this will help you understand the complete concept of what, why an how we measure averages. kind regards

1. Measure of
Central
Tendency
PROF. HAJRAH MUGHAL

2. Measure of Central Tendency or
Averages
Introduction:
A measure of central tendency is a single value which represents the
observations to clusters in the centre of the distribution.
 It indicates the location or general position of the distribution, it is also known
as a measure of location or position.
The measure of central tendency or measure of location are generally known as
averages.

3. Points to be Noted:
Two points should be noted while calculating averages (measure of
central tendency):
First, A measure of central tendency should be somewhere within
the range of the given data.
Secondly, It should remain unchanged by a rearrangement of the
observations in the different order.

4. Criteria Of A Satisfactory Average:
A satisfactory average should be
Rigidly defined.
Based on all observations.
Simple to understand and easy to calculate.
Easy to interpret.
Capable of further mathematical treatments.
Relatively stable in repeated sampling experiments.
It should not be effected by extreme values.

5. Types of Averages:
Arithmetic Mean
Geometric Mean
Harmonic Mean
Median
Mode

6. The Arithmetic Mean:
The sum of all the values in the data divided by their total
number is called arithmetic mean
Mean=
Sum of all the observations
. Number of the observations

7. Arithmetic Mean:
For Ungrouped Data:
For population:
𝝁 =
𝑿
𝑵
=
𝑿 𝟏+𝑿 𝟐+⋯+𝑿 𝒏
𝑵
For Sample:
𝑿 =
𝒙
𝒏
=
𝒙 𝟏+𝒙 𝟐+⋯+𝒙 𝒏
𝒏
For Grouped Data:
For population:
𝝁 =
𝒇𝑿
𝑓
=
𝑓1 𝑿 𝟏+𝑓2 𝑿 𝟐+⋯+𝑓𝑛 𝑿 𝒏
𝑓1+ 𝑓2+⋯+𝑓𝑛
For Sample:
𝑿 =
𝒇𝒙
𝑓
=
𝑓1 𝒙 𝟏+𝑓2 𝒙 𝟐+⋯+𝑓𝑛 𝒙 𝒏
𝑓1+ 𝑓2+⋯+𝑓𝑛

8. Numerical of A.M for Ungrouped data:
The Quick oil company has a number of outlets in the Pakistan. The numbers of changes at the islamabad
outlet in the past 20 days are:
65, 98, 55, 62, 79, 79, 59, 51, 90, 72, 56, 70, 62, 66, 80, 94, 63, 73, 71, 85
Calculate the AM from the following data.
Solution:
𝑿 =
𝒙
𝒏
=
𝟔𝟓 + 𝟗𝟖 + 𝟓𝟓 + 𝟔𝟐 + 𝟕𝟗 + 𝟕𝟗 + 𝟓𝟗 + 𝟓𝟏 + 𝟗𝟎 + 𝟕𝟐 + 𝟓𝟔 + 𝟕𝟎 + 𝟔𝟐 + 𝟔𝟔 + 𝟖𝟎 + 𝟗𝟒 + 𝟔𝟑 + 𝟕𝟑 + 𝟕𝟏 + 𝟖𝟓
𝟐𝟎
𝑿 =
𝟏𝟒𝟑𝟎
𝟐𝟎
𝑿 =71.5

9. Numerical of A.M for grouped data:
A sample of fifteen University of Lahore undergraduate students gives the frequency distribution.
Obtain the mean credit hours for this sample of students.
Solution:
𝑿 =
𝒇𝒙
𝑓
𝑿 =
𝟏𝟖𝟔
𝟏𝟓
𝑿 = 12.5
No. of credit hours 3 9 12 14 15 17
No. of Students 1 3 4 1 4 2
x f fx
3
9
12
14
15
17
1
3
4
1
4
2
3
27
48
14
60
34
total 𝒇=15 𝒇𝒙=186

10. Geometric Mean:
The geometric mean is defined as the nth root of the product of n
positive values.
The geometric mean must be used when working with percentages
which are derived from values, while the arithmetic mean works with the
values themselves.

11. Geometric Mean:
For Ungrouped Data:
G.M= 𝑛
𝑥 𝟏 × 𝒙 𝟐 × … 𝒙 𝒏
G.M=antilog
𝒍𝒐𝒈𝒙
𝒏
For Grouped Data:
G.M=antilog
𝒇𝒍𝒐𝒈𝒙
𝒇

12. Calculating G.M for ungrouped
data:
Find the geometric mean of the following
values:
15, 12, 13, 19, 10
Solution:
G.M=antilog
𝒍𝒐𝒈𝒙
𝒏
G.M=antilog
𝟓.𝟔𝟒𝟖
𝟓
G.M=antilog 𝟏. 𝟏𝟐𝟗𝟔
G.M=13.48
x Log x
15
12
13
19
10
1.1761
1.0792
1.1139
1.2788
1.0000
Total 𝒍𝒐𝒈𝒙=5.648

13. Harmonic Mean:
Harmonic mean is defined as the recipirocal of the mean of the
reciprocals of the values.
he harmonic mean is best used for fractions such as rates(speed) or
multiples.

