WELCOME TO THE LECTURE
SERIES ON BUSINESS
STATISTICS
10/2/2021
Afzal Hossain (Assistant Professor)
1
COURSE NAME: BUSINESS STATISTICS
COURSE CODE: BUS-2409
10/2/2021
Afzal Hossain (Assistant Professor)
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Course Teacher: Afzal Hossain
Assistant Professor
Department of CSE
Army IBA
E-mail:
afzal.armyiba@gmail.com
Contact: 01920979195
10/2/2021
Afzal Hossain (Assistant Professor)
3
SUBMITTED BY
IDs
Section 1, 2, 3, 9, 13, 14, 18, 19, 21, 23, 27, 29, 30, 31, 35, 37,
A
41, 43, 45, 51, 52, 53, 55, 57, 59, 61, 63, 65, 67, 68,
69,71, 75, 77, 81, 85, 70(BBA-5)
Section 4, 5, 6, 8, 10, 11, 14, 15, 17, 20, 22, 24, 26, 28, 32, 33,
B
34, 36, 40, 44, 46, 47, 48, 49, 50, 56, 60, 64, 66, 76,
80, 82, 83
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Afzal Hossain (Assistant Professor)
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BUSINESS STATISTICS
MEAN , MEDIAN AND MODE
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MEAN
Definition: The "mean" is the "average" you're used
to, where you add up all the numbers and then divide
by the number of numbers.
For example: The Mean of 13, 18, 13, 14, 13, 16, 14,
21, 13 are : (13 + 18 + 13 + 14 + 13 + 16 + 14 + 21 +
13) ÷ 9 = 15
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ARITHMETIC MEAN
Example: Suppose we have the following observations: 10, 15,30, 7, 42, 79
and 83. Find out the Arithmetic Mean.
Solutions: These are seven observations. Symbolically, the arithmetic mean,
𝑛
𝑖=1 𝑥𝑖
also called simply mean 𝑥 =
Now, 𝑥 =
𝑛
𝑖=1 𝑥𝑖
𝑛
=
Here 𝑥 is simple mean.
𝑛
𝑥1 +𝑥2 +𝑥3 +𝑥4 +𝑥5 +𝑥6 +𝑥7
7
=
10+15+30+7+42+79+83
7
=
266
7
= 38
It may be noted that the Greek letter μ is used to denote the mean of the
population and n to denote the total number of observations in a population.
Thus the population mean μ =
10/2/2021
𝑛
𝑖=1 𝑥𝑖
𝑛
.
Afzal Hossain (Assistant Professor)
7
ARITHMETIC MEAN
Ungrouped Data-weighted Average: Suppose we have the following
observations:
Type of Test
Mid-term
Laboratory
Final
Relative Weight (w)
2
3
5
Marks (x)
30
25
20
Find out the Arithmetic Mean.
Solutions:μ =
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𝑛
𝑖=1 𝑤𝑖 𝑥𝑖
𝑛 𝑤
𝑖=1 𝑖
=
𝑤1 𝑥1 +𝑤2 𝑥2 +𝑤3 𝑥3
𝑤1 +𝑤2 +𝑤3
=
Afzal Hossain (Assistant Professor)
2×30+3×25+5×20
2+3+5
=
235
10
= 23.5
8
ARITHMETIC MEAN
Ungrouped Data-weighted Average: An investor is fond of investing in
equity shares. During a period of falling prices in the stock exchange, a stock
is sold at BDT 120 per share on one day, BDT 105 on the next and BDT 90
on the third day. The investor has purchased 50 shares on the first day, 80
shares on the second day and 100 shares on the third' day. What average price
per share did the investor pay?
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ARITHMETIC MEAN
Solutions:
Day
1
2
3
Price per Share (BDT) (x)
120
105
90
No of Shares Purchased (w)
50
80
100
𝑤1 𝑥1 +𝑤2 𝑥2 +𝑤3 𝑥3
120 × 50 + 105 × 80 + 90 × 100
μ=
=
=
𝑤1 +𝑤2 +𝑤3
50 + 80 + 100
6000 + 8400 + 9000 23400
=
=
= 101.7
230
230
𝑛
𝑖=1 𝑤𝑖 𝑥𝑖
𝑛
𝑖=1 𝑤𝑖
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ARITHMETIC MEAN
Grouped Data-arithmetic Mean: For grouped data, arithmetic mean may
be calculated by applying any of the following methods:
(i) Direct method
(ii) Short-cut method
(iii) Step-deviation method
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ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Direct
Method): In the case of direct method, the
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
formula μ =
is used. Here m is
𝑛
mid-point of various classes, f is the
frequency of each class and n is the total
number of frequencies.
The given table gives the marks of 58
students in Statistics. Calculate the average
marks of this group.
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Afzal Hossain (Assistant Professor)
Marks No. of
Students
0-10 4
10-20 8
20-30 11
30-40 15
40-50 12
50-60 6
60-70 2
Total 58
12
ARITHMETIC MEAN
Grouped Data-arithmetic
Mean (Direct Method):
Solution: Here n=58
𝑛
1940
𝑖=1 𝑓𝑖 𝑚𝑖
μ=
=
𝑛
58
Or, μ =33.45
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Marks
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Mid-point
(𝒎𝒊 )
5
15
25
35
45
55
65
Afzal Hossain (Assistant Professor)
No. of
𝒇𝒊 𝒎𝒊
Students (𝒇𝒊 )
4
20
8
120
11
275
15
525
12
540
6
330
2
130
n
i=1 fi mi = 1940
13
DIRECT METHOD
SOME EXAMPLES
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Afzal Hossain (Assistant Professor)
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ARITHMETIC MEAN
Question 1: Marks obtained by the 30 students of BBA-06 in
Business Statistics course-
60, 65, 76, 73, 73, 90, 78, 81, 62, 92, 75, 73, 66, 66, 68, 71, 74,
83, 78, 80, 86, 61, 87, 92, 87, 86, 81, 77, 74, 83
Now, Find out the mean using Direct method
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ARITHMETIC MEAN
Solution: Arithmetic mean using Direct method
Marks
Mid-point (𝒎𝒊 )
No. of Students (𝒇𝒊 )
𝒇𝒊 𝒎𝒊
60-64
62
3
186
65-69
67
4
268
70-74
72
6
432
75-79
77
5
385
80-84
82
5
410
85-89
87
4
348
90-94
92
3
276
Σfi =30
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Afzal Hossain (Assistant Professor)
n
i=1 fi mi =2305
16
ARITHMETIC MEAN
Here n=30
μ=
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
𝑛
=
2305
30
Or, μ = 76.833
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Afzal Hossain (Assistant Professor)
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ARITHMETIC MEAN
Question 2: The monthly income of 10 employees working in a
firm is as follows:
4493 4502 4446 4475 4492 4572 4516 4468 4489 4520
Calculating average monthly income by direct method.
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Afzal Hossain (Assistant Professor)
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ARITHMETIC MEAN
Marks
No. of Students (𝒇𝒊 )
Mid-point (𝒎𝒊 )
4440-4459
4460-4479
4480-4499
4500-4519
4520-4539
4540-4559
4560-4579
1
2
3
2
1
0
1
n=10
4449.5
4469.5
4489.5
4509.5
4529.5
4549.5
4569.5
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Afzal Hossain (Assistant Professor)
𝒇𝒊 𝒎𝒊
4449.5
8939
13468.5
9019
4529.5
0
4569.5
n
i=1 fi mi = 44975
19
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Direct Method) Solution:
Here n=10
Known that,μ =
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
𝑛
44975
=
10
Or, μ =4497.5
So, Average monthly income by using direct method is = 4497.5
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Afzal Hossain (Assistant Professor)
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ARITHMETIC MEAN
Question 3: A contractor employs three types of works- male, female
and children. To a male he pays Rs. 200 per day, to a female worker Rs.
150 per day and to child worker Rs. 100 per day.
Now, Find out the mean by using Direct method
Answer:
𝒇𝒊 𝒎𝒊
Salary
Mid-point (𝒎𝒊 )
No. of people (𝒇𝒊 )
200
150
100
100
75
50
20
15
5
2000
1125
250
Total =
-
40
3375
ARITHMETIC MEAN
Now 𝜇 =
=
𝑛
𝑓𝑚
𝑖=1 𝑖 𝑖
𝑛
3375
40
= 84.375 (approx.)
ARITHMETIC MEAN
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Find out the mean using
direct method
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Age (yrs.)
No. of
workers
Age (yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
Afzal Hossain (Assistant Professor)
23
ARITHMETIC MEAN
Solutions-Arithmetic mean
using direct method:
Age (yrs.)
Mid-point (𝒎𝒊 )
No. of workers (𝒇𝒊 )
𝒇𝒊 𝒎𝒊
18-22
20
120
2400
22-26
24
125
3000
26-30
28
280
7840
30-34
32
260
8320
34-38
36
155
5580
38-42
40
184
7360
42-46
44
162
7128
46-50
48
86
4128
50-54
52
75
3900
54-58
56
53
2968
N= 1500
n
i=1 fi mi =
52624
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Afzal Hossain (Assistant Professor)
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24
ARITHMETIC MEAN
Here,
m= Mid-point of various classes,
f= Frequency of each class,
N= Total number of frequencies =1500
Class intervene= 4 and we know
μ=
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
𝑛
52624
=
= 35.08
1500
So, the average age of the workers is 35.08
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Afzal Hossain (Assistant Professor)
25
ARITHMETIC MEAN
Question 5: Given below is the
frequency distribution of the profit
obtained by 250 companies. Find out
arithmetic mean using Direct
Method.
10/2/2021
Profits (in BDT.)
10000-20000
20000-30000
30000-40000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000-100000
Total
No. of companies
15
35
47
68
32
22
12
11
8
250
26
ARITHMETIC MEAN
Mid Point
(𝒎𝒊 )
10000-20000
15000
15
225000
20000-30000
30000-40000
25000
35000
35
47
875000
1645000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000100000
45000
55000
65000
75000
85000
95000
10/2/2021
No of companies
(𝒇𝒊 )
𝒇𝒊 𝒎𝒊
Profits
in(BDT)
68
32
22
12
11
8
Afzal Hossain (Assistant Professor)
n=250
3060000
1760000
1430000
900000
935000
760000
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖 = 11590000
27
ARITHMETIC MEAN
Grouped Data Arithmetic Mean(Direct Method):
Solution: Here, n = 250
µ=
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
𝑛
11590000
=
250
=46360
⸫Arithmetic mean = 46360
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Afzal Hossain (Assistant Professor)
28
ARITHMETIC MEAN
Question 6: Here given the
following Incomplete frequency
distribution.
Find
out
the
arithmetic mean using Direct
Method.
10/2/2021
Sales
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
Total
Afzal Hossain (Assistant Professor)
No. of Companies
5
17
80
468
326
7
88
9
1000
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Direct Method): Solution:
Sales
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
10/2/2021
Mid-point (𝒎𝒊 )
No. of Companies (𝒇𝒊 )
312.5
337.5
362.5
387.5
412.5
437.5
462.5
487.5
5
17
80
468
326
7
88
9
Afzal Hossain (Assistant Professor)
𝒇𝒊 𝒎𝒊
1562.5
5737.5
29000
181350
134475
3062.5
40700
4387.5
n
i=1 fi mi = 400275
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Direct Method):
Solution:
Here n=1000
μ=
10/2/2021
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
𝑛
400275
=
= 400.275
1000
Afzal Hossain (Assistant Professor)
ARITHMETIC MEAN
Question 8: Find the
arithmetic mean from the
given data using Direct
method
Weekly Rent (in
Rs)
200-400
No. of Persons
Paying the Rent
6
400-600
600-800
800-1000
9
11
14
1000-1200
1200-1400
20
15
1400-1600
1600-1800
1800-2000
10
8
7
ARITHMETIC MEAN
Weekly
Rent
200-400
400-600
600-800
Mid-Point (𝒎𝒊 ) No. of Persons Paying
the Rent(𝒇𝒊 )
300
6
500
9
700
11
800-1000
1000-1200
1200-1400
900
1100
1300
14
20
15
1400-1600
1600-1800
1800-2000
1500
1700
1900
10
8
7
N=100
𝒇𝒊 𝒎𝒊
1800
4500
7700
12600
22000
19500
15000
13600
13300
n
i=1 fi mi =110,000
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Direct Method):
Solution:
Here, n=100
𝜇=
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
𝑛
=
110,000
100
= 1100
ARITHMETIC MEAN
Question 9: Consider the following data and find the
arithmetic mean using shortcut Direct Method.
10/2/2021
Class
10-20
20-30
30-40 40-50
50-60
60-70
70-80
Frequenc
y
12
30
33
45
25
18
65
Afzal Hossain (Assistant Professor)
35
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Direct Method):
Class
Mid-point
No. of frequency
Solution:
(𝒇𝒊 )
𝒇𝒊 𝒎𝒊
(𝒎𝒊 )
10/2/2021
10-20
15
12
180
20-30
25
30
750
30-40
35
33
1155
40-50
45
65
2925
50-60
55
45
2475
60-70
65
25
1625
70-80
75
18
1350
N=228
10460
Afzal Hossain (Assistant Professor)
36
ARITHMETIC MEAN
Grouped Data-arithmetic
Method): Solution:
Here n=228
µ=
𝑛
𝑖=1 𝑓𝑖𝑚𝑖
𝑛
=
10460
=
228
Mean
(Direct
45.88
⸫ Arithmetic mean = 45.88
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Afzal Hossain (Assistant Professor)
37
ARITHMETIC MEAN
QUESTION 10: Given below
is the frequency distribution of
the marks obtained by 90 student.
Find out arithmetic mean using
direct method.
MARKS
20-29
30-39
NO. OF THE STUDENTS
2
12
40-49
50-59
60-69
15
20
18
70-79
80-89
10
9
90-99
4
Afzal Hossain (Assistant Professor)
ARITHMETIC MEAN
Grouped Data-arithmetic
Mean (Direct Method):
Solution:
Marks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
Mid-point
(𝒎𝒊 )
24.5
No. of the
Students (𝒇𝒊 )
2
34.5
44.5
54.5
12
15
20
64.5
74.5
84.5
18
10
9
94.5
4
n = 90
Afzal Hossain (Assistant Professor)
(𝒇𝒊 𝒎𝒊 )
49
414
667.5
1090
1161
745
760.5
378
𝑛
𝑖=1 fimi= 5265
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Direct Method):
Solution:
Solution:
Here, n = 90
𝑛
5265
𝑖=1 fimi
μ=
=
= 58.5
𝑛
90
∴ Arithmetic mean = 58.5
Afzal Hossain (Assistant Professor)
SHORTCUT METHOD
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Afzal Hossain (Assistant Professor)
41
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Shortcut
Method): The formula for calculation of the
arithmetic mean by the short-cut method is given
below:
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
μ=𝐴+
is used. Here, d is deviation from
𝑛
the arbitrary or assumed mean, f is the frequency of
each class, A is arbitrary or assumed mean and n is
the total number of frequencies.
The given table gives the marks of 58 students in
Statistics. Calculate the average marks of this group.
10/2/2021
Afzal Hossain (Assistant Professor)
Marks No. of
Students
0-10 4
10-20 8
20-30 11
30-40 15
40-50 12
50-60 6
60-70 2
Total 58
42
ARITHMETIC MEAN
Grouped Data-arithmetic
Mean (Shortcut Method):
Solution: Here n=58, A=35
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
μ=𝐴+
𝑛
−90
= 35+
58
Or, μ =35-1.55
Or, μ=33.45
10/2/2021
Marks
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Mid-point
(𝒎𝒊 )
5
15
25
35
45
55
65
Afzal Hossain (Assistant Professor)
No. of
Students
(𝒇𝒊 )
4
8
11
15
12
6
2
Deviation 𝒇𝒊 𝒅𝒊
(𝒅𝒊 )
-30
-20
-10
0
10
20
30
-120
-160
-110
0
120
120
60
n
i=1 fi 𝑑i = -90
43
SHORTCUT METHOD
SOME EXAMPLES
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Afzal Hossain (Assistant Professor)
44
ARITHMETIC MEAN
Question 1: Marks obtained by the 30 students of
BBA-06 in Business Statistics course-
60, 65, 76, 73, 73, 90, 78, 81, 62, 92, 75, 73, 66, 66,
68, 71, 74, 83, 78, 80, 86, 61, 87, 92, 87, 86, 81, 77,
74, 83
Now, Find out the mean using shortcut method
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Afzal Hossain (Assistant Professor)
45
ARITHMETIC MEAN
Solution-Arithmetic mean using short-cut method:
Marks
60-64
65-69
70-74
75-79
80-84
85-89
90-94
10/2/2021
Mid-point (mi)
62
67
72
77
82
87
92
No. of people (fi)
3
4
6
5
5
4
3
Σfi =30
Afzal Hossain (Assistant Professor)
Deviation (di )
-15
-10
-5
0
5
10
15
fi di
-45
-40
-30
0
25
40
45
Σfi di =-5
46
ARITHMETIC MEAN
We know,
−5
30
μ= 77 +
= 76.833
Ans: 76.84 (approx.)
10/2/2021
Here,
μ = Mean
f = Frequency
d = Deviation
n = Total number of frequency=30
A= Arbitrary or assumed mean=77
Afzal Hossain (Assistant Professor)
47
ARITHMETIC MEAN
Question 2: The monthly income 0f 10 employees
working in a firm is as follows (Average short cut) 4487
4493 4502 4446 4475 4492 4572 4516 4468 4489
Calculating average monthly income by short cut method.
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Afzal Hossain (Assistant Professor)
48
ARITHMETIC MEAN
Solution:
Here n = 10, A = 4520
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
−126
=4509+
10
μ=𝐴+
𝑛
=4496.4
average monthly income by
short cut method is 4496.4
10/2/2021
Monthly
Income
4446-4488
4488-4530
4530-4572
No. of
Deviation Mid-point fidi
Employ (di)
(mi)
(fi)
4
-42
4467
-168
5
0
4509
0
1
42
4551
42
Σfi =10
Σfi di = -126
Afzal Hossain (Assistant Professor)
49
ARITHMETIC MEAN
Question 3: A contractor employs three types of works- male,
female and children. To a male he pays Rs. 200 per day, to a
female worker Rs. 150 per day and to child worker Rs. 100 per
day. Now, Find out the mean by using shortcut method
10/2/2021
Afzal Hossain (Assistant Professor)
50
ARITHMETIC MEAN
Solutions-Arithmetic mean using shortcut method:
Salary
200
150
100
10/2/2021
Mid-point (mi)
100
75
50
No. of
people
(fi)
20
15
5
n=40
Deviati
on (di )
-1
0
1
Afzal Hossain (Assistant Professor)
fi di
-20
0
5
Σfi di =-15
51
ARITHMETIC MEAN
We know,
Here,
μ = Mean
f = Frequency
d = Deviation
n = Total number of frequency
A= Arbitrary or assumed mean
−15
40
μ= 75 +
= 74.625
Ans: 74.625 (approx.)
10/2/2021
Afzal Hossain (Assistant Professor)
52
ARITHMETIC MEAN
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Age
(yrs.)
Find out the mean using
shortcut method
10/2/2021
No. of
workers
Age
(yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
Afzal Hossain (Assistant Professor)
53
ARITHMETIC MEAN
Solutions-Arithmetic mean
using shortcut method:
Age (yrs.)
Mid-point
(mi)
No. of workers
(fi)
18-22
20
120
22-26
24
125
-12
-1500
280
-8
-2240
26-30
28
fidi
-1920
30-34
32
260
-4
-1040
34-38
36
155
0
0
38-42
40
184
4
736
42-46
44
162
8
1296
46-50
48
86
12
1032
50-54
52
75
16
1200
54-58
56
53
20
1060
N= 1500
10/2/2021
Deviation
(di)
-16
Afzal Hossain (Assistant Professor)
Σ fi𝑑 ′ i= -1376
54
54
ARITHMETIC MEAN
Here,
m= Mid-point of various classes, d= Deviation from the arbitrary or assumed
mean, f= Frequency of each class, A= Arbitrary or assumed mean =36, N= Total
number of frequencies =1500 and Class intervene= 4 and we know
μ=𝐴+
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
𝑛
−1376
μ= 36+
1500
Or,
Or, μ=35.083
So, the average age of the workers is 36.08
10/2/2021
Afzal Hossain (Assistant Professor)
55
ARITHMETIC MEAN
Question 5: Given below is
the frequency distribution of
the profit obtained by 250
companies.
