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St. Philomena’s College, Mysore-15
MODULE VI: MEASURES OF CENTRAL TENDENCY:
By,
Prof. Susamma Chacko
J.S.S. College,
Nanjanagud.
2 Marks questions:
1. Which are the different measures of central tendency?
2. Mention any two measures of central tendency and state the formula used to compute
them.
3. What are the uses of an average?
4. Distinguish between mean and median.
5. What is the difference between simple arithmetic mean and weighted arithmetic mean?
6. What are the uses of weighted arithmetic mean?
7. What are the properties of a good average?
8. What is meant by grouped data and ungrouped data?
9. What is the difference between individual Series, discrete Series and continuous series?
10. Define Mean. Give an example.
11. Define Median. Give an example.
12. Define Mode. Give an example.
13. Define weighted arithmetic mean. Give an example.
14. Define Harmonic mean. Give an example.
15. Define Geometric mean. Give an example.
16. Mention any two advantages of arithmetic mean.
17. Mention any two advantages of weighted arithmetic mean.
18. What are the uses of geometric mean?
19. What are the uses of harmonic mean?
20. State the formula to Calculate weighted arithmetic mean.
Q 21. Calculate the mean for the following data.
40, 50, 55, 78, 58, 60, 73, 35, 43, 48
Solution : Formula = x = x
= 540 = 54
n
10
File: ECO WORKSHOP-2005
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St. Philomena’s College, Mysore-15
Q . 22. The following are the marks scored by 10 students in a test paper. Calculate
the arithmetic average (mean).
75, 47, 43, 75, 50, 25, 75, 43, 42, 25
Solution: Formula = x = x
n
= 500 = 50
10
Q 23. Following are the height measurement of 8 persons in centimeters. Find out the
mean height.
159, 161, 163, 165, 167, 169, 171, 173.
Solution : Formula = x = x
= 1028 = 128.5
n
8
Q 24. Daily cash earnings of 10 workers working in different industries are as
follows:
Calculate the average daily earnings.
Rs. 50, 70, 80, 90, 100, 75, 40, 65, 85, 120
Solution: Formula = x = x
= 770 = Rs. 77
n
10
Q 25. Calculate the simple arithmetic average of the following items.
Size of items:
20
50
72
28
53
74
34
54
75
39
59
78
42
64
79
Solution : Formula = x = x
n
Alternative method:
x
20
28
34
39
42
50
53
54
59
64
72
74
75
78
79
File: ECO WORKSHOP-2005
= 821
15
= 54.73
d (x-A)
- 30
- 28
- 60
- 11
-8
0
3
4
9
14
22
24
25
28
29
d = 71
2
St. Philomena’s College, Mysore-15
Solution : = x = A+d
n
= 50 + 71 = 50 + 4.73 = 54.73
15
N.B : A = Assumed mean (Average)
d = Deviation from the assumed mean
n = No. of items
26. Find the harmonic mean for the following data, using the following formula
3, 5, 6, 6, 7, 10, 12.
n
H.M=
1
  x 
Answer:
7
n
 5.98
=
 1  1.17
  x 
27. Estimate the median for the following observations.
1, 2, 5, 7, 9, 10, 12
Answer: M = 7
NB: M=(N+1)th item=(7+1)th= 8 =4th item. 4th item is 7
2
2
2
28. Estimate the median for the following even observations.
1, 2, 5, 7, 9, 10, 12, 14
Answer: M = (N+1)th item
2
=(8+1)th item on 9 = 4.5th item
2
2
th
th
Taking the average of 4 and 5 item.
M = 7+9 =16 = 8
2
2
29. Locate the mode for the following ungrouped data.
3, 7, 3, 3, 5, 1, 7
Answer: Z = 3
Because 3 is the most repeated number.
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St. Philomena’s College, Mysore-15
30. Locate the mode for the following discrete data.
Marks
40
50
60
70
80
90
Frequency
10
8
20
10
8
9
Answer: Z = 60
NB: Because maximum frequency (i.e. 20) is for 60 marks
31. Compute the geometric mean of 2 and 8 by using the formula G.M= n x1  x2  x3, ... xn
Answer:- GM= 2 8 = 16 = 4
32. Compute the Geometric mean of 2, 4, 8 by using the formula GM= n x1  x2  x3, ... xn
Answer:GM=
3
2 4 8 =
3
2 4 8 = 4
Mean, Median and Mode
Mean:
5 Marks questions:
1. a) Define arithmetic mean. What are its uses?
b) The following table given the monthly income of 12 families in a town.
Calculate the mean.
Sl No.
1
2
3
4
5
6
7
8
9
10
11
12
Monthly
Income
280
180
96
98
104
75
80
84
100
76
600
20
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St. Philomena’s College, Mysore-15
Solution:
Sl N0.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
N =12
Monthly Income
(in Rs.) (X)
280
180
96
98
104
75
80
84
100
76
600
20
x = 1793
x =  x = 1793 = Rs. 149 . 42
n
12
2. The following table gives the number of children born per family in 735 families. Calculate
the average number (mean) of children born per family.
No of Children
born per family
No of families
0
1
96
108
2
3
154 126
4
5
6
7
8
9
10
11
12
13
95
62
45
20
11
6
5
5
1
1
Solution :No of Children born
per family (x)
0.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
File: ECO WORKSHOP-2005
Number of Families
(f)
96
108
154
126
95
62
45
20
11
6
5
5
1
1
f = 735
fx
0
108
308
378
380
310
270
140
88
54
50
55
12
13
fx = 2166
5
St. Philomena’s College, Mysore-15
x = f x = 2166 = 2.9 or 3
f
735
The average no of Children born per family is approximately 3.
2. a) Mention the merits and demerits of arithmetic mean.
b) A candidate obtains the following percentages of marks in an examination.
Calculate the arithmetic mean.
Subjects
Sanskrit
Mathematics
Economics
English
Political Science
History
Geography
Marks %
75
84
86
68
70
75
80
Solution:-
N.B. :- n - No of subjects
 x - Total no of Marks.
x =x
n
= 538
7 = 76. 85
3. a) Define arithmetic mean. What are its advantages and disadvantages?
b) For the grouped discrete data given below obtain the mean.
Group
Marks (X)
Frequency (f)
1
75
3
2
50
1
3
47
1
4
43
3
5
25
2
Solution:
Formula: x = fx or fx
f
n
Group
No
1
2
3
4
5
Total
X = fx
f
= 501
10
File: ECO WORKSHOP-2005
Marks scored
(x)
75
50
45
43
25
-----
f
fx
3
1
1
3
2
10
225
50
47
129
50
501
= 50.1
6
St. Philomena’s College, Mysore-15
5. Compute the arithmetic mean for the following data:
Marks Scored
75
50
47
43
25
No. of Students
30
10
10
30
20
Solution:
x
75
50
47
43
25
x
f
fx
30
10
10
30
20
 f 100
2250
500
470
1290
500
 fx  50110
 fx 50110

