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INTRODUCTION TO STATISTICS
Definitions “Statistics is a numerical statement of facts in any department of
enquiry placed in relation to each other’. -Bowley
“Statistics are the classified facts representing the conditions of the people in a State specially those facts which can be stated in numbers or any tabular or classified arrangement”. -Webster
“Statistics can be defined as the aggregate of facts affected to a marked extent by multiplicity of causes, numerically expressed, enumerated or estimated according to a reasonable standard of accuracy, collected in in a systematic manner, for a pre-determined purpose and placed in relation to each other”. -Secrist
Statistics is the science of collecting, organizing , analyzing, interpreting and presenting data.
SCOPE OF STATISTICS1.Social Sciences -Man Power Planning
-Crime Rates
-Income & Wealth Analysis of Society
-In studying Pricing, Production, Consumption, Investments & Profits etc.
2.Planning -Agriculture
-Industry
-Textiles
-Education etc.
For ex. Five Year Plans in India.
n
SCOPE OF STATISTICS contd…
3. Mathematics
-Extensive use of Differentiation, Algebra, Trigonometry, Matrices etc in modern business analysis.
-Statistics now treated as Applied Mathematics.
4. Economics
- Family Budgeting
-Applied in solving economic problems related to production, consumption, distribution of products as per income & wealth related patterns, wages, prices, profits & individual savings, investments, unemployment & poverty etc.
SCOPE OF STATISTICS contd…
5. Business Management - Trend Analysis
- Market Research & Analysis
- Product Life Cycle
i) Marketing Marketing Policy Decisions depend on forecasting, demand
analysis, time & motion studies, inventory control, investments & analysis of consumer data for production & sales.
SCOPE OF STATISTICS contd…
ii) Production - Designs
- Methods of Production
- Technology Selection
- Quality Control Mechanisms
- Product Mix
- Quantities
- Time Schedules for Manufacturing & Distribution
SCOPE OF STATISTICS contd…iii) Finance -Correlation Analysis of profits & dividends, assets & liabilities
-Analysis of income & expenditure
- Financial forecasts, break-even analysis, investment & risk analysis
iv) Sales -Demand Analysis
-Sales Forecasts
v) Personnel - Wage plans, Incentive plans, Cost of living, Labor turnover ratio,
Employment trends, Accidental Rates, Performance Appraisals
etc.
SCOPE OF STATISTICS contd…
vi) Accounting & Auditing -Analysis of Income, Expenditure, Investment, Profits and
Optimization of Production etc
- Forecasting costs of production & price
vii) Other Areas
-Insurance, Astronomy, Social Sciences, Medical Sciences, Psychology, Education etc.
LIMITATIONS OF STATISTICS
Does not study individual items, deals with aggregates.
Statistical laws are not exact. Not suitable for the study of qualitative
phenomenon. Statistical methods are only means and not
end for solving problems.
ROLE OF STATISTICS IN MANAGEMENT DECISIONS
A. Marketing & Sales
- Product selection & competence strategies
- Utilization of resources including territory control
- Advertising decisions for cost & time effectiveness
- Forecasting & trend analysis
- Pricing & market research
ROLE OF STATISTICS IN MANAGEMENT DECISIONS contd…
B. Production Management
- Product mix & product positioning
- Facility & production planning
- Distribution management
- Material handling & facility planning
- Maintenance policies
- Activity planning & resources allocation
- Quality control decisions
ROLE OF STATISTICS IN MANAGEMENT DECISIONS contd… C. Materials Management - Buying policy- Sourcing & Procurement - Material Planning & Lead Times D. Finance, Investments & Budgeting - Profit planning - Cash Flow Analysis - Investment decisions - Dividend policy decisions - Risk Analysis - Portfolio Analysis
ROLE OF STATISTICS IN MANAGEMENT DECISIONS contd…
E. Personnel Management
- Optimum organization level
- Job evaluation & assignment analysis
- Social / habit analysis
- Salary / wage policies
- Recruitment & Training
ROLE OF STATISTICS IN MANAGEMENT DECISIONS contd…
F. Research & Development
- Area of thrust – Analysis & Planning
- Project Selection Criteria\
- Alternatives analysis
- Trade – off analysis - cost & revenue
ROLE OF STATISTICS IN MANAGEMENT DECISIONS contd…
G. Defense
- Optimization of weapon system
- Force deployment
- Transportation Cost Analysis
- Assignment Suitabilities
Definitions Continued
Observations: Numerical quantities that measure specific characteristics. Examples include height, weight, gross sales, net profit, etc.