Find
out
arithmetic
mean
using
Short-cut Method.
10/2/2021
Profits (in BDT.)
10000-20000
20000-30000
30000-40000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000-100000
Afzal Hossain (Assistant Professor)
No. of companies
15
35
47
68
32
22
12
11
8
56
ARITHMETIC MEAN
Solutions-Arithmetic
mean using shortcut
method:
Profits
in(BDT)
Mid Point
(mi)
No of
companies
(fi)
f id i
10000-20000 15000
15
-40000
-600000
20000-30000 25000
35
-30000
-1050000
30000-40000 35000
47
-20000
-940000
40000-50000 45000
68
-10000
-680000
50000-60000 55000
32
0
0
60000-70000 65000
22
10000
220000
70000-80000 75000
12
20000
240000
80000-90000 85000
11
30000
330000
90000100000 95000
8
40000
320000
n=250
10/2/2021
Deviation
(di)
Afzal Hossain (Assistant Professor)
𝑓𝑖𝑑𝑖 = 2160000
57
ARITHMETIC MEAN
Here, No. of companies (n) =250 and A=55000
We know, µ = A+
𝑛
𝑖=1 𝑓𝑖𝑑𝑖
𝑛
=55000 +
−2160000
250
= 55000 – 8640
= 46360
∴ Arithmetic mean (short-cut method)= 46360
10/2/2021
Afzal Hossain (Assistant Professor)
58
ARITHMETIC MEAN
Question 6: Here given the
following Incomplete frequency
distribution.
Find
out
the
arithmetic mean using short cut
method.
10/2/2021
Sales
320-325
325-330
330-335
335-340
340-345
345-350
350-355
355-360
Total
Afzal Hossain (Assistant Professor)
No. of Companies
5
17
80
468
326
7
88
9
1000
59
2
ARITHMETIC MEAN
Solutions-Arithmetic mean
using shortcut method:
Here n=1000, A=337.5
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
μ=𝐴+
𝑛
2555
μ = 337.5+
1000
Or, μ=340.05
10/2/2021
Sales
320-325
325-330
330-335
335-340
340-345
345-350
350-355
355-360
Mid-point
No. of
Deviation
fidi
(mi)
Companies
(di)
(fi)
322.5
5
-15
-75
327.5
17
-10
-170
332.5
80
-5
-400
337.5
468
0
0
342.5
326
5
1630
347.5
7
10
70
352.5
88
15
1320
357.5
9
20
180
n
n = 1000
i=1 fi 𝑑i = 2555
Afzal Hossain (Assistant Professor)
60
ARITHMETIC MEAN
Question 7: Find the
Weekly Income(BDT) No. of employees
arithmetic mean from the
given data using short cut
method
500-550
550-600
600-650
650-700
700-750
750-800
800-850
10/2/2021
Afzal Hossain (Assistant Professor)
6
10
22
30
16
12
15
61
ARITHMETIC MEAN
Solutions-Arithmetic mean
using shortcut method:
H
ere n=111, A=675
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
μ=𝐴+
𝑛
1250
Or, μ=675+
111
Or, μ =675+11.3
Weekly
No. of
Income(BDT) employees(fi)
Deviation
(di)
f id i
500- 550
6
525
-150
-900
550-600
10
575
-100
-1000
600-650
22
625
-50
-1100
650-700
30
675
0
0
700-750
16
725
50
800
750-800
12
775
100
1200
800-850
15
825
150
2250
n=111
n
fi 𝑑i = 1250
Or, μ=686.3
10/2/2021
mi
i=1
Afzal Hossain (Assistant Professor)
62
ARITHMETIC MEAN
Question 8: Find the
arithmetic mean from the
given data using short cut
method
10/2/2021
Weekly Rent (in
Rs)
200-400
400-600
No. of Persons
Paying the Rent
6
9
600-800
11
800-1000
1000-1200
14
20
1200-1400
1400-1600
15
10
1600-1800
1800-2000
8
7
Afzal Hossain (Assistant Professor)
63
ARITHMETIC MEAN
Solutions-Arithmetic mean
using shortcut method:
Here, Total frequency n=100, and
A=1100
We know, µ = A+
=1100+
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
𝑛
0
100
= 1100
∴ Arithmetic mean =1100
10/2/2021
Weekly
Rent
(in BDT)
No. of Persons
Paying the Rent
(𝒇𝒊 )
mi
Deviation
(𝒅𝒊 )
𝒇𝒊 𝒅𝒊
200-400
6
300
-800
-4800
400-600
9
500
-600
-5400
600-800
11
700
-400
-4400
800-1000
14
900
-200
-2800
1000-1200
20
1100
0
0
1200-1400
15
1300
200
3000
1400-1600
10
1500
400
4000
1600-1800
8
1700
600
4800
1800-2000
7
1900
800
5600
n=100
Afzal Hossain (Assistant Professor)
∑ 𝑓𝑖 𝑑𝑖 =0
64
ARITHMETIC MEAN
Question 9: Consider the following data and Find the arithmetic
mean using Shortcut Method.
Class
10-20 20-30
Frequency 12
30
10/2/2021
30-40
33
40-50
65
Afzal Hossain (Assistant Professor)
50-60
45
60-70
25
70-80
18
65
ARITHMETIC MEAN
Solutions-Arithmetic
mean
using
shortcut
method:
Here, Total frequency
n =228, A=45.
We know,
µ = A+
= 45 +
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
𝑛
200
=
228
45.877
variable Mid-point
(𝑚𝑖 )
Deviation
(𝒅𝒊 )
𝒇𝒊 𝒅𝒊
10-20
15
12
-30
-360
20-30
25
30
-20
-600
30-40
35
33
-10
-330
40-50
45
65
0
0
50-60
55
45
10
450
60-70
65
25
20
500
70-80
75
18
30
540
N=228
∴ Arithmetic mean = 46.75
10/2/2021
No. of
frequency
(fi)
Afzal Hossain (Assistant Professor)
∑ 𝑓𝑖 𝑑𝑖 =200
66
ARITHMETIC MEAN
Question10: Given below is the
frequency distribution of the marks
obtained by 90 student. Find out
arithmetic mean in, 1. Short-cut
Method & 2. Step-deviation Method.
10/2/2021
Afzal Hossain (Assistant Professor)
Marks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
No. of the students
2
12
15
20
18
10
9
4
67
ARITHMETIC MEAN
Solutions-Arithmetic mean
using shortcut method:
Here, No. of students n =90
and A=54.5, We know,
µ = A+
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
µ = 54.5+
𝑛
360
=54.5+4=58.5
90
∴ Arithmetic mean (short-cut
method)= 58.5
10/2/2021
Marks
20-29
30-39
MidNo. of
Deviation
point(mi) students(fi)
(di)
24.5
2
-30
34.5
12
-20
fidi
-60
-240
40-49
50-59
60-69
44.5
54.5
64.5
15
20
18
-10
0
10
-150
0
180
70-79
80-89
74.5
84.5
10
9
20
30
200
270
90-99
94.5
4
40
160
N=90
Afzal Hossain (Assistant Professor)
∑fidi=360
68
SETP DEVIATION METHOD
10/2/2021
Afzal Hossain (Assistant Professor)
69
ARITHMETIC MEAN
Grouped Data-arithmetic Mean (Step-deviation
Method): The formula for calculation of the
arithmetic mean by the Step-deviation Method is
given below:
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
μ=𝐴+
×C is used. Here m is mid-point of
𝑛
various classes, d is deviation from the arbitrary or
assumed mean, f is the frequency of each class, A is
arbitrary or assumed mean, C is class interval and n is
the total number of frequencies.
The given table gives the marks of 58 students in
Statistics. Calculate the average marks of this group.
10/2/2021
Afzal Hossain (Assistant Professor)
Marks No. of
Students
0-10 4
10-20 8
20-30 11
30-40 15
40-50 12
50-60 6
60-70 2
Total 58
70
ARITHMETIC MEAN
Grouped
Dataarithmetic Mean(Stepdeviation Method):
Solution: Here n=58,
A=35
μ=𝐴+
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
𝑛
×𝐶
−9
𝑂𝑟, μ =35+
×10
58
Or, μ =35-.155×10
Or, μ =35-1.55
Or, μ=33.45
10/2/2021
Marks
0-10
10-20
20-30
30-40
40-50
50-60
60-70
Midpoint (mi)
5
15
25
35
45
55
65
No. of
Students (fi)
4
8
11
15
12
6
2
Afzal Hossain (Assistant Professor)
𝒅i
-3
-2
-1
0
1
2
3
f i𝒅 i
-12
-16
-11
0
12
12
6
n
i=1 fi 𝑑i = -9
71
SETP DEVIATION METHOD
SOME EXAMPLES
10/2/2021
Afzal Hossain (Assistant Professor)
72
ARITHMETIC MEAN
Question 1: Marks obtained by the 30 students of BBA-06
in Business Statistics course60, 65, 76, 73, 73, 90, 78, 81, 62, 92, 75, 73, 66, 66, 68, 71,
74, 83, 78, 80, 86, 61, 87, 92, 87, 86, 81, 77, 74, 83
Now, Find out the mean using step-deviation method
10/2/2021
Afzal Hossain (Assistant Professor)
73
ARITHMETIC MEAN
Solution-Arithmetic mean using step-deviation method:
Marks
Mid-point (mi)
No. of people
(fi)
Step deviation
(di)
60-64
65-69
70-74
75-79
80-84
85-89
90-94
62
67
72
77
82
87
92
3
4
6
5
5
4
3
-3
-2
-1
0
1
2
3
N=30
10/2/2021
Afzal Hossain (Assistant Professor)
f i di
-9
-8
-6
0
5
8
9
Σfi d′i =-1
74
ARITHMETIC MEAN
We know,
Here,
μ = Mean
f = Frequency
d′ = Step deviation
n = Total number of frequency
A= Arbitrary or assumed mean
C = Class Interval
′
μ= 77 +
−1
30
× 5 = 76.833
Ans: 76.84 (approx.)
10/2/2021
Afzal Hossain (Assistant Professor)
75
ARITHMETIC MEAN
Question 2: The monthly income of 10 employees
working in a firm is as follows:
4493 4502 4446 4475 4492 4572 4516 4468 4489
Calculating average monthly income by step-deviation
method.
10/2/2021
Afzal Hossain (Assistant Professor)
76
ARITHMETIC MEAN
Solution:
Here n= 10, d = 340 and
A= 4460
μ=A+
n
i=1 fi di
n
−3
+
10
= 4509
=4496.4
×C
× 42
Monthly
Income
4446-4488
4488-4530
4530-4572
No. of
Employ (fi)
4
5
1
n=10
𝒅i
-1
0
1
Midfidi
point (mi)
4467
-4
4509
0
4551
1
Σfi di =-3
average monthly income by
using step-deviation method
is 4496.4
10/2/2021
Afzal Hossain (Assistant Professor)
77
ARITHMETIC MEAN
Question 3: A contractor employs three types of works- male,
female and children. To a male he pays Rs. 200 per day, to a
female worker Rs. 150 per day and to child worker Rs. 100 per
day.
Now, Find out the mean by using step-deviation method
10/2/2021
Afzal Hossain (Assistant Professor)
78
ARITHMETIC MEAN
Solutions: Arithmetic mean using step-deviation method:
Salary
200
150
100
10/2/2021
Mid-point (mi) No. of people Deviation (di )
(fi)
100
20
-1
75
15
0
50
5
1
n=31
Afzal Hossain (Assistant Professor)
fi di
-20
0
5
Σfi d′i =-15
79
ARITHMETIC MEAN
We know,
μ= 75 +
=56.25
−15
×
40
Here,
μ = Mean
f = Frequency
d′ = Step deviation
n = Total number of frequency
A= Arbitrary or assumed mean
C = Class Interval
50
Ans: 𝟓𝟔. 𝟐𝟓(approx.)
10/2/2021
Afzal Hossain (Assistant Professor)
80
ARITHMETIC MEAN
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Find out the mean using
step-deviation method.
10/2/2021
Age (yrs.)
No. of
workers
Age (yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
Afzal Hossain (Assistant Professor)
81
ARITHMETIC MEAN
Solutions:
Arithmetic mean
using step-deviation
method:
Mid-point
(mi)
No. of workers
(fi)
18-22
20
120
22-26
24
125
-3
-375
26-30
28
280
-2
-560
30-34
32
260
-1
-260
34-38
36
155
0
0
38-42
40
184
1
184
42-46
44
162
2
324
46-50
48
86
3
258
50-54
52
75
4
300
54-58
56
53
5
265
N= 1500
10/2/2021
Afzal Hossain (Assistant Professor)
Deviation
(di)
-4
fi𝒅i
Age
(yrs.)
-480
Σ fi𝑑 ′ i= -344
82
ARITHMETIC MEAN
Here,
A=36,
N=1500,
C= 4
n
i=1 fi 𝑑i = -344
We know,
𝑛
𝑖=1 𝑓𝑖 𝑚𝑖
μ=𝐴+
×𝐶
𝑛
−344
𝑂𝑟, μ =36+
×4
1500
Or, μ=35.08
So, the average age of the workers is 35.08
10/2/2021
Afzal Hossain (Assistant Professor)
83
ARITHMETIC MEAN
Question 5: Given below is
the frequency distribution of the
profit
obtained
by
250
companies. Find out arithmetic
mean
using
Step-deviation
Method.
10/2/2021
Profits (in BDT.)
No. of companies
10000-20000
20000-30000
30000-40000
15
35
47
40000-50000
50000-60000
60000-70000
68
32
22
70000-80000
80000-90000
90000-100000
12
11
8
Afzal Hossain (Assistant Professor)
84
ARITHMETIC MEAN
Solutions:
Arithmetic
mean
using step-deviation
method:
Profits
in(BDT)
Mid Point
(mi)
No of
companies (fi)
f id i
10000-20000
15000
15
-4
-60
20000-30000
25000
35
-3
-105
30000-40000
35000
47
-2
-94
40000-50000
45000
68
-1
-68
50000-60000
55000
32
0
0
60000-70000
65000
22
1
22
70000-80000
75000
12
2
24
80000-90000
85000
11
3
33
90000-100000 95000
8
4
32
n=250
10/2/2021
Deviation
(di)
Afzal Hossain (Assistant Professor)
𝑓𝑖𝑑𝑖 = -216
85
ARITHMETIC MEAN
Here, No. of companies (n) =250 and A=55000
We know,
µ = A+
𝑛
𝑖=1 𝑓𝑖𝑑𝑖
𝑛
× C=55000 +
−216
250
× 10000
= 55000 – 0.864 × 10000= 55000 – 8640 = 46360
∴ Arithmetic mean (short-cut method)= 46360
10/2/2021
Afzal Hossain (Assistant Professor)
86
ARITHMETIC MEAN
Question
6:
Here given the
following
Incomplete
frequency
distribution. Find out arithmetic mean
using Step-deviation Method.
10/2/2021
Sales
No. of Companies
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
5
17
80
468
326
7
88
9
Total
1000
Afzal Hossain (Assistant Professor)
87
2
ARITHMETIC MEAN
Solutions:
Arithmetic
mean
using
stepdeviation method:
Here n=1000, A=468
μ=𝐴+
n
′
i=1 fi 𝒅 i
𝑛
×𝐶
511
𝑂𝑟, μ = 387.5+
×25
1000
Or, μ=400.2
10/2/2021
Sales
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
Mid-point
No. of
(mi)
Companies (fi)
312.5
5
337.5
17
362.5
80
387.5
468
412.5
326
437.5
7
462.5
88
487.5
9
n = 1000
Afzal Hossain (Assistant Professor)
di
fi𝒅′ i
-3
-2
-1
0
1
2
3
4
-15
-34
-80
0
326
14
264
36
𝑛
′
𝑓
𝑑
𝑖=1 𝑖 𝑖 =511
88
ARITHMETIC MEAN
Question 7: Find the
Weekly Income(BDT)
No. of employees
arithmetic mean from the
given data using stepdeviation method.
500-550
550-600
600-650
650-700
700-750
750-800
800-850
6
10
22
30
16
12
15
10/2/2021
Afzal Hossain (Assistant Professor)
89
ARITHMETIC MEAN
Solutions:
Arithmetic
mean using step-deviation
method:
Here
n=111,
A=675
μ=𝐴+
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
𝑛
25
675+
111
×𝐶
𝑂𝑟, μ =
× 50
Or, μ = 675+11.3
Or, μ=686.3
10/2/2021
Weekly
Income(BDT)
mi
No. of
employees(fi)
𝒅𝒊
fi𝒅𝒊
500- 550
550-600
600-650
525
575
625
6
10
22
-3
-2
-1
-18
-20
-22
650-700
700-750
750-800
675
725
775
30
16
12
0
1
2
0
16
24
800-850
825
15
n=111
3
45
n
fi 𝑑i = 𝟐𝟓
Afzal Hossain (Assistant Professor)
i=1
90
ARITHMETIC MEAN
Question 8: Find the
arithmetic mean from the
given data using stepdeviation method.
10/2/2021
Weekly Rent (in
BDT)
200-400
No. of Persons
Paying the Rent
6
400-600
600-800
800-1000
9
11
14
1000-1200
1200-1400
20
15
1400-1600
1600-1800
1800-2000
10
8
7
Afzal Hossain (Assistant Professor)
91
ARITHMETIC MEAN
Solutions: Arithmetic mean
using step-deviation method:
Here, Total frequency n =100,
A=1100, C=200
We know, µ = A+
𝑛
′
𝑖=1 𝑓𝑖 𝑑𝑖
=1100+
𝑛
0
100
= 1100
∴ Arithmetic
mean=1100
10/2/2021
xC
x 200
Weekly Rent
(in Rs)
200-400
400-600
600-800
800-1000
1000-1200
1200-1400
1400-1600
1600-1800
1800-2000
No. of Persons
Paying the Rent (𝑓𝑖 )
6
9
11
14
20
15
10
8
7
n=100
Afzal Hossain (Assistant Professor)
Deviation
(𝑑𝑖 )
-8
-6
-4
-2
0
2
4
6
8
𝑓𝑖 𝑑𝑖
-48
-54
-44
-28
0
30
40
48
56
∑ 𝑓𝑖 𝑑𝑖′ =0
92
ARITHMETIC MEAN
Question 9: Consider the following data and Find the arithmetic mean
using Step Deviation Method.
Class
10-20 20-30
Frequency 12
30
10/2/2021
30-40
33
40-50
65
Afzal Hossain (Assistant Professor)
50-60
45
60-70
25
70-80
18
93
ARITHMETIC MEAN
Solutions: Arithmetic mean variab Mid-point
le
(𝑚𝑖 )
using step-deviation method:
15
Here, Total frequency n =228, 10-20
20-30
25
A=400, C=10
We know, µ = A+
= 45 +
20
228
𝑛
′
𝑓
𝑑
𝑖
𝑖
𝑖=1
𝑛
× 10 = 45.87
×C
𝒅𝒊
𝒇 𝒊 𝒅𝒊
-3
-36
-2
-1
0
-60
-33
0
1
2
3
45
50
54
30-40
40-50
35
45
30
33
65
50-60
60-70
70-80
55
65
75
45
25
18
∴ Arithmetic mean= 45.87
10/2/2021
No. of
frequency (fi)
12
N=228
Afzal Hossain (Assistant Professor)
∑ 𝑓𝑖 𝑑𝑖 =20
94
ARITHMETIC MEAN
Question 10: Given below is the
frequency distribution of the marks
obtained by 90 students. Find out
arithmetic mean using Step-deviation
Method.