14. Harmonic Mean:
For Ungrouped Data:
H.M=
𝒏
𝟏
𝒙
For Grouped Data:
H.M=
𝒇
𝒇
𝒙

15. Example for Continuous Frequency Distribution:
Calculate the harmonic mean for the given below:
Solution:
𝐌𝐢𝐝 𝐩𝐨𝐢𝐧𝐭 = 𝒙 =
𝒍𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕=𝒖𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕
𝟐
=
𝟑𝟎+𝟑𝟗
𝟐
= 𝟑𝟒. 𝟓
H.M=
𝒇
𝒇
𝒙
H.M=
𝟏𝟎𝟎
𝟏.𝟒𝟑𝟕
H.M=69.60
marks 30-39 40-49 50-59 60-69 70-79 80-89 90-99
No. of students 2 3 11 20 32 25 7
Marks f Mid points= x 𝒇
𝒙
30-39
40-49
50-59
60-69
70-79
80-89
90-99
2
3
11
20
32
25
7
34.5
44.5
54.5
64.5
74.5
84.5
94.5
0.0580
0.0674
0.2018
0.3101
0.4295
0.2959
0.0741
𝛴𝑓 = 100 𝑓
𝑥
=1.437

16. Median:
The median is a measure of central tendency, which
denotes the value of the middle-most observation in the
data
Median is defined as the central value of the arranged
data. It is positional average denoted by 𝑋.

17. Median:
For Ungrouped or Discrete
Frequency Distribution:
Median= 𝑿 =
n+1
𝟐
th value of
array data
For Grouped Data:
For Continuous Frequency Distribution:
Median= 𝑿 = 𝒍 +
𝒉
𝒇
𝒏
𝟐
− 𝑪
l = lower class boundary of median class
n = no. of observations(total frequency)
C = cumulative frequency of the class preceding the
median class
f = frequency of median class
h = class size = U.C.B – L.C.B

18. How to calculate Median step by step?
For ungrouped data:
Step 1. Arrange the given values in the ascending order.
Step 2. Find the number of observations in the given set of data. It is denoted by n.
Step 3. If n is odd, the median equals the [(n+1)/2]th observation.
Step 4. If n is even, then the median is given by the mean of (n/2)th observation and
[(n/2)+1]th observation.

19. Example for Ungrouped Data:
The heights (in cm) of 11 players of a team are as follows:
160, 158, 158, 159, 160, 160, 162, 165, 166, 167, 170.
Solution:
Arranging the variates in the ascending order, we get
157, 158, 158, 159, 160, 160, 162, 165, 166, 167, 170.
The number of variates = 11, which is odd.
Therefore,
Median= 𝑿 =
n+1
𝟐 th value=
11+1
𝟐
th value=
12
𝟐 th value=(6)th value
Median= 𝑿 =160

20. Example for Discrete Frequency Distribution:
The following table gives the number of the petal on the flowers on varies branches
of the trees. Compute the median
Solution:
Median= 𝑿 =
n+1
𝟐
th value
=
25+1
𝟐
th value
=
26
𝟐
th value
= 𝟏𝟑 th value
Median= 𝑿 = 4
No. of petals 1 2 3 4 5 6 7 8 9
No. of branches 3 4 4 3 2 4 3 1 1
No. of petals(x) f C.F
1
2
3
4
5
6
7
8
9
3
4
4
3
2
4
3
1
1
3
7
11
14
16
20
23
24
25
𝛴𝑓 = 25

21. How to calculate Median step by step?
For grouped data:
Step 1. Make a table with 3 columns. First column for the class interval, second column for frequency, f, and the third
column for cumulative frequency, cf.
Step 2. Write the class intervals and the corresponding frequency in the respective columns.
Step 3. Write the cumulative frequency in the column cf. It is done by adding the frequency in each step.
Step 4. Find the sum of frequencies, ∑f. It will be the same as the last number in the cumulative frequency column.
Step 5. Find (n/2)th value. Then find the class whose cumulative frequency is greater than and nearest to n/2. This is
the median class.
Step 6. Now we use the formula Median
𝑿 = 𝒍 +
𝒉
𝒇
𝒏
𝟐
− 𝑪