 50.11
x
100
6. Calculate the arithmetic mean for the following data.:
Wages
5
7
9
11
13
15
No. of Workers
2
4
5
6
2
1
Solution:
x
5
7
9
11
13
15
f
2
4
5
6
2
1
 f  20
fx
10
28
45
66
26
15
 fx  190
 fx
x
190

 9 .5
20

x
File: ECO WORKSHOP-2005
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St. Philomena’s College, Mysore-15
Median:
7. From the following data of the wages of 7 labourers, compute the median wage.
Wage (in Rs.)
1100, 1150, 1080, 1120, 1160, 1400.
Solution:
Step 1. Arrange the data in the ascending order.
N  1 th
Step 2. Apply the formula
item
2
Wages (Arranged in
the ascending order)
1080
1100
1120
1150
1160
1200
1400
Sl. No.
1
2
3
4
5
6
7
n=7
m
N 1
2
th
item
7 1 8
  4 th item
2
2
The 4th item is 1150.
Hence the median wage is Rs. 1150.

8. The following are the marks scored by 7 students; find out the median marks.
Roll No.
Marks
1
45
2
32
3
18
4
57
5
65
6
28
7
46
Solution:
m
N 1
2
th
item.
Sl. No.
Roll No.
1
2
3
4
5
6
7
3
6
2
1
4
7
5
File: ECO WORKSHOP-2005
Marks (Arranged in
ascending order)
18
28
32
45
57
58
65
8
St. Philomena’s College, Mysore-15
m
7 1
2
th
item. 
8
 4th item.
2
4th item is 45
Therefore the median marks is 45
9. a) Define Median. What are its uses?
b) Obtain the value of median from the following data.
391, 384, 591, 407, 672, 522, 777, 753, 2488, 1490.
Solution : Step 1. Arrange the data in the ascending (descending) order.
2. Apply the formula (N+1)th item.
2
3. Take the mean of the middle two values.
Data arranged in the
Sl. No.
ascending order
1
384
2
391
3
405
4
522
5
591
6
672
7
753
8
777
9
1490
10
2488
M = (N+1)th item = 10 + 1 = 11 = 5.5th item.
2
2
2
Take the mean of the 5th and 6th item.
i.e., 591+672 = 1263 = 631.5
2
2
10. Find out the median from the following.
57, 58, 61, 42, 38, 65, 72, 60.
Solution :
Sl. No.
1
2
3
4
6
7
8
File: ECO WORKSHOP-2005
Values
(Arranged in the
ascending order)
38
42
57
58
61
65
66
72
9
St. Philomena’s College, Mysore-15
M = (N+1)th item = 8 + 1
2
2
th
Size of the 4.5 item is
i.e., 58 +61 = 59.5
2
= 9 = 4.5th item.
2
Mode:
11. Locate the mode for the following continues frequency distribution.
Class
interval
40-50
50-60
60-70
70-80
80-90
Frequency
10
8
25
10
12
Answer: Z = lo + (f0 - f1)X(l1-l0)
2f0-f1-f2
= 60 + 25 - 8 X 70-60
(2X25)-8-10
= 60 +
17
X 10
50-8-10
= 60 + 17 X 10
32
= 60 + 170 = 60+5.31
32
= 65.31
NB: First locate the modal class i.e. class having maximum frequency.
12. From the following continuous data locate the mode by using the interpolation method.
Class interval
4-6
6-8
8-10
10-12
12-14
Frequency
4
5
20
15
5
Answer: Z = 8+20-5 X 10-8
2X20-5-15
=8+15X2 = 8 +1.5
20
= 9.5
13. Define Mean, Median and Mode and mention their uses.
14. Discuss the merits and demerits of Mean and Median.
File: ECO WORKSHOP-2005
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St. Philomena’s College, Mysore-15
Weighted Arithmetic Mean:
5 Marks questions:
1. A Contractor employs three types of workers -- male, female and children.
To a male worker he pays Rs. 100/- per day, to a female worker Rs. 80/per day and to a child Rs. 30/- per day. He employs 10 workers in each
category. What is the average wage per day paid by the contractor? Apply
the weighted average method.
Solution:
Wx
Xw =
W
= (10X100)+(10X80)+(10X30)
10+10+10
=
(1000+800+300)
30
=
2100
30
= Rs.70
2. A contractor employs three types of workers male, female and children. To
a male worker he pays Rs.80/day, to a female worker he pays Rs.60/day