10/2/2021
Afzal Hossain (Assistant Professor)
Marks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
No. of the students
2
12
15
20
18
10
9
4
95
ARITHMETIC MEAN
Solutions:
Arithmetic
mean using step-deviation
method: Here,
No. of
students n =90
µ = A+
= 55.5 +
𝑛
′
𝑓
𝑑
𝑖=1 𝑖 𝑖
𝑛
400
228
×C
× 10 = 73.04
Marks
Midpoint(mi)
No. of
students(fi)
di
fidi
20-29
24.5
2
-3
-6
30-39
34.5
12
-2
-24
40-49
44.5
15
-1
-15
50-59
55.5
20
0
0
60-69
65.5
18
1
18
70-79
74.5
10
2
20
80-89
84.5
9
3
27
90-99
94.5
4
4
16
N=90
10/2/2021
Afzal Hossain (Assistant Professor)
∑ fidi=36
96
BUSINESS STATISTICS
GEOMETRIC MEAN, HARMONIC MEAN &
QUADRATIC MEAN
8/14/2020
Afzal Hossain (Assistant Professor)
97
GEOMETRIC MEAN
Definition : There are two other means that are used sometimes in business
and economics. These are the geometric mean and the harmonic mean.
Geometric mean is defined at the nth root of the product of n observations of
a distribution. Symbolically,
(
1/𝑁
𝑁
𝑖=1 𝑋𝑖 )
=
𝑁
𝑋1 . 𝑋2 … … … . . 𝑋𝑁 , If we
have only two observations, say, 4 and 16 then GM = √(4 ×16) =
64 =8
have to calculate the cube root of the product of these three observations; and
so on. When the number of items is large, it becomes extremely difficult to
multiply the numbers and to calculate the root.
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98
GEOMETRIC MEAN
Example 1: What is the geometric mean of 2,3,and 6?
Solution: First, multiply the numbers together and then take the
cubed root (because there are three numbers) = (2×3×6)1/3 = 3.30
Example 2: What is the geometric mean of 4,8,3,9 and 17?
Solution: First, multiply the numbers together and then take the 5th
root (because there are 5 numbers) = (4*8*3*9*17)(1/5) = 6.81
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Afzal Hossain (Assistant Professor)
99
GEOMETRIC MEAN
Example: If we have to
find out the geometric mean
of 2, 4 and 8, then we find.
10/2/2021
log 𝑥𝑖
𝑁
Solution : log 𝐺𝑀 =
log 2 + log 4 + log 8
=
3
.3010 + .6021 + .9031
=
3
1.8062
=
3
Or, log 𝐺𝑀 = .60206
Or, GM= 𝐴𝑛𝑡𝑖 𝐿𝑜𝑔(.60206)=4
Afzal Hossain (Assistant Professor)
100
GEOMETRIC MEAN
Geometric
Mean
For Group
Data:
Geometric mean can
𝑓𝑖 . 𝐿𝑜𝑔 𝑥𝑖
𝐿𝑜𝑔 𝐺𝑀 =
𝑓1 + 𝑓2 + ⋯ … … … … + 𝑓𝑛
𝐿𝑜𝑔 𝐺𝑀
be obtained by the
formula:
𝑓1 . 𝐿𝑜𝑔 𝑥1 + 𝑓2 . 𝐿𝑜𝑔 𝑥2 + ⋯ … … … … + 𝑓𝑛 . 𝐿𝑜𝑔 𝑥𝑛
=
𝑓1 + 𝑓2 + ⋯ … … … … + 𝑓𝑛
Then, GM
10/2/2021
= Antilog
Afzal Hossain (Assistant Professor)
𝑓𝑖 .𝐿𝑜𝑔 𝑥𝑖
𝑓1 +𝑓2 +⋯…………+𝑓𝑛
101
GEOMETRIC MEAN
Example : A person has invested Rs
5,000 in the stock market. At the end
of the first year the amount has
grown to Rs 6,250; he has had a 25
percent profit. If at the end of the
second year his principal has grown
to Rs 8,750, the rate of increase is 40
percent for the year. What is the
average rate of increase of his
investment during the two years?
10/2/2021
Solution :
GM = .25 × .40 = .316
The average rate of increase as .316
or 31.6 percent.
Afzal Hossain (Assistant Professor)
102
GEOMETRIC MEAN
Example: An economy has grown at 5 percent in the first year, 6 percent
in the second year, 4.5 percent in the third year, 3 percent in the fourth
year and 7.5 percent in the fifth year. What is the average rate of growth
of the economy during the five years?
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Afzal Hossain (Assistant Professor)
103
GEOMETRIC MEAN
GM = Antilog
= Antilog
Rate of
Growth
(present)
Value at the end of
the -Year x (in BDT)
(xi )
Log xi
Log 𝑥𝑖
𝑛
Year
10.10987
1
5
105
2.02119
5
2
6
106
2.02531
3
4.5
104.5
2.01912
4
3
103
2.01284
5
7.5
107.5
2.03141
= Antilog 2.021974
= 105.19
The average rate of growth during the fiveyear period is 105.19 - 100 = 5.19 percent
per annum. In case of a simple arithmetic
average, the corresponding rate of growth
would have been 5.2 percent per annum.
10/2/2021
Afzal Hossain (Assistant Professor)
Log 𝑥𝑖
= 10.10987
104
GEOMETRIC MEAN
Example: An economy has grown at 5 percent in the last 2 years, 6
percent in the next 3 Years , 4.5 percent in the next 4 years , 3 percent
in the next 5 years and 7.5 percent in the next 6 years. What is the
average rate of growth of the economy during the 20 years?
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Afzal Hossain (Assistant Professor)
105
GEOMETRIC MEAN
Solution :
Rate of Growth
(present)
5
6
4.5
3
7.5
Value at the end of the -Year x
(in BDT) (xi )
105
106
104.5
103
107.5
Year
(fi )
2
3
4
5
6
Log xi
2.02119
2.02531
2.01912
2.01284
2.03141
𝑓𝑖 . Log 𝑥𝑖
10/2/2021
Afzal Hossain (Assistant Professor)
fi . Log xi
4.0424
6.0759
8.0765
10.064
12.188
40.447
106
GEOMETRIC MEAN
GM
= Antilog
= Antilog
𝑓𝑖 .Log 𝑥𝑖
40.447
= Antilog
𝑓1 +𝑓2 +𝑓3 +𝑓4 +𝑓5
2+3+4+5+6
40.447
=105.281
20
The average rate of growth during the five-year period is 105.281 - 100 =
5.281 percent per annum.
10/2/2021
Afzal Hossain (Assistant Professor)
107
GEOMETRIC MEAN
Example: Given below is the
frequency distribution of the marks
obtained by 90 students. Find out
Geometric mean.
10/2/2021
Afzal Hossain (Assistant Professor)
Marks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
No. of the students
2
12
15
20
18
10
9
4
108
GEOMETRIC MEAN
Solution:
Marks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
No. of the
students (𝒇𝒊 )
2
12
15
20
18
10
9
4
Mid Point (𝒙𝒊 )
𝐋𝒐𝒈 𝒙𝒊
𝒇𝒊 . 𝐋𝒐𝒈 𝒙𝒊
24.5
34.5
44.5
54.5
64.5
74.5
84.5
94.5
1.389
1.537
1.648
1.736
1.809
1.874
1.926
1.975
2.778
18.444
24.72
34.72
32.562
18.72
17.34
7.9
157.178
𝑓𝑖 . Log 𝑥𝑖
10/2/2021
Afzal Hossain (Assistant Professor)
109
GEOMETRIC MEAN
Solution:
GM
𝑓𝑖 .Log 𝑥𝑖
= Antilog
𝑓1 +𝑓2 +𝑓3 +𝑓4 +𝑓5
157.178
=Antilog
90
= Antilog 1.746 =55.71
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Afzal Hossain (Assistant Professor)
110
GEOMETRIC MEAN
ADVANTAGES OF G. M.
•
•
•
LIMITATIONS OF G. M.
Geometric mean is based on each
and every observation in the data
set.
It is rigidly defined.
It is more suitable while averaging
ratios and percentages as also in
calculating growth rates.
10/2/2021
•
•
•
As compared to the arithmetic mean,
geometric mean is difficult to
understand.
Both computation of the geometric
mean and its interpretation are rather
difficult.
When there is a negative item in a
series or one or more observations
have zero value, then the geometric
mean cannot be calculated.
Afzal Hossain (Assistant Professor)
111
BUSINESS STATISTICS
EXAMPLES OF GEOMETRIC MEAN
8/14/2020
Afzal Hossain (Assistant Professor)
112
GEOMETRIC MEAN
Question 1: Marks obtained by the 30 students of BBA-06 in
Business Statistics course- 60, 65, 76, 73, 73, 90, 78, 81, 62,
92, 75, 73, 66, 66, 68, 71, 74, 83, 78, 80, 86, 61, 87, 92, 87,
86, 81, 77, 74, 83. Now, Find out the Geometric mean.
8/15/2020
Afzal Hossain (Assistant Professor)
113
GEOMETRIC MEAN
Solution: Geometric Mean Calculation
We know, 𝑮𝑴 = (𝒙𝟏 . 𝒙𝟐 . 𝒙𝟑 … . . … … 𝒙𝑵 )
Here,
𝑥 = Individual Terms
N = Number of numbers or terms
𝟏
𝑵
G= (60. 65. 76. 73. 73. 90. 78. 81. 62. 92. 75. 73. 66. 66. 68. 71.
1
30
74. 83. 78. 80. 86. 61. 87. 92. 87. 86. 81. 77. 74. 83)
= 𝟕𝟔. 𝟎𝟔
8/15/2020
Afzal Hossain (Assistant Professor)
114
GEOMETRIC MEAN
Question 2: The monthly income of 10 employees working in a
firm is as follows:
4487 4493 4502 4446 4475 4492 4572 4516 4468 4489.
Calculate the GEOMETRIC MEAN.
GEOMETRIC MEAN
Solution: Here n= 10
Log GM=
Log
Log
log 𝑥𝑖
𝑛
log 4487+𝑙𝑜𝑔4493+𝑙𝑜𝑔4502+𝑙𝑜𝑔4446+𝑙𝑔4475+𝑙𝑜𝑔4492+𝑙𝑜𝑔4572+𝑙𝑜𝑔4516+𝑙𝑜𝑔4468+𝑙𝑜𝑔4489
GM=
10
3.65+3.667+3.64+3.650+3.652+3.660+3.654+3.650+3.652
GM==
10
32.87
GM= =
10
Log
GM=Antilog 3.28
=1905.46 (approx.)
GEOMETRIC MEAN
Question 3: A contractor employs three types of worksmale, female and children. To a male he pays Rs. 200 per
day, to a female worker Rs. 150 per day and to child
worker Rs. 100 per day. What is the average wage per day
paid by the contractor?
10/2/2021
Afzal Hossain (Assistant Professor)
GEOMETRIC MEAN
Solution: Calculation of GEOMETRIC MEAN
Here n= 3
Log GM =
Log GM
log 𝑥𝑖
𝑛
log 200+𝑙𝑜𝑔150+𝑙𝑜𝑔100
=
3
2.30+2.17+2
=
3
Log GM
Log GM =2.1567
GM= Antilog (2.1567) = 143.44 (approx.)
10/2/2021
Afzal Hossain (Assistant Professor)
Illustration 9(A)
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Age
(yrs.)
Find out the geometric
mean for this group data.
No. of
workers
Age
(yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
119
GEOMETRIC MEAN
Solution : Calculation
of Geometric Mean
through Group Data
Age (yrs.)
(𝒙𝒊 )
(fi)
𝐋𝒐𝒈 𝒙𝒊
𝒇𝒊 . 𝐋𝒐𝒈 𝒙𝒊
18-22
20
120
1.301
156.12
22-26
24
125
1.380
172.5
26-30
28
280
1.447
405.16
30-34
32
260
1.505
391.3
34-38
36
155
1.556
241.18
38-42
40
184
1.602
294.768
42-46
44
162
1.643
266,166
46-50
48
86
1.681
144.566
50-54
52
75
1.716
128.7
54-58
56
53
1.748
97.888
𝒏
𝒊=𝟏 𝒇𝒊 =1500
𝑓𝑖 . Log 𝑥𝑖 =
2298.348
120
GEOMETRIC MEAN
GM
= Antilog
= Antilog
𝑓𝑖 .Log 𝑥𝑖
𝒏 𝒇
𝒊=𝟏 𝒊
2298.348
1500
= Antilog 1.5322
=34.05
121
GEOMETRIC MEAN
Question
5:
Given
below is the frequency
distribution of the profit
obtained
by
250
companies. Find out the
Geometric mean.
Profits (in BDT.)
10000-20000
20000-30000
No. of companies
15
35
30000-40000
40000-50000
50000-60000
47
68
32
60000-70000
70000-80000
80000-90000
22
12
11
90000-100000
8
GEOMETRIC MEAN
Profits in
(BDT )
We know,
10000-20000
15000
15
4.176
62.64
20000-30000
25000
35
4.398
153.93
30000-40000
35000
47
4.544
213.57
40000-50000
45000
68
4.653
316.40
50000-60000
55000
32
4.740
151.68
60000-70000
65000
22
4.813
105.89
70000-80000
75000
12
4.875
58.50
80000-90000
85000
11
4.929
54.22
90000-100000 95000
8
4.978
32.82
GM=Antilog
∑fi logxi
𝒏 𝒇
𝒊=𝟏 𝒊
1149.65
GM= Antilog
250
= 39719.154
Mid Point No of companies
(xi)
(fi)
𝒏
𝒊=𝟏 𝒇𝒊 =250
Log xi
𝒇𝒊 . 𝐋𝒐𝒈 𝒙𝒊
Solution:
∑𝒇𝒊 . 𝐋𝒐𝒈 𝒙𝒊 = 1149.65
GEOMETRIC MEAN
Question 6: Here given the
following Incomplete frequency
distribution. It is known as the total
frequency is 1,000 . Find out the
Geometric Mean.
10/2/2021
Sales
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
Total
No. of Companies
5
17
80
468
326
7
88
9
1000
2
GEOMETRIC MEAN
Solution:
Sales
Here, Geometric mean,
GM=Antilog
log(𝑥𝑖)
𝑓𝑖
2601.348
GM=Antilog(
100
GM=1.023 (approx.)
10/2/2021
)
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
Mid-point
(mi)
312.5
337.5
362.5
387.5
412.5
437.5
462.5
487.5
No. of
Companies (fi)
5
17
8
46
32
7
8
9
𝑓𝑖= 100
𝒇𝒊 𝐋𝒐𝒈 𝒙𝒊
12.474
42.981
204.745
1211.311
852.628
18.487
234.53
24.19
log(𝑥𝑖)=2601.348
125
GEOMETRIC MEAN
Question 7: Find out the
Mean by using direct data
from the grouped data.
Find out the Geometric
mean
Weekly Wage (Rs)
1200 - 1400
1400 - 1600
1600 - 1800
1800 - 2000
2000 - 2200
2200 - 2400
2400 - 2600
2600 - 2800
2800 - 3000
3000- 3200
No of workers (fi)
8
12
20
30
40
32
18
7
6
4
GEOMETRIC MEAN
Weekly
Wage (Rs)
Solution:
Mid point
(mi)
Log (xi)
fi log xi
1200 - 1400
8
1300
3.1139
24.9112
fi log 𝑥𝑖
1400 - 1600
12
1500
3.1760
38.112
1600 - 1800
20
1700
3.2304
64.608
𝑓𝑖
586.3749
177
1800 - 2000
30
1900
3.2787
98.361
2000 - 2200
40
2100
3.3222
132.888
2200 - 2400
32
2300
3.3617
107.5744
2400 - 2600
18
2500
3.3979
61.1622
2600 - 2800
7
2700
3.4313
24.0191
2800 - 3000
6
2900
3.4623
20.7738
3000- 3200
4
3100
3.4913
13.9652
We Know,
Log GM=
No of workers
(fi)
Log GM =
GM= Antilog(3.31)
GM
= 2041.73
𝑓𝑖 = 177
Σfi log xi = 586.3749
127
GEOMETRIC MEAN
Question 8: An incomplete distribution is given below :
Variable : 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency : 12
30
33
65
45
25
18
Find the Geometric Mean
GEOMETRIC MEAN
Solution: Here we know,
GM
= Antilog
= Antilog
= 42.73
𝑓𝑖 .Log 𝑥𝑖
𝑓𝑖
371.769
228
variable
Midpoint
xi
10-20
20-30
30-40
40-50
50-60
60-70
70-80
15
25
35
45
55
65
75
No. of
frequency
(fi)
Log 𝒙𝒊
(fi) log xi
12
1.176
14.112
30
1.397
41.91
33
1.544
50.952
65
1.653
107.445
45
1.740
78.3
25
1.812
45.3
18
1.875
33.75
𝑓𝑖=228 Σfi log xi =371.769
GEOMETRIC MEAN
Question
9:
From the
following distribution of travel
time to work of a firm’s
employees, find the Geometric
mean and Harmonic mean.
Travel time (in
Minutes)
Frequency
70-79
60-69
50-59
218
215
195
40-49
30-39
20-29
156
85
50
10-19
18
0-9
2
GEOMETRIC MEAN
GM
= Antilog
= Antilog
= 45.763
𝑓𝑖 .Log 𝑥𝑖
𝑓𝑖
1559.224
939
Travel time (in
Minutes)
Mid-point
𝑥𝑖
Frequency
(𝑓𝑖 )
Log 𝑥𝑖
𝑓𝑖 . Log 𝑥𝑖
70-79
74.5
218
1.872
408.1
60-69
64.5
215
1.80
387
50-59
54.5
195
1.736
338.52
40-49
44.5=A
156
1.648
257.1
30-39
34.5
85
1.537
76.85
20-29
24.5
50
1.389
69.45
10-19
14.5
18
1.161
20.898
0-9
4.5
2
0.653
1.306
N=939
𝑓𝑖 . Log 𝑥𝑖 = 1559.224
GEOMETRIC MEAN
Question 10: From the
following distribution of
sales of companies , find the
Geometric mean.
Sale’s (BDT)
300-325
No. of Companies
5
325-350
350-375
375-400
17
80
468
400-425
425-450
450-475
326
7
88
475-500
9
GEOMETRIC MEAN
Solution:
GM = Antilog
= Antilog
=399.04
𝑓𝑖 .Log 𝑥𝑖
𝑓𝑖
2601.018
1000
sale’s
(BDT)
Mid-point
(𝑥𝑖 )
Frequency
(𝒇𝒊 )
Log 𝑥𝑖
300-325
312.5
5
2.494
12.47
325-350
337.5
17
2.528
42.976
350-375
362.5
80
2.559
204.72
375-400
387.5=A
468
2.588
1211.184
400-425
412.5
326
2.615
852.49
425-450
437.5
7
2.641
18.487
450-475
462.5
88
2.665
234.5
475-500
487.5
9
2.687
24.191
N=1000
𝒇𝑖 log𝑥𝑖
∑flogx =
2601.018
BUSINESS STATISTICS
GEOMETRIC MEAN, HARMONIC MEAN &
QUADRATIC MEAN
8/14/2020
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134
HARMONIC MEAN
Definition : The harmonic mean is defined as the reciprocal of the
arithmetic
mean
of
Symbolically, 𝑯𝑴 =
the
𝒏
𝒏 𝟏
𝒊=𝟏𝒙
𝒊
reciprocals
=
of
individual
observations.
𝒏
𝟏
𝟏
𝟏
𝟏
+
+
+⋯+
𝒙𝟏 𝒙𝟐 𝒙𝟑
𝒙𝒏
The calculation of harmonic mean becomes very tedious when a
distribution has a large number of observations.
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135
HARMONIC MEAN
Formula for Ungrouped Data
𝒏
𝒏
𝑯𝑴 =
=
𝟏
𝟏
𝟏
𝟏
𝒏 𝟏
+ +
+ ⋯+
𝒊=𝟏 𝒙
𝒙𝟏 𝒙𝟐 𝒙𝟑
𝒙𝒏
𝒊
Here, n= Total number of numbers or terms
𝑥1 , 𝑥2 , 𝑥3 … 𝑥𝑛 = Individual terms or individual values
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HARMONIC MEAN
Example:
Suppose we have three observations
4, 8 and 16. We are required to
calculate the harmonic mean.
1 1 1
Reciprocals of 4,8 and 16 are- , ,
Solution:
Since, 𝐻𝑀 =
4 8 16
𝑛
1
1
1
1
+
+
+⋯+
𝑥1 𝑥2 𝑥3
𝑥𝑛
=1
3
1
1
+ +
4 8 16
3
0.25+0.125+0.0625
=
= 6.857 𝑎𝑝𝑝𝑟𝑥.
