The document discusses different measures of central tendency including arithmetic mean. It provides formulas and examples for calculating arithmetic mean using direct, discrete and continuous series. It also discusses methods like step deviation and shot-cut for calculating mean and provides examples. It further discusses concepts like finding correct mean when initial mean is incorrect, calculating combined mean of two series and finding missing frequency from the data where mean is given.

1. Measure of Central
tendency
(Airthmetic mean)
By-
himanshu Malhotra,
Assistant professor
m.Sc (Statistics)
Department of Management
Dev bhoomi group of institutions
Dehradun,uttarakhand
E-mail- himmalhotra27@gmail.com

2. Average are generally the central part of the distribution and therefore they are also
called as Measure of Central tendency. There are five types of measure of Central
Tendency or averages .
Airthmetic mean
Median
Mode
Geometric Mean
Harmonic Mean

3. Airthmetic mean
Sum of Observations divided by number of observations.
Airthmetic Mean is denoted with a symbol “ 𝑿”
Airthmetic Mean can be solved by using three methods
Step-Deviation Method
Shot-cut Method
Direct Method

4. Direct Method

5. Before Starting Direct Method let us Understand Some basics first of all
There are three kind of Series in Statistics
Individual Series
Discrete Series
Continous Series
Continous is further divided into
Exclusive Series and Inclusive Series

6. Individual Series Discrete Series Continous Series
• Only Observations will be
given in the question like
Find AM
1,2,3,4
• “n” represent no of
observations
There will be two Column
one is for variable and
another one is for
Frequency like
Find AM
“N” represent Sum of
frequency
In this series marks will be
given in class intervals like
Find AM
“N” represent Sum of
frequency
Marks No of
Students
5 3
10 2
15 1
20 2
Marks No of
Students
0-10 3
10-20 2
20-30 1
30-40 2

7. Individual Series
Find the AM form the following data
1,2,3,4
Simply add the values and divide by number of obervation.
So, its
1+2+3+4
4
=
10
4
=
𝑋 =
𝑥
𝑛
Example
Solution

8. Discrete Series
In discrete Series we have variable with its frequency .Hence the formula and
Calculation process is liitle different
Find the AM form the following data
Marks 52 58 60 65 68 70 75
No of Students 7 5 4 6 3 3 2
𝑋 =
𝑓𝑥
𝑁
Example

9. Marks(X) No of Students(F) F*X
52 7 360
58 5 290
60 4 240
65 6 390
68 3 204
70 3 210
75 2 150
30 1848
In discrete Series “N” is the not the sum of observations its Sum of Frequency
column.Hence N=30
So first of all we have to multiply Marks with No of Students(Frequency)Solution
𝑋 =
𝑓𝑥
𝑁
𝑋 =
1848
30
= 61.6

10. Continous Series
In Continous Series the variable presented is in the form of Class.So again the
formula and Calculation process is liitle different
Find the AM form the following data
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No of Students 2 6 9 7 4 2
So first of all we will find the mid points for the marks column and than that mid
point column is our X value.After that the procedure remain same as in the case of
Discrete Series
𝑋 =
𝑓𝑥
𝑁
Example
Solution

11. Marks(X) No of
Students(F)
0-10 2
10-20 6
20-30 9
30-40 7
40-50 4
50-60 2
30
• The working rule for calculating mid points is
• Simply you have to add the upper limit and lower limit of every class intervals
Like
𝑈𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡+𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡
2
=
0+10
2
= 5 and so on……..
Mid-point(x)
5
15
20
25
30
35
F*x
10
90
225
245
180
110
860
𝑋 =
𝑓𝑥
𝑁
𝑋 =
860
30
= 28.67

12. Age No of Patients
10-19 7
20-29 1
30-39 4
40-49 3
As we discussed Continous Series is divided into Inclusive and Exclusive
Conversion of Inclusive into Exclusive Series
The data here is in the form of Inclusive
Series.We have to change the inclusive
series into Exclusive Series like
We find the distance between the lower limit of
the Second class Interval and the upper limit of
the first class –Interval.This is equal to 20 minus
19 =1.We Subtract
1
2
of this i.e, 0.5 form the lower limit and add it to
the upper limt.So the new classes will be
Age No of Patients
9.5-19.5 7
19.5-29.5 1
29.5-39.5 4
39.5-49.5 3
Inclusive Series
Exclusive Series

13. Shot-cut Method

14. Individual Series Discrete Series Continous Series
“n”=No of observations
“d” =deviation of the vaiate
x from “a”
“a” = assumed mean
“N”= sum of frequencies
“d” =deviation of the
vaiate x from “a”
“a” = assumed mean
“N”= sum of frequencies
“d” =deviation of the
vaiate x from “a”
“a” = assumed mean
𝑋 = 𝑎 +
𝑑
𝑛
𝑋 = 𝑎 +
𝑓𝑑
𝑁
𝑋 = 𝑎 +
𝑓𝑑
𝑁
Formula for Calculating AM by using Shot- cut
method in individual,discrete and continous
series