and to a child Rs.40/day. He employs 20 male workers, 15 female workers
and 10 child workers. Calculate the weighted arithmetic mean wage.
Solution:
Wages /day
(X)
80
60
40
File: ECO WORKSHOP-2005
No. of workers
(W)
20
15
10
W=45
WX
1600
900
400
WX=2900
11
St. Philomena’s College, Mysore-15
Xw =
Wx
W
=
2900
45
=
64.4
3. The following table shows the speed and time taken by 3 different types of
trains. Calculate the weighted average speed.
Speed
(Miles /hour)
30
40
10
24
Time taken
50
75
6
60
Solution:
Speed
(Miles /hour) (X)
30
40
10
24
Average speed
WX
1500
3000
60
1440
WX=6000
= 6000/191
=
File: ECO WORKSHOP-2005
Time taken
(W)
50
75
6
60
W=191
31.41miles/hour
12
St. Philomena’s College, Mysore-15
4. The following table indicates the increase in cost of living for a working
class family and the weights assigned to each item. Find out the weighted
average of the increase in cost of living.
Items
Food
Rent
Clothing
Fuel and light
Other items
Percentage Increase
29
54
97.5
75
75
Weights
7.5
2.0
1.5
1.0
0.5
Solution:
Items
Food
Rent
Clothing
Fuel and light
Other items
Percentage
Increase(X)
29
54
97.5
75
75
Weights
(W)
7.5
2.0
1.5
1.0
0.5
W=12.5
WX
217.50
108.00
146.25
75.00
37.50
WX=584.25
Wx
Xw =
W
= 584.25
12.5
=
46.74 %
5. The following table gives the total number of labour force and the ratio of
unemployment in 5 states. Find the weighted arithmetic mean.
States
Karnataka
Tamil Nadu
Andra pradesh
Maharashtra
Kerala
File: ECO WORKSHOP-2005
Unemployment ratio
203
195
250
300
200
Total No. of labour
force (in thousands)
183
198
201
107
210
13
St. Philomena’s College, Mysore-15
Solution:
States
Karnataka
Tamil Nadu
Andra pradesh
Maharashtra
Kerala
Xw =
=
Unemployment
ratio ( X)
203
195
250
300
200
Total No. of
labour force (in
thousands) (W)
183
198
201
107
210
W=899
WX
37149
38610
50250
32100
42000
WX=200109
Wx
W
=
200109
899
222.5
=
22.25%
6. From the following data find out the academic performance of a student,
by using the weighted arithmetic mean method.
Work
Seminar
Test
Assignment
Marks/100
45
62
52
Weights
4
2
3
Solution:
Work
Seminar
Test
Assignment
Marks/100
(X)
45
62
52
Weights
(W)
4
2
3
W=9
WX
180
124
156
WX=460
Wx
Xw = W
= 460/9
= 51.1
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St. Philomena’s College, Mysore-15
7. The following table shows the results of two colleges. Which is better on
the average?
Colleges
College A
College B
Courses
No.of
No.of
Marks %
Marks %
students
students
I degree
200
70
150
80
II degree
150
60
100
60
III degree
100
80
50
80
Solution:
College A :
Xw A =
Wx
W
= (200X70)+(150X60)+(100X80)
200+150+100
= 31000/450
= 68.9
College B :
Xw B =
Wx
W
= (150X80)+(100X60)+(50X80)
150+100+50
= 22000/300
= 73.3
College B is better because the average is more.
File: ECO WORKSHOP-2005
15
St. Philomena’s College, Mysore-15
Alternative method
Colleges
Courses
College B
College A
I degree
No.of
students
(X)
200
II degree
150
60
9000
100
60
6000
III degree
100
80
8000
50
80
4000
W=450
WX=
31000
W=300
WX=
22000
Xw A =
Marks %
(W)
WX
70
14000
No.of
students
(X)
150
Marks %
(W)
WX
80
12000
Wx
W
= 31000/450
= 68.9
Wx
W
= 22000/300
Xw B =
= 73.3
Harmonic Mean
5 Marks Questions:
1. The monthly income of 10 families in a certain village are given below.
Calculate the Harmonic Mean by using the following formula HM=
Family
Income (in RS)
File: ECO WORKSHOP-2005
1
85
2
70
3
10
4
75
5
500
6
8
7
42
8
250
9
40
n