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HARMONIC MEAN
Formula for Grouped Data
𝒏
𝒊=𝟏 𝒇𝒊
𝒇𝟏 + 𝒇𝟐 + 𝒇𝟑 … … . . +𝒇𝒏
𝑯𝑴 =
=
𝒇𝟏 𝒇𝟐 𝒇𝟑
𝒇𝒏
𝒏 𝒇𝒊
+ +
+ ⋯+
𝒊=𝟏 𝒙
𝒙𝟏 𝒙𝟐 𝒙𝟑
𝒙𝒏
𝒊
Here, f= Frequency
𝑥1 , 𝑥2 , 𝑥3 … 𝑥𝑛 = Individual terms or individual values
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HARMONIC MEAN
Example: Consider the following series:
and calculate the Harmonic Mean.
Class-Interval
Frequency
10/2/2021
2-4
20
4-6
40
Afzal Hossain (Assistant Professor)
6-8
30
8-10
10
139
HARMONIC MEAN
Solution: Let us set up the table as follows:
Class
-Interval
2-4
4-6
6-8
Mid-value
(xi)
3
5
7
Frequency
(fi )
20
40
30
8-10
9
10
n
i=1 fi =100
10/2/2021
Afzal Hossain (Assistant Professor)
fi
xi
6.6660
8.0000
4.2870
1.1111
n fi
i=1 x =20.0641
i
140
HARMONIC MEAN
Finally,
𝐻𝑀 =
𝑛
𝑖=1 𝑓𝑖
𝑛 𝑓𝑖
𝑖=1 𝑥
𝑖
100
20.0641
=
= 4.984
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141
HARMONIC MEAN
Example of the Harmonic Mean
As an example, take two firms. One has a market capitalization of $100
billion and earnings of $4 billion (P/E of 25) and one with a market
capitalization of $1 billion and earnings of $4 million (P/E of 250). In an
index made of the two stocks, with 10% invested in the first and 90%
invested in the second, the P/E ratio of the index is:
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142
HARMONIC MEAN
Solutions:
The WAM = (0.1×25+0.9×250)/(.1+.9) = 227.5
The W𝐻𝑀 =
𝑛
𝑖=1 𝑓𝑖
𝑛 𝑓𝑖
𝑖=1𝑥
𝑖
=
.1+.9
.1
.9
+
25 250
≈ 131.6
Where,
WAM=Weighted Arithmetic Mean
P/E=Price-to-Earnings Ratio
WHM=Weighted Harmonic Mean​
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143
HARMONIC MEAN
Advantages
Disadvantages
The main advantage of the harmonic mean
is that it is based on all observations in a
distribution and is amenable to further
algebraic treatment. When we desire to
give greater weight to smaller observations
and less weight to the larger observations,
then the use of harmonic mean will be
more suitable.
 It is difficult to understand as well as
difficult to compute
 It cannot be calculated if any of the
observations is zero or negative
 it is only a summary figure, which may
not be an actual observation in the
distribution
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144
BUSINESS STATISTICS
EXAMPLES OF HARMONIC MEAN
8/14/2020
Afzal Hossain (Assistant Professor)
145
HARMONIC MEAN
Question 1: Marks obtained by the 30 students of BBA-06 in
Business Statistics course- 60, 65, 76, 73, 73, 90, 78, 81, 62, 92, 75,
73, 66, 66, 68, 71, 74, 83, 78, 80, 86, 61, 87, 92, 87, 86, 81, 77, 74, 83.
Now, Find out the Harmonic mean.
8/15/2020
Afzal Hossain (Assistant Professor)
146
HARMONIC MEAN
Solution(Harmonic Mean Calculation):
We know,
𝑛
𝐻𝑀 =
1
1
1
1
+ + + ⋯+
𝑥1 𝑥2 𝑥3
𝑥𝑛
Here,
𝑥 = Individual Terms
n = Number of numbers or
terms
30
𝐻 =
1
1
1
1
1
1
1
1
1
1
+
+
+
+
+
+
+
+
+
+
60 65 76 73 73 90 78 81 62 92
1
1
1
1
1
1
1
1
1
1
+
+
+
+
+
+
+
+
+
+
75 73 66 66 68 71 74 83 78 80
1
1
1
1
1
1
1
1
1
1
+
+
+
+
+
+
+
+
+
86 61 87 92 87 86 81 77 74 83
=80
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Afzal Hossain (Assistant Professor)
147
HARMONIC MEAN
Question 2: The monthly income of 10 employees working in a firm is as
follows:
4487 4493 4502 4446 4475 4492 4572 4516 4468 4489. Calculate the
Harmonic Mean
HARMONIC MEAN
Solution:
Here n= 10
𝑛
HM= 1 1 1
1
+ + +⋯+
𝑥1 𝑥2 𝑥3
𝑛
=
10
1
1
1
1
1
1
1
1
1
1
+
+
+
+
+
+
+
+
+
4487 4493 4502 4446 4475 4492 4572 4516 4468 4489
=4493.45 (approx.)
HARMONIC MEAN
Question 3: A contractor employs three types of worksmale, female and children. To a male he pays Rs. 200 per
day, to a female worker Rs. 150 per day and to child
worker Rs. 100 per day. What is the average wage per day
paid by the contractor?
Calculate the Harmonic Mean
10/2/2021
Afzal Hossain (Assistant Professor)
HARMONIC MEAN
Solution:
Here n= 3
HM= 1 1
=
𝑛
1
1
+ + +⋯+𝑛
𝑥1 𝑥2 𝑥3
3
1
1
1
+
+
200 150 100
3
−
=
03
5×10 +6.67×10 03 +0.01
−
=138.440 (approx.)
10/2/2021
Afzal Hossain (Assistant Professor)
HARMONIC MEAN
Question 4: 1,500 workers are working in an industrial establishment. Their
age is classified as follows.
Age
(yrs.)
Find out the harmonic
mean using group
data.
No. of
workers
Age
(yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
152
HARMONIC MEAN
Class-Interval
Mid-value
(𝑥𝑖 )
Frequency
(𝒇𝑖 )
Recripocal of
MV
𝒇𝒊
𝒙𝒊
18-22
20
120
156.12
6
22-26
24
125
172.5
5.208
280
405.16
10
Solution : Calculation of
Harmonic Mean through
Group Data
26-30
28
30-34
32
260
391.3
8.125
34-38
36
155
241.18
4.306
38-42
40
184
294.768
4.6
42-46
44
162
266.166
3.682
46-50
48
86
144.566
1.792
50-54
52
75
128.7
1.44
54-58
56
53
97.888
0.946
𝒏
1500
𝒏
𝒇𝒊
𝒊=𝟏
𝒊=𝟏
𝒇𝒊
𝒙𝒊
46.099
153
HARMONIC MEAN
Finally,
𝑛
𝑖=1 𝑓𝑖
1500
𝐻𝑀 =
=
= 32.539
46.099
𝑛 𝑓𝑖
𝑖=1 𝑥
𝑖
So, the Geometric Mean of the workers is 32.539
154
HARMONIC MEAN
Profits (in BDT.)
Question 5: Given below is the 10000-20000
frequency distribution of the profit
20000-30000
obtained by 250 companies. Find
30000-40000
out the Harmonic mean.
No. of companies
15
35
40000-50000
50000-60000
47
68
32
60000-70000
22
70000-80000
80000-90000
12
11
90000-100000
8
HARMONIC MEAN
Solution:
HM =
=
𝑁
𝑓𝑖
𝑥𝑖
250
0.0063
= 39,682.53
Profits in(Rs.)
Mid Point (𝐱 𝒊 )
𝒇𝒊
𝒙𝒊
No of companies
(𝒇𝒊 )
10000-20000
15000
15
0.003
20000-30000
25000
35
0.009
30000-40000
35000
47
0.015
40000-50000
45000
68
0.002
50000-60000
55000
32
0.005
60000-70000
65000
22
0.02
70000-80000
75000
12
0.0005
80000-90000
85000
11
0.0004
90000-100000
95000
8
0.0001
N=250
𝑓𝑖
𝑥𝑖
=0.0063
HARMONIC MEAN
Question 6: Here given the
following Incomplete frequency
distribution. It is known as the total
frequency is 1,000 and that the
median is 413.11. Estimate by
calculation the missing frequencies
and find the value of the mode.
10/2/2021
Sales
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
Total
No. of Companies
5
17
80
468
326
7
88
9
1000
2
HARMONIC MEAN
Solution:
Harmonic mean of X
(
=
𝑛
𝑓𝑖
𝑥𝑖
)
1000
2.5095
398.486 (approx.)
10/2/2021
Sales
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
Mid-point
(𝒙𝒊 )
No. of
Companies (𝒇𝒊 )
312.5
337.5
362.5
387.5
412.5
437.5
462.5
487.5
5
17
8
46
32
7
8
9
n = 1000
𝒇𝒊
𝒙𝒊
0.016
0.050
0.221
1.208
0.790
0.016
0.190
0.018
2.5095
158
HARMONIC MEAN
Weekly Wage (BDT)
Question 7: Find out the Mean by
using direct data from the grouped
data.
Find out the Harmonic mean
No of workers (fi)
1200 - 1400
1400 - 1600
1600 - 1800
8
12
20
1800 - 2000
2000 - 2200
2200 - 2400
30
40
32
2400 - 2600
18
2600 - 2800
2800 - 3000
7
6
3000- 3200
4
HARMONIC MEAN
Weekly
Wage
No of workers
(fi)
Mid point
(xi)
𝒙𝒊
fi
𝒙𝒊
1200 - 1400
8
1300
3.1139
2.5691
𝒏
𝒊=𝟏 i
1400 - 1600
12
1500
3.1760
3.7783
1600 - 1800
20
1700
3.2304
6.1911
𝒏
𝒊=𝟏𝑥
𝑖
1800 - 2000
30
1900
3.2787
9.1499
2000 - 2200
40
2100
3.3222
12.0402
2200 - 2400
32
2300
3.3617
9.5189
2400 - 2600
18
2500
3.3979
5.2973
2600 - 2800
7
2700
3.4313
2.0400
2800 - 3000
6
2900
3.4623
1.7329
3000- 3200
4
3100
3.4913
1.1457
𝒏 fi
=
𝒊=𝟏
Solution:
HM=
f
fi
177
53.4634
=
HM= 3.3106
𝒏
fi = 𝟏𝟕𝟕
𝒊=𝟏
𝑥𝑖
53.4634
160
HARMONIC MEAN
Question 8: An incomplete distribution is given below :
Variable : 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total
Frequency : 12
30
33
65
45
25
18
228
Find the Harmonic Mean
HARMONIC MEAN
Solution:
𝑛
𝐻𝑀 =
𝑓𝑖
𝑥𝑖
111
=
= 41.75
0.162
variable
Mid-point
(xi)
No. of
frequency
(fi)
𝟏
𝒙𝒊
fi
𝒙𝒊
10-20
15
12
0.066
0.792
20-30
25
30
0.04
1.2
30-40
35
33
0.028
0.924
40-50
45
65
0.022
1.43
50-60
55
45
0.018
0.81
60-70
65
25
0.015
0.375
70-80
75
18
0.013
0.234
228
∑ =5.675
HARMONIC MEAN
Question 9: From
the
following
distribution of travel
time to work of a
firm’s employees, find
the Geometric mean
and Harmonic mean.
Travel time (in Minutes)
70-79
60-69
Frequency
218
215
50-59
40-49
30-39
195
156
85
20-29
10-19
50
18
0-9
2
HARMONIC MEAN
Solution:
𝐻𝑀 =
𝑛
𝑓𝑖
𝑥𝑖
= 939 / 20.896
= 44.936
Travel time (in
Minutes)
Mid-point (xi) Frequency (fi)
fi
𝑥𝑖
70-79
60-69
50-59
74.5
64.5
54.5
218
215
195
2.926
3.33
3.577
40-49
30-39
20-29
44.5
34.5
24.5
156
85
50
3.505
2.463
2.040
10-19
0-9
14.5
4.5
18
2
n=939
0.805
2.25
𝑓𝑖
=20.896
𝑥𝑖
HARMONIC MEAN
Question 10: From the
following distribution of
sales of companies , find the
Harmonic mean.
Sale's (Rs)
300-325
325-350
No. of companies
5
17
350-375
375-400
400-425
80
468
326
425-450
450-475
475-500
7
88
9
HARMONIC MEAN
Mid-point
(xi)
Frequency
(fi)
300-325
325-350
350-375
312.5
337.5
362.5
5
17
80
0.016
0.051
0.221
375-400
400-425
425-450
387.5=A
412.5
437.5
468
326
7
1.207
0.877
0.016
450-475
462.5
88
0.191
475-500
487.5
9
N=1000
0.018
Solution:
𝐻𝑀 =
=
𝑛
𝑓𝑖
𝑥𝑖
1000 / 2.597
= 38.059
𝒇𝒊
𝒙𝒊
Sale's
𝑓𝑖
=2.597
𝑥𝑖
BUSINESS STATISTICS
GEOMETRIC MEAN, HARMONIC MEAN &
QUADRATIC MEAN
8/14/2020
Afzal Hossain (Assistant Professor)
167
QUADRATIC MEAN
Definition: The quadratic mean (Q) is the square root of the arithmetic
mean of the squares.
Symbolically,
Q=
𝑛
2
𝑥
𝑖
𝑖=1
𝑛
=
2
𝑥12 +𝑥22 +….+𝑥𝑛
𝑛
Instead of using original values, the quadratic mean can be used while
averaging deviations when the standard deviation is to be calculated.
This will be used in the next chapter on dispersion.
8/14/2020
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168
QUADRATIC MEAN
Formula for Ungrouped Data
Q=
𝒏 𝒙 𝟐
𝒊=𝟏 𝒊
𝒏
=
𝒙𝟐𝟏 +𝒙𝟐𝟐 +….+𝒙𝟐𝒏
𝒏
Here, n= Total number of numbers or terms
𝑥1 , 𝑥2 , 𝑥3 … 𝑥𝑛 = Individual terms or individual values
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169
QUADRATIC MEAN
Solution:
Example:
Suppose we have three observations
4, 8 and 16. We are required to
calculate the QUADRATIC mean.
Since,
Q=
=
=
𝑛 𝑥 2
𝑖=1 𝑖
𝑛
2
𝑥12 +𝑥22 +….+𝑥𝑛
𝑛
42 +82 +162
3
16 + 64 + 256
=
3
=
10/2/2021
Afzal Hossain (Assistant Professor)
336
= 10.58
3
170
QUADRATIC MEAN
Formula for Grouped Data
Q=
𝒏 𝒇 𝒙 𝟐
𝒊=𝟏 𝒊 𝒊
𝒏 𝒇
𝒊=𝟏 𝒊
=
𝒇𝟏 𝒙𝟐𝟏 +𝒇𝟐 𝒙𝟐𝟐 +….+𝒇𝒏 𝒙𝟐𝒏
𝒇𝟏 +𝒇𝟐 +𝒇𝟑 …….+𝒇𝒏
Here, n= Total number of numbers or terms and f is the frequency
of individual values. 𝑥1 , 𝑥2 , 𝑥3 … 𝑥𝑛 = Individual terms or
individual values
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171
QUADRATIC MEAN
Example: Consider the following series:
and calculate the Quadratic Mean.
Class-Interval
Frequency
10/2/2021
2-4
20
4-6
40
Afzal Hossain (Assistant Professor)
6-8
30
8-10
10
172
QUADRATIC MEAN
Solution: Let us set up the table as follows:
Class-Interval
2-4
4-6
6-8
8-10
Mid-value
(xi)
3
5
7
9
Frequency
(𝑓𝑖 )
20
40
30
10
𝑛
𝑛
𝑓𝑖
𝑓𝑖 𝑥𝑖 2
9
25
49
81
180
1000
1470
810
𝑓𝑖 𝑥𝑖 2
3460
100
𝑖=1
10/2/2021
𝑥𝑖 2
𝑖=1
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QUADRATIC MEAN
Solution:
Q =
=
𝑛
2
𝑖=1 𝑓𝑖 𝑥𝑖
𝑛
𝑖=1 𝑓𝑖
3460
100
= 5.89
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QUADRATIC MEAN
Relative Position of Different Means:
The relative position of different means will always be:
Q> X >G>H provided that all the individual observations in a series are
positive and all of them are not the same.
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175
BUSINESS STATISTICS
EXAMPLES OF QUADRATIC MEAN
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176
QUADRATIC MEAN
Question 1: Marks obtained by the 30 students of BBA-06 in
Business Statistics course- 60, 65, 76, 73, 73, 90, 78, 81, 62,
92, 75, 73, 66, 66, 68, 71, 74, 83, 78, 80, 86, 61, 87, 92, 87,
86, 81, 77, 74, 83. Now, Find out the Quadratic mean.
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QUADRATIC MEAN
Solution(Quadratic Mean Calculation):
We know,
Q=
𝒏 𝒙 𝟐
𝒊=𝟏 𝒊
𝒏
Here,
n= Total number of numbers
or terms
𝑥1 , 𝑥2 , 𝑥3 … 𝑥𝑛 = Individual
terms or individual values
8/15/2020
Q=
602 +652 +762 + 732 +732 +902 + 782 +812 +622
+ 922 +752 +732 + 662 +662 +682 + 712 +74 2 +832
+782 +802 +862 +612 +872 + 922 +872 +862
+ 812 +772 + 742 +832
30
=
178462
30
= 77.128
Afzal Hossain (Assistant Professor)
178
QUADRATIC MEAN
Question 2: The monthly income of 10 employees working in a firm is as
follows:
4487 4493 4502 4446 4475 4492 4572 4516 4468 4489.
Calculate the Quadratic Mean.
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179
QUADRATIC MEAN
Solution:
Here n= 10
Q=
𝒙𝟐𝟏 +𝒙𝟐𝟐 +….+𝒙𝟐𝒏
𝒏
𝟐
=
=
8/15/2020
𝟐
𝟐
𝟐
𝟐
𝟐
4487 +4493 +4502 +4446 +4475 +4492 +4572
𝟐
𝟐
𝟐
4516 +4468 +4486
𝟐
+
10
201970412
10
= 4494.11
Afzal Hossain (Assistant Professor)
180
QUADRATIC MEAN
Question 3: A contractor employs three types of worksmale, female and children. To a male he pays Rs. 200 per
day, to a female worker Rs. 150 per day and to child
worker Rs. 100 per day. What is the average wage per day
paid by the contractor?
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181
QUADRATIC MEAN
Solution:
Here n= 3
Q=
=
=
8/15/2020
𝑛 𝑥 2
𝑖=1 𝑖
𝑛
=
2
𝑥12 +𝑥22 +….+𝑥𝑛
𝑛
2002 +1502 +1002
3
72500
3
=155.45
Afzal Hossain (Assistant Professor)
182
QUADRATIC MEAN
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Find out the Quadratic
mean using group
data.
8/15/2020
Age (yrs.)
No. of
workers
Age (yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
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183
QUADRATIC MEAN
Solution : Calculation of
Quadratic Mean through
Group Data.
Q=
=
𝑛 𝑓 𝑥 2
𝑖=1 𝑖 𝑖
𝑛
𝑖=1 𝑓𝑖
1981824
= 36.34
1500
𝟐
𝒇𝒊 𝒙𝒊 𝟐
120
400
48000
24
125
576
72000
26-30
28
280
784
219520
30-34
32
260
1024
266240
34-38
36
155
1296
200880
38-42
40
184
1600
294400
42-46
44
162
1936
313632
46-50
48
86
2304
198144
50-54
52
75
2704
202800
54-58
56
53
3136
166208
Class-Interval
Mid-value
(𝒙𝒊 )
Frequency
(𝒇𝒊 )
18-22
20
22-26
𝒏
𝒏
𝒇𝒊
8/15/2020
𝒙𝒊
𝒊=𝟏
Afzal Hossain (Assistant Professor)
𝒇𝒊 𝒙 𝒊 𝟐
1500
1981824
𝒊=𝟏
184
QUADRATIC MEAN
Profits (in BDT.)
Question 5: Given below is the 10000-20000
frequency distribution of the profit 20000-30000
obtained by 250 companies. Find 30000-40000
out the Quadratic mean.
