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STUDY NOTE –6

DATA
6.1 DATA : A statistician begins the work with the collection of data i.e. numerical facts. The data so collected are called raw materials (or raw data). It is from these raw materials, a statistician analysis after proper classification and tabulation, for the final decision or conclusion. Therefore it is undoubtedly important that the raw data collected should be clear, accurate and reliable. Before the collection of data, every enquiry must have a definite object and certain scope, that is to say, what information will be collected for whom it will be collected, how often or at what periodically it will be collected and so on. If the object and the scope of enqiry are not clearly determined before hand, difficulties may arise at the time of collection which will be simply a wastage of time and money. Statistical Units : The unit of measurement applied to the data in any particular problem is the statistical unit. Physical units of the measurement like quintal, kilogramme, metre, hour and year, etc. do not need any explanation or definition. But in some cases statistician has to give some proper definition regarding the unit. For examples, the wholesale price of commodity. Now what does the form ‘wholesale price’ signify? Does it stand for the price at which the producer sells the goods concrened to the stockist, or the price at which the stockist sells to a wholesaler? Is it the price at which the market opened at the day of enquiry? Many such problems may arise as stated. It is thus essential that a statistician should define the units of data before he starts the work of collection. Types of Methods of Collection of Data : Statistical data are usually of two types : (i) Primary, (ii) Secondary

Data which are collected for the first time, for a specific purpose are known as primary data, while those used in an investigation, which have been originally collected by some one else, are known as secondary data. For example, data relating to national income collected by government are primary data, but the same data will be secondary while those will be used by a different concern.
1.5.1.1.1.1.1.1.1 Let us take another example, known to everyone. In our country after every ten years of Let us take another example, known to everyone. In our country after every ten years counting counting of is done, which is commonlycommonly known asFor this data this collected by the Government population population is done, which is known as Census. Census. For are data are collected by the Government of India. The data collected areprimaryas primary data. Now in the data, except population of India. The data collected are known as known data. Now in the data, except population information information about age of persons, income etc. are available. available. Now a separate about age of persons, education, education, income etc. are Now a separate

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department of the government or any other private concern use these related data for any purpose, then the data will be known as secondary data to them. Data are primary to the collector, but secondary to the user. Example. For primary data : (i) Reserve Bank of India Bulletin (monthly) (ii) Jute Bulletin (monthly), (published by Govt. of India). (iii) ndian Textile Bulletin (monthly). (vi) Statement of Railway Board (yearly), (published by Ministry of Railway, Govt. of India). For secondary data : (i) Statistical Abstract of the Indian Union (ii) Monthly Abstract of Statictics. (iii) Monthly Statistical Digest. (iv) International Labour Bulletin (monthly). Distinction between Primary and secondary Data : 1. Primary data are those data which are collected for the first time and thus original in character. Secondary data are those data that have already been collected earlier by some other persons. 2. Primary data are in the form of raw materials to which statistical methods are applied for them purpose of analysis. On the other hand, secondary data are in form of finished products as they have been already statistically applied. 3. Primary data are collected directly from the people to which enquiry is related. Secondary data are collected from published materials. 4. f observed closely the difference is one of degree only. Data are primary to an institution collecting it, while they secondary for all others. Thus data which are primary in the hands of one, are secondary in the hands of other. Primary Method : The following methods are common in use : (i) Direct Personal Observation : Under this method, the investigator collects the data personally. He has to go to the spot for conducting enquiry has to meet the persons concerned. It is essential that the investigator should be polite, tactful and have a sense of observation. This method is applicable when the field of enquiry is small and there is an intention of greater accuracy. This method however, gives satisfactory result provided the investigator is fully dependable.
1.5.1.1.1.1.1.1.1 Oral Investigation : In Oral method data are collected through are collected through (ii) Indirect this Investigation : In this method data indirect sources (ii) Indirect indirect sources. Persons having some knowledge regarding the enquiry are cross-examined and the desired Persons. having some knowledge regarding the enquiry are cross-examined and the desired