22. Example for Continuous Frequency Distribution:
The following table gives marks obtained by 50 students in statistics. Find median
Solution:
Median= 𝑿 = 𝒍 +
𝒉
𝒇
𝒏
𝟐
− 𝑪
For median class =
𝒏
𝟐
th value
=
𝟓𝟎
𝟐
th value 𝑿 𝑪𝒍𝒂𝒔𝒔
= 𝟐𝟓 th value
𝑿 = 24.5+
𝟓
𝟓
𝟐𝟓 − 𝟐𝟎
𝑿 = 29.5
marks 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49
No. of students 4 6 10 5 7 3 9 6
Marks f Class boundaries c.f
10-14
15-19
20-24
25-29
30-34
35-39
40-44
45-49
4
6
10
5
7
3
9
6
9.5 – 14.5
14.5 – 19.5
19.5 – 24.5
24.5 – 29.5
29.5 – 34.5
34.5 – 39.5
39.5 – 44.5
44.5 – 49.5
4
10
20
25
32
35
44
50
𝛴𝑓 = 50

23. Quartiles:
Quartiles are the values which divides the arranged data into four equal parts
For Ungrouped Data:
For Ungrouped or Discrete Frequency
Distribution:
𝑸 𝒓 = 𝒓
n+1
𝟒
th value of array data
For Grouped Data:
For Continuous Frequency Distribution:
𝑸 𝒓 =𝒍 +
𝒉
𝒇
𝒓𝒏
𝟒
− 𝑪
l = lower class boundary of quartile class
r = number of quartile
n = no. of observations(total frequency)
C = cumulative frequency of the class preceding the quartile class
f = frequency of quartile class
h = class size = U.C.B – L.C.B

24. Example for Discrete Frequency Distribution:
The following table gives the number of the petal on the flowers on varies
branches of the trees. Compute the 𝑄1, 𝑄 𝟐 and 𝑄 𝟑
Solution:
Q1 Class
Q3 Class
No. of petals 1 2 3 4 5 6 7 8 9
No. of branches 3 4 4 3 2 4 3 1 1
No. of petals(x) f C.F
1
2
3
4
5
6
7
8
9
3
4
4
3
2
4
3
1
1
3
7
11
14
16
20
23
24
25
𝛴𝑓 = 25

25. Conti…
First quartile= 𝑄1 = 𝟏
n+1
𝟒
th value
= 1
25+1
𝟒
th value
= 𝟏
26
𝟒
th value
=𝟏 𝟔. 𝟓 th value
𝑄1 = 2
𝑄 𝟐 = 𝑴𝒆𝒅𝒊𝒂𝒏
Third quartile= 𝑄 𝟑 = 𝟑
n+1
𝟒
th value
= 𝟑
25+1
𝟒
th value
= 𝟑
26
𝟒
th value
=3 𝟔. 𝟓 th value
= 19.5th value
𝑄 𝟑 = 6

26. Example for Continuous Frequency Distribution:
The following table gives marks obtained by 50 students in statistics. Find third quartile
Solution:
𝑄3 = 𝒍 +
𝒉
𝒇
𝟑𝒏
𝟒
− 𝑪
For 𝑄3 class =
𝟑𝒏
𝟒
th value
=
𝟑(𝟓𝟎)
𝟒
th value
=
𝟏𝟓𝟎
𝟒
th value
= 𝟑𝟕. 𝟓 th value
𝑄3 = 39.5+
𝟓
𝟗
𝟑𝟕. 𝟓 − 𝟑𝟓
𝑄3 = 40.9
marks 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49
No. of students 4 6 10 5 7 3 9 6
Marks f Class boundaries c.f
10-14
15-19
20-24
25-29
30-34
35-39
40-44
45-49
4
6
10
5
7
3
9
6
9.5 – 14.5
14.5 – 19.5
19.5 – 24.5
24.5 – 29.5
29.5 – 34.5
34.5 – 39.5
39.5 – 44.5
44.5 – 49.5
4
10
20
25
32
35
44
50
𝛴𝑓 = 50

27. Deciles:
Deciles are the values which divides the arranged data into ten equal parts
For Ungrouped Data:
For Ungrouped or Discrete Frequency
Distribution:
𝑫 𝒓 = 𝒓
n+1
𝟏𝟎
th value of array data
For Grouped Data:
For Continuous Frequency Distribution:
𝑫 𝒓 =𝒍 +
𝒉
𝒇
𝒓𝒏
𝟏𝟎
− 𝑪
l = lower class boundary of decile class
r = number of deciles
n = no. of observations(total frequency)
C = cumulative frequency of the class preceding the decile class
f = frequency of decile class
h = class size = U.C.B – L.C.B