Some More Definitions Raw Data: Data collected in original form.
Classes / class intervals: Subgroups within aset of collected data. Ex.10-20,20-30 etc
Width of class-interval = upper limit – lower limit
Mid – Value = (U.L + L.L)/2
Frequency: The number of times a certainvalue or class of values occurs.
Frequency Distribution Table: The organization ofraw data into table form using classes andfrequencies.
More Definitions
Cumulative Frequency of a class is the sum of the frequency of that class and the frequencies of all the preceding or succeeding classes which are listed in some sensible order (numerical order, alphabetical order, etc.)
Illustration – Individual Series
Marks of ten students of a class in Statistics
15, 35, 55, 67, 78, 84, 79, 90, 89, 94
Illustration – Discrete Frequency Distribution
Height(in
inches)
No. of Students
60 12
62 18
64 10
66 6
68 4
Illustration – Grouped or Continuous Frequency Distribution
Exclusive Type Class – Intervals
Class-Intervals
Frequency
20-25 8
25-30 2
30-35 40
35-40 23
40-45 9
Illustration – Grouped or Continuous Frequency Distribution contd…
Inclusive Type Class - Intervals
Class-Intervals
Frequency
1-10 2
11-20 6
21-30 10
31-40 15
41-50 12
CONVERSION OF INCLUSIVE TYPE CLASS-INTERVALS TO EXCLUSIVE TYPE CLASS INTERVALS1. Calculate ADJUSTMENT FACTOR as follows:
A.F= Lower Limit of Next C.I – Upper Limit of Previous C.I
2
using the given inclusive type class intervals.
2. Obtain new class intervals as follows:
New Lower Limit = Old Lower limit – A.F
New Upper Limit = Old Upper Limit + A.F
CONVERSION OF INCLUSIVE TYPE CLASS-INTERVALS TO EXCLUSIVE TYPE CLASS INTERVALS contd…
Class-Intervals
Frequency
1-10 2
11-20 6
21-30 10
31-40 15
41-50 12
A.F = (11 – 10)/2
= 0.5
For 1st C.I i.e 1-10
New L.L = 1(old L.L) – 0.5
= 0.5
New U.L=10(old U.L) +0.5
= 10.5
And so on.
CONVERSION OF INCLUSIVE TYPE CLASS-INTERVALS TO EXCLUSIVE TYPE CLASS INTERVALS contd…
Class-Intervals
Frequency
0.5-10.5 2
10.5-20.5 6
20.5-30.5 10
30.5-40.5 15
40.5-50.5 12
Now calculations
can be made.
Obtaining Cumulative Frequency Distribution
Class -Intervals
Frequency Less than type More than type
Cum.frequency cum.frequency
20-25 15 15 60 + 15 = 75
25-30 34 15 +34 =49 26 + 34 = 60
30-35 6 49 + 6 =55 20 + 6 = 26
35-40 10 55 + 10 = 65 10 + 10 = 20
40-45 8 65 + 8 = 73 2 + 8 = 10
45-50 2 73 + 2 = 75 2
Introduction to Measures of Central Tendency
Also known as averages. Values show a distinct tendency to cluster or
group around a value. This behavior is central tendency of data. The value around which the data clusters is
the measure of central tendency which represents the whole set of data.
Objectives of Averages
To find out one value that represents the whole mass of data.
To enable comparison. To establish relationship. To derive inferences about universe to which
sample belongs. To aid decision – making.
