[n+1]/2}th.term
M=((n/2) th+(n/2+1)th)/2 Term
Case II: When
data is grouped data.
I] when data is Discrete
N= Total Frequency
Note: Find
cumulative frequency and find nearest greatest value of Mth term in
cumulative frequency, so its corresponding observation is median value.
II] When data is Continuous
Where as l=is lower limit of median lies Class Interval.
f=Frequency of median lies Class Interval.
C= Preceding cumulative frequency of median lies Class
Interval.
N= Total Frequency
h = is
width of Class Interval.
Merits and Demerits of Median:
Merits of Median:
• It is easy to understand and calculate.
· • It is rigidly defined.
• It is not affected by extreme values.
· • It is calculated in case of open-end interval.
· • It can be located graphically.
Demerits of Median:
· • It is not based on all observations.
· • It is least affected by sampling fluctuation.
· • It is not capable of further algebraic treatment.
Case I : When data is Ungrouped data/Raw data
Example 1: Calculate median from following data
10, 17, 18, 12, 23, 22, 25, 21,11
Solution:
Let’s arrange given data in ascending order
10, 11, 12, 17, 18, 21, 22, 23, 25
Here n=9 which is odd number so formula is
M={[n+1]/2}th.tern
M={[9+1]/2}th.tern
M={[10]+1/2}th.tern
M= 5 th term (5 th Term value in ascending order data)
M = 18
So Median of given data is 18.
Example 2: Calculate median from following data
56, 34, 31, 87, 23, 45, 25, 49
Solution:
Let’s arrange given data in ascending order
23, 25, 31, 34, 45, 49, 56, 87.
Here n=8 which is Even number so formula is
Case II : When data is Grouped data.
Example 2: Calculate median of following frequency distribution data
|
Marks |
40 |
42 |
44 |
45 |
47 |
50 |
|
Number
of students |
08 |
18 |
12 |
09 |
07 |
06 |
Solution:
|
Marks (Xi) |
Number of students (fi) |
Cumulative Frequency (cf) |
|
40 |
08 |
08 |
|
42 |
18 |
26 |
|
44 |
12 |
38 > 30.5th |
|
45 |
09 |
47 |
|
47 |
07 |
54 |
|
50 |
06 |
60 |
|
Total |
N=60 |
|
|
Size of items: |
0-4 |
4-8 |
8-12 |
12-16 |
16-20 |
20-24 |
24-28 |
28-32 |
32-36 |
36-40 |
|
Frequency |
5 |
7 |
9 |
17 |
12 |
10 |
6 |
3 |
2 |
1 |
|
Size of items |
Number of students (fi) |
Cummulative Fequency (cf) |
|
0-4 |
5 |
5 |
|
4-8 |
7 |
12 |
|
8-12 |
9 |
21 (c) |
|
(l)
12-16 |
17 (f) |
38 >36 |
|
16-20 |
12 |
50 |
|
20-24 |
10 |
60 |
|
24-28 |
6 |
66 |
|
28-32 |
3 |
69 |
|
32-36 |
2 |
71 |
|
36-40 |
1 |
72 |
|
Total |
N=72 |
|
(So 36th term nearest greatest value in cumulative frequency is 38, so its corresponding Class interval is median lies class interval i.e. 12-16).
|
Marks |
50 |
60 |
70 |
80 |
90 |
100 |
|
Number
of students |
4 |
12 |
28 |
33 |
11 |
7 |
Solution:
In above given frequency distribution data the maximum frequency is 33.
Therefore, its corresponding observation is 80 marks
So Mode = 80 marks
|
Class
Interval: |
0-20 |
20-40 |
40-60 |
60-80 |
80-100 |
100-120 |
12-140 |
140-160 |
|
Frequency |
2 |
5 |
10 |
23 |
21 |
14 |
08 |
03 |
Class Interval | Frequency (fi) |
0-20 | 2 |
20-40 | 5 |
40-60 | 10 (f1) |
(l) 60-80 | 23(fm) |
80-100 | 21 (f2) |
100-120 | 14 |
120-140 | 08 |
140-160 | 03 |
Total | N=72 |