40000-50000
8/15/2020
No. of companies
15
35
50000-60000
47
68
32
60000-70000
22
70000-80000
80000-90000
12
11
90000-100000
8
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185
QUADRATIC MEAN
Solution:
Q=
=
𝒏 𝒇 𝒙 𝟐
𝒊=𝟏 𝒊 𝒊
𝒏 𝒇
𝒊=𝟏 𝒊
6.2945e+11
250
𝒙𝒊 𝟐
𝒇𝒊 𝒙𝒊 𝟐
15
225000000
3375000000
25000
35
625000000
21875000000
30000-40000
35000
47
1225000000
57575000000
40000-50000
45000
68
2025000000
137700000000
50000-60000
55000
32
3025000000
96800000000
60000-70000
65000
22
4225000000
92950000000
70000-80000
75000
12
5625000000
67500000000
80000-90000
85000
11
7225000000
79475000000
90000-100000
95000
8
9025000000
72200000000
𝒏
𝟐
𝒊=𝟏 𝒇𝒊 𝒙𝒊 = 6.2945e+11
Profits in(Rs.)
Mid Point (𝒙𝒊 )
10000-20000
15000
20000-30000
No of companies
(𝒇𝒊 )
𝒏
𝒊=𝟏 𝒇𝒊 =250
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QUADRATIC MEAN
Question 6: Here given the
following Incomplete frequency
distribution. It is known as the total
frequency is 1,000 and that the
median is 413.11. Estimate by
calculation the missing frequencies
and find the value of the mode.
8/15/2020
Sales
300-325
325-350
350-375
375-400
400-425
425-450
450-475
475-500
Total
Afzal Hossain (Assistant Professor)
No. of Companies
5
17
80
468
326
7
88
9
1000
187
QUADRATIC MEAN
Solution:
Quadratic mean of X
Q=
=
𝒏 𝒇 𝒙 𝟐
𝒊=𝟏 𝒊 𝒊
𝒏 𝒇
𝒊=𝟏 𝒊
21018125
1000
=144.97
8/15/2020
Sales
Mid-point
No. of
𝒙𝒊 𝟐
(xi)
Companies (𝒇𝒊 )
300-325
312.5
5
97656.25
325-350
337.5
17
113906.25
𝒇𝒊 𝒙𝒊 𝟐
488281.25
1936406.25
350-375
362.5
8
131406.25
1051250
375-400
387.5
46
150156.25
6907187.5
400-425
412.5
32
170156.25
5445000
425-450
437.5
7
191406.25
1339843.75
450-475
462.5
8
213906.25
1711250
475-500
487.5
9
237656.25
2138906.25
𝒏
𝒊=𝟏 𝒇𝒊
Afzal Hossain (Assistant Professor)
= 1000
𝒏
𝟐
𝒊=𝟏 𝒇𝒊 𝒙𝒊 =21018125
188
QUADRATIC MEAN
Weekly Wage (BDT)
Question 7: Find out the Mean by
using direct data from the grouped
data. Find out the Quadratic mean.
8/15/2020
No of workers (fi)
1200 - 1400
1400 - 1600
1600 - 1800
8
12
20
1800 - 2000
2000 - 2200
2200 - 2400
30
40
32
2400 - 2600
18
2600 - 2800
2800 - 3000
7
6
3000- 3200
4
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189
QUADRATIC MEAN
Solution:
Q=
=
𝒏 𝒇 𝒙 𝟐
𝒊=𝟏 𝒊 𝒊
𝒏 𝒇
𝒊=𝟏 𝒊
804730000
177
= 2132.25
Weekly Wage
No of workers
(fi)
Mid point
(xi)
𝒇𝒊 𝒙𝒊 𝟐
1200 - 1400
8
1300
1690000
13520000
1400 - 1600
12
1500
2250000
27000000
1600 - 1800
20
1700
2890000
57800000
1800 - 2000
30
1900
3610000
108300000
2000 - 2200
40
2100
4410000
176400000
2200 - 2400
32
2300
5290000
169280000
2400 - 2600
18
2500
6250000
112500000
2600 - 2800
7
2700
7290000
51030000
2800 - 3000
6
2900
8410000
50460000
3000- 3200
4
3100
9610000
38440000
𝒏
𝟐
𝒊=𝟏 𝒇𝒊 𝒙𝒊 =804730000
𝒏
fi = 𝟏𝟕𝟕
8/15/2020
𝒙𝒊 𝟐
𝒊=𝟏
Afzal Hossain (Assistant Professor)
190
QUADRATIC MEAN
Question 8: An incomplete distribution is given below :
Variable : 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total
Frequency : 12
30
33
65
45
25
18
228
Find the Quadratic Mean
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191
QUADRATIC MEAN
Solution:
Q=
=
𝑛
2
𝑖=1 𝑓𝑖 𝑥𝑖
𝑛
𝑖=1 𝑓𝑖
536500
228
= 48.50
8/15/2020
No. of
frequency
(fi)
𝒙𝒊 𝟐
𝒇𝒊 𝒙𝒊 𝟐
10-20
20-30
Midpoint
(xi)
15
25
12
30
225
625
2700
18750
30-40
40-50
50-60
35
45
55
33
65
45
1225
2025
3025
40425
131625
136125
60-70
65
25
4225
105625
70-80
75
5625
101250
𝒏
𝟐
𝒊=𝟏 𝒇𝒊 𝒙𝒊 =
536500
192
Variabl
e
Afzal Hossain (Assistant Professor)
18
𝒏
𝒊=𝟏 𝒇𝒊 =
228
QUADRATIC MEAN
Question 9: From the
following distribution of
travel time to work of a
firm’s employees, find
the Quadratic mean.
8/15/2020
Travel time (in Minutes)
70-79
60-69
Frequency
218
215
50-59
40-49
30-39
195
156
85
20-29
10-19
50
18
0-9
2
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193
QUADRATIC MEAN
Solution:
Q=
=
𝒏 𝒇 𝒙 𝟐
𝒊=𝟏 𝒊 𝒊
𝒏 𝒇
𝒊=𝟏 𝒊
3127534.75
939
= 57.71
Travel time
(in Minutes)
Midpoint (xi)
Frequency
(fi)
𝒙𝒊 𝟐
𝒇𝒊 𝒙𝒊 𝟐
70-79
74.5
218
5550.25
1209955
60-69
64.5
215
4160.25
894453.8
50-59
54.5
195
2970.25
579198.8
40-49
44.5
156
1980.25
308919
30-39
34.5
85
1190.25
101171.3
20-29
24.5
50
600.25
30012.5
10-19
14.5
18
210.25
3784.5
0-9
4.5
2
20.25
40.5
𝒏
𝟐
𝒊=𝟏 𝒇𝒊 𝒙𝒊 =
3127534.75
𝒏
𝒊=𝟏 𝒇𝒊
8/15/2020
=939
Afzal Hossain (Assistant Professor)
194
QUADRATIC MEAN
Question 10: From the
following distribution of
sales of companies , find the
Quadratic mean.
8/15/2020
Sale's (Rs)
300-325
325-350
No. of companies
5
17
350-375
375-400
400-425
80
468
326
425-450
450-475
475-500
7
88
9
Afzal Hossain (Assistant Professor)
195
QUADRATIC MEAN
Sale's
Mid-point
(xi)
Frequency
(fi)
𝒙𝒊 𝟐
𝒇𝒊 𝒙𝒊 𝟐
300-325
312.5
5
97656.25
488281.25
325-350
337.5
17
113906.25
1936406.25
350-375
362.5
80
131406.25
10512500
375-400
387.5
468
150156.25
70273125
400-425
412.5
326
170156.25
55470937.5
425-450
437.5
7
191406.25
1339843.75
450-475
462.5
88
213906.25
18823750
475-500
487.5
9
237656.25
Solution:
Q=
=
𝒏 𝒇 𝒙 𝟐
𝒊=𝟏 𝒊 𝒊
𝒏 𝒇
𝒊=𝟏 𝒊
160983750
1000
= 401.22
2138906.25
𝒏
𝟐
𝒊=𝟏 𝒇𝒊 𝒙𝒊 =16098375
𝒏
𝒊=𝟏 𝒇𝒊 =100
0
8/15/2020
Afzal Hossain (Assistant Professor)
0
196
BUSINESS STATISTICS
MEAN , MEDIAN AND MODE
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197
MEDIAN
Definition:The "median" is the "middle" value in the list of numbers. To
find the median, your numbers have to be listed in numerical order from
smallest to largest, so you may have to rewrite your list before you can find
the median.
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198
MEDIAN
Characteristics Of The Median : Unlike the arithmetic mean, the
median can be computed from open-ended distributions. This is
because it is located in the median class-interval, which would not be
an open-ended class.
The median can also be determined graphically whereas the arithmetic
mean cannot be ascertained in this manner.
As it is not influenced by the extreme values, it is preferred in case of
a distribution having extreme values
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199
MEDIAN
Median for Ungrouped Data: The median will in (N+1)/2-th
Position for ungrouped data but if the result is given fraction value
the median will be average prior and posterior values of fraction. For
example, The Median of 13, 18, 13, 14, 13, 16, 14, 21, 13 are :
Here N=9
Step-1: Sort the Value-13 , 13, 13, 13, 14, 14, 16,18,21
Step-2: Median Position: (N+1)/2 -th Position= (9+1)/2 -th Position=
10/2 -th Position= 5 -th Position=14
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Afzal Hossain (Assistant Professor)
200
MEDIAN
Median for Ungrouped Data:
Example-1: The Median of 13, 18, 13, 14, 13, 16, 14, 21, 13,16 are :
Here N=10
Step-1: Sort the Value-13 , 13, 13, 13, 14, 14, 16,16,18,21
Step-2: Median Position: (N+1)/2 -th Position= (10+1)/2 -th Position=
11/2 -th Position= 5.5 -th Position.
Since Position of median is fraction. So Prior value of fraction =5 and
Posterior value of fraction =6.
th
th
Median=(5 Position Value+6 Position Value)/2 =(14+14)/2=14.
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201
MEDIAN
Median for Grouped Data: In
the case of a grouped series, the
median is calculated by linear
interpolation with the help of the
following formula:
𝑳𝟐 − 𝑳𝟏
𝑴 = 𝑳𝟏 +
𝒇
10/2/2021
𝒎−𝒄
Where M = the median
L1 = the lower limit of the class in which the
median lies.
L2 = the upper limit of the class in which the
median lies
f = the frequency of the class in which the
median lies
m = the middle item or (n + 1)/2th, where n
stands for total number of items
c = the cumulative frequency of the class
preceding the one in which the median lies
Afzal Hossain (Assistant Professor)
202
MEDIAN
Example : Find the Median Value
of the Given Data
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
No. of
Workers
18
25
30
34
26
10
Total
143
Afzal Hossain (Assistant Professor)
203
MEDIAN
Solution : We Know
𝐿2 − 𝐿1
𝑀 = 𝐿1 +
𝑓
Monthly Wages
(BDT)
𝑚−𝑐
𝑛+1 143+1
=
=72
2
2
Here, m=
So, Median Lies in 1200-1400
L1 =1200 and L2 = 1400, f=30,
c=43.
10/2/2021
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
204
MEDIAN
Solution : Now,
𝐿2 − 𝐿1
𝑀 = 𝐿1 +
𝑓
Monthly Wages
(BDT)
𝑚−𝑐
1400 − 1200
= 1200 +
72 − 43
30
200
= 1200 +
29
30
= 1200 + 6.67 × 29
= 1200 + 193.43 = 1393.43
10/2/2021
Frequency
(f)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
205
MEDIAN
Median for Grouped Data: In
the case of a grouped series, the
median is calculated by linear
interpolation with the help of the
following formula:
𝒏
−𝒄
𝑴 = 𝑳𝟏 + 𝟐
×𝒊
𝒇
10/2/2021
Where M = the median
L1 = the lower limit of the class in which the
median lies.
f = the frequency of the class in which the
median lies
c = the cumulative frequency of the class
preceding the one in which the median lies
i=Class Interval of median Class.
Afzal Hossain (Assistant Professor)
206
MEDIAN
Example : Find the Median Value
of the Given Data.
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
No. of
Workers
18
25
30
34
26
10
Total
143
Afzal Hossain (Assistant Professor)
207
MEDIAN
Solution : We Know
𝒏
−𝒄
𝟐
𝑴 = 𝑳𝟏 +
×𝒊
𝒇
𝑛 143
Here, m= =
=71.5
2
2
So, Median Lies in 1200-1400
L1 =1200
f=30
c=43
i=1400-1200=200.
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
208
MEDIAN
Solution : 𝑵𝒐𝒘,
𝑛
−𝑐
𝑀 = 𝐿1 + 2
×𝑖
𝑓
71.5 − 43
= 1200 +
× 200
30
= 1200 + .95 × 200
=1200+190 =1390
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
209
MEDIAN
Median for Grouped Data at Quartiles, Deciles and Percentiles : Quartiles,
Deciles and Percentiles are special case of quantities.
Quartiles: Quartiles are those values of the variate which divide the frequencies
into four equals parts.
Deciles: Deciles are those values of the variate which divide the frequencies into
ten equals parts.
Percentiles: Percentiles are those values of the variate which divide the
frequencies into hundred equals parts.
10/2/2021
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210
MEDIAN
Median for Grouped Data
Median for Grouped Data
(Percentiles)
(Quartiles)
𝑀𝑛
𝑗𝑛
−𝑐
−𝑐
100
𝑃𝑀 = 𝐿1 +
×𝑖
𝑄𝐽 = 𝐿1 + 4
×𝑖
𝑓
𝑓
Where M=1,2,3,…99
Where J=1,2,3
Median for Grouped Data
(Deciles)
𝐾𝑛
−𝑐
𝐷𝐾 = 𝐿1 + 10
×𝑖
𝑓
Where K=1,2,3……..9
10/2/2021
Afzal Hossain (Assistant Professor)
211
MEDIAN
Example : Find the Quartiles
(Q2), Deciles (D2), Percentiles
(P3) Value of the Given Data
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
No. of
Workers
18
25
30
34
26
10
Total
143
Afzal Hossain (Assistant Professor)
212
MEDIAN
Solution : Quartiles Calculation
We Know,
𝑗𝑛
−𝑐
𝑄𝐽 = 𝐿1 + 4
×𝑖
𝑓
Here, For 𝑄2 ,
𝐽𝑛 2×143 286
m= =
=
=71.5
4
4
4
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
So, Median Lies in 1200-1400 and
L1 =1200 , f=30, c=43 and i=14001200=200.
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213
MEDIAN
Solution : Quartiles Calculation
𝑗𝑛
−𝑐
𝑄𝐽 = 𝐿1 + 4
×𝑖
𝑓
71.5 − 43
𝑄2 = 1200 +
× 200
30
𝑄2 = 1200 + .95 × 200
𝑄2 =1200+190
𝑄2 =1390
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
71.5% Worker get wages 1390
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214
MEDIAN
Solution : Deciles Calculation
We Know,
𝐾𝑛
−𝑐
𝐷𝐾 = 𝐿1 + 10
×𝑖
𝑓
For, 𝐷2 , K=2
𝐾𝑛 2×143
286
Here, m= =
=
=28.6
10
10
10
So, Median Lies in 1000-1200 and
L1 =1000 , f=25, c=18 and i=12001000=200.
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
215
MEDIAN
Solution : Deciles Calculation
𝐾𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
28.6 − 18
𝐷2 = 1000 +
× 200
25
𝐷2 = 1000 + .424 × 200
𝐷2 =1200+84.8
𝐷2 =1284.8
28.6% Worker get wages 1284.8
BDT
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
216
MEDIAN
Solution : Percentiles Calculation
We Know,
𝑀𝑛
−𝑐
𝑃𝑀 = 𝐿1 + 100
×𝑖
𝑓
For 𝑃3 , 𝑀 = 3
𝑀𝑛 3×143
429
Here, m=
=
=
= 4.29
100
100
100
So, Median Lies in 800-1000
and L1 =800 , f=18, c=0 and i=1000800=200.
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
217
MEDIAN
Solution : Percentiles Calculation
𝐾𝑛
−𝑐
𝑃𝑀 = 𝐿1 + 10
×𝑖
𝑓
4.29 − 0
𝑃3 = 800 +
× 200
18
𝑃3 = 800 + .2383 × 200
𝑃3 =800+47.667
𝑃3 =847.667
4.29% Worker get wages 847.667
BDT
10/2/2021
Monthly Wages
(BDT)
800-1,000
1,000-1,200
1,200-1,400
1,400-1,600
1,600-1,800
1,800-2,000
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
18
25
30
34
26
10
143
Cumulative
Frequency
(c)
18
43
73
107
133
143
218
QUARTILES
SOME EXAMPLES
10/2/2021
Afzal Hossain (Assistant Professor)
219
MEDIAN
Question 1: Marks obtained by the 30 students of
BBA-06 in Business Statistics course-
60, 65, 76, 73, 73, 90, 78, 81, 62, 92, 75, 73, 66, 66,
68, 71, 74, 83, 78, 80, 86, 61, 87, 92, 87, 86, 81, 77,
74, 83
Now, Find out median in Quartile.
10/2/2021
Afzal Hossain (Assistant Professor)
220
MEDIAN
Solution :
We Know, 𝑄𝐽 = 𝐿1 +
𝑗𝑛
−𝑐
4
𝑓
×𝑖
Here, For 𝑄2 ,
m=
𝐽𝑛
4
=( 2 ∗ 30)/4 = 60/4 = 15
So, Median Lies in 75-79 and L1 =75 , f=5,
c=13 and i=79-75=4
Marks
Frequency
(f)
60-64
65-69
70-74
75-79
80-84
85-89
90-94
Total
3
4
6
5
5
4
3
30
Cumulative
Frequency
(c)
3
7
13
18
23
27
30
MEDIAN
Solution :
𝑗𝑛
−𝑐
𝑄𝐽 = 𝐿1 + 4
×𝑖
𝑓
Q2 = 75+
15−13
5
Q2 = 75 +
2
5
Q2 = 75 +
8
5
Q2 = 75 + 1.6
Q2 = 76.5
×4
×4
Marks
Frequency
(f)
60-64
65-69
70-74
75-79
80-84
85-89
90-94
Total
3
4
6
5
5
4
3
30
Afzal Hossain (Assistant Professor)
Cumulative
Frequency
(c)
3
7
13
18
23
27
30
MEDIAN
Question 2:
The monthly income 0f 10 employees working in a firm is as
follows (Average short cut)
4487 4493 4502 4446 4475 4492 4572 4516 4468 4489.
Find the Quartiles.
Afzal Hossain (Assistant Professor)
MEDIAN
Solution:
𝑗𝑛
−𝑐
4
𝑄𝐽 = 𝐿1 +
×𝑖
𝑓
Here,
𝑗𝑛 2×10
m= =
=5
4
4
MONTHLY INCOME
NO. OF EMPLOY
(f)
4446 - 4488
4488 - 4530
4530 - 4572
4
5
1
Σf =10
So, the median lies in 4488-4530.
L1 = 4488, f = 5, c = 4 and i = 4530-4488 = 42.
Afzal Hossain (Assistant Professor)
CUMULATIVE
FREQUENCY (c)
4
9
10
MEDIAN
Solution:
𝑗𝑛
−𝑐
4
𝑄𝐽 = 𝐿1 +
×𝑖
𝑓
5−4
𝑄2 = 4488 +
× 42
5
42
+
5
𝑄2 = 4488
𝑄2 = 4488+8.4
𝑄2 = 4496.4
5% Worker get wages 4496.4
MONTHLY INCOME
NO. OF EMPLOY
(f)
4446 - 4488
4488 - 4530
4530 - 4572
4
5
1
Σf =10
Afzal Hossain (Assistant Professor)
CUMULATIVE
FREQUENCY (c)
4
9
10
MEDIAN
Question 3: A contractor employs three types of works- male,
female and children. To a male he pays Rs. 200 per day, to a
female worker Rs. 150 per day and to child worker Rs. 100 per
day. Now, Find out the mean by using quartile.
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226
MEDIAN
Solution:
Here N=3
Step-1: Sort the value- 100,150,200
(𝑁+1)
(3+1)
4
Step-2: Median position:
th position =
th position = th
2
2
2
position = 2nd position
150
∴ 𝑀𝑒𝑑𝑖𝑎𝑛 =
= 75.