28. Example for Discrete Frequency Distribution:
The following table gives the number of the petal on the flowers on varies
branches of the trees. Compute the 𝑫 𝟓 and 𝑄 𝟕
Solution:
D5 Class
D7 Class
No. of petals 1 2 3 4 5 6 7 8 9
No. of branches 3 4 4 3 2 4 3 1 1
No. of petals(x) f C.F
1
2
3
4
5
6
7
8
9
3
4
4
3
2
4
3
1
1
3
7
11
14
16
20
23
24
25
𝛴𝑓 = 25

29. Conti…
Seventh decile= 𝑫 𝟕 = 𝟕
n+1
𝟏𝟎
th value
= 𝟕
25+1
𝟏𝟎
th value
= 𝟕
26
𝟏𝟎
th value
= 7 𝟐. 𝟔 th value
= 18.2th value
𝑫 𝟕 = 6
Fifth decile = 𝑫 𝟓 = Median = 4

30. Example for Continuous Frequency Distribution:
The following table gives marks obtained by 50 students in statistics. Find ninth decile.
Solution:
𝑫 𝟗 = 𝒍 +
𝒉
𝒇
𝟗𝒏
𝟏𝟎
− 𝑪
For 𝑫 𝟗 class =
𝟗𝒏
𝟏𝟎
th value
=
𝟗(𝟓𝟎)
𝟏𝟎
th value
=
𝟒𝟓𝟎
𝟏𝟎
th value
= 𝟒𝟓 th value
𝑫 𝟗= 44.5+
𝟓
𝟔
𝟒𝟓 − 𝟒𝟒
𝑫 𝟗 = 43.3 𝑫 𝟗 Class
marks 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49
No. of students 4 6 10 5 7 3 9 6
Marks f Class boundaries c.f
10-14
15-19
20-24
25-29
30-34
35-39
40-44
45-49
4
6
10
5
7
3
9
6
9.5 – 14.5
14.5 – 19.5
19.5 – 24.5
24.5 – 29.5
29.5 – 34.5
34.5 – 39.5
39.5 – 44.5
44.5 – 49.5
4
10
20
25
32
35
44
50
𝛴𝑓 = 50

31. Percentile:
percentiles are the values which divides the arranged data into hundred equal parts
For Ungrouped Data:
For Ungrouped or Discrete Frequency
Distribution:
𝑷 𝒓 = 𝒓
n+1
𝟏𝟎𝟎
th value of array data
For Grouped Data:
For Continuous Frequency Distribution:
𝑷 𝒓 =𝒍 +
𝒉
𝒇
𝒓𝒏
𝟏𝟎𝟎
− 𝑪
l = lower class boundary of percentile class
r = number of deciles
n = no. of observations(total frequency)
C = cumulative frequency of the class preceding the percentile class
f = frequency of percentile class
h = class size = U.C.B – L.C.B

32. Example for Discrete Frequency Distribution:
The following table gives the number of the petal on the flowers on varies branches of the trees.
Compute the 85th percentile
Solution:
85th Percentile = 𝑷 𝟖𝟓 = 𝟖𝟓
n+1
𝟏𝟎𝟎
th value
= 85
25+1
𝟏𝟎𝟎
th value
= 𝟖𝟓
26
𝟏𝟎𝟎
th value
= 8𝟓 𝟎. 𝟐𝟔 th value
= 22.1th value
𝑷 𝟖𝟓= 7
No. of petals 1 2 3 4 5 6 7 8 9
No. of branches 3 4 4 3 2 4 3 1 1
No. of petals(x) f C.F
1
2
3
4
5
6
7
8
9
3
4
4
3
2
4
3
1
1
3
7
11
14
16
20
23
24
25
𝛴𝑓 = 25

33. Example for Continuous Frequency Distribution:
The following table gives marks obtained by 50 students in statistics. Find ninth percentile.
Solution:
𝑷 𝟗 = 𝒍 +
𝒉
𝒇
𝟗𝒏
𝟏𝟎𝟎
− 𝑪
For 𝑷 𝟗 class =
𝟗𝒏
𝟏𝟎𝟎
th value
=
𝟗(𝟓𝟎)
𝟏𝟎𝟎
th value 𝑷 𝟗 Class
=
𝟒𝟓𝟎
𝟏𝟎𝟎
th value
= 𝟒. 𝟓 th value
𝑷 𝟗= 14.5+
𝟓
𝟔
𝟒. 𝟓 − 𝟒
𝑷 𝟗 = 14.9
marks 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49
No. of students 4 6 10 5 7 3 9 6
Marks f Class boundaries c.f
10-14
15-19
20-24
25-29
30-34
35-39
40-44
45-49
4
6
10
5
7
3
9
6
9.5 – 14.5
14.5 – 19.5
19.5 – 24.5
24.5 – 29.5
29.5 – 34.5
34.5 – 39.5
39.5 – 44.5
44.5 – 49.5
4
10
20
25
32
35
44
50
𝛴𝑓 = 50