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information Evidence of Evidence should person should a number of views a number information is collected.is collected. one person of one bot be relied, butbot be relied, but should be of views should be position. This method is usually adopted by is usually adopted by taken to find out realtaken to find out real position. This methodenquiry committees or enquiry committees or commissions appointed by governments private government commissions appointed by governments or semi- government oror semi- institutions. or private institutions. to be taken here. Firstly it should here. Firstly it the informant Certain precautions are Certain precautions are to be takenbe seen whether should be seen whether the informant knows under investigations. Secondly should be considered that knows full facts of the problem full facts of the problem underitinvestigations. Secondly it should be considered that the person questioned is not prejudiced and also not motivated the person questioned is not prejudiced and also not motivated to colour the facts. Of to colour the facts. Of cource, due allowance should be made for optimism and pessimism. cource, due allowance should be made for optimism and pessimism.
(iii) Schedules and Questionnaires : A list of questions regarding the enquiry is prepared and printed. Data are collected in any of the following ways : (a) By sending the questionnaire to the persons concerned with a request to answer the questions and return the questionnaire. Success in this method depends entirely on the co-operation of the informants. The advantage in this method is that it is less costly, as no enumerators are required and investigations can be completed within a short time. The disadvantages are – many individuals do not return the forms in time and some of the individuals make mistake in filling up the forms. (b) By sending the questionnaire through enumerators for helping the informants. In this method, enumerators go to the informants to help them in filling the answers. This method is useful for extensive enquiries. It is expensive. Population census is conducted by this method. It is essential enumerators should be polite, and have proper training. The implications and scope of each question, to be asked to the informants, should be explained clearly to the enumerators. They should be instructed how to check up apparently wrong replies. They should have intelligence and capacity to cross examine the informants for finding out the true result. (iv) Local Reports : This method does not imply a formal collection of data. Only local agents or correspondents are requested to supply the estimate required. This method gives only approximate results, of course at a low cost. Questionaires : In a statistical enquiry, the necessary information is generally collected in a printed sheet in the form of a questionnaire. This sheet contains a set of questions which the investigator asks to the informant, and the answers are noted down against the respective questions on the sheet. Choice of questions is a a very important part of the enquiry whatever be its nature. For satisfactory investigation a questionnaire should possess the following points :
1.5.1.1.1.1.1.1.1 (i) The scheduling of questions must not be lengthy. Many questions may arise during (i) The scheduling of questions must not are lengthy. Many questions maypersons who prepreparations of questionnaire. If all of them be included, the result is that the arise during parations of questionnaire. If all of them are included, the result is that the persons who

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are interviewed bored and reluctant to answer all the the questions. only the important are interviewed may fell may fell bored and reluctant to answer all questions. So So only the important questions are to be included. questions are to be included.

(ii) It should be simple and clear. The questions should be understandable even by the most uneducated people so that informants do not find any difficulty in furnishing the answers. The factors os simplicity and clarity also imply that the questions should be few so that the informant may not be confused. If possible, the questions should be so set up that require brief answers viz, ‘yes’, ‘no’ or a ‘number’, etc. (iii) Each questions should be brief and must aim to some particular information necessary for the investigation of the problem. Lengthy questions may be spilt up into smaller parts, which will be easily grasped by the informants. (iv) Questions on personal matter like income or property should be avoided as far as possible, as people are generally reluctant to disclose the truth. In such cases, the information may be collected on guess work. (v) The questions should be arranged in a logical sequence. The first part may contain questionnaire so that the informant may answer them when he feels easy with the interviewer. (vi) The units of information should be clearly shown in the schedule. For example. State your age, years ….. months… What is your weight? Kg….. Example : The following form was used in census of population India of 1961, for having a census of Scientific and Technical Personnel. CENSUS OF INDIA 1961 ; SCIENTIFIC& TECHNICAL PERSONAL Only a person with a reconginised degree or diploma in Science, Engineering, Technology or Medicine should fill in this card. READ CAREFULLY BEFORE FILLING IN TICK (3 ) WITHHIN BRACKETS PROVIDED WHERE APPLICABLE 1. NAME ………………… 3. DESIGNATION & OFFICE ADDRESS …………… (if employed) 4. PERMANENT ADDRESS ……………. 5. (a) Male 6. (a) Never married 7.On Feb. 1st, 1961 were you If so, monthly total income Rs. ………… () () (b) Female (b) Married (a) Employed? () () () CENSUS LOCATION CODE 2. DATE OF BIRTH

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(b) Full time Student? If so, how long? … yrs. … months. 8. ACADEMIC QUESTIONS (ANSWER FULLY) Degree/Diploma Subject taken

() ( )

(c) Unemployed?