1
x
10
36
16
St. Philomena’s College, Mysore-15
Solution: Family
1
2
3
4
5
6
7
8
9
10
N=10
Income (x)
85
70
10
75
500
8
42
250
40
36
Harmonic Mean =
Harmonic Mean
=
1/x
0.1176
0.1426
0.1000
0.1333
0.0020
0.1250
0.2318
0.0040
0.0250
0.2778
(1/x) = 0.3463

n
OR
(1/x1 +1/x2 + 1/x3-----1/xn)
10
0.346
n
 (1/x)
= 28.87
2. A truck company has 5 trucks to bring red soil from a pit of 5kms away from the
brickyard. The following table shows the time taken per load of all the 5 trucks.
Obtain the harmonic mean by using the formula HM=
Truck no
Minutes per hour
1
48
2
40
n

3
40
1
x
4
48
5
32
Solution: Truck no
1
2
3
4
5
n=5
Minutes per hour
48
40
40
48
32
Harmonic Mean =
=
File: ECO WORKSHOP-2005
n
 (1/x)
40.7

1/x
0.0208
0.0250
0.0250
0.0208
0.0312
x =0.1228
= 5/(0.1228)
17
St. Philomena’s College, Mysore-15
3. Calculate the harmonic Mean for the following data by using the formula
Size of Items
Frequency
6
4
7
6
8
9
9
5
10
2

n
1
f 
 x
11
8
Solution: -
x
6
7
8
9
10
11
F
4
6
9
5
2
8

1/x
0.167
0.143
0.125
0.111
0.100
0.090
f (1/x)
0.6668
0.8574
1.1250
0.5555
0.2000
0.7272
 f(1/x)=4.1319
f = 34
Harmonic Mean =
n
 f(1/x)
= 34
4.1319
= 8.23
4. From the following data Compute the value of harmonic Mean
Marks
No of Students
10
20
Solution:
Marks (x)
10
20
25
40
50
f
20
30
50
15
5
Harmonic Mean =
25
50
1/x
0.100
0.050
0.040
0.025
0.020
(1/x)=0.235
N
f
  x 
= 120
5.97
File: ECO WORKSHOP-2005
20
30
or
=

f
f
  x 
40
15
50
5
F(1/x)
2.00
1.50
2.00
0.37
0.10
 f(1/x)=5.97
20.08
18
St. Philomena’s College, Mysore-15
5. From the following data compute the value of harmonic mean. Use the
formula HM=
f
f 
  m 
Class
10-20
interval
Frequency 4
20-30
30-40
40-50
50-60
6
10
7
3
f
4
6
10
7
3
f=30
f/m
0.267
0.240
0.286
0.156
0.055
(f/m)=1.004
Solution
Class interval
10-20
20-30
30-40
40-50
50-60
Midpoints
15
25
35
45
55
Harmonic Mean =

N
f 
  m 
or

f
f 
  m 
30
=
29.88
=
1.004
6. Calculate harmonic mean of the following data.
Use the formula HM=
Marks
Frequency
Solution:
Marks
30-40
40-50
50-60
60-70
70-80
80-90
90-100
File: ECO WORKSHOP-2005
30-40 40-50
15
13
N
f 
  m 
50-60
8
Mid value (m)
35
45
55
65
75
85
95
60-70
6
70-80
15
Frequency (f)
15
13
8
6
15
7
6
 f=70
80-90 90-100
7
6
1/m
0.02857
0.02222
0.01818
0.01534
0.01333
0.01176
0.01053