34. Mode:
The most frequent or repeated value in the data is called mode.
Unimodal Distribution:
The distribution having only one mode is called unimodal distribution.
Bimodal Distribution:
The distribution having exactly two modes is called bimodal distribution.
Multimodal Distribution:
The distribution having more the two modes is called multimodal distribution.

35. Mode:
For Ungrouped:
Most frequent value of the data.
Discrete Frequency Distribution:
Mode= 𝑋 = 𝑣𝑎𝑙𝑢𝑒 ℎ𝑎𝑣𝑖𝑛𝑔 ℎ𝑖𝑔ℎ𝑒𝑠𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
For Grouped Data:
For Continuous Frequency Distribution:
Mode= 𝑋 = l +
(fm−f1)
(fm−f1)+(fm−f2)
× h
l = lower class boundary of median class
fm = maximum frequency( frequency of modal class)
f1 =frequency of the class preceding the modal class
f2 =frequency of the class following the modal class
h = class size = U.C.B – L.C.B

36. Calculating mode for Ungrouped data:
The heights (in cm) of 11 players of a team are as follows:
160, 158, 158, 159, 160, 160, 162, 165, 166, 167, 170.
Find the mode.
Solution:
Arranging the variates in the ascending order, we get
157, 158, 158, 159, 160, 160, 162, 165, 166, 167, 170.
In the given data, the observation 158 and 160 occurs maximum number of times (2)
So,
Mode= 𝑋 =158, 160
This is a bimodal distribution.

37. Calculating Mode for Discrete data:
A sample of fifteen University of Lahore undergraduate students gives the frequency distribution.
Obtain the mode for this sample of students.
Solution:
Mode= 𝑋 = value having maximum frequency
𝑋 =12
No. of credit hours 3 9 12 14 15 17
No. of Students 1 3 4 1 3 2
x f
3
9
12
14
15
17
1
3
4
1
3
2
total 𝒇=14

38. Example for Continuous Frequency Distribution:
The following table gives marks obtained by 50 students in statistics. Find median
Solution:
Mode= 𝑋 = l +
(fm−f1)
(fm−f1)+(fm−f2)
× h
𝑋 = 19.5 +
(10−6)
10−6 +(10−5)
× 5
𝑋 = 19.5 +(0.44)5
𝑋 = 19.5 +2.22
𝑋 = 21.72
marks 10-14 15-19 20-24 25-29 30-34 35-39 40-44 45-49
No. of students 4 6 10 5 7 3 9 6
Marks f Class boundaries
10-14
15-19
20-24
25-29
30-34
35-39
40-44
45-49
4
6 𝐟 𝟏
10 𝐟 𝐦
5 𝐟 𝟐
7
3
9
6
9.5 – 14.5
14.5 – 19.5
19.5 – 24.5
24.5 – 29.5
29.5 – 34.5
34.5 – 39.5
39.5 – 44.5
44.5 – 49.5
𝛴𝑓 = 50

39. Relationship Between Mean, Median
and Mode:
We will understand the empirical relationship between mean, median, and mode by means of a
frequency distribution graph. We can divide the relationship into four different cases:
Symmetrical Distribution
Positively Skewed Distribution
Negatively Skewed Distribution
Moderately Skewed Distribution

40. Symmetrical Distribution
In the case of a frequency
distribution which has a
symmetrical frequency curve. The
empirical relation states that
Mean=Median=Mode

41. Positively Skewed Distribution
In the case of positively skewed
distribution curve,
mean > median > mode.

42. Negatively Skewed Distribution
In the case of negatively
skewed frequency distribution
mean < median < mode

43. Moderately Skewed Distribution
For moderately skewed distribution median divides the distance between mean
and mode by ration 1:2
𝑚𝑒𝑎𝑛 −𝑚𝑒𝑑𝑖𝑎𝑛
𝑚𝑒𝑑𝑖𝑎𝑛 −𝑚𝑜𝑑𝑒
=
1
2
2(mean - median)=1(median - mode)
2mean - 2median = median - mode
Mode = median + 2median - 2mean
Mode = 3median – 2mean