ARITHMETIC MEAN
Case 1:
Ungrouped data
(Individual data)
Direct
method
Short-cut /
Assumed
mean method
𝑥
𝑛
𝑑
𝑛
𝑥=
𝑥 =A+
(where, n = number of observations
d=x–A
A = assumed mean
d' = d ÷ c
c = common factor)
Step
deviation
method
𝑥 =A+
𝑑′
𝑛
×c
CASE 1: Ungrouped Data (Individual data)
a) Direct method
During the COVID-19 induced lockdown, 10 families engaged in the family
game of ‘Housie’ online everyday. The virtual game also had cash prizes for the
winning family. The weekly earnings of cash prize (in ₹) of the 10 families are:
1400 ; 1500 ; 1560 ; 1440 ; 1475 ; 1490 ; 1575 ; 1520 ; 1470 ; 1480
Find the average weekly cash prize searnings.
x (Weekly earnings)
1400
1500
Number of observations (n) = 10
Sum ( 𝒙) = 14,910
1560
1440
1475
1490
1575
1520
̅)=
Mean ( 𝒙
=
𝒙
𝑛
14,910
10
̅ = 1491
.`. 𝒙
1470
1480
𝒙 = 14,910
.`. Average weekly earnings = ₹ 1491
b) Short-cut / Assumed mean method
During the COVID-19 induced lockdown, 10 families engaged in the family
game of ‘Housie’ online everyday. The virtual game also had cash prizes for the
winning family. The weekly earnings of cash prize (in ₹) of the 10 families are:
1400 ; 1500 ; 1560 ; 1440 ; 1475 ; 1490 ; 1575 ; 1520 ; 1470 ; 1480
Find the average weekly cash prize earnings.
x (Weekly earnings)
d=x–A
1400
- 100
1500
0
1560
60
1440
- 60
1475
- 25
1490
- 10
1575
75
1520
20
1470
- 30
1480
- 20
𝒅 = - 90
(Note: Here, d = x – A
A = assumed mean)
Here, A = 1500
Number of observations (n) = 10
Sum of deviations ( 𝒅) = - 90
̅)=A
Mean ( 𝒙
+
𝒅
𝑛
= 1500 +
−90
10
= 1500 – 9
̅ = 1491
.`. 𝒙
.`. Average weekly earnings = ₹ 1491
c) Step deviation method
During the COVID-19 induced lockdown, 10 families engaged in the family
game of ‘Housie’ online everyday. The virtual game also had cash prizes for the
winning family. The weekly earnings of cash prize (in ₹) of the 10 families are:
1400 ; 1500 ; 1560 ; 1440 ; 1475 ; 1490 ; 1575 ; 1520 ; 1470 ; 1480
Find the average weekly cash prize earnings.
x (Weekly
d=x–A
d' =
earnings)
𝒅
Here, A = 1500 ; c = 10
𝒄
Number of observations (n) = 10
1400
- 100
- 10
1500
0
0
1560
60
6
1440
- 60
-6
1475
- 25
- 2.5
1490
- 10
-1
Sum of deviations ( 𝒅 ′) = - 9
̅)=A
Mean ( 𝒙
+
= 1500 +
𝒅′
𝑛
−9
10
1575
75
7.5
1520
20
2
= 1500 – 9
1470
- 30
-3
̅ = 1491
.`. 𝒙
1480
- 20
-2
𝒅 = - 90
𝐝′= -9
(Note: Here, d = x – A
d' = d ÷ c
A = assumed mean
c = common factor)
×c
× 10
.`. Average weekly earnings = ₹ 1491
ARITHMETIC MEAN
Case 2:
Grouped data
(Discrete data)
Direct
method
𝑥=
𝑓𝑥
𝑓
Short-cut /
Assumed
mean method
𝑥 = A+
(where, n = number of observations
f = frequency
d=x–A
A = assumed mean
d' = d ÷ c
c = common factor)
𝑓𝑑
𝑓
Step
deviation
method
𝑥 =A+
𝑓𝑑′
𝑓
×c
CASE 2: Grouped Data ~ Discrete data
a) Direct method
The following data gives the diameters of screws (in mm) obtained in a sample
enquiry in a hardware shop. Calculate the arithmetic mean.
x
f
(Diameter
in mm)
(Number
of screws)
fx
130
1
130
135
4
540
140
5
700
145
10
1450
150
4
600
155
6
930
𝒇 = 30
̅)=
Arithmetic Mean ( 𝒙
=
𝒇𝒙
𝒇
4350
30
̅ = 145
.`. 𝒙
.`. Average diameter of screws = 145 mm
𝒇𝒙 = 4350
Here,
f = frequency
b) Short-cut / Assumed mean method
The following data gives the diameters of screws (in mm) obtained in a sample
enquiry in a hardware shop. Calculate the arithmetic mean.
x
f
(Diameter
in mm)
(Number
of screws)
d=x–A
fd
130
1
- 10
- 10
135
4
-5
- 20
140
5
0
0
d=x–A
145
10
5
50
150
4
10
40
A = assumed
mean
155
6
15
90
𝒇 = 30
Here, A = 140
̅)=A
Arithmetic Mean ( 𝒙
𝒇𝒅
+
= 140 +
𝒇
150
30
= 140 + 5
̅ = 145
.`. 𝒙
.`. Average diameter of screws = 145 mm
𝒇𝒅 = 150
Here,
f = frequency
c) Step deviation method
The following data gives the diameters of screws (in mm) obtained in a sample
enquiry in a hardware shop. Calculate the arithmetic mean.