Requisites of a Good Average
Should be rigidly defined. Should be mathematically expressed. Should be readily comprehensible & easy to
calculate. Should be calculated on the basis of all the
observations. Should be least affected by extreme values and
sampling fluctuations. Should be suitable for further mathematical
treatment.
Common Measures of Central Tendency
Arithmetic Mean Geometric Mean Harmonic Mean Median Mode Partition Values like Deciles ,Quartiles &
Percentiles.
Averages
Mathematical Averages Positional Averages
A.M G.M H.M Median Mode
Arithmetic Mean
Individual Series
μ = x1 + x2 +…… + xn
n
For ex. A.M of 3, 6, 24 and 48
μ = 3 + 6 + 24 + 48
4
= 81/4 = 20.25 Ans.
Arithmetic Mean contd…
Discrete Frequency Distribution
μ = f1x1 + f2x2 + …..fnxn =Σfx
N Σf
Where N = f1 +f2 +…+fn
n = no. of observations
X Freq. fx
x1 f1 f1x1
x2 f2 f2x2
x3 f3 f3x3
x4 f4 f4x4
Height
(in inches)
X
No. of Students
f fX
60 12 60 x 12 = 720
62 18 1116
64 10 640
66 6 396
68 4 272
50 = N 3144 = Σ fx
μ = 3144 / 50 = 62.88 Ans.
Illustration
Arithmetic Mean contd…
Continuous Frequency Distribution
- Direct Method
- Assumed Mean Method
- Step Deviation Method
Arithmetic Mean Formulae
Direct Method
μ = f1x1 + f2x2 + …..fnxn = Σfx N Σf Where N = f1 +f2 +…+fn
x = mid value of a C.I = (U.L + L.L) 2
Arithmetic Mean Formulae contd… Assumed Mean Method
μ = A + Σ fd
N
Where A = assumed mean
N = Σ f
d = x – A
x = mid - value
Arithmetic Mean Formulae contd…
Step Deviation Method
μ = A + Σ fd x i
N
where A = assumed mean
N = Σ f
d = x – A
i
x = mid – value
i = width of C.I = U.L – L.L
Illustration – Direct Method
C.I Freq
f
Mid-Value
X
fX
4-6 6 5 30
6-8 12 7 84
8-10 17 9 153
10-12 10 11 110
12-14 5 13 65
Total 50 = Σf
442 = Σfx
= 442/50
= 8.84 Ans.
μ = Σ fx Σ f
Illustration – Assumed Mean Method
C.I Freq.
f
Mid Values (x)
d =(x-A)
fd
10-15 2 12.5 -10 -20
15-20 7 17.5 -5 -35
20-25 9 22.5 = A 0 0
25-30 8 27.5 5 40
30-35 6 32.5 10 60
35-40 4 37.5 15 60
Σf= 36
Σfd = 105
= 22.5 + 105
36
= 22.5 + 2.916
= 25.416 Ans.
μ = A + Σ fd Σf
Illustration- Step Deviation Method
C.I Freq.(f) MidValues (x)
d= (x-A)
I
(i= 5)
fd
10-15 200 12.5 -2 -400
15-20 700 17.5 -1 -700
20-25 900 22.5 = A 0 0
25-30 800 27.5 1 800
30-35 600 32.5 2 1200
35-40 400 37.5 3 1200
Σf= 3600
Σfd = 2100
= 22.5 + 2100 x 5
3600
= 22.5 + 2.916
= 25.416 Ans.
μ = A + Σ fd x i Σf
Illustration
Marks X or more
Cum.
Freq.C.I Freq.
10 140 10-20 140-133= 7
20 133 20-30 133-118=15
30 118 30-40 118-100=18
40 100 40-50 100-75=25
50 75 50-60 75-45=30
60 45 60-70 45-25=20
70 25 70-80 25-9=16
80 9 80-90 9-2=7
90 2 90-100 2-0=2
100 0
Proceed as usual
What if…
C.I Frequency
50-59 1
40-49 3
30-39 8
20-29 10
10-19 15
0-9 3
Total N=40
?