2
10/2/2021
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227
MEDIAN
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Find out the mean using
quartile method.
10/2/2021
Age (yrs.)
No. of
workers
Age (yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
Afzal Hossain (Assistant Professor)
228
MEDIAN
Age (yrs.)
No. of workers
(fi)
18-22
20
120
22-26
24
125
-12
-1500
280
-8
-2240
Solution:
26-30
28
-1920
30-34
32
260
-4
-1040
34-38
36
155
0
0
38-42
40
184
4
736
42-46
44
162
8
1296
46-50
48
86
12
1032
50-54
52
75
16
1200
54-58
56
53
20
1060
N= 1500
10/2/2021
Deviation
(di)
-16
fi𝒅′ i
Mid-point (mi)
Afzal Hossain (Assistant Professor)
Σ fi𝑑 ′ i= -1376
229
229
MEDIAN
Here,
𝑗𝑛
−𝑐
= 𝐿1 + 4
×𝑖
𝑓
750−525
Or, 𝑄4 = 30+
×4
260
Or, 𝑄4 = 33.46
So, the average age of the workers is 33.46
10/2/2021
Afzal Hossain (Assistant Professor)
230
MEDIAN
Question 5: Given below is
the frequency distribution of the
profit
obtained
by
250
companies. Find out quartile.
10/2/2021
Profits (in BDT.)
10000-20000
20000-30000
15
35
30000-40000
40000-50000
50000-60000
47
68
32
60000-70000
70000-80000
80000-90000
22
12
11
90000-100000
8
Afzal Hossain (Assistant Professor)
No. of companies
231
MEDIAN
Solution :
We Know,
𝑗𝑛
−𝑐
4
𝑄𝐽 = 𝐿1 +
×𝑖
𝑓
Here, For 𝑄2 ,
𝐽𝑛
m= = (2×250)/4= (500/4) =125
4
So, Median Lies in 40000-50000
and L1 =40000, f=68, c=97 and
i=50000-40000 =10000
Monthly Wages
(BDT)
10000-20000
20000-30000
30000-40000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000-100000
Frequency
(f)
15
35
47
68
32
22
12
11
8
Cumulative
Frequency
(c)
15
50
97
165
197
219
231
242
250
MEDIAN
Solution:
𝑗𝑛
−𝑐
𝑄𝐽 = 𝐿1 + 4
×𝑖
𝑓
Q2
= 40000 + (125 − 97)/68
× 10000
Q2 = 40000 + 0.411 × 10000
𝑄2 = 40000 + 4117.64
𝑄2 = 44117.64
125% Worker get wages 44117.64
Monthly Wages
(BDT)
10000-20000
20000-30000
30000-40000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000-100000
Frequency
(f)
15
35
47
68
32
22
12
11
8
Cumulative
Frequency
(c)
15
50
97
165
197
219
231
242
250
MEDIAN
Question 6: Find the
Median from the given data
using Quartiles 2.
Sales
320-325
325-330
No. of companies
5
17
330-335
335-340
340-345
80
468
326
345-350
350-355
355-360
7
88
9
N=1000
MEDIAN
Solution:
Sales
No. of companies
(f)
Cumulative frequency
(c)
320-325
325-330
5
17
5
22
330-335
335-340
340-345
80
468(f)
326
102(c)
570
896
345-350
350-355
355-360
7
88
9
903
991
1000
N=1000
MEDIAN
Solution:
We Know,
𝑗𝑛
−𝑐
4
𝑄𝐽 = 𝐿1 +
×𝑖
𝑓
Here,
𝑗𝑛
m=
4
2×1000
=
4
=500
MEDIAN
𝐿1 = 335, 𝑓 = 468, 𝑖 = 340 − 335 = 5, 𝑐 = 102
𝑗𝑛
−𝑐
𝑄2 = 𝐿1 + 4
×𝑖
𝑓
=335+
2×1000
−102
4
468
×5
= 335+4.252 = 339.252
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237
MEDIAN
Question 7: Find the
Weekly Income(BDT) No. of employees
Quartiles Value from the
given data.
500-550
550-600
600-650
650-700
700-750
750-800
800-850
10/2/2021
Afzal Hossain (Assistant Professor)
6
10
22
30
16
12
15
238
MEDIAN
Solution:
10/2/2021
Weekly
Income(BDT)
500-550
550-600
600-650
650-700
700-750
750-800
800-850
No. of employees
6
10
22
30
16
12
15
Afzal Hossain (Assistant Professor)
Cumulative
frequency(c)
6
16
38
68
84
96
111
239
MEDIAN
Solution:
We know, 𝑄𝑗 = 𝐿1 +
𝐽𝑛
−𝐶
4
𝑓
𝑗𝑛
4
×𝑖
2×111
4
For, P2,M=2. Here, j= =
= 55.5
So, median lies in 500-550 and L1=500,f=6,c=0,i=550-500=50
55.5 − 0
∴ 𝑄𝑗 = 500 +
× 50
6
Or, 𝑄𝑗 = 500 + 9.25 × 50
Or, 𝑄𝑗 = 500 + 462.5 = 962.5
We can say, 55.5% employees set weekly income is 962.5 BDT.
Afzal Hossain (Assistant Professor)
240
MEDIAN
Question
quartile.
10/2/2021
8:
Find
using
Weekly Rent (in Rs)
No. of Persons
Paying the Rent
200-400
400-600
600-800
6
9
11
800-1000
1000-1200
1200-1400
14
20
15
1400-1600
1600-1800
1800-2000
10
8
7
Afzal Hossain (Assistant Professor)
241
MEDIAN
Solution:
We Know,
WeeklyRent
(RS)
𝑗𝑛
−𝑐
4
𝑄𝐽 = 𝐿1 +
×𝑖
𝑓
Here, For 𝑄2 ,
𝐽𝑛 (2×100) 200
m= =
=
=50
4
4
4
So, Median Lies in 1000-1200 and
L1 =1000, f=20, c=40 and
i=1200-1000 =200.
10/2/2021
200-400
400-600
600-800
800-1000
1000-1200
1200-1400
1400-1600
1600-1800
1800-2000
Afzal Hossain (Assistant Professor)
No. Of person
paying the rent
(f)
6
9
11
14
20
15
10
8
7
(c)
6
15
26
40
60
75
85
93
100
242
MEDIAN
Solution:
𝑗𝑛
−𝑐
𝑄𝐽 = 𝐿1 + 4
×𝑖
𝑓
(50−40)
Q2 = 1000 +
× 200
20
10
Q2 = 1000 + ×200
20
𝑄2 = 1000 + 0.5 × 200
𝑄2 = 1100
50% Worker get rent 1100
10/2/2021
WeeklyRent
(RS)
200-400
400-600
600-800
800-1000
1000-1200
1200-1400
1400-1600
1600-1800
1800-2000
Afzal Hossain (Assistant Professor)
No. Of person
paying the rent
(f)
6
9
11
14
20
15
10
8
7
(c)
6
15
26
40
60
75
85
93
100
243
MEDIAN
Question 9: Consider the following data and Find the quartile.
Class
10-20 20-30
Frequency 12
30
10/2/2021
30-40
33
40-50
65
Afzal Hossain (Assistant Professor)
50-60
45
60-70
25
70-80
18
244
MEDIAN
Solution:
We Know,
𝑗𝑛
−𝑐
4
𝑄𝐽 = 𝐿1 +
×𝑖
𝑓
Here, For 𝑄2 ,
𝐽𝑛 (2×228) 456
m= =
=
=114
Monthly Wages
(BDT)
10-20
20-30
30-40
40-50
50-60
4
4
4
60-70
So, Median Lies in 40-50 and L1 =40 70-80
Frequency
(f)
12
30
33
65
45
25
18
Cumulative
Frequency
(c)
12
42
75
140
185
210
228
, f=65, c=75 and i=50-40 =10.
10/2/2021
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245
MEDIAN
Solution:
Monthly Wages
(BDT)
𝑗𝑛
−𝑐
4
10-20
𝑄𝐽 = 𝐿1 +
×𝑖
20-30
𝑓
Q2 = 40 + (114 − 75)/65 × 10 30-40
40-50
Q2 = 40 + 0.6 × 10
50-60
𝑄2 = 40 + 6 = 46
60-70
70-80
Frequency
(f)
12
30
33
65
45
25
18
Cumulative
Frequency
(c)
12
42
75
140
185
210
228
114% Worker get wages 46
10/2/2021
Afzal Hossain (Assistant Professor)
246
MEDIAN
Marks
Question 10: Given below is the
frequency distribution of the marks
obtained by 90 student.
Find out Quartile.
10/2/2021
20-29
30-39
40-49
50-59
60-69
70-79
80-89
Afzal Hossain (Assistant Professor)
Frequency
(f)
2
12
15
20
18
10
9
Cumulative
Frequency
(c)
18
30
45
65
83
93
102
247
MEDIAN
Solution:
𝑄𝑗 = 𝑙1 + (
Marks
𝑗𝑛−𝑐
4
𝑓
)× 𝑖
Here, l = 50
f=20
c=65
i= (59-50) = 9
10/2/2021
20-29
30-39
40-49
50-59
60-69
70-79
80-89
Frequency
(f)
2
12
15
20
18
10
9
Afzal Hossain (Assistant Professor)
Cumulative
Frequency (c)
18
30
45
65
83
93
102
248
MEDIAN
So the value of 𝑄𝑗 is,
2
4
𝑄𝑗 = 50+ ( -65) × 9
20
1.5−65
= 50+ (
)×
20
−63.5
=50+(
)×9
20
9
=50+ (-3.175)× 9
=50 - 28.575 = 21.425
Afzal Hossain (Assistant Professor)
249
DECILES
SOME EXAMPLES
10/2/2021
Afzal Hossain (Assistant Professor)
250
MEDIAN
Question 1: Marks obtained by the 30 students of
BBA-06 in Business Statistics course-
60, 65, 76, 73, 73, 90, 78, 81, 62, 92, 75, 73, 66, 66,
68, 71, 74, 83, 78, 80, 86, 61, 87, 92, 87, 86, 81, 77,
74, 83
Now, Find out the mean in decile.
Afzal Hossain (Assistant Professor)
251
MEDIAN
Solution :
We Know,
Marks
𝐾𝑛
−𝑐
𝐷𝐾 = 𝐿1 + 10
×𝑖
𝑓
For, 𝐷2 , K=2
𝐾𝑛 (2×30)
Here, m= =
=6
10
10
So, Median Lies in 65-69 and
L1 =65 , f=4, c=3 and i=69-65=4.
60-64
65-69
70-74
75-79
80-84
85-89
90-94
Afzal Hossain (Assistant Professor)
Frequency
(f)
3
4
6
5
5
4
3
Cumulative
Frequency
(c)
3
7
13
18
23
27
30
252
MEDIAN
Solution:
𝐾𝑛
−𝑐
𝐷𝐾 = 𝐿1 + 10
×𝑖
𝑓
𝐷2 = 65 + (6-3)/4 × 4
𝐷2 = 65 + (3/4) × 4
𝐷2 = 65 + 0.75 × 4
𝐷2 = 65 + 3
𝐷2 = 68
Marks
60-64
65-69
70-74
75-79
80-84
85-89
90-94
Frequency
(f)
3
4
6
5
5
4
3
Cumulative
Frequency
(c)
3
7
13
18
23
27
30
6% student get 68 marks
Afzal Hossain (Assistant Professor)
253
MEDIAN
Question 2: The monthly income 0f 10 employees working in a
firm is as follows
4487 4493 4502 4446 4475 4492 4572 4516 4468 4489
Calculating monthly income by Deciles Calculation method.
10/2/2021
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254
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Solution:
We Know,
Monthly income
𝐾𝑛
−𝑐
𝐷𝐾 = 𝐿1 + 10
×𝑖
𝑓
For, 𝐷2 , K=2
𝐾𝑛 2×10
20
Here, m= =
= =2
10
10
10
So, Median Lies in 4446-4488 and
L1 =4446 , f=4, c=0 and
i= 4446-4488 =42
10/2/2021
4446-4488
4488-4530
4530-4572
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
4
5
1
10
Cumulative
Frequency
(c)
4
9
10
255
MEDIAN
Monthly income
2−0
𝐷2 = 4446 +
× 42
4
𝐷2 = 4446 +
𝐷2 = 4467
10/2/2021
1
2
× 42
4446-4488
4488-4530
4530-4572
Total
Afzal Hossain (Assistant Professor)
Frequency
(f)
4
5
1
10
Cumulative
Frequency
(c)
4
9
10
256
MEDIAN
Question 3: A contractor employs three types of works- male,
female and children. To a male he pays Rs. 200 per day, to a
female worker Rs. 150 per day and to child worker Rs. 100 per
day. Now, Find out the mean by using decile.
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257
MEDIAN
Solution:
Here N=3
Step-1: Sort the value- 100,150,200
(𝑁+1)
(3+1)
4
Step-2: Median position:
th position =
th position = th
2
2
2
position = 2nd position
∴ 𝑀𝑒𝑑𝑖𝑎𝑛 =
150
2
= 75.
Afzal Hossain (Assistant Professor)
258
MEDIAN
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Age (yrs.)
Find out the median using
Deciles calculation.
10/2/2021
No. of
workers
Age (yrs.)
No. of
workers
18-22
120
38-42
184
22-26
125
42-46
162
26-30
280
46-50
86
30-34
260
50-54
75
34-38
155
54-58
53
Afzal Hossain (Assistant Professor)
259
MEDIAN
Solutions:
We know
Age (yrs.)
𝐷𝑘 = 𝐿1 +(
𝐾𝑛
−𝑐
10
𝑓
)×i
For 𝐷2 ,k=2
𝐾𝑛
2×1500
Here , m=
=
3000
=
10
10
10
= 300
So median lies in 26-30
and 𝐿1 =26, f=280, c=245,
i=(30-26)=4
10/2/2021
No. of workers
(fi)
Cumulative Frequency
(c)
18-22
120
120
22-26
125
245
26-30
280
525
30-34
260
785
34-38
155
940
38-42
184
1124
42-46
162
1286
46-50
86
1372
50-54
75
1447
54-58
53
1500
N= 1500
Afzal Hossain (Assistant Professor)
260
260
MEDIAN
Age (yrs.)
No. of workers
(fi)
Cumulative Frequency
(c)
)* I
18-22
120
120
22-26
125
245
∗4
26-30
280
525
30-34
260
785
34-38
155
940
38-42
184
1124
42-46
162
1286
46-50
86
1372
50-54
75
1447
54-58
53
1500
We know
𝐷𝑘 = 𝐿1 +(
𝐾𝑛
−𝑐
10
𝑓
300−245
=26+
280
55
=26+( )*4
280
=26+.78571429
=26.78
Ans: 26.78
N= 1500
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261
MEDIAN
Question 5: Given below is
the frequency distribution of the
profit
obtained
by
250
companies. Find out decile.
10/2/2021
Profits (in BDT.)
10000-20000
20000-30000
30000-40000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000-100000
Afzal Hossain (Assistant Professor)
No. of companies
15
35
47
68
32
22
12
11
8
262
MEDIAN
Solution:
Profits
10,000-20,000
20,000-30,000
30,000-40,000
40,000-50,000
50,000-60,000
60,000-70,000
70,000-80,000
Total
10/2/2021
Frequency
(f)
15
35
47
68
32
22
12
231
Afzal Hossain (Assistant Professor)
Cumulative Frequency
(c)
15
50
97
165
197
219
231
263
MEDIAN
Here,
m= 46.2, So, Median lies in 20000-30000 L1 = 20000, f=35, C=15,
i=(30000-20000)=10000
𝑘𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
46.2−15
Or, 𝐷2 = 20000+
× 10000
35
Or, 𝐷2 = 28914.29
So, the solution is 28914.29.
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MEDIAN
Question
6:
Here given the
following
Incomplete
frequency
distribution. Find out decile.
10/2/2021
Sales
320-325
325-330
330-335
335-340
340-345
345-350
350-355
355-360
Total
Afzal Hossain (Assistant Professor)
No. of Companies
5
17
80
468
326
7
88
9
1000
265
2
MEDIAN
Solution:
We Know,
𝐾𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
For, 𝐷2 , K=2
𝐾𝑛 2×1000
2000
Here, m= =
=
=200
10
10
10
So, Median Lies in 335-340 and
L1 =335 , f=468, c=102 and
i= 340-335 =5
10/2/2021
Marks
Frequency
(f)
320-325
325-330
330-335
335-340
340-345
345-350
350-355
355-360
5
17
80
468
326
7
88
9
Afzal Hossain (Assistant Professor)
Cumulative
Frequency
(c)
5
22
102
570
896
903
991
1000
266
MEDIAN
Solution:
𝐾𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
200 − 102
𝐷2 = 335 +
×5
468
49
𝐷2 = 335+
×
234
245
𝐷2 = 335 +
234
𝐷2 = 336.05
10/2/2021
5
Marks
Frequency
(f)
320-325
325-330
330-335
335-340
340-345
345-350
350-355
355-360
5
17
80
468
326
7
88
9
Afzal Hossain (Assistant Professor)
Cumulative
Frequency
(c)
5
22
102
570
896
903
991
1000
267
MEDIAN
Question 7: Find the
Weekly Income(BDT)
No. of employees
Deciles Value from the
given data.
500-550
550-600
600-650
650-700
700-750
750-800
800-850
6
10
22
30
16
12
15
10/2/2021
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268
MEDIAN
Solutions:
We know,
𝐾𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
For, 𝐷2 , K=2
𝐾𝑛 2×111
222
Here, m= =
=
=22.2
10
10
10
So, Median Lies in 600-650 and
L1 =600 , f=22, c=16 and i=600650=50.
10/2/2021
Weekly
Income(BDT)
No. of employees(f)
Cumulative
frequency(c)
500-550
6
6
550-600
10
16
600-650
22
38
650-700
30
68
700-750
16
84
750-800
12
96
800-850
15
111
Afzal Hossain (Assistant Professor)
269
MEDIAN
Solution:
𝐾𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
22.2 − 16
𝐷2 = 600 +
× 50
22
𝐷2 = 600 + .282 × 50
𝐷2 =600+14.1
𝐷2 =614.1
22.2% Employees weekly income
is 614.1 BDT.
10/2/2021
Weekly
Income(BDT)
No. of
employees(f)
Cumulative
frequency(c)
500-550
6
6
550-600
10
16
600-650
22
38
650-700
30
68
700-750
16
84
750-800
12
96
800-850
15
111
Afzal Hossain (Assistant Professor)
270
MEDIAN
Question 8: An incomplete distribution is given below :
Variable : 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total
Frequency : 12
30
33
65
45
25
18
228
Find the median using deciles.
MEDIAN
Solution:
Variable
No. of
frequency (fi)
Cumulative
frequency(c)
10-20
20-30
30-40
40-50
50-60
60-70
70-80
12
30
33
65
45
25
18
12
42
75
140
185
210
228
Afzal Hossain (Assistant Professor)
272
MEDIAN
Solution:
We know, 𝐷𝑘 = 𝐿1 +
𝑘𝑛
−𝐶
10
𝑓
𝑘𝑛
10
×𝑖
2×228
10
For, D2,M=2. Here, 𝑚 = =
= 45.6
So, median lies in 600-800 and L1=600,f=11,c=15,i=800-600=200
45.6 − 15
∴ 𝐷𝑘 = 600 +
× 200
11
Or, 𝐷2 = 600 + 2.78 × 200
Or, 𝐷2 = 600 + 556.36 = 1156.36
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MEDIAN
Question 9: Consider the following data and calculate the deciles
value?
