Properties of arithmetic mean

Properties of arithmetic mean

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 2
x kx k
k
k
c
nnn
xnxnxn
x
+++
+++
=
......
...
21
2211 

=
n
xn
x
c

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 xy3= 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
1
st

2
nd


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
1
st

2
nd

3
rd
10
15
20
60
50
68
600
900
1360
Sum 45 2860

Kg



n x xn 56
25
1400
n
xn
x
c ===

 n x xn 56.63
45
2860
n
xn
x
c ===



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 =−=