A.F = (L.L of 1st C.I – U.L of 2nd C.I)/2
= (50-49)/2
= 0.5
New C.I
L.L of new C.I = L.L of original C.I – A.F
U.L of new C.I= U.L of original C.I + A.F
For ex. For 1st C.I,new L.L = 50-0.5
= 49.5
new U.L = 59 +0.5
= 59.5 and so on.
Now Continue as usual.
Determining missing frequency when A.M is known – Illustration Mean = 16.82
Marks Freq. M.V (x) d=
(x –A)/i
fd
0-5 10 2.5 -3 -30
5-10 12 7.5 -2 -24
10-15 16 12.5 -1 -16
15-20 ? = f4 17.5 = A 0 0
20-25 14 22.5 1 14
25-30 10 27.5 2 20
30-35 8 32.5 3 24
N = 70 + f4 Σfd = -12
Determining missing frequency when A.M is known - Illustration
Soln. μ = A + Σ fd x I
Σf
μ = 16.82 (given) , I = 5
Hence 16.82 = 17.5 + ( -12 ) x 5
70 + f4
- 0.68 = - 60
70 + f4
- 0.68 (70 + f4) = - 60
f4 = 12.4/0.68 = 18 approx. Ans.
Some More Applications of A.M
Q1.The avg. marks secured by 50 students was 44.Later on it was discovered that a score 36 was misread as 56. Find the correct average marks secured by the students.
Soln. Given N = 50 and mean μ = 44
μ = ΣX
N
ΣX = 44N
i.e ΣX = 44x55
ΣX = 2200
Since 36 was misread as 56
Hence correct ΣX = 2200 – 56 + 36 = 2180
Correct mean = 2180/50 = 43.6 Ans.
Combined A.M
Suppose for k different series with n1,n2……nk observations each, the respective A.M s are μ1,μ2,….μk. Then the A.M of the new series obtained on combining all the n1,n2,…nk observations is obtained using the formula:
μ = n1μ1+n2μ2+….+nkμk
n1+n2+….+nk
Illustration- Combined A.M
There are two branches of a Co. employing 100 and 80employees respectively .If A.Ms of the monthly salaries paid by the two branches are Rs.4570 and
Rs.6750 respectively, find the A.M of the salaries of the employees of the Co. as a whole.
Soln. Given No. of employees in 1st factory, n1 = 100 Avg. Salary of employees in 1st factory, μ1 = Rs. 4750
No. of employees in 2nd factory, n2 = 80
Avg. Salary of employees in 2nd factory, μ2 = Rs.6750
Avg salary of the employees of the Co. as a whole
= 100 x 4750 + 80 x 6750 = 997000 = Rs. 5538.89
100 + 80 180
Practice Questions- Arithmetic Mean
Q1.
Q2
Q3
Weekly Income
(in Rs.)
20-25 25-30 30-35 35-40 40-45 45-50
No.of workers 200 700 900 800 600 400
Weight (in kgs) 30-34 35-39 40-44 45-49 50-54 55-59 60-64
No.of Students 3 5 12 18 14 6 2
Wages(in Rs.)
125-175 175-225 225-275 275-325 325-375 375-425 425-475
No.of workers
8 10 25 35 12 10 4
Practice Questions- Arithmetic Mean contd…
Q4 Lifetime (in hrs.) No. of tubes
Less than 300 0
Less than 400 20
Less than 500 60
Less than 600 116
Less than 700 194
Less than 800 265
Less than 900 324
Less than 1000 374
Less than 1100 392
Less than 1200 400
Merits of A.M
Is rigidly defined and has a definite value. Is based on all the observations. Is capable of algebraic treatments for further
data analysis & interpretation. Easy to calculate & simple to understand. For a large no. of observations, A.M provides
a good basis of comparison.
