STATISTICS

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.


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.


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.


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


E. Personnel Management

- Optimum organization level

- Job evaluation & assignment analysis

- Social / habit analysis

- Salary / wage policies

- Recruitment & Training


F. Research & Development

- Area of thrust – Analysis & Planning

- Project Selection Criteria\

- Alternatives analysis

- Trade – off analysis - cost & revenue


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


(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


G = (x1f1.x2

f2…….xnfn)1/N

log G = 1( f1logx1 + f2logx2 +……+xnlogfn)

N

G = antilog ( 1 Σfilogxi)

N

Geometric Mean


- 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


- Formula same as that of Discrete Frequency Distribution with x (as observations) replaced by x (as mid values).

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