()

(d) Retired

()

Division

Year of Passing

If employed fill in Qs. 9–12 9. Nature of employed (b) Teaching in College (d) Technical Outside 10. Any Research Assignment 11. Where employed (b) Private sector 12. How employment? (b) Temporary (d) Research Scholar ( ) ( ) Date Secondary Method : The main sources from which secondary data are collected are given below– (i) Official publications by the Central and State Government, District Boards, (ii) Reports of Committees, Commissions. (iii) Publications by Research Institutions, Universities, (iv) Economic and Commercial Journals. (v) Publications of Trade Associations, Chambers of Commerce, etc. (vi) Market reports, individual research works of Statisticians. Secondary data are also available from unpublished records of government offices, chambers of commerce, labour bureaus, etc. Editing and Scrutiny : Secondary data Secondary data should be used only after and with due and with It criticism. It is 1.5.1.1.1.1.1.1.1 should be used only after careful enquirycareful enquiry criticism.dueis advisable not to
advisable not their face value. Scrutinyvalue. Scrutiny is essential because the data might beunsuitable and take them at to take them at their face is essential because the data might be inaccurate, inaccurate, unsuitable and

(a) Teaching in School ( ) ( ) (c) Technical inindustry (e) Non-Technical Yes ( ) No ( ) (a) Public Sector ( ) (c) Self employment (a) Permanent (c) On contract (e) Otherwise Signature

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

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inadequate. According to Bowley, “It is never safe to take published statistics at their face value
STATISTICAL METHODS

without knowing their meanings and limitations ….” Secondary data may, however, be used provided they possess the attributes (i.e. qualities) shown below– 1. Data should be reliable : The reliability of data depends on the following queries– (a) The sources of original collector’s informations. (b) Original compiler’s reference. (c) Method of collection including instructions given to the enumerators. (d) Period of collection of data. (e) Degree of accuracy desired and achieved by the complier. 2. Data should be suitable : For the purpose of investigation, even the reliable data should be avoided if they are found to be not suitable for the purpose concerned. Data suitable for one enquiry may be unsuitable for the other. 3. They should be adequate : Even the reliable and also suitable data may become inadequate sometimes for enquiry. The original data may refer to a certain market price during disturbed period ; for a normal period the above reference will be inadequate. Universe or Population : Statistics is taken in relation to a large data. Single and unconnected data is not statistics. In the field of any statistical enquiry there may be persons, items or any other similar units. The aggregate of all such similar units under consideration is called Universe or Population. Example L For collecting the data regarding height, weight or age of the male candidates who appeared in the last H.S. Examination, the aggregate of such candidates is universe. Universe may be aggregate of items or any other similar things other than persons. The books in your college library or produced goods in a factory may be taken as Universe. Population may be finite or infinite according to finite or infinite number of members. In the field of enquiry if the number of units is finite, then Populationor Universe is finite. For Example, first class cricket or football players in India is finite. But the temperature in any day at Calcutta is infinite, although temperature lies between two finite limits. Within these two finite limits it takes up an infinity of values. Sample : If a part is selected out of the Universe then the selected part (or portion) is sample. It means sample is a part of the Universe.
1.5.1.1.1.1.1.1.1 Examplescrews or bulbs produced in produced in a factorytested. be tested. The of al Example : Suppose the : Suppose the screws or bulbs a factory are to be are to The aggregate aggregate of all such items is universe but it is not possible to test every item. So in such case, a part of the such items is universe but it is not possible to test every item. So in such case, a part of the whol whole i.e., universe is taken and then tested. Now this part is known as sample.

i.e., universe is taken and then tested. Now this part is known as sample.

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Note. While collecting primary data (discussed before) it should be decided at first whether the purpose will be solved if collection is made from universe or sample.