Drawbacks of A.M Being based on all the observations, is considerably
affected by abnormal observations. For ex. A.M of 1000, 25, 35 & 40 will be (1000+25+35+40)/4 = 275 which is not at all a representative figure.
Cannot be calculated even if a single observation is missing.
Cannot be obtained just by inspection as in case of median & mode.
May give absurd results. For ex. If avg. no. of children per family is to be calculated and the result is 3.4 children per family, how would you interpret it?
Weighted Arithmetic Mean
Formula Used
μw = x1w1+ x2w2 +…….+xnwn
w1+ w2 +…….+wn
Illustration – Weighted A.MDesignation Monthly
Salary
(in Rs.) (X)
No. of employees
(w)
wX
Class I Officers
1500 10 15000
Class II officers
800 20 16000
Subordinate Staff
500 70 35000
Clerical Staff 250 100 25000
Lower Staff 100 150 15000
350 = Σw 106000 = ΣwX
Illustration – Weighted A.M
Weighted A.M = Σ wX
Σw
= 106000
350
= Rs 302.857 Ans.
Median – Positional Average
The value of the middle term of a series arranged in ascending or descending order of magnitude.
Its value is the value of the middle item irrespective of all other values.
Calculation of Median Individual Series N = no. of observations or items in the series - Arrange all the items in ascending or
descending order of magnitude.Case I N = Odd Median = Value at (N+1) th position in 2 the arranged series.Case II N = Even Median = A.M of values at (N, N+1)th 2 2 position.
Calculation of Median – Illustration (Individual Series)
Ex.1 Find the median 5, 7, 9, 12, 10, 8, 7, 15,21
Solution: Arranging in ascending order we get
5, 7, 7, 8, 9, 10, 12, 15, 21
Here N = 9 i.e odd
Hence Md = (N+1) th item in the arranged order
2
= (9 +1) th item
2
= 5 th item
= 9 Ans.
Calculation of Median – Illustration (Individual Series)
Ex 2. Find the median 10, 18, 9, 17, 15, 24, 30, 11Solution Arranging in ascending order we get 9, 10, 11, 15, 17, 18, 24, 30 Here N = 8 i.e even Hence Md = A.M of the ( N , N+1)th items in the 2 2 arranged order. = A.M of (4th, 5th) items = (15 + 17) 2 = 16 Ans.
Calculation of Median
Discrete Frequency Distribution
(i) Find less than type cum.frequency.
(ii) Find N/2.( N = Σf)
(iii) Find the cum.freq. just greater than N/2. Suppose it is C.
(iv) Find the corresponding value of X. (the item) This is median.
Calculation of Median-Illustration (Discrete Freq. Distribution)
Height
(in inches)
No. of students
Cum.
Freq.
60 12
12
62 18
30
64 10 40
66 6
46
68 4 50
N = 50
Here N = 50(i) N/2 = 25(ii) Cum. Frequency just greater than N/2 = 30(iii)Corresponding value of item is 62.Median = 60 Ans.
Calculation of Median Grouped Frequency Distribution
(i) Find less than type cum.frequency.
(ii) Find N/2.( N = Σf)
(iii) Find the cum.freq. just greater than N/2. Suppose it is X.
(iv) Look for the cum.freq. preceding X. Find the corresponding class interval.This is median class
Formula Used
Where L1 = L.L of median class
L2 = U.L of median class
C =cum.freq. of class preceding the median class.
f = frequency of median class.
Md = L1 + N/2 - C (L2 – L1) f
C.I Freq.(f) Cum. Freq
10-15 200 200
15-20 700 900
20-25 900 1800
25-30 800 2600
30-35 600 3200
35-40 400 3600
Σf= 3600
Calculation of Median-Illustration(Grouped Freq. Distribution)
N/2 = 3600/2 = 1800
Cum.freq. just greater than 1800 is 2600. Hence median class is 25-30.Hence L1 = 25 L2 = 30 C = 1800 f = 800
Md = 25 + 1800 - 1800 (30 – 25 ) 800 = 25 Ans.
