The document defines properties of arithmetic mean and provides examples to illustrate these properties. The key properties are: 1) The sum of deviations from the mean is zero. 2) The sum of squares of deviations from the mean is minimum. 3) The mean is unaffected by a constant change of scale or origin. 4) The mean of multiple data sets can be calculated as a weighted average based on sample sizes. The examples demonstrate calculating means and verifying properties for various data sets.

2. Properties of Arithmetic Mean:
1) The sum of deviation from mean is zero i.e. .
2) The sum of the square of deviation from mean is minimum i.e.
where A is a assumed mean.
3) The mean of constant is constant.
4) The mean is effected by change of origin (Addition or Subtraction)
i.e. A.M where a is any constant.
5) The mean is effected by change of scale (Multiplication or
Division)
i.e. A.M or A.M
where a is any constant.
6) The mean is effected by change of origin and scale i.e.
A.M where a and b are any constant.
7) Combined Mean: If n1 values have mean , n2 values have mean
, ……, nk values have mean the mean of all the values is
given as:
( ) 0xx =−
( ) ( ) −−
22
Axxx
( ) axax =
( ) xaxa =
a
x
a
x
=





( ) bxabxa =
1x
2x kx
k
kk
c
nnn
xnxnxn
x
+++
+++
=
......
...
21
2211

=
n
xn
xc

3. Example-7:
Find the mean and show that the sum of deviations from mean is zero
for the following data.
3, 7, 9, 11, 10
Solution:
3
7
9
11
10
−5
−1
1
3
2
Sum 40 0
First, find the mean:
Verification:
= Zero
Proved
x 8x −
5
40
n
x
x == 
8x =
( ) ( )  −=− 8xxx

4. Example-8:
For the data given below, verify that the sum of the square of deviation
from mean is minimum i.e. is minimum.
10, 20, 30, 40, 50
Solution:
10
20
30
40
50
−20
−10
0
10
20
400
100
0
100
400
−30
−20
−10
0
10
900
400
100
0
100
Sum 150 0 1000 −50 1500
Verification:
Taking L.H.S:
Hence
Taking R.H.S:
If then A select any value within and outside set of data. We
choose
A is 40.
Hence,
So,
is the minimum or least.
( ) −
2
xx
x 30x − ( )2
30x − 40x − ( )2
40x −
( ) ( ) −=−
22
Axxx
30
5
150
n
x
x === 
( ) ( ) =−=− 100030xxx
22
( ) ( ) =−=− 150040xAx
22
xA 
15001000
( ) −
2
xx

5. Example-9:
Show that the mean of constant value is constant i.e. .
10, 10, 10, 10, 10
Solution:
x
10
10
10
10
10
Sum 50
We know that:
Hence, the mean of constant value is constant i.e.
Proved
( ) aaM.A =
10
5
50
n
x
x === 
( ) 1010aM.A ==

6. Example-10:
A student obtained the following grades on six examination:
84, 91, 72, 68, 87 and 48.
Find: (a) Average grade
(b) if (i) (ii) (iii)
(iv) (v)
Solution:
x
84
91
72
68
87
48
Sum 450
(a)
(b) (i)
By property of mean:
(ii)
By property of mean:
(iii)
By property of mean:
(iv)
By property of mean:
y 50xy −= 60xy += x2y =
5
x
y = 50x2y −=
75
6
450
n
x
x === 
50xy −=
25507550xy =−=−=
60xy +=
135607560xy =+=+=
x2y =
150752x2y ===
5
x
y =

7. (v)
By property of mean:
Example-11:
Calculate the arithmetic mean for the following:
20, 25, 15, 10, 30
Verify:
(i) (ii) (iii) (iv)
Solution:
X
20
25
15
10
30
60
65
55
50
70
60
75
45
30
90
4
5
3
2
6
−10
0
−20
−30
10
Sum 100 300 300 20 −50
First, find the mean of :
(i)
Taking L.H.S
15
5
75
5
x
y ===
50x2y −=
5075250x2y −=−=
10050150y =−=
40xy += x3y =
5
x
y = 50x2y −=
40+= xy xy 3=
5
x
y = 502 −= xy
x
20
5
100
n
x
x === 
40xy +=

8. where
Taking R.H.S
Hence or 60 = 60
Mean is dependent by change of origin or affected by addition.
(ii)
Taking L.H.S
where
Taking R.H.S
Hence or 60 = 60
Mean is dependent by change of scale or affected by
multiplication.
(iii)
Taking L.H.S
n
y
y = 40xy +=
60
5
300
y ==
60402040x =+=+
40xy +=
x3y =
n
y
y = x3y =
60
5
300
y ==
60203x3 ==
x3y =
5
x
y =

9. where
Taking R.H.S
Hence or 4 = 4
Mean is dependent by change of scale or affected by division.
(iv)
Taking L.H.S
where
Taking R.H.S
Hence or −10 = −10
Mean is dependent by change of origin and scale or affected by
multiplication and subtraction.
n
y
y =
5
x
y =
4
5
20
y ==
4
5
20
5
x
==
5
x
y =
50x2y −=
n
y
y = 50x2y −=
10
5
50
y −=−=
1050405020250x2 −=−=−=−
50x2y −=

10. Example-12:
The average wage of 10 men is Rs.50 per hour and the average wage of
further 15 men is Rs.60 per hour. Find the average wage of all men.
Solution:
Groups
1st
2nd
10
15
50
60
500
900
Sum 25 1400
Example-13:
The average wage of 10, 15 and 20 students is 60 kg, 50 kg and 68 kg
respectively. Find the average weight of the 45 students.
Solution:
Groups
1st
2nd
3rd
10
15
20
60
50
68
600
900
1360
Sum 45 2860
Kg
n x xn
56
25
1400
n
xn
xc ===


n x xn
56.63
45
2860
n
xn
xc ===



11. Example-14:
The mean of the values 3, 8, A, 10 and 4 is 6. Find the value of A.
Solution:
x
3
8
A
10
4
Sum 25+A
Given that:
We know that:
6x =
6
n
x
x == 
6
5
25
=
+ A
56A25 =+
30A25 =+
52530A =−=