f/m
0.42855
0.28886
0.14544
0.09264
0.19995
0.08232
0.06318
(f/m)=1.30034
19
St. Philomena’s College, Mysore-15
f
Harmonic Mean = 
f 
  m 
= 70
1.300
=
or
N
f 
  m 
53.83
GEOMETRIC MEAN
5 Marks questions:
1. Calculate Geometric mean of the following by using the formula
GM=antilog of
 log x
n
Solution:
Log of x
1.6990
1.8573
1.7324
1.9138
1.9685
x
50
72
54
82
93
 log x  9.1710
GM= 50 72548293
Or
 log x
N
9.1710
 Anti log 1.8342  68.26
= Antilog
5
GM = Antilog
2. Daily income of ten families of a particular place in given below. Find out
the Geometric mean by using the formula GM= Antilog of
 log x
N
Income in Rs. 85, 70, 15, 75, 500, 8, 45, 250, 40, 36
x
85
70
15
75
500
8
45
250
40
36
File: ECO WORKSHOP-2005
log x
1.9294
1.8451
1.1761
1.8751
2.6990
0.9031
1.6532
2.3979
1.6021
1.5563
 log x  17.6373
20
St. Philomena’s College, Mysore-15
GM = Antilog of
= Antilog
 log x
N
17.6373
10
= 58.03
3. For the grouped data given below obtain the geometric mean by using the
formula GM = Antilog
 f log x
.
N
10
2
x
f
100
3
1000
2
10000
3
Solution:
f
2
3
2
3
x
10
100
1000
10000
 f 10
GM = Antilog
Log x
1
2
3
4
f log x
2
6
6
12
 f log x  26
 f log x
N
Here N= c.f.
= Antilog of
26
10
= 398.1
File: ECO WORKSHOP-2005
21
St. Philomena’s College, Mysore-15
Discrete series
10 Marks questions:
4. The following table gives the weight of 31 persons in a sample survey.
Calculate Geometric mean by using the formula GM = Antilog
 f log x
N
Solution:
Weight (in lbs)
No. of persons.
Size of item
130
135
140
145
146
148
149
150
157
130 135 140 145 146 148 149 150 157
3
4
6
6
3
5
2
1
1
Frequency
3
4
6
6
3
5
2
1
1
N=31
Log x
2.1139
2.1303
2.1461
2.1614
2.1644
2.1703
2.1732
2.1761
2.1959
f log x
6.3417
8.5212
12.8766
12.9684
6.4932
10.8515
4.3464
2.1761
2.1959
 f log x  66.7710
 f log x
N
66.7710
 Anti log of 2.1539
= Antilog
31
GM = Antilog
G.M. Weight = 142.5 lbs
5. Find out the Geometric mean of the following data by using the formula
GM = Antilog
 f log m
N
Field of Wheat
7.5-10.5
10.5-13.5
13.5-16.5
16.5-19.5
19.5-22.5
22.5-25.5
25.5-28.5
File: ECO WORKSHOP-2005
No. of farms
5
9
19
23
7
4
1
22
St. Philomena’s College, Mysore-15
Solution:
Mid Value
9
12
15
18
21
24
27
log m
0.9542
1.0792
1.1761
1.2553
1.3222
1.3802
1.4314
f
5
9
19
23
7
4
1
N=68
f log m
4.7710
9.7128
22.3459
28.8719
9.2554
5.5208
1.4314
 f log m  81.9092
 f log m
N
81.9092
= Antilog of
68
GM = Antilog
= 16.02
6. Compute the Geometric mean of the following data. Use the formula GM
= Antilog of
Marks
No. of
Students
 f log m
N
0-10
10-20
20-30
30-40
40-50
5
7
15
25
8
Solution:
Marks
0-10
10-20
20-30
30-40
40-50
Midpoints
(m)
5
15
25
35
45
GM = Antilog
Log m
Frequency
f. log m
0.6990
1.1761
1.3979
1.5441
1.6532
5
7
15
25
8
3.4950
8.2327
20.9685
38.6025
13.2256
  60
  84.5243
84.5243
60
= Antilog 1.40874
= 25.64
File: ECO WORKSHOP-2005
23
St. Philomena’s College, Mysore-15
Mean
10 Marks Questions:
1. a) Define mean. What are its advantages and disadvantages?
b) The following table gives the monthly income of 10 employees in an office.
Calculate the arithmetic mean of income.
Income (In Rs.)
1780, 1760, 1690, 1750, 1840, 1920, 1100, 1810, 1050, 1950.
Solution : Direct Method :
x = x = 16650
n
10
= Rs. 1665
Shortcut Method:
Employee
Income
1
2
3
4
5
6
7
8
9
10
n = 10
1780
1760
1690
1750
1840
1920
1100
1810
1050
1950
x = A + d
n
= 1800 + - 1350
10
Deviations from
assumed mean
(x-1800)
- 20
- 40
- 110
- 50
+ 40
+ 120
- 700
+ 10
- 750
+ 150
d = - 1350
= 1800 + (-1350)
1800 – 135 = 1665
The average income is Rs. 1665
NB: A= 1800
2. From the following data of the marks obtained by 60 students of a class, calculate the
arithmetic mean by using Direct Method & Short cut method.
Marks
No. of Students
File: ECO WORKSHOP-2005
20
8
30
12
40
20
50
10
60
6
70
4
24
St. Philomena’s College, Mysore-15
Solution : Direct Method :
Marks
20
30
40
50
60
70
x = fx
n
Shortcut Method:
x = fx
f
No. of Students
8
12
20
10
6
4
f = 60
= 2460
60
x
20
30
40
50
60
70
fx
160
360
800
500
360
280
fx = 2460
= 41
f
8
12
20
10
6
4
f = 60
d (x-40)
- 20
- 10
0
10
20
30
fd
- 160
- 120
0
100
120
120
fd = 60
x = fd
= 40 + 60
= 41
f
60
N.B A – Assumed mean
D – Deviations from the assumed mean. Here 40 is the assumed mean.
3. Calculate the mean for the following data by using direct method and short cut method.
1
21
Value
Frequency
2
30
Solution: Direct Method:
x
1
2
3
4
5
6
7
8
9
10
File: ECO WORKSHOP-2005
3
28
x = fx
f
f
21
30
28
40
26
34
40
9
15
57
f = 300
4
40
5
26
6
34
7
40
8
9
9
15
10
57
fx
21
60
84
160
130
204
280
72
135
570
fx = 1716
25
St. Philomena’s College, Mysore-15
x = fx
f
= 1716
300
= 5.72
Shortcut Method:
x
1
2
3
4
5
6
7
8