Class
10-20
Frequency 12
10/2/2021
20-30
30
30-40
33
40-50
65
Afzal Hossain (Assistant Professor)
50-60
45
60-70
25
70-80
18
274
MEDIAN
Class
No. of
frequency (f)
𝑪𝒖𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆
frequency
10-20
20-30
30-40
12
30
33
12
42
75
So,median lies in 30-40 and 𝐿1 =30,
40-50
50-60
60-70
65
45
25
140
185
210
f =33
70-80
18
N=228
228
Solution:
We know, 𝐷k = 𝐿1 +(
𝑘𝑛
−𝐶
10
𝑓
)𝑥𝑖
For, 𝐷2 , K=2
Here,m=
𝑘𝑛
10
=
2 𝑥 228
10
=
456
10
= 45.6
C=42 and i=40-30=10
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275
MEDIAN
Solution:
10-20
No. of frequency
(f)
12
𝑪𝒖𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆
frequency
12
20-30
30-40
40-50
30
33
65
42
75
140
50-60
60-70
70-80
45
25
18
185
210
228
Class
We know, 𝐷k = 𝐿1 +
𝐷2 = 30+(
𝑘𝑛
−𝐶
10
𝑓
× 𝑖
45.6−42
)
33
𝑥 10
𝐷2 = 30 + (0.109 𝑥 10)
𝐷2 = 30 + 1.09
𝐷2 = 31.09
N=228
So,45.6% of class will get the frequency of 31.09
10/2/2021
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MEDIAN
Question 10: Given below is the
frequency distribution of the marks
obtained by 90 student. Find out the
Deciles (D2) of the Given Data
10/2/2021
Afzal Hossain (Assistant Professor)
Marks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
No. of the students
2
12
15
20
18
10
9
4
277
MEDIAN
Solution:
Marks
No. of the
students (f)
Cumulative
Frequency (c)
20-29
2
2
30-39
12
14
40-49
15
29
50-59
20
49
60-69
18
67
70-79
10
77
80-89
9
86
90-99
4
90
Total
90
We Know,
𝐾𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
For, 𝐷2 , K=2
𝐾𝑛 2×90
180
Here, m= =
=
=18
10
10
10
So, Median Lies in 30-39 and
L1 =30 ,
f=12,
c=14 and
i=39-30=9.
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278
MEDIAN
Solution:
𝐾𝑛
−𝑐
10
𝐷𝐾 = 𝐿1 +
×𝑖
𝑓
18 − 14
𝐷2 = 30 +
×9
12
𝐷2 = 30 + .333 × 9
𝐷2 =30 + 2.997
𝐷2 =32.997
𝐷2 =33 (Approx)
So, 33 students get 18% marks.
10/2/2021
Marks
No. of the
students (f)
Cumulative
Frequency (c)
20-29
2
2
30-39
12
14
40-49
15
29
50-59
20
49
60-69
18
67
70-79
10
77
80-89
9
86
90-99
4
90
Total
90
Afzal Hossain (Assistant Professor)
279
PERCENTILE
SOME EXAMPLES
10/2/2021
Afzal Hossain (Assistant Professor)
280
MEDIAN
Question 1: Marks obtained by the 30 students of bba 06 in
Business Statistics course.
60,65,76,73,73,90,78,81,62,92,75,73,66,66,68,71,74,83,78,80,86,6
1,87,92,87,86,81,77,74,83
Find out the 15th percentile or 𝑃15 of this frequency.
10/2/2021
Afzal Hossain (Assistant Professor)
281
MEDIAN
Solution:
Marks
Here, for 𝑃15
𝑚𝑛
M=
60-64
65-69
70-74
75-79
80-84
85-89
90-94
=
100
15×30
100
𝑃15 = 𝐿1 +
= 65+
= 4.5
𝑀𝑛
−𝐶
100
4.5−3
4
10/2/2021
𝑓
×𝑖
× 4 = 66.5
No of
people (f)
3
4
6
5
5
4
3
Afzal Hossain (Assistant Professor)
Cumulitive(c)
3
7
13
18
23
27
30
MEDIAN
Question 2: The monthly income 0f 10 employees
working in a firm is as follows 4487 4493 4502 4446
4475 4492 4572 4516 4468 4489
Find out the 5th percentile or 𝑃5 of this frequency.
10/2/2021
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283
MEDIAN
Solution:
We know,
Income
Employ (fi)
4446-4488
4
Cumulative
(C)
4
Here,
𝑚𝑛
M=
4488-4530
5
9
1.5
So median lies in (4446-4488)
So we know 𝐿1 =4446, f=4, c=0,
i=(4488-4446)=42
45304572
1
N=10
10
𝑃15 = 𝐿1 +
100
15∗10
=
=
100
10/2/2021
𝑀𝑛
−𝐶
100
𝑓
×𝑖
Afzal Hossain (Assistant Professor)
284
MEDIAN
Solution:
We know,
𝑃15 = 𝐿1 +
𝑀𝑛
−𝐶
100
𝑓
×𝑖
1.5−0
=4446+(
)*42
4
=4446+15.75
4461.75
Income
Employ (f)
4446-4488
4
Comulative
(C)
4
4488-4530
5
9
45304572
1
N=10
10
Ans: 4461.75
10/2/2021
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285
MEDIAN
Question 3: A contractor employs three types of works- male, female
and children. To a male he pays Rs. 200 per day, to a female worker Rs.
150 per day and to child worker Rs. 100 per day. Now, Find out the
mean by using percentilies.
10/2/2021
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286
MEDIAN
Solution:
here N=3
Step-1: sort the value- 100,150,200
(𝑁+1)
(3+1)
4
Step-2: Median position:
th position =
th position = th
2
2
2
position = 2nd position
∴ 𝑀𝑒𝑑𝑖𝑎𝑛 =
150
2
= 75.
Afzal Hossain (Assistant Professor)
287
MEDIAN
Question 4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows.
Age (yrs.)
Find out the 3rd percentile
or 𝑷𝟑 of this frequency
using percentiles method.
10/2/2021
No. of
workers
Age (yrs.)
No. of
workers
18-22
22-26
26-30
120
125
280
38-42
42-46
46-50
184
162
86
30-34
34-38
260
155
50-54
54-58
75
53
Afzal Hossain (Assistant Professor)
288
MEDIAN
Age(yrs.)
Solution:
We know,
𝑃𝑀 =𝐿1 +
𝑀𝑛
−𝐶
100
𝑓
×𝑖
For 𝑃3 , M=3
𝑀𝑛
3×1500
Here,m=
=
=45
100
100
So,median lies in 18-22 and
𝐿1 =18,f=120,C=0 and i=22-18=4.
Frequency(f)
Cumulative
Frequency(c)
18-22
120
120
22-26
125
245
26-30
280
525
30-34
260
785
34-38
155
940
38-42
184
1124
42-46
162
1286
46-50
86
1372
50-54
75
1447
54-58
53
1500
Total
10/2/2021
Afzal Hossain (Assistant Professor)
1500
289
289
MEDIAN
Age(yrs.)
Frequency(f)
Cumulative
Frequency(c)
×i
18-22
120
120
22-26
125
245
×4
26-30
280
525
30-34
260
785
𝑃3 =18+.375×4
𝑃3 =27
34-38
155
940
38-42
184
1124
42-46
162
1286
⸫45% worker’s age is 27 yrs.
46-50
86
1372
50-54
75
1447
54-58
53
1500
Solution:
𝑃𝑀 =𝐿1 +
𝑃3 = 18+
𝑘𝑛
−𝑐
10
𝑓
45−0
120
Total
10/2/2021
Afzal Hossain (Assistant Professor)
1500
290
290
Question 5: Given below is the
frequency distribution of the
profit obtained by 250 companies.
Find out the 2nd percentile or
𝑷𝟐 of this frequency using
percentiles method.
10/2/2021
Profits (in BDT.)
10000-20000
20000-30000
30000-40000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000-100000
Afzal Hossain (Assistant Professor)
No. of companies
15
35
47
68
32
22
12
11
8
291
MEDIAN
Solution:
Profits (in BDT.)
10000-20000
20000-30000
30000-40000
40000-50000
50000-60000
60000-70000
70000-80000
80000-90000
90000-100000
No. of companies
15
35
47
68
32
22
12
11
Afzal Hossain (Assistant Professor)
8
Cumulative
frequency (C)
15
50
97
165
197
219
231
242
250
292
MEDIAN
Solution:
We know, 𝑃𝑀 = 𝐿1 +
𝑀𝑛
−𝐶
100
𝑓
𝑀𝑛
100
×𝑖
2×250
100
For 𝑃2 , M=2. Here, 𝑚 =
=
=5
So, median lies in 10000-20000 and 𝐿1 =10000,f=15,c=0,i=2000010000=10000
5−0
∴ 𝑃𝑀 = 10000 +
× 10000
15
Or, 𝑃2 = 10000 + 0.33 × 10000
Or, 𝑃2 = 10000 + 3300 = 13300
We can say, 5% company’s earned profit is 13300 BDT.
Afzal Hossain (Assistant Professor)
293
MEDIAN
Sales
No. of Companies
Question 6: Here given the following
320-325
5
Incomplete frequency distribution.
325-330
17
330-335
80
335-340
468
340-345
326
345-350
7
350-355
88
355-360
9
Total
1000
Find out the 53th percentile or P53 of this
frequency.
10/2/2021
Afzal Hossain (Assistant Professor)
294
294
MEDIAN
Solution:
Here we have to calculate P53
Mn
c
PM  L1  ( 100
)i
f
Here M = 53
53000
m= 53×1000=
= 530
100
So the median lies here in fourth row.
L1 = 335, m= 530, c = 102, i= 5
10/2/2021
Class
Interval
320-325
Frequency
(f)
5
Cumulative
Frequency
5
325-330
330-335
335-340
17
80
468
22
102
570
340-345
345-350
350-355
326
7
88
896
903
991
355-360
Total
9
1000
1000
Afzal Hossain (Assistant Professor)
295
295
MEDIAN
Mn
c
PM  L1  ( 100
)i
f
P53 = 335+ (428÷468)×5
= 335+4.573
= 339.572
So the answer is 339.572 .
10/2/2021
Class
Interval
Frequency
(f)
Cumulative
Frequency
320-325
325-330
5
17
5
22
330-335
335-340
340-345
80
468
326
102
570
896
345-350
350-355
355-360
7
88
9
903
991
1000
Total
1000
Afzal Hossain (Assistant Professor)
296
MEDIAN
Question 7: Here given
the following frequency
distribution.
th
Find out the 30
Percentiles or P30 of this
frequency.
10/2/2021
Weekly Income(BDT)
No. of employees
500-550
550-600
600-650
6
10
22
650-700
700-750
750-800
30
16
12
800-850
15
Afzal Hossain (Assistant Professor)
297
MEDIAN
Solution:
Here, we have to calculate 𝑃30
So, Median lies in 600-650 and
𝐿1 = 600, f= 22,c=16 and
i=650-600 = 50.
10/2/2021
Weekly
Income(BDT)
500-550
550-600
600-650
650-700
700-750
750-800
800-850
Afzal Hossain (Assistant Professor)
No. of
Cumulative
employees(f) Frequency (c)
6
6
10
16
22
38
30
68
16
84
12
96
15
111
298
MEDIAN
Solution:
Here,
= 600 + 0.78636
= 600+ 39.31818
= 639.318
So, the answer is 639.318
10/2/2021
Weekly
Income(BDT)
500-550
550-600
600-650
650-700
700-750
750-800
800-850
Afzal Hossain (Assistant Professor)
No. of
Cumulative
employees(f) Frequency (c)
6
6
10
16
22
38
30
68
16
84
12
96
15
111
299
MEDIAN
Weekly Rent (in Rs)
Question 8: Find the
median from the given data
using percentile calculation.
rd
Find out the 3 Percentiles
or 𝑷𝟑 of this frequency.
10/2/2021
200-400
400-600
No. of Persons Paying
the Rent
6
9
600-800
11
800-1000
1000-1200
14
20
1200-1400
1400-1600
15
10
1600-1800
1800-2000
8
7
Afzal Hossain (Assistant Professor)
300
MEDIAN
Solutions:
Weekly Rent
(in BDT)
No. of Persons
Paying the Rent
(f)
Cumulative
Frequency
(c)
200-400
6
6
400-600
9
15
600-800
11
26
800-1000
14
40
1000-12000
20
60
1200-1400
15
75
1400-1600
10
85
1600-1800
8
93
1800-2000
7
100
Total
100
We know,
𝑃𝑀 =𝐿1 +
𝑃3 =200+
𝐾𝑛
−𝐶
10
𝑓
3−0
6
×𝑖
×200
𝑃3 =200+100
𝑃3 =300
3% workers weekly rent is 300 BDT.
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301
MEDIAN
Solutions:
Weekly Rent
(in BDT)
We know,
No. of Persons
Paying the Rent
(f)
Cumulative
Frequency
(c)
200-400
6
6
Here, m =3
400-600
9
15
600-800
11
26
800-1000
14
40
1000-12000
20
60
f=6
1200-1400
15
75
1400-1600
10
85
C=0
1600-1800
8
93
1800-2000
7
100
Total
100
So, Median lies in 200-400 and
𝐿1 =200
i=400-200=200.
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Afzal Hossain (Assistant Professor)
302
MEDIAN
Question 9: Find the
median from the given data
using percentile calculation.
rd
Find out the 3
Percentiles or 𝑷𝟑 of this
frequency.
10/2/2021
Class
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Afzal Hossain (Assistant Professor)
Frequency
12
30
33
65
45
25
18
303
MEDIAN
Solutions:
Class
Frequency
(f)
Cumulative
Frequency
(c)
10-20
12
12
20-30
30
42
30-40
33
75
40-50
65
140
50-600
45
185
60-70
25
210
70-80
18
228
Total
228
We know,
For ,M=3 ; m = 6.84
So,Median lies in 10-20 and
𝐿1 =10
f=12
C=0
i=20-10=10.
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Afzal Hossain (Assistant Professor)
304
MEDIAN
Solutions :
Class
Frequency
(f)
Cumulative
Frequency
(c)
10-20
12
12
20-30
30
42
30-40
33
75
40-50
65
140
50-600
45
185
60-70
25
210
𝑃3 =10+2.5
70-80
18
228
𝑃3 =12.5
Total
228
We know,
𝐾𝑛
−𝐶
10
𝑃𝑀 =𝐿1 +
𝑃3 =10+
𝑓
3−0
12
×𝑖
×10
Answer: 3% Classes’ Frequency is 12.5
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305
MEDIAN
Question 10: Given below is the
frequency distribution of the marks
obtained by 90 student.
Find out the 3rd Percentiles or 𝑷𝟑 of this
frequency.
10/2/2021
Afzal Hossain (Assistant Professor)
Marks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
No. of the students
2
12
15
20
18
10
9
4
306
MEDIAN
Marks
Solution:
We know,
Pm = 𝐿1 +
𝑀𝑛
−𝐶
100
𝑓
×i
For, 𝑃3 , M = 3
Here , m =
𝑀𝑛 3×90
=
=
100 100
2.7
So median 30-39
And 𝐿1 =30, f =12,C=14
i=39-30 =9
10/2/2021
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
Afzal Hossain (Assistant Professor)
Total
Frequency
2
12
15
20
18
10
9
4
90
Cumulative
Frequency
2
14
29
49
67
77
86
90
307
MEDIAN
Solution:
𝑃𝑚 = 𝐿1 +
𝑃3 =30 +
𝐾𝑛
(
10
-c)×𝑖
2.7−14
(
12
)×9
𝑃3 = 30 + (- 0.94167) × 9
𝑃3 = 21.52497
2.7 % students get marks 21.524
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308
BUSINESS STATISTICS
MEAN , MEDIAN AND MODE
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309
MODE
Definition: The "mode" is the value that
occurs most often. If no number in the
list is repeated, then there is no mode for
the list.
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310
MODE
Mode for Ungroup Data: For
example: The Mode of 13, 18,
13, 14, 13, 16, 14, 21, 13 are :
The mode is the number that is
repeated more often than any
other, so 13 is the mode.
Marks
Obtain
13
14
16
18
21
Tally
Frequency
IIII
II
4
2
I
I
I
1
1
1
Total 9
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311
MODE
Mode for Group Data: Mode is
the value that has the highest
frequency in a data set. •For
grouped data, class mode (or,
modal class) is the class with the
highest frequency. •To find mode
for grouped data, use the
following formula:
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
10/2/2021
Where, 𝐿1 = The lower value of the class
in which the mode lies
fl = The frequency of the class in which
the mode lies
fo = The frequency of the class preceding
the modal class
f2 = The frequency of the class
succeeding the modal class
i = The class-interval of the modal class
Afzal Hossain (Assistant Professor)
312
MODE
Example: Let us take the given
frequency distribution and Find out
the mode.
10/2/2021
Class Intervals
30-40
40-50
50-60
60-70
70-80
80-90
90-100
Afzal Hossain (Assistant Professor)
Frequency
4
6
8
12
9
7
4
313
MODE
Solution: We can see from Column
(2) of the table that the maximum
frequency of 12 lies in the classinterval of 60-70. This suggests that
the mode lies in this class- interval.
Applying the formula given earlier,
we get: 𝐿1 = 60, 𝑓1 = 12,
𝑓0 = 8, 𝑓2 = 9 and i=70-60=10
10/2/2021
Class Intervals
(i)
30-40
40-50
50-60
60-70
70-80
80-90
90-100
Afzal Hossain (Assistant Professor)
Frequency
(f)
4
6
8
12
9
7
4
314
MODE
Solution: Now
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
12 − 8
𝑀 = 60 +
× 10
2 × 12 − 8 − 9
4
𝑀 = 60 +
× 10
24 − 17
40
𝑀 = 60 +
7
M=60+5.71=65.71
10/2/2021
Class Intervals
(i)
30-40
40-50
50-60
60-70
70-80
80-90
90-100
Afzal Hossain (Assistant Professor)
Frequency
(f)
4
6
8
12
9
7
4
315
MODE
Mode for Group Data Using Method of Grouping: In several cases,
just by inspection one can identify the class-interval in which the mode
lies. One should see which the highest frequency is and then identify to
which class-interval this frequency belongs. Having done this, the
formula given for calculating the mode in a grouped frequency
distribution can be applied. At times, it is not possible to identify by
inspection the class where the mode lies. In such cases, it becomes
necessary to use the method of grouping. This method consists of two
parts: Preparation of a grouping table and Preparation of an analysis
table.
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316
MODE
Preparation of a Grouping Table: A grouping table has six columns, the
first column showing the frequencies as given in the problem. Column 2
shows frequencies grouped in two's, starting from the top. Leaving the
first frequency, column 3 shows frequencies grouped in two's. Column 4
shows the frequencies of the first three items, then second to fourth item
and so on. Column 5 leaves the first frequency and groups the remaining
items in three's. Column 6 leaves the first two frequencies and then
groups the remaining in three's. Now, the maximum total in each column
is marked and shown either in a circle or in a bold type.
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317
MODE
Preparation of an Analysis Table: After having prepared a grouping
table, an analysis table is prepared. On the left-hand side, provide the first
column for column numbers and on the right-hand side the different
possible values of mode. The highest values marked in the grouping table
are shown here by a bar or by simply entering 1 in the relevant cell
corresponding to the values they represent. The last row of this table will
show the number of times a particular value has occurred in the grouping
table. The highest value in the analysis table will indicate the classinterval in which the mode lies.
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318
MODE
Example1: Find out the mode of given
Data.
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Afzal Hossain (Assistant Professor)
Size of Item
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
10
18
25
26
17
4
319
MODE
Solution: Preparation of a Grouping Table
Size of Item
10-20
20-30
30-40
40-50
50-60
60-70
10/2/2021
1
10
18
25
26
17
4
2
28
Column Number
3
4
43
5
6
53
69
51
43
68
47
21
Afzal Hossain (Assistant Professor)
320
MODE
Solution: Preparation of an Analysis Table
Col No
1
2
3
4
5
6
Total
10/2/2021
10-20
1
1
20-30
Size of Item
30-40
40-50
1
50-60
1
1
1
1
1
1
1
1
1
1
1
1
1
5
5
2
3
Afzal Hossain (Assistant Professor)
60-70
1
321
MODE
Solution: This is a bi-modal series as is evident from the analysis table,
which shows that the two classes 30-40 and 40-50 have occurred five
times each in the grouping (If the repetition of 5 is not occurred i.e.
Only one 5 is occurred in analysis table then we will apply normal
formula of Mode ). In such a situation, we may have to determine mode
indirectly by applying the following formula:
Mode = 3 ×median - 2 ×mean
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Afzal Hossain (Assistant Professor)
322
MEDIAN
Solution (Median Calculation) :
We Know
𝐿2 − 𝐿1
𝑀 = 𝐿1 +
𝑚−𝑐
𝑓
𝑛+1 100+1
=
=50.5
2
2
Here, m=
So, Median Lies in 30-40
L1 =30 and L2 = 40, f=25, c=28.
