Post on 24-Apr-2023
SYLLABUS
INRODUCTION OF BUSINESS STATISTICS-DEFINITION-VARIABLES-QUALITATIVE AND
QUANTITATIVE DATA’S-PRIMARY AND SECONDARY DATA’S-COLLECTION OF DATA’S-CENSUS
METHODS-SAMPLING METHODS-PRECAUTIOUS BY USING SECONDARY DATA.
CLASSIFICATION AND PRESENTATION OF DATA’S-TABULATION-FREQUENCY DISTRIBUTION-
DIAGRAMS AND GRAPHICAL REPRESENTATION OF DATA’S-BAR DIAGRAMS-PIE DIAGRAMS-FREQUENCY
CURVES-OGIVE CURVES-HISTOGRAM-POLYGON-LORENZ CURVE.
MEASURES OF CENTRAL TENDENCY-MEAN-MEDIAN AND-MODE-MEASURE OF VARIATION-
RANGE-QUARTILE DEVIATION, STANDARD DEVIATION, MEAN DEVIATION AND COEFFICIENT-
CHARACTERISTICS-USES AND LIMITATIONS ADVANTAGES- DISADVANTAGES.
UNIT – 1
UNIT – 2
UNIT - 3
SYLLABUS
SKEWNESS TYPES AND MEASUREMENTS-KARL PEARSON-BOWLEY
TIME SERIES-COMPONENTS-MODELS, TREND MEASUREMENTS (GRAPHIC AND STRAIGHT LINE
METHOD) SEASONAL INDEX MEASUREMENTS-SIMPLE AVERAGE AND MOVING AVERAGE METHOD
REFERENCE:
S.B. GUPTA – SULTAN CHAND
V.R. VICTOR – MARGHAM PUBLICATIONS
UNIT – 4
UNIT – 5
BUSINESS STATISTICS
FOR EXAMPLE:
Statistics is an imposing form of mathematics whereas suggest tables,
charts, and figures which is found in newspapers, journal's, books,
radio and lectures etc.,
BUSINESS STATISTICS
There are 932 females for every 1000 males in india.
Whereas in russia, there are 1170 females per 1000 males. These statements contains
figures and they may be called as numerical statements of facts. They are highly
convenient form of communication. At the same time they are clear precise and
meaningful.
Facts and figures related to population, production, national income, profit, sales, births
and deaths etc., are all connected with statistics.
Statistics refers to quantitative information.
MEASURES OF CENTRAL TENDENCY OR
AVERAGES
CENTRAL TENDENCY OR AVERAGES
Objectives of statistical analysis is to get one single value that describes the
characteristics of the entire mass of data's. Such value is called central value or
average.
FOR EXAMPLE:
Average height or life of an indian, average income, average marks
and average sales etc.,
MEAN OR ARITHEMATIC MEAN
MEAN OR ARITHEMATIC MEAN
Count the number of times a particular value
is repeated is called the frequency of that
class
a) Individual observation
b) Discrete series
c) Continuous series
FREQUENCY DISTRIBUTION
SUM NO:1- THE MARKS OBTAINED BY 25 STUDENTS A CLASS
10 20 20 30 40 25 25 30 40 20 25 25 50 15 25 30 40
50 40 50 30 25 25 15 40
MARKS (X) TALLY BARS FREQUENCY
10
15
20
25
30
40
50
I
II
III
IIII II
IIII
IIII
III
1
2
3
7
4
5
3
N = 25
SUM NO: 2 – THE WEEKLY WAGES PAID TO 30 WORKERS IN A FACTORY GIVEN
BELOW. CALCULATE FREQUENCY TABLE.