9
10
x = fd
f
f
21
30
28
40
26
34
40
9
15
57
f = 300
= 216
300
d (x-A) (x-5)
-4
-3
-2
-1
0
1
2
3
4
5
= 5 + 0.72
fd
-84
-90
-56
-40
0
34
80
27
60
285
fd = 216
= 5.72
4. From the following data find out the mean profits.
Profits per shop
(In Rs.)
100-200
200-300
300-400
400-500
500-600
600-700
700-800
No. of Shops
10
18
20
26
30
28
18
Solution: Direct Method:
x = fm
f
Profits per shop
(Class interval)
100-200
200-300
300-400
400-500
500-600
600-700
700-800
File: ECO WORKSHOP-2005
No. of Shops
10
18
20
26
30
28
18
f = 150
Mid Points of
class intervals (m)
150
250
350
450
550
650
750
fm
1500
4500
7000
11700
16500
18200
13500
fm = 72900
26
St. Philomena’s College, Mysore-15
x = fm = 72900
= 486
f
150
The average profit is Rs. 486.
Shortcut Method:
x = A + fd or = A + fd
f
n
Class interval
f
m
100-200
200-300
300-400
400-500
500-600
600-700
700-800
10
18
20
26
30
28
18
f = 150
150
250
350
450
550
650
750
x = A + fd = 450 + 5400 = 450 + 36
f
150
d
(m-450)
-300
-200
-100
0
100
200
300
fd
-3000
-3600
-2000
0
3000
5600
5400
fd = 5400
= 486
N.B : A – Assumed Mean. Here we have taken 450 as the assumed mean.
5. For the data given below calculate the mean.
Strength
60-65
65-70
70-75
75-80
80-85
85-90
90-95
95-100
100-105
105-110
110-115
115-120
No. of lots
1
3
10
18
20
16
14
14
6
4
2
1
x = fm
f
(m = Mid point of class intervals)
Solution:
File: ECO WORKSHOP-2005
27
St. Philomena’s College, Mysore-15
No. of Lots
(f)
1
3
10
18
20
16
14
14
6
4
2
1
100
Strength
60-65
65-70
70-75
75-80
80-85
85-90
90-95
95-100
100-105
105-110
110-115
115-120
Total
Solution:
x = fm
f
Mid value
(m)
62.5
67.5
72.5
77.5
82.5
87.5
92.5
97.5
102.5
107.5
112.5
117.5
fm
62.5
202.5
365.5
775.0
1485.0
1750.0
1480.0
1365.0
615.0
430.0
225.0
117.5
8870
= 8870 = 88.70
100
6. From the following data calculate arithmetic mean.
Marks
0-10
10-20
20-30
30-40
40-50
50-60
Solution:
Class interval
0-10
10-20
20-30
30-40
40-50
50-60
Solution:
x = fm
f
File: ECO WORKSHOP-2005
f
5
10
25
30
20
10
f = 100
No. of
students
5
10
25
30
20
10
m
5
15
25
35
45
55
fm
25
150
625
1050
900
550
fd = 3300
= 3300 = 33
100
28
St. Philomena’s College, Mysore-15
7. Comment on the performance of the students of 3 universities given below by
using the weighted average method.
Universities
Bombay
No. of
Pass % Student
s
71
3
83
4
73
5
74
2
65
3
66
3
Course of
Study
M.A
M.Com
BA
B.com
BSC
MSC
Calcutta
No. of
Pass % Student
s
82
2
76
3
73
6
76
7
65
3
60
7
Madras
No. of
Pass % Student
s
81
2
76
3
74
4
58
2
70
7
73
2
Solution:
Universities
Course of
Study
M.A
M.Com
BA
B.com
BSC
MSC
Bombay
Calcutta
Madras
x
w
wx
x
w
wx
x
w
wx
71
83
73
74
65
66
3
4
5
2
3
3
w= 20
213
332
365
148
195
198
wx= 1451
82
76
73
76
65
60
2
3
6
7
3
7
w= 28
164
228
438
532
195
420
wx= 1977
81
76
74
58
70
73
2
3
4
2
7
2
w= 20
162
228
296
116
490
146
wx= 1438
xw = wx
w
Bombay - 1451
= 72.55
20
Calcutta - 1977
= 70.61
28
Madras – 1438
= 71.9
20
Answer: Bombay University has the best performance.
Weighted Arithmetic Mean
8. A candidate obtains the following percentages of marks in an examinations.
Sanskrit
Mathematics
Economics
English
Political Science
History
Geography
File: ECO WORKSHOP-2005
75
84
56
78
57
54
47
29
St. Philomena’s College, Mysore-15
It is agreed to give double weight to marks in English, Mathematics and Economics. What is
the weighted means?
Solution:
Subjects
Marks (x)
Sanskrit
Mathematics
Economics
English
Political Science
History
Geography
75
84
56
78
57
54
47
x w= wx
w
Weights
(w)
1
2
2
2
1
1
1
w = 10
Weighted
marks (wx)
75
168
112
156
57
54
47
wx = 649
= 649 = 64.9
10
10. Calculate the mean, median and mode for the following data.
x
f
10-20
3
20-30
4
30-40
10
40-50
5
50-60
2
Solution:
x = fm
f
x
10-20
20-30
30-40
40-50
50-60
f
3
4
10
5
2
f =24
m
15
25
35
45
55
Mean
Median
fm
cf
45
3
100
7
350
17
225
22
110
24
fm=830
x = 830 = 34.58
24
Median:
M=
( n  1) th
N
item or   thitem
2
2
=(24+1) 25 = 12.5 th item or 12th item.
2
2
Median comes in 30-40 class interval.
File: ECO WORKSHOP-2005
30
St. Philomena’s College, Mysore-15
N
 c. f
2
M 
i
f
 24 
 7
2
30   
 10
10
Mode
Z = 3 Median-2 mean. on 3M-2x
=(3X35)-(2X34.58)
= 105-69.16
= 39.84
11.Calculate the mean, median and mode for the following data.
Age
55-60
50-55
45-50
40-45
35-40
30-35
25-30
20-25
No of People
7
12
15
20
30
33
28
14
Solution:
NB: Step1. Arrange the data in the ascending order to calculate median.
20-25
25-30
30-35
35-40
40-45
45-50
50-55
55-60
f
14
28
33
30
20
15
13
7
f=160
x = fm
f
5892.5 = 36.8
160
Median:
n
M =   th item =
2
m
22.5
27.5
32.5
37.5
42.5
47.5
52.5
57.5
fm
315
770
1072.5
1125
850
675
682.5
402.5
5892.5
c.f
14
42
75
165
125
140
153
160
 160 