Complete enumeration : If detail information regarding every individual person or items of a given universe is collected, then the enquiry will be complete enumeration. Another common name of complete enumeration is census. Example : If it is required to compute the average height or weight of all the employees working under the Government of West Bengal by the complete enumeration, then the heights or weights of all such employees are to be counted. (No one should be excluded). Since this methods requires time, expenditure, strength of working person, etc., application of the method is less. But for the interest of accurate observation of a particular individual item of the universe or if universe is small, them this method may be applied. In case of cesus of any country, detail enquiries of age, education, religion, occupation, income etc. of every individual (man of woman) are collected. In our counry census is made after every ten years. In certain cases complete enumeration is impossible. For export purpose it is not possible to test the quality of every grain of rice or wheat in a bag. Example 1. Part I Dear friend, The academic session of your college is going to be over. After few months you may go to a different Institution for further higher studies. You must have experienced some problems in your college. This survey is conducted to collect these informations. Of course, the main aim is to collect suggestions and hence to improve your college so that future students may get the advantage. It may be noted that the present survey is without any prejudice to any individua, group or Institution. Your are requested to fill the form in free mind and in the spirit of helping your Institution only. Thanking you, yours faithfully, ……………. Part II A survey of social, economic and educational problems experienced by students reading in a degree college. QUESIONNAIRE [Use tick mark wherever applicable.] 1. Personal Bio-data : (a) Name ………… (d) Stream ……….
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(b) Sex …….. (e) Year ……

(c) Age …… (f) Sec ……. (g) Roll No ….
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6. Savings : (in Rs.) yes/no If yes, average monthly saving Saving in current month Saving in previous month 7. Indebtedness : (in Rs.) yes/no If yes, source (b) Private money-lender ( ), 8. Living Status : (a) Own house : yes/no (c) Any arrear in rent/tax : yes ( ) /no (d) Area of residence : ( ) sq. m. (d) No. of rooms ( ) (b) Rent/tax (Rs.) (a) friend/relative (c) bank ( ), (d) any other ( ). ( ) ( ) ( )

(f) Electricity : yes/no If yes, monthly bill (Rs.) (f) Do you have : fan/radio/bicycle/T.V./scooter, 9. Conveyance : (a) Have you any vehicle : yes/noIf yes, mention…. (b) Conveyance charge for school/college going students : (Rs.) (d) Miscellaneous : (Rs.) 10. Amenities : (a) Running water-tap : yes/no (c) Dispensary : yes/no (b) Primary School : yes/no (d) Play-ground : yes/no

……….. Signature with date 6.2. CLASSIFICATION AND TABULATION : Classification : It is the process of arranging data into different classes or group according to resemblance and similarities. An ideal classification should be unambiguous, stable and flexible. Type of Classification : There are two types of classification depending upon the nature of data. (i) Classification according to attribute – if the data is of a descriptive nature having several qualifications i.e. males, female, illiterate, etc.
1.5.1.1.1.1.1.1.1 (ii) Classification according to class-interval if the data are expressed in numerical quantities i.e… ages of person vary and so do their heights and weights.

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Classification according to Attributes : (i) Simple classification is that when one attribute is present i.e. classification of persons according to sex–– males or female. (ii) Manifold classification is that when more than one attributes are present simultaneously two attributes – deafness and sex. A person may be either deaf or not deaf, further a person may be a male or a female. The data, thus are to be divided into four classes :(a) males who are deaf, (b) males who are not deaf, (c) females who are deaf, (d) females who are not deaf. The study can be further continued, if we find another attribute, say religion. Classification according to Class-intervals : The type arises when direct measurements of data is possible. Data relating to height, weight, production etc. comes under this category. For instance persons having weight, say 100-110 Lbs, can form one group, 110-120 lbs. Another group and so on. In this way data are divided into different classes ; each of which is known as class interval. Number of items which fall in any class-interval is known as class frequency. In the class-intervals mentioned above, the first figures in each of them are the lower limits, while the second figure are the upper limits. The difference between the limits of a class interval is known as magnitude of the class interval. If for each class intervals the frequencies given are aggregates of the preceding frequencies, they are known as cumulative frequencies. The frequencies may be cumulated either from top or from below. Method of forming Class-intervals : The class-intervals i.e. 100-110, 110-120, 120-130, etc. are overlapping. Difficulty arises when placing an item say 110 in the above class-interval. Whether 110 lbs should be placed in the classintervals 100-110 or 110-120. Now in this method, known as Exclusive method, an item which is identical to the upper limit of a class-interval is excluded from that class-interval, and is included in the next class-interval. So the item 110 lbs. will belong to the class interval of 110-120. For all practical uses, 100-110 means 100 and less than 110 again, 110-120 means 110 and less than 120, and so on. Again the class-intervals may be formed as 100-109, 110-119, 120-129 etc. In this method known as Inclusive method, also difficulty arises when there is an item lying between the upper limit of a class and lower limit of the next class. The above class-intervals may also be arranged as 100-109.5, 110119.5 and so on.
1.5.1.1.1.1.1.1.1 Class-intervals with Cumulative Frequencies : Class-intervals with Cumulative Frequencies :