Calculation of Missing Frequencies when median is known : Illustration : Median = 50
Expenditure No. of Families Cumulative Freq.
0-20 14 14
20-40 ? = f1 14 + f1
40-60 27 41 + f1
60-80 ? = f2 41+ f1+f2
80-100 15 56 + f1 + f2
N = 100
Calculation of Missing Frequencies when median is known : Illustration
Here median = 50 L1 = 40
N = 100 L2 = 60
N/2 = 50 f = 27
Hence median class 40-60 C = 14 + f1Md = L1 + N/2 - C (L2 – L1) f
50 = 40 + 50 – (14 + f1)(60 – 40)
27
10 = 720 – 20 f1
27
f1 = 450/20 = 22.5 = 23 families approx.
N = 56 + f1 + f2
100 = 56 + 23 + f2
f2 = 21 Ans. f1 = 23 and f2 = 21
Practice Numericals - Median
Q1. Q2.Age No. of Persons
20-25 14
25-30 28
30-35 33
35-40 30
40-45 20
45-50 15
50-55 13
55-60 7
Value Frequency
Less than 10 4
Less than 20 16
Less than 30 40
Less than 40 76
Less than 50 96
Less than 60 112
Less than 70 120
Less than 80 125
Practice Problems- Median
Q3. Determine the missing frequencies. The median is 46.Also determine the A.M.
Class-Intervals Frequency
10-20 12
20-30 30
30-40 ?
40-50 65
50-60 ?
60-70 25
70-80 18
229 = N
Merits - Median
Is rigidly defined. Can be easily calculated. Not affected by extreme values. Can be located merely by inspection.
Demerits - Median
May not represent the entire series in many cases.
Not suitable for further algebraic treatment. More likely to be affected by sampling
fluctuations.
Mode
The value occurring the largest no. of times in a series. That is the value having the maximum frequency.
Is calculated for discrete and continuous frequency distributions only.
For ex. How to obtain the mode for 1,2,3,4,5 ?
as the maximum frequency is 1 and each observation has frequency 1.
Mode – Discrete Frequency Distribution
The value corresponding to maximum frequency is the mode.
For ex. The weight 132 pounds has the maximum frequency 3. Hence 130 pounds is the mode for this frequency distribution.
Wt. in pounds
No.of students
120 1
130 3
132 2
135 2
140 1
141 1
Total 10
Mode – Continuous Frequency Distribution
1.Look for the class-interval with maximum frequency. This is the modal class.
2. Note down the following:
L1 = lower limit of the modal class. i = width of class-interval
f0 = frequency of class preceding the modal class.
f1 = frequency of modal class.
f2 = frequency of class succeeding the modal class.
Mode: Formula for Continuous Frequency Distribution
Mode = L1 + h(f1 – f0) 2f1-f0-f2
Empirical Relationship between Mean, Median & Mode
Mode = 3 Median – 2 Mean
Geometric Mean
Individual Series
G = (x1.x2.x3……xn)1/n
log G = 1 (logx1 + log x2 +….+ logxn)
n
G = antilog ( 1 Σ log x)
n
Geometric Mean
Discrete Frequency Distribution
G = (x1f1.x2
f2…….xnfn)1/N
log G = 1( f1logx1 + f2logx2 +……+xnlogfn)
N
G = antilog ( 1 Σfilogxi)
N
Geometric Mean
Continuous Frequency Distribution
- Formula same as in case of discrete frequency distribution with x (as observations) replaced by x (as mid-values)
Harmonic Mean Reciprocal of A.M of reciprocals - Individual Series H = 1 1( 1 + 1 +…..+ 1 ) n x1 x2 xn
H = n Σ(1 ) x
Harmonic Mean
-Discrete Frequency DistributionH = 1 1( f1 + f2+…..+ fn ) N x1 x2 xn
H = N Σ(fi ) xi
Harmonic Mean
Continuous Frequency Distribution
- Formula same as that of Discrete Frequency Distribution with x (as observations) replaced by x (as mid values).