 = 80th item
2


80th item is in between 35 and 40
File: ECO WORKSHOP-2005
31
St. Philomena’s College, Mysore-15
 c. f .
i
f
# It can also be done in the descending order.
Median = L +
N
2
= 35+ 80-75 X 5
30
= 35+ 25 =35 + 0.83 = 35.83
30
Mode
Z = 3 median - 2 mean
=(3 X 35.83)-(2 X 36.8)
=107.49-73.6
=33.89
12. Calculate the mean, median and mode for the following data.
Wage (Rs)
90
70
50
30
20
10
No. of workers
2
4
5
6
2
1
Solution:
Mean (X) = fx
f
X
90
70
50
30
20
10
f
2
4
5
6
2
1
f=20
Mean:-
Median:-
File: ECO WORKSHOP-2005
fx
180
280
250
180
40
10
fx=940
(X) = 940
20
= 47
n  1  th
M= 
 item
 2 
32
St. Philomena’s College, Mysore-15
x
90
70
50
30
20
10
f
2
4
5
6
2
1
cf
2
6
11
17
19
20
M=(20+1)/2
= 21/2
= 10.5th item
10.5th item comes in 50.
Therefore median value is 50.
Mode:
Mode(Z)=3M-2X
=(3X50)-(2X47)
= 150 – 92
= 58
Note: if the data is in the irregular form, arrange them in the ascending order to find out the
median.
13. Calculate the mean, median and mode from the following frequency distribution of marks
at a test in statistics.
Marks
No. of Students
Solution: Direct Method:
Marks
5
10
15
20
25
30
40
45
50
5
20
15
75
x = fx
f
No. of Students
(f)
20
43
75
76
72
45
9
8
50
f = 398
x = fx
= 7295
n
398
Median m = (n+1) th item
2
File: ECO WORKSHOP-2005
10
43
20
76
25
72
30
45
40
9
45
8
50
50
fx
cf
100
430
1125
1520
1800
1350
360
360
250
fx = 7295
20
63
138
214
286
331
340
348
398
= 18.3
33
St. Philomena’s College, Mysore-15
= (398 + 1) = 399 = 199.5th item
2
2
th
= 199.5 item belong to 20
= M = 20
Mode:
Z = 3M = 2x or 3 Median –2 Mean.
(3x20) – (2x18.3)
60 – 36.6
23.4
Median
10 Marks questions:
1. a) Define median. What are its uses?
b) From the following data find the value of the median:
Income
(in Rs.)
Marks
1000
1500
800
2000
2500
1800
24
26
16
20
6
30
Solution:
Steps: 1. Arrange the data in the ascending order
2. Find out the cumulative frequencies
N  1 th
3. Apply the formula
item
2
Income (Arranged in
the ascending order)
800
1080
1500
1800
2000
2500
M 
N 1
2
th
No. of Persons
16
24
26
30
20
6
Cumulative frequency
c.f.
16
40
66
96
116
122
item.
NB: Here N= Cumulative frequency
122  1 123
M 