If the class-frequencies are given as cumulative class-frequencies then the class-intervals also are expressed only the upper limit preceded by the word 'below' (or less than) or 'above' (or more than)

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according as the frequencies are cumulated from the top or bottom. Before treating with such data for any statistical purpose, it is necessary to convert it into usual class-intervals with their corresponding class-frequencies. From the following examples, the idea of converting the cumulative frequencies to usual frequencies will be clear. (a) Class-frequencies Weights Below110 ” 120 ” 130 ” 140 ” 150 cumulated from top Persons 10 15 17 21 27 (b) Class-frequencies Weights (lb) Above 100 ” ” ” ” 110 120 130 140 cumulated from bottom Persons 27 17 12 10 6

Now the usual type of class-intervals having class-frequencies will be as follows––

Weights (lb.) 100 – 110 110 – 120 120 – 130 130 – 140 140 – 150

Persons 10 5 2 4 6

DISCRETE AND CONTINUOUS SERIES : Statistical series may be either discrete or continuous. A discrete series is formed from items which are exactly measurable, Every unit of data is separate, complete and not capable of divisions. For instance, the number of students obtaining marks exactly 10, 14, 18, 29, can easily be counted. But phenomenon like height or weight cannot be measured exactly or with absolute accuracy. So the number of students (or individuals) having height exactly 5¢2¢¢ cannot be counted. Exact height may be either 5¢2¢¢ by a hundredth part of an inch. In such cases, we are to count the number of students whose heights lie between 5¢0¢¢ to 5¢2¢¢. Such series are known as ‘continuous’ series. Example : Discrete Series Marks 10 14 18 20 No. of Students 12 16 15 7 Continuous Series Height (inch) 58 – 60 60 – 62 62 – 64 64 – 66 No. of students 6 10 13 11

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TABULATION :
STATISTICAL METHODS

Tabulation is a systematic and scientific presentation of data in a suitable form for analysis and interpretation. After the data have been collected, they are tabulated i.e. put in a tabular form of columns and rows. The function of tabulation is to arrange the classified data in an orderly manner suitable for analysis and interpretation. Tabulation is the last stage in collection and compilation of data, and is a kind of stepping-stone to the analysis and interpretation. A table broadly consists of five parts – (i) Number and Title indicating the serial number of the table and subject mater of the table. (ii) stub i.e. the column indicating the headings or rows. (iii) Caption i.e. the headings of the column (other than stub) (iv) Body i.e. figures to be entered in the table (v) Foot-note is source from which the data have been obtained. Thus table should be arranged as follows :-

Table No. Title Stub Caption Total

Body

Total

Types of Tabulation : Mainly there are two types of tables –– Simple and Complex. Simple tabulation reveals information regarding one or more groups of independent question, while complex table gives information about one or more interrelated questions. One-way table is one that answers one or more independent questions. So it is a simple tabulation. The following table will explain the point :

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Table 1. Daily wages in Rs. obtained by 50 workers in a factory Wages (Rs.) 4–6 6–8 8 – 10 10 – 12 12 – 14 Total No. of Workers 20 9 10 7 4 50

The table shows the number of workers belonging to each class-interval of wages. We can now easily say that there are 20 workers, obtain wages between 4 and 6 (the minimum range) and there are 4 workers, obtain wages between 12 and 14 (the maximum range). So this table reveals information regarding only one characteristic of data i.e. wages of workers. Two-way table shows subdivision of a total and is able of answering two mutually dependent questions. In the above table (no. 1), if the workers are divided into sex-wise, then we would get a table as follows. :-

Table 2 No. of Workers Wages (Rs.) 4–6 6–8 8 – 10 10 – 12 12 – 14 Total Male 12 6 6 4 4 32 Female 8 3 4 3 0 18 Total 20 9 10 7 4 50

The above table shows the wages obtained by workers and sex-wise distribution of workers i question.