10/2/2021
Size of
Item
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
(f)
10
18
25
26
17
4
Afzal Hossain (Assistant Professor)
Cumulative
Frequency (c)
10
28
53
79
96
100
323
MEDIAN
Solution (Median Calculation) :
We Know,
𝐿2 − 𝐿1
𝑀 = 𝐿1 +
𝑚−𝑐
𝑓
40 − 30
𝑀 = 30 +
50.5 − 28
25
10
𝑀 = 30 +
× 22.5
25
𝑀 = 30 + .4 × 22.5
𝑀 = 30 + 9 = 39
10/2/2021
Size of
Item
10-20
20-30
30-40
40-50
50-60
60-70
Afzal Hossain (Assistant Professor)
Frequenc
y (f)
10
18
25
26
17
4
Cumulative
Frequency (c)
10
28
53
79
96
100
324
MEDIAN
Solution (Mean Calculation) :
We Know
Here n=100, A=35, Class Interval
C=10
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
μ=𝐴+
×𝐶
𝑛
34
𝑂𝑟, μ = 35+
× 10
100
Or, μ = 35 + .34 × 10
Or, μ = 35+3.4
Or, μ=38.4
10/2/2021
Size
of
Item
10-20
20-30
30-40
40-50
50-60
60-70
Freque Mid-point Deviation
ncy (fi)
(mi)
(di)
10
18
25
26
17
4
Afzal Hossain (Assistant Professor)
15
25
35
45
55
65
fidi
-2
-1
0
1
2
3
-20
-18
0
26
34
12
n
i=1 fi 𝑑i = 34
325
MEDIAN
Solution (Mode Calculation) :
Finally,
Mode = 3 ×median - 2 ×mean
= 3 ×39 - 2 ×38.4
= 117- 76.8
= 40.2
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326
MODE
Example 2: Find out the mode of given
Data.
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Afzal Hossain (Assistant Professor)
Size of Item
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
13
20
23
23
17
4
327
MODE
Solution: Preparation of a Grouping Table
Size of Item
10-20
20-30
30-40
40-50
50-60
60-70
10/2/2021
1
13
20
23
23
17
4
2
33
Column Number
3
4
43
5
6
56
66
46
40
63
44
21
Afzal Hossain (Assistant Professor)
328
MODE
Solution: Preparation of an Analysis Table
Col No
1
2
3
4
5
6
Total
10/2/2021
10-20
1
1
20-30
Size of Item
30-40
40-50
1
1
50-60
1
1
1
1
1
1
1
1
1
1
1
6
4
1
3
60-70
1
Afzal Hossain (Assistant Professor)
329
MODE
Solution: From the analysis table we have fund
that maximum value is 6 which lies in class
interval 30-40. This suggests that the mode lies
in this class- interval. Applying the formula
given earlier, we get: 𝐿1 = 30, 𝑓1 = 23 ,
𝑓0 = 20, 𝑓2 = 23 and i=40-30=10.
Since we have maximum values. So we apply
following formula.
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
10/2/2021
Afzal Hossain (Assistant Professor)
Size of Item
10-20
20-30
30-40
40-50
50-60
60-70
Frequency
13
20
23
23
17
4
330
MODE
Solution: Now
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
23 − 20
𝑀 = 30 +
× 10
2 × 23 − 20 − 23
3
𝑀 = 30 +
× 10
46 − 43
30
𝑀 = 30 +
3
M=30+10=40
10/2/2021
Size of Item
10-20
20-30
30-40
40-50
50-60
60-70
Afzal Hossain (Assistant Professor)
Frequency
13
20
23
23
17
4
331
MODE
Question-1: Marks obtained by the 30 students of BBA-06 in Business
Statistics course60, 65, 76, 73, 73, 90, 78, 81, 62, 92, 75, 73, 66, 66,
68, 71, 74, 83, 78, 80, 86, 61, 87, 92, 87, 86, 81, 77,
74, 83
332
MODE
Solutions- Mode for group data
Here
L1= 60 ,f1=5 ,i=5 ,f0=6 ,f2=5
We know,
M = L1+
𝑓1−𝑓0
∗𝑖
2𝑓1−𝑓0−𝑓2
= 60 +
5−6
2∗5−6−5
= 60 (Ans)
∗5
Marks (i)
60-65
No. of people
(f)
3
65-69
70-74
75-79
4
6
5
80-84
85-89
90-94
5
4
3
Total =30
MODE SOLUTION
QUESTION-2 : The monthly income 0f 10 employees working in a firm is as
follows (Average short cut) 4487 4493 4502 4446 4475 4492 4572 4516
4468 4489 Calculating average monthly income by Mode.
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334
MODE SOLUTION
Solution (Median Calculation) :
We Know,
𝐿2 −𝐿1
𝑀=𝐿1 +
𝑚−𝑐
Size of Item
Frequency (f)
Cumulative
Frequency (c)
𝑓
4446-4478
2
2
= 5.5
4478-4510
4510-4542
3
4
5
9
So, Median Lies in 4510-4542
4542-4574
1
10
Here, m=
n+1
2
=
10+1
2
𝐿1 =4510 and 𝐿2 =4542, f=4, c=5
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335
MODE SOLUTION
Solution (Median Calculation) :
We Know,
𝑀=𝐿1 +
𝐿2−𝐿1
𝑀=4510+
𝑓
𝑚−𝑐
4542−4510
4
𝑀=4510+8×0.5
(5.5-5)
Size of Item
Frequency (f) Cumulative
Frequency (c)
4446-4478
2
2
4478-4510
3
5
4510-4542
4
9
4542-4574
1
10
𝑀= 4514
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336
MODE SOLUTION
Solution (Mean Calculation) :
We Know
Here n=10, A=4526, Class Interval C=32
μ=𝐴+
𝑛
𝑖=1 𝑓𝑖 𝑑𝑖
×𝐶
𝑛
−128
𝑂𝑟, μ = 4526+
× 32
10
Or, μ = 4526 − 12.8 × 32
Or, μ = 4526−409.6
Or, μ=4116.4
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337
MODE SOLUTION
Size of Item Frequency
(f)
Mid-point
(mi)
Deviation
(di)
4446-4478
2
4462
-32
-64
4478-4510
3
4494
-32
-96
4510-4542
4
4526
0
0
4542-4574
1
4558
32
32
n
i=1 fi 𝑑i =
10/2/2021
fidi
Afzal Hossain (Assistant Professor)
-128
338
MODE SOLUTION
Solution (Mode Calculation) :
Finally,
Mode = 4 ×median - 5 ×mean
= 4 ×4514 - 5 ×4116.4
= 18056- 20582
= -2526
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339
MODE
Question 3: A contractor
employs three types of worksmale, female and children. To
a male he pays Rs. 200 per
day, to a female worker Rs.
150 per day and to child
worker Rs. 100 per day. Now,
Find out the mean by using
Mode method .
10/2/2021
Salary
No. of people
200
20
150
15
100
5
Afzal Hossain (Assistant Professor)
340
MODE
Solutions- Mode for group data
Here
L1= 200 ,f1=20 ,i=50 ,f0=0 ,f2=15
We know,
M= L1+
𝑓1−𝑓0
2𝑓1−𝑓0−𝑓2
∗𝑖
=200 +40
=240 ( Ans )
10/2/2021
Afzal Hossain (Assistant Professor)
341
MODE
Question-4: 1,500 workers are working in an industrial establishment. Their age
is classified as follows. Find out the mode of given data.
Age (yrs.)
10/2/2021
No. of workers
18-22
120
22-26
125
26-30
280
30-34
260
34-38
155
38-42
184
42-46
162
46-50
86
50-54
75
54-58
53
MODE
Solution: Preparation of a Grouping Table
Age (yrs.)
1
18-22
120
22-26
125
26-30
280
30-34
260
34-38
155
38-42
184
42-46
162
46-50
86
50-54
75
54-58
53
Column Number
2
245
3
4
5
6
665
695
405
525
540
501
415
339
346
161
599
432
248
323
128
53
214
MODE
Solution: Preparation of an Analysis Table
Size of Item
Col No 18-22
22-26
26-30
1
1
2
1
30-34
34-38
1
1
4
1
1
1
6
Total
10/2/2021
1
46-50
50-54
54-58
1
3
5
38-42 42-46
1
1
1
1
1
4
5
3
1
1
Afzal Hossain (Assistant Professor)
344
MODE
Solution: This analysis table
shows that the (30-34) class has
occurred 260 times in the grouping. In such a situation, we may have
to determine mode directly by applying the traditional mode
calculating formula.
10/2/2021
Afzal Hossain (Assistant Professor)
345
MODE
Age (yrs.)
Solution: Mode
calculationHere ,
𝐿1 = 30
𝑓1 = 260
𝑓0 = 280
𝑓2 = 155
i=30-34=4
18-22
22-26
26-30
30-34
34-38
38-42
42-46
46-50
50-54
54-58
No. of workers
120
125
280
260
155
184
162
86
75
53
MODE
Age (yrs.)
Solution: Now
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
260 − 280
𝑀 = 30 +
×4
2 × 260 − 280 − 155
−20
𝑀 = 30 +
×4
85
𝑀 = 30 − 0.2353 × 4
M =29.0588
No. of workers
18-22
22-26
120
125
26-30
280
30-34
34-38
260
155
38-42
42-46
46-50
184
162
86
50-54
54-58
75
53
ARITHMETIC MEAN
Question-5:
Find out the mode of given data.
Profits (in BDT.)
No. of companies
10000-20000
15
20000-30000
35
30000-40000
47
40000-50000
68
50000-60000
32
60000-70000
22
70000-80000
12
80000-90000
11
90000-100000
8
MODE
Solution: Preparation of a Grouping Table
Profits
10000-20000
1
15
20000-30000
35
30000-40000
47
40000-50000
68
50000-60000
32
60000-70000
22
70000-80000
12
80000-90000
11
90000-100000
8
Column Number
3
4
97
2
50
82
5
6
150
115
147
100
122
54
66
34
23
45
31
19
MODE
Solution: Preparation of an Analysis Table
Profits
10000- 20000- 30000- 40000- 50000- 60000- 70000- 80000- 90000Col
20000 30000 40000 50000 60000 70000 80000 90000 100000
No
1
2
3
4
5
6
Total
10/2/2021
1
1
1
1
1
3
1
1
1
1
1
1
1
1
1
6
3
Afzal Hossain (Assistant Professor)
1
1
350
MODE
Solution: This analysis table shows that the (40000-50000) class has
occurred six times in the grouping. In such a situation, we may have to
determine mode directly by applying the traditional mode calculating
method.
10/2/2021
Afzal Hossain (Assistant Professor)
351
MODE
Solution: We can see from Column
(2) of the table that the maximum
frequency of 68 lies in the classinterval of 40000-50000. This
suggests that the mode lies in this
class- interval. Applying the formula,
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
Profits (in BDT.)
No. of companies
10000-20000
15
20000-30000
35
30000-40000
47
40000-50000
68
50000-60000
32
60000-70000
22
70000-80000
12
80000-90000
11
90000-100000
8
MODE
Solution:
we get: 𝐿1 = 40000, 𝑓1 = 68, 𝑓0 = 47, 𝑓2 = 32 and i=50000-40000=10000
Now put these value in the formula,
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
68 − 47
𝑀 = 40000 +
× 10000
2 × 68 − 47 − 32
21
𝑀 = 40000 +
× 10000
136 − 79
𝑀 = 40000 + 3684.2105
M=43,684.2105
MODE
sales
320-325
Question6:
Find out the mode of given data. 325-330
companies
5
17
330-335
80
335-340
468
340-345
326
345-350
7
350-355
88
355-360
9
354
MODE
Solution: Preparation of grouping table
Column Number
sales
1
2
320-325
5
22
325-330
17
330-335
80
335-340
468
340-345
326
345-350
7
350-355
88
355-360
9
3
4
97
102
5
6
565
548
875
794
801
333
95
104
355
MODE
Sales
Solution:
Col
No
Preparation of an
1
Analyzing table
2
320325
325330
330335
335340
340345
1
355360
1
1
1
4
1
1
1
6
Total
350355
1
3
5
345350
1
1
1
1
1
1
3
6
3
1
1
356
MODE
Solution:
Here,
𝐿1 =335, 𝑓1 =468, 𝑓0 = 80, 𝑓2 =326
And i =340-335=5
sales
No. of companies
320-325
5
325-330
17
330-335
80
335-340
468
340-345
326
345-350
7
350-355
355-360
88
9
357
MODE
We know,
𝑓1 −𝑓0
M=𝐿1 +
2𝑓1 −𝑓0 −𝑓2
×𝑖
M=335+
468−80
2×468−80−326
M=335+
388
530
×5
×5
1940
M=335+
530
M=335+3.66
M=338.6
358
ARITHMETIC MEAN
Question-7: Find out the mode of
given Data.
10/2/2021
Weekly
Income(BDT)
500-550
550-600
600-650
650-700
700-750
750-800
800-850
Afzal Hossain (Assistant Professor)
No. of employees
6
10
22
30
16
12
15
359
MODE
Solution: Preparation of a Grouping Table
Weekly
Income(BDT)
500-550
550-600
1
6
10
600-650
650-700
700-750
22
30
16
750-800
800-850
12
15
10/2/2021
2
Column Number
3
4
5
6
16
32
38
62
52
68
46
58
28
43
Afzal Hossain (Assistant Professor)
360
MODE
Solution: Preparation of an Analysis Table
2
3
Size of Item
500-550 550-600 600-650 650-700 700-750 750-800 800-850
1
1
1
1
1
4
5
6
1
1
1
1
1
1
1
Total
1
3
6
3
Col No
1
10/2/2021
1
1
Afzal Hossain (Assistant Professor)
1
361
MODE
Solution: From the analysis table we have fund
that maximum value is 6 which lies in class
interval 650-700. This suggests that the mode
lies in this class- interval. Applying the formula
given earlier, we get: 𝐿1 = 650, 𝑓1 = 30 ,
𝑓0 = 23, 𝑓2 = 16 and i=550-500=50.
Since we have maximum values. So we apply
following formula.
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
10/2/2021
Afzal Hossain (Assistant Professor)
Weekly
Income(BDT)
No. of employees
500-550
550-600
6
10
600-650
650-700
700-750
22
30
16
750-800
800-850
12
15
362
MODE
Solution: Now
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
30 − 23
𝑀 = 30 +
× 10
2 × 30 − 23 − 16
7
𝑀 = 30 +
× 10
60 − 39
70
𝑀 = 30 +
21
M=30+3.33=33.33
10/2/2021
Weekly
Income(BDT)
No. of employees
500-550
550-600
6
10
600-650
650-700
700-750
22
30
16
750-800
800-850
12
15
Afzal Hossain (Assistant Professor)
363
ARITHMETIC MEAN
Question-8: Find the arithmetic
mean from the given data using
short cut method
10/2/2021
Weekly Rent (in Rs)
No. of Persons Paying
the Rent
200-400
400-600
600-800
6
9
11
800-1000
1000-1200
1200-1400
14
20
15
1400-1600
1600-1800
10
8
1800-2000
7
Afzal Hossain (Assistant Professor)
364
MODE
Solution: Grouping tableWeekly Rent (in Rs)
1
200-400
6
400-600
9
600-800
11
800-1000
14
1000-1200
20
1200-1400
15
1400-1600
10
1600-1800
8
1800-2000
7
10/2/2021
2
Column Number
3
4
5
6
15
20
26
25
34
34
35
45
49
25
45
18
33
15
25
Afzal Hossain (Assistant Professor)
365
MODE
Solution: Preparation of an Analysis Table
Col No
200400
400600
Marks
600800800
1000
12001400
1
10001200
1
2
1
1
14001600
3
1
1
4
1
1
1
1
1
1
3
1
5
6
1
1
1
Total
1
3
6
10/2/2021
Afzal Hossain (Assistant Professor)
16001800
18002000
366
MODE
Solution: 𝐿1 = 1000, 𝑓1 = 20,
𝑓0 = 14, 𝑓2 = 15 and i=1200-1000=200
Weekly Rent (in
Rs)
No. of Persons
Paying the Rent
200-400
400-600
600-800
6
9
11
800-1000
1000-1200
1200-1400
14
20
15
1400-1600
1600-1800
10
8
1800-2000
7
Now,
𝑓1 − 𝑓0
𝑀 = 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
20 − 14
𝑀 = 1000 +
× 200
2 × 20 − 14 − 15
6
𝑀 = 1000 +
× 200
40 − 29
1200
𝑀 = 1000 +
11
𝑀 = 1109.09
10/2/2021
Afzal Hossain (Assistant Professor)
367
MODE
Question-9: Consider the following data and Find the mode
Class
10-20 20-30
Frequency 12
30
10/2/2021
30-40
33
40-50
65
Afzal Hossain (Assistant Professor)
50-60
45
60-70
25
70-80
18
368
MODE
Solution: Grouping tableMarks
10-20
20-30
30-40
40-50
50-60
60-70
70-80
10/2/2021
1
12
30
33
65
45
25
18
Column Number
3
4
2
5
6
42
63
75
98
128
110
70
143
135
43
Afzal Hossain (Assistant Professor)
88
369
MODE
Solution: Preparation of an Analysis Table
Marks
Col No
10-20 20-30 30-40 40-50
1
1
2
1
1
3
1
4
1
5
1
1
1
6
1
1
Total
1
3
6
10/2/2021
Afzal Hossain (Assistant Professor)
50-60
60-70 70-80
1
1
1
1
3
1
370
MODE
Solution: We can see from Column
(2) of the table that the maximum
frequency of 65 lies in the classinterval of 40-50. This suggests that
the mode lies in this class- interval.
Applying the formula given earlier,
we get: 𝐿1 = 40, 𝑓1 = 65 ,
𝑓0 = 33, 𝑓2 = 45 and i=40-50=10
10/2/2021
Class
(i)
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Afzal Hossain (Assistant Professor)
Frequency
(f)
12
30
33
65
45
25
18
371
MODE
Solution: 𝑀 = 𝐿 +
1
𝑓1 −𝑓0
2𝑓1 −𝑓0 −𝑓2
×𝑖
65 − 33
𝑀 = 40 +
× 10
2 × 65 − 33 − 45
32
𝑀 = 40 +
× 10
130 − 78
320
𝑀 = 40 +
52
Class Intervals
(i)
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Frequency
(f)
12
30
33
65
45
25
18
M=40+6.15=46.15
10/2/2021
Afzal Hossain (Assistant Professor)
372
MODE
Question 10: Given below is the
frequency distribution of the marks
obtained by 90 students. Find out
mode.
10/2/2021
Afzal Hossain (Assistant Professor)
373
MODE
Solution: We can see from Column (2)
of the table that the maximum frequency
of 20 lies in the marks of 50-59. This
suggests that the mode lies in this classinterval. Applying the formula given
earlier, we get:
𝐿1 = 50, 𝑓1 = 20, 𝑓0 = 15, 𝑓2 = 18 and
i=59-50=9
10/2/2021
Afzal Hossain (Assistant Professor)
374
MODE
Solution: Grouping tableMarks
20-29
30-39
40-49
50-59
60-69
70-79
80-89
90-99
10/2/2021
1
2
12
15
20
18
10
9
4
2
14
Column Number
3
4
27
5
6
29
47
35
38
53
48
28
13
19
Afzal Hossain (Assistant Professor)
37
23
375
MODE
Solution: Preparation of an Analysis Table
Marks
Col No
20-29 30-39 40-49 50-59
1
1
2
1
1
3
1
4
1
5
1
1
1
6
1
1
Total
1
3
6
10/2/2021
Afzal Hossain (Assistant Professor)
60-69
70-79 80-89 90-99
1
1
1
1
3
1
376
MODE
Solution: Now,
𝑀
𝑀
𝑀
𝑀
𝑀
𝑓1 − 𝑓0
= 𝐿1 +
×𝑖
2𝑓1 − 𝑓0 − 𝑓2
20 − 15
= 50 +
×9
2 × 20 − 15 − 18
5
= 50 +
×9
40 − 33
45
= 50 +
7
= 56.4286
10/2/2021
Afzal Hossain (Assistant Professor)
377
THANK YOU
10/2/2021
Afzal Hossain (Assistant Professor)
378