3000 3500 2000 3000 2250 2000 2000 3000 2250 2000 3000 2250
3000 2250 2000 3000 2250 2000 3000 3000 2250 3000 2000 2000
3000 2250 3500 3000 2000 2250
WAGES (X) TALLY BARS FREQUENCY
2000
2250
3000
3500
IIII IIII
IIII III
IIII IIII I
II
9
8
11
2
N = 30
I. MEAN OR ARITHMETIC MEAN
a) INDIVIDUAL OBSERVATION
N = Total Frequency
b) DISCRETE SERIES
C) CONTINUOUS SERIES
𝑋 =𝛴𝑓𝑚
𝑁
𝑋 = 𝐴 +𝛴𝑓𝑑
𝛴𝑓 SHORT CUT METHOD
SUM NO: 1- FOLLOWING TABLES GIVES A MARKS OBTAINED BY 10
STUDENTS IN A CLASS. CALCULATE ARITHMETIC MEAN (AM)
X 40 50 30 60 70 80 40 50 60 90
X
40
50
30
60
70
80
40
50
60
90
Σx =570
𝑋 =570
10= 57
𝑋 = 57
a) INDIVIDUAL OBSERVATION
SUM NO: 2- CALCULATE ARITHMETIC MEAN FROM THE FOLLOWING DATA.
FIND THE ARITHMETIC MEAN .
MARKS 20 30 40 50 60 70
X
20
30
40
50
60
70
Σx = 270
𝑋 =𝛴𝑋
𝑁
𝑋 = 270
6= 45
𝑋 = 45
SUM NO: 1- FROM THE FOLLOWING DATA OF MARKS, OBTAIN BY 60 STUDENTS
OF A CLASS. CALCULATE THE AVERAGE MARKS.
b) DISCRETE SERIES
MARKS 20 30 40 50 60 70
NO.OF.
STUDENTS
8 12 20 10 6 4
MARKS
(X)
NO. OF.
STUDENT (F)
FX
20 8 160
30 12 360
40 20 800
50 10 500
60 6 360
70 4 280
Σ(F) = 60 ΣF(X) = 2460
𝑋 = 𝛴𝑓(𝑋)
𝑁 =
2460
60 = 41
𝑋 = 41
SUM NO: 1 - SHORT CUT METHOD:
MARKS
(X)
NO. OF.
STUDENT
(F)
dA = 40
(X-A) fd
20 8 -20 -160
30 12 -10 -120
40 20 0 0
50 10 10 100
60 6 20 120
70 4 30 120
Σ(F) = 60 Σfd= 60
𝑋 = 40 + 60
60 = 41
SUM NO: 1 – FROM THE FOLLOWING DATA OF MARKS. OBTAIN CALCULATE
ARITHMETIC UNDER DIRECT METHOD. FIND M.
c) CONTINUOUS SERIES
MARKS (X) N0. OF.
STUDENTS
(f)
m fm
0-10 5 5 25
10-20 10 15 150
20-30 25 25 625
30-40 30 35 1050
40-50 20 45 900
50-60 10 55 550
Σ(f) = 100 Σfm=3300
MARKS 0-10 10-20 20-30 30-40 40-50 50-60
NO. OF.
STUDENTS
5 10 25 30 20 10
𝑋 = 3300
100 = 33
𝑋 = 33
SUM NO: 1- SHORT CUT METHOD
MARKS
(X)
N0. OF.
STUDENTS
(f)
m M-AA = 25
fd
0-10 5 5 -20 -100
10-20 10 15 -10 -100
20-30 25 25 0 0
30-40 30 35 10 300
40-50 20 45 20 400
50-60 10 55 30 300
Σ(f) = 100 Σfd = 800
𝑋 = 25 + 800
100 = 33
𝑋 = 33
SUM NO: 2 – CALCULATE ARITHMETIC MEAN FROM THE FOLLOWING DATA.
FIND M.
MARKS 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
FREQUENCY 4 6 5 10 5 5 10 20
MARKS (F) m fm0-10 4 5 20
10-20 6 15 90
20-30 5 25 125
30-40 10 35 350
40-50 5 45 225
50-60 5 55 275
60-70 10 65 650
70-80 20 75 1500
Σ(f)= 65 Σ(fm)=3235
𝑋 = 3235
65= 49.76
𝑋 = 49.76
SUM NO: 3 – CALCULATE ARITHMETIC MEAN UNDER STEP DEVIATION
METHOD.
CLASS
LIMITS
0-10 10-20 20-30 30-40 40-50 50-60
FREQUENCY 5 10 25 30 20 10
CLASS
LIMITS f m d
(m-a/c)
fd
0-10 5 5 -2 -10
10-20 10 15 -1 -10
20-30 25 25 0 0
30-40 30 35 1 30
40-50 20 45 2 40
50-60 10 55 3 50
N= 100 Σ(fd)= 80
𝑋 = 25 + 80
100 × 10
𝑋 = 25 + 8 𝑋 = 33
𝑋 = A + 𝛴𝑓𝑑
Σf × 𝐶
SUM NO: 4 – FROM THE FOLLOWING DATA CALCULATE THE ARITHMETIC MEAN
DATA.