x
f
(Diameter
in mm)
(Number
of screws)
d=x–A
130
1
- 10
-2
-2
135
4
-5
-1
-4
140
5
0
0
0
d=x–A
145
10
5
1
10
150
4
10
2
8
A = assumed
mean
155
6
15
3
18
d' =
𝒇 = 30
𝒄
f d'
𝒇𝒅′ = 30
Here, A = 140 ; c = 5
̅)=A
Arithmetic Mean ( 𝒙
𝒅
𝒇𝒅′
+
= 140 +
𝒇
30
30
×c
×5
= 140 + 5
̅ = 145
.`. 𝒙
.`. Average diameter of screws = 145 mm
Here,
f = frequency
d' = d ÷ c
c = common
factor
ARITHMETIC MEAN
Case 3:
Grouped data
(Continuous data)
Direct
method
𝑥=
𝑓𝑚
𝑓
Short-cut /
Assumed
mean method
𝑥 = A+
(where, n = number of observations ;
𝑓𝑑
𝑓
f = frequency
d=m–A
m = Mid-value = (Lower Limit + Upper limit) ÷ 2
A = assumed mean
d' = d ÷ c
c = common factor)
Step
deviation
method
𝑥 =A+
𝑓𝑑′
𝑓
×c
CASE 3: Grouped Data ~ Continuous data
a) Direct method
A survey was conducted among BCom students regarding the monthly usage of
smartphones during COVID-19 induced lockdown. Ascertain the average
monthly usage of smartphones (in hours).
Monthly usage
(hours)
No. of students
120 &
more
100
130 &
more
88
140 &
more
65
150 &
more
36
160 &
more
28
170 &
more
17
180 &
more
5
Solution:
CI
f
(Monthly usage
in hours)
(No. of
students)
120 – 130
m
(Mid-value)
fm
12 (100 – 88)
125
1500
130 – 140
23 (88 – 65)
135
3105
140 – 150
29
145
4205
150 – 160
8
155
1240
160 – 170
11
165
1815
170 – 180
12
175
2100
180 – 190
5
185
925
𝒇 = 100
̅)=
Arithmetic Mean ( 𝒙
=
Here,
f = frequency
m = mid-value
= ( l1 + l2 ) ÷ 2
l1 = lower limit
l2 = upper limit
𝒇𝒎 = 14890
𝒇𝒎
𝒇
14890
100
̅ = 148.9 hours
.`. 𝒙
= 148 hours 54 minutes
Note: The question doesn’t
directly give CIs and frequencies.
More-than cumulative frequencies
(MCF) are given, and using that,
simple frequencies are calculated
for the solution. Example:
f for class 120 – 130 is
(100 – 88 = 12) and so on….
.`. Average monthly usage of smartphones = 148 hrs 54 mins
b) Short-cut / Assumed mean method
A survey was conducted among BCom students regarding the monthly usage of smartphones
during COVID-19 induced lockdown. Ascertain the average monthly usage of smartphones
(in hours).
Monthly usage
(hours)
No. of students
120 &
more
100
130 &
more
88
140 &
more
65
150 &
more
36
160 &
more
28
170 &
more
17
180 &
more
5
Solution:
CI
f
m
(Monthly usage
in hours)
(No. of
students)
(Midvalue)
d=m–A
fd
120 – 130
12
125
- 30
- 360
130 – 140
23
135
- 20
- 460
140 – 150
29
145
- 10
- 290
150 – 160
8
155
0
0
160 – 170
11
165
10
110
170 – 180
12
175
20
240
d=m–A
180 – 190
5
185
30
150
A = assumed
mean
𝒇 = 100
𝒇𝒅 = - 610
Here,
f = frequency
m = mid-value
= ( l1 + l2 ) ÷ 2
l1 = lower limit
l2 = upper limit
Here, A = 155
̅)=
Arithmetic Mean ( 𝒙
A+
= 155 +
𝒇𝒅
𝒇
−610
100
= 155 – 6.1
Note: The question doesn’t directly
give CIs and frequencies. More-than
cumulative frequencies (MCF) are
given, and using that, simple
frequencies are calculated for the
solution.
̅ = 148.9 hours
.`. 𝒙
.`. Average monthly usage of smartphones = 148 hrs 54 mins
c) Step deviation method
A survey was conducted among BCom students regarding the monthly usage of smartphones
during COVID-19 induced lockdown. Ascertain the average monthly usage of smartphones
(in hours).
120 &
more
100
Monthly usage
(hours)
No. of students
130 &
more
88
140 &
more
65
150 &
more
36
160 &
more
28
170 &
more
17
180 &
more
5
Solution:
CI
f
m
(Monthly usage
in hours)
(No. of
students)
(Midvalue)
120 – 130
12
125
- 30
130 – 140
23
135
140 – 150
29
150 – 160
d = m – A d' =
𝒅
Here,
f d'
f = frequency
-3
- 36
- 20
-2
- 46
m = mid-value
= ( l1 + l2 ) ÷ 2
145
- 10
-1
- 29
8
155
0
0
0
160 – 170
11
165
10
1
11
d=m–A
170 – 180
12
175
20
2
24
A = assumed
mean
180 – 190
5
185
30
3
15
d' = d ÷ c
𝒇 = 100
𝒄
𝒇𝒅′ = - 61
l1 = lower limit
l2 = upper limit
c = common
factor
Here, A = 155 ; c = 10
̅)=
Arithmetic Mean ( 𝒙
A+
= 155 +
𝒇𝒅′
𝒇
−61
100
= 155 – 6.1
×c
× 10
Note: The question doesn’t directly
give CIs and frequencies. More-than
cumulative frequencies (MCF) are
given, and using that, simple
frequencies are calculated for the
solution.
̅ = 148.9 hours
.`. 𝒙
.`. Average monthly usage of smartphones = 148 hrs 54 mins