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Example 1. Construct a blank table in which could be shown at different dates and in five industries
STATISTICAL METHODS

the average wages of the four groups, males and females, eighteen years and over, and under eithteen years.

Average Wages of Employees in 5 Industries :

As on … (date) Industry Under 18 yes. 18 yrs. and over

As on … (date) Under 18 yes. male female total 18 yrs. and over male female total

Male female total male female total 1 2 3 4 5 Total

Example 2. Draw up a blank table to show the number of students reading in 1st, 2nd and 3rd year class (Pass and Honours) of a certain college in a faculties of Arts, Science and Commerce in the year 1983. Number of Students reading in different years (Pass or Hons.) with different streams in a college

Year Course® Stream¯ Arts Science Commerce Total

1st Year

2nd Year

3rd Year

Total number Pass Hons. Total

Pass Hons.Total Pass Hons.Total Pass Hons. Total

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Example 3 : Draw a blank table to show the number of students sexwise admitted in each of a 3 streams Arts. Science and commerce in the years 2000 and 2001 in a college of Kolkata showing totals in each stream, sex and year. [ICWA (F) Dec. 2006]

Table 3. Number of students according to sex, stream and year

Year ® Sex ® Stream¯ Arts Science Commerce Total male

2000 female total male

2001 female total

6.3 FREQUENCY DISTRIBUTION : Frequency of a value of a variable is the number of times it occurs in a given series of observations. A tally-sheet may be used to calculate the frequencies from the raw data (primary data not arranged in the Tabular form). A tally-mark (/) is put against the value when it occurs in the raw data. The following example shows how raw-data can be represented by a tally-sheer : Example : Raw data Marks in Mathematics of 50 students. (selected from among candidates in ICWA Examination)

37 21 50 37 44 30 47

47 41 45 45 48 40 37

32 38 52 31 46 36 47

26 41 46 40 16 32 50

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Table-4 Tally-sheet of the given raw data Marks (x) 16 21 26 30 31 32 33 36 37 38 39 Total / / // / / /// / // //// //// / Tally-Marks Frequencies (f) 1 1 2 1 1 3 1 2 4 5 1 22 Marks (x) 40 41 43 44 45 46 47 48 50 51 52 Total Total Frequency //// //// / / /// // /// // //// / // Tally-Marks Frequencies (f) 4 5 1 1 3 2 3 2 4 1 2 28 50

Such a representation of the data is known as the Frequency Distribution. The number of classes should neither be too large nor too small. It should not exceed 20 but should not be less than 5, normally, depending on the number of observations in the raw data.

GROUP FREQUENCY DISTRIBUTION :

When large masses of raw data are to be summarised and the identity of the individual observation or the order in which observations arise are not relevant for the analysis, we distribute the data into classes or categories and determine the number of individuals belonging to each class, called the classfrequency.
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A tabular arrangement of raw data by classes where the corresponding class-frequencies are indicated is known as Grouped Frequency distribution.

Table No. 5 : Grouped Frequency Distribution of Marks of 50 students in Mathematics Serial No. 1 2 3 4 5 6 7 8 Total Marks 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 No. of Students 1 1 3 5 16 10 11 3 50

Few Terms (associated with grouped frequency distribution) :(a) Class-interval (b) Class-frequency, total frequency (c) Class-limits (upper and lower) (d) Class boundaries (upper and lower) (e) Mid-value of class interval (or class mark) (f) Width of class interval (g) Frequency denisty (h) Percentage Frequency. (a) Class-interval : In the above table, class intervals are 16-20, 21-25 …. etc. In all there are eight class-intervals. If, however, one end of class-interval is not given then it is known as open-end class. For example, less than 10, 10-20, 20-30, 30 and above. The class-interval having zero frequency is know as empty class.