MARKS BELOW 10 BELOW 20 BELOW 30 BELOW 40 BELOW 50 BELOW 60 BELOW 70 BELOW 80
NO. OF.
STUDENTS
15 35 60 84 96 127 198 250
MARKS NO. OF.
STUDENTSsf m d
(m-a/c)
fd
0-10 15 15 5 -3 -45
10-20 35 20 15 -2 -40
20-30 60 25 25 -1 -25
30-40 84 24 35 0 0
40-50 96 12 45 1 12
50-60 127 31 55 2 62
60-70 198 71 65 3 213
70-80 250 52 75 4 208
Σ(f)=250 Σ(d)= 4 Σ(fd=385
𝑋 = A + 𝛴𝑓𝑑
Σf × 𝐶
𝑋 = 35 + 385
250 × 10
𝑋 = 35 + 15.4 𝑋 = 50 . 4
SUM NO: 1 – CALCULATE THE MEDIAN FROM THE FOLLOWING DATA, THE CLASS
INTERVALS ARE UNEQUAL.
MARKS 0-10 10-30 30-60 60-80 80-90
FREQUENCY 5 15 30 8 2
MARKS FREQUENCY CUMULATIVE
FREQUENCY
(CF)
0-10 5 5
10-20 7.5 12.5
20-30 7.5 20.0 (PCF)
30-40 10 30.0
40-50 10 40.0
50-60 10 50.0
60-70 4 54.0
70-80 4 58.0
80-90 2 60.0
N = 60
20 N = 30
Median = 30 + 30 − 20.0
10 × 10
Median = 30 + 10
10 × 10 Median = 40
SUM NO: 2 – FIND THE MEDIAN FROM THE FOLLOWING CLASS INTERVALUES.
WAGES 60-69 70-79 80-89 90-99 100-109 109-119
FREQUENCY 5 15 20 30 20 8
WAGES C.I FREQUENCY CUMULATIVE
FREQUENCY
60-69 59.5-69.5 5 5
70-79 69.5-79.5 15 20
80-89 79.5-89.5 20 40 (PCF)
90-99 89.5-99.5 30 70
100-109 99.5-100.5 20 90
110-119 100.5-109.5 8 98
N = 98 N = 323
𝑁
2 =
98
2 = 49
Median = 89 . 5 + 49 − 40
30 × 10 = 92 . 5
Median = 92 . 5
III. MODE
MODE Mode is the measure of central value. It is the value which
has the maximum frequency. In case of frequency
distribution of an continuous variable mode is obtained by
giving weights to the frequency of modal class, then pre
modal and post modal class.
SUM NO: 1 – FIND MODE FROM THE FOLLOWING GIVEN DATA.
110 120 130 120 110 140 130 120 140 120
a) INDIVIDUAL OBSERVATION
X TALLY BAR f
110 I I 2
120 I I I I 4
130 I I 2
140 I I 2
Mode = 120
SUM NO: 1 – CALCULATE MODE FROM THE FOLLOWING DATA.
b) DISCRETE SERIES
SIZE 28 29 30 31 32 33
NO. OF.
PERSONS
10 20 40 65 50 15
SIZE NO. OF.
PERSONS
28 10
29 20
30 40
31 65
32 50
33 15
Mode = 31
SUM NO: 1 – CALCULATE MODE FROM THE FOLLOWING DATA.
c) CONTINUOUS SERIES
SALES BELOW - 60 BELOW - 62 BELOW - 64 BELOW - 66 BELOW - 68 BELOW- 70 BELOW - 72
FREQUENCY 12 18 25 30 10 3 2
SALES (X) f
58-60 12
60-62 18
62-64 25 (f0)
64-66 30 (f1)
66-68 10 (f2)
68-70 3
70-72 2
Mode = 64 + 30 − 25
2 30 − ( 25 + 10 ) × 2
Mode = 64 . 4