15. Individual Series
Find the AM form the following data
1,2,3,4,5
Let a=3 ,i.e, assumed mean(this you have to let from the
above observations)
Example
Solution
𝑋 = 𝑎 +
𝑑
𝑛
(X) d=x-a
1 1-3=-2
2 2-3=-1
3 3-3=0
4 4-3=1
5 5-3=2
𝑑=0
𝑋 = 𝑎 +
𝑑
𝑛
𝑋 = 3 +
0
5
𝑋 =3

16. Discrete Series
In discrete Series we have variable with its frequency .Hence the formula and
Calculation process is liitle different
Find the AM form the following data
X 0 1 2 3 4 5 6 7 8 9 10
F 2 8 43 133 207 260 213 120 54 9 1
Example
𝑋 = 𝑎 +
𝑓𝑑
𝑁

17. (X) (F) d=x-a F*d
0 2 0-5=-5 -10
1 8 1-5=-4 -32
2 43 2-5=-3 -129
3 133 3-5=-4 -266
4 207 4-5=-1 -207
5 260 5-5=0 0
6 213 6-5=1 213
7 120 7-5=2 240
8 54 8-5=3 162
9 9 9-5=1 36
10 1 10-5= 5 5
𝑓=1050 𝑓𝑑=12
Let a=5 ,i.e, assumed mean(this you have to let from the above observations)Solution
= 5.011𝑋 = 𝑎 +
𝑓𝑑
𝑁
𝑋 = 5 +
12
1050

18. Continous Series
In Continous Series the variable presented is in the form of Class.So again the
formula and Calculation process is liitle different
Find the AM form the following data
Class 100-120 120-140 140-160 160-180 180-200 200-220 220-240
Frequency 10 8 4 4 3 1 2
So first of all we will find the mid points for the marks column and than that mid
point column is our X value.After that the procedure remain same as in the case of
Discrete Series
𝑋 =
𝑓𝑥
𝑁
Example
Solution

19. (X) (F)
100-120 10
120-140 8
140-160 4
160-180 4
180-200 3
200-220 1
220-240 2
𝑓=32
Mid-point(x)
110
130
150
170
190
210
230
D=x-a F*d
-60 -600
-40 -320
-20 -80
0 0
20 60
40 40
60 120
𝑓𝑑=760
= 145.62
Assumed
mean
𝑋 = 𝑎 +
𝑓𝑑
𝑁
𝑋 = 170 +
−(760)
32

20. Step-Deviation Method

21. Formula for Calculating AM by using Step-
deviation method in individual,discrete and
continous series
𝑋 = 𝑎 +
𝑓𝑑
𝑁
*i
The procedure of calculating AM is almost same as in the case of Shot-cut
method.The only difference is you have to divide the deviation column
i.e,(d=x-a) column by “i” which stands for width between the class
interval.
𝑑 =
𝑥 − 𝑎
𝑖
where

22. HOW TO FIND THE CORRECT
MEAN FROM INCORRECT
MEAN

23. Formula for calculating Correct Mean
𝐶𝑜𝑟𝑟𝑒𝑐𝑡 𝑀𝑒𝑎𝑛 =
incorrect ∑x – incorrect value+ correct value
𝑛

24. Average height is given in the question i.e, x̄ = 166 cmSolution
The average height of 10 students in a class was calculated as
166 cm. On verification it was found that one reading was
wrongly recorded as 160 cm instead of 150 cm. Find the
correct mean height
Example
But x̄ = 166 cm is incorrect mean as it was found that one
reading was wrongly recorded as 160 cm instead of 150 cm
𝑋 =
𝑥
𝑛
166=
𝑥
10
𝑥 =1660
Hence Incorrect 𝑥 =1660

25. Now by applying the formula
𝐶𝑜𝑟𝑟𝑒𝑐𝑡 𝑀𝑒𝑎𝑛 =
incorrect ∑x – incorrect value+ correct value
𝑛
𝐶𝑜𝑟𝑟𝑒𝑐𝑡 𝑀𝑒𝑎𝑛 =
1660– 160+150
10
Hence the correct mean height is 165 cm

26. Combined mean

27. This formula is used when the mean of the
two series are given in the question
where
= Combined mean of the two series
= Size of the first Series
= Mean of the first series
= Size of the second series
= Mean of the Second series

28. Total number of studets in the class =50Solution
There are 50 students in a class of which 40 are boys and rest
girls.The Average weight of the class is 44 kg and the average
weight of the girls is 40 kg.Find the average weight of the
boys
Example
= 44 (No of Boys in class)= 40
(No of girls in class)= 10
(Average weight of the boys)= ?
(Average weight of the girls)= 40

29. Now by applying the formula
44 =
40 ∗ 𝑋1 + 40 ∗ 40
40 + 10
= 45

30. Missing frequency

31. Find the value of k from the following data whose mean is 16.6
X 8 12 15 K 20 25 30
F 12 16 20 24 16 8 4
𝑋 =
𝑓𝑥
𝑁
Example
Solution
X F F*x
8 12 96
12 16 192
15 20 300
K 24 24k
20 16 320
25 8 200
30 4 120
𝑓𝑥=24k+1228
16.6 =
24𝑘 + 1228
𝑁
24𝑘 = 1660 − 1228 = 432
𝑘 = 18