 61.5 th item.
2
2
61.5th item is 1500
The median value is 1500
NB: Same procedure for both even number and odd number of observations.
File: ECO WORKSHOP-2005
34
St. Philomena’s College, Mysore-15
2. a) what are the merits and demerits of median?
b) Locate median from the following:
Size of Shoes
5
5.5
6
6.5
7
7.5
8
Frequency
10
16
28
15
30
40
34
Solution:
Size of Shoes
5
5.5
6
6.5
7
7.5
8
Median 
173  1
2
th
F
10
16
28
15
30
40
34
c.f.
10
26
54
69
99
139
173
item.
= 87th item.
Size of 87th item is 7
Therefore Median size of shoe is 7
3. Locate the median for the following data
Weekly Wage
100
200
150
250
300
No. of Workers
10
25
12
33
20
Solution:
x
100
200
150
250
300
File: ECO WORKSHOP-2005
f
10
25
12
33
20
cf
10
22
47
80
100
35
St. Philomena’s College, Mysore-15
M 
N 1
2
=
or
M
cf  1
2
100  1 101

 50.5 th item
2
2
50.5th item comes in 250
Therefore Median is 250
4. Calculate the median for the following
Marks
Frequency
40
7
50
3
70
5
75
6
90
4
Solution:
Group
Marks (x)
Frequency (f)
1
2
3
4
5
40
50
70
75
90
7
3
5
6
4
M= Value of
=
N 1
2
25  1
2
th
th
Cumulative frequency
c.f.
7
10
15
21
25
item.
item. =
26
 13
2
th
item.
The 13th item belongs to the third group and the value of x corresponding to the 3rd group is 70
Hence the Median M=70
5. Calculate the median for the following frequency distributions.
Marks
5-10
10-15
15-20
20-25
25-30
30-35
35-40
40-45
45-50
File: ECO WORKSHOP-2005
No. of Students
7
15
24
31
42
30
26
15
10
36
St. Philomena’s College, Mysore-15
Solution:
Marks
(Class interval)
5-10
10-15
15-20
20-25
25-30
30-35
35-40
40-45
45-50
Median=
f
cf
7
15
24
31
42
30
26
15
10
7
22
46
77
119
149
175
190
200
N 200

 100 The 100th item lies in 25-30 marks group L  25
2
2
N
 c. f
M  L 2
i
f
200
 77
= 25  2
5
42
= 25 
100  77
 5  25  (.54  5)
42
= 27.3
NB:
L- Lower limit of the median class
f- frequency of median class
cf- cumulative frequency of the class proceeding the median class
i- class interval of median class
N- cumulative frequency
6. Calculate the median from the following table:
Marks
10-25
25-40
40-55
55-70
70-85
85-100
File: ECO WORKSHOP-2005
Frequency
6
20
44
26
3
1
37
St. Philomena’s College, Mysore-15
Solution:
Marks
10-25
25-40
40-55
55-70
70-85
85-100
Median=
N
2
th
f
6
20
44
26
3
1
c.f.
6
26
70
96
99
100
item
100
 50 th item
2
N
 cf
2
M=L+
i
f
=
100
 26
2
= 40+
15
44
50  26
 15
44
24
= 40+  15
44
= 40+
= 40+0.54x15
= 40+8.18
= 48.18
7. Find out the median for the following data:
Class interval
0-10
10-20
20-30
30-40
40-50
No. of persons
5
11
19
21
16
Solution:
x
0-10
10-20
20-30
30-40
40-50
File: ECO WORKSHOP-2005
f
5
11
19
21
16
cf
5
16
35
56
72
38
St. Philomena’s College, Mysore-15
N
2
Median=
th
item
72
 36
2
36 comes in 30-40 class interval L  30
=
N
 c. f
2
M=L+
i
f
36  35
10
21
1
=30+ 10
21
=30.5
=30+
8. The following table shows age distribution of persons in a particular region. Find the
median age.
Age
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
No. of persons
2
3
4
3
2
1
0.5
0.1
Solution:
Class interval
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
f
2
3
4
3
2
1
0.5
0.1
c.f.
2
5
9
12
14
15
15.5
15.6
Median less in the 20-30 age group
N
 cf
2
M=L+
i
f
File: ECO WORKSHOP-2005
39
St. Philomena’s College, Mysore-15
15.6
5
=20 + 2
10  27 years
4
NB: Find the median class first, and then apply the formula
N 15.6

 7.8
2
2
7.8 Comes in 20-30-class interval.
File: ECO WORKSHOP-2005
40