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(b)Class frequency : The number of observations (frequency) in a particular class-interval is known as class-frequency. In the table, for the class-interval 26-30, class frequency is 3 and so on. The sum of all frequencies is total frequency. Here in the table total frequency is 20. (c) Class limits : The two ends of a class-interval are called class-limits. (d) Class boundaries : The class boundaries may be obtained from the class limits as follows : Lower class-boundary = lower class limit – ½ d Upper class-boundary = upper class limit + ½ d Where d = common difference between upper class of any class-interval with the lower class of the next class-interval. In the table d = 1. Lower class boundary = 16 Upper class boundary = 20 +
1 ´ 1 = 16 – .5 = 15.5 2 1 ´ 1 = 20 + .5 = 20.5 2

Again, for the next class-interval, boundary = 25.5 and so on. (c)

lower class-boundary = 20.5, upper class

Mid value : (or class mark). It is calculated by adding the two class limits divided by 2. In the above table : for the first class-interval Mid-value =
16 + 20 36 = = 18 2 2

For the next one, mid value =

21 + 25 = 23 and so on. 2

(d) Width : The width (or size) of a class interval is the difference between the classboundaries (not class limits) \ Width = Upper class boundary – lower class boundary For the first class, width = 20.5 – 15.5 = 5 For the second class width = 25.5 – 20.5 = 5, so on. (g) Frequency density : It is the ratio of the class frequency to the width of that class-interval i.e. frequency density = For the first class frequency density = For the third class frequency density =
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Class frequency width of the class
1 = 0.2 5 3 = 0.6 5
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(h)Percentage frequency : It is the ratio of class-frequency to total frequency expressed as percentage. i.e. percentage frequency = Class frequency ´ 100 Total frequency

In the table for the frequency 5, % frequency = 5 / 50 ´ 100 = 10 and so on. All the above terms have been shown in the following summary table :

Summary Illustration of class-boundaries, mid-value, width etc. (Data : Table 1.3) Class Class Class-limits Class-boundaries Mid value Width Frequency Percentage of class Lower Upper 1 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Total 2 1 1 3 5 16 10 11 3 50 3 16 21 26 31 36 41 46 51 – 4 20 25 30 35 40 45 50 55 – Lower 5 15.5 20.5 25.5 30.5 35.5 40.5 45.5 50.5 – Upper 6 20.5 25.5 30.5 35.5 40.5 45.5 50.5 55.5 – 7 18 23 28 33 38 43 48 53 – 8 5 5 5 5 5 5 5 5 – 9 0.2 0.2 0.6 1.0 3.2 2.0 2.2 0.6 – 10 2 2 6 10 32 20 22 6 100 density frequency

Interval frequency

1.5.1.1.1.1.1.1.1

(i) Cumulative Frequency distribution : As the name suggests,

in this distribution, the

frequencies are cumulated. This is prepared from a grouped frequency distribution showing the class boundaries by adding each frequency to the total of the previous one, or those following it. The former is termed as Cumulative frequency of less than type and the latter, the cumulative frequency of greater than type. Numerical examples will be given while doing Graphical representation of a statistical data. Example : The following is an array of 65 marks obtained by students in a certain examination :– 26 32 48 34 27
1.5.1.1.1.1.1.1.2 33 35

45 36 27 42 49
37

27 41 46 45 48
47

50 31 47 31 47
28

45 41 31 28 32

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26 33 42 39 55 36 61

46 42 44 51 39 37 65

31 31 41 54 52 38 64

35 41 36 53 38 56 72 45 37 38 54 59 64

1.5.1.1.1.1.1.1.1

Draw up a frequency distribution table classified on the basis of marks with class-intervals of 5.

Class-intervals Of marks 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74 Total
1.5.1.1.1.1.1.1.2

Tally marks

Frequency

//// // //// //// //// //// /// //// /// //// //// /// //// / /// /// / /

7 10 13 8 13 6 3 3 1 1 65

Now the required frequency distribution is shown below : Frequency distribution of marks obtained by 65 students Marks 25-29 30-34 35-39 40-44 45-49 50-54 Frequency 7 10 13 8 13 6

6.22

MATHS

55-59 60-64 65-69 70-74 Total
1.5.1.1.1.1.1.1.1

3 3 1 1 65

CUMULATIVE FREQUENCY DISTRIBUTION : It is a form of frequency distribution in which each frequency beginning with the second from the top is added with the total of the previous ones, the class-intervals being adjusted accordingly. Example : Cumulative frequency distribution showing the marks (Data, Reference Table above) Marks 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74 Total Frequency 7 10 13 8 13 6 3 3 1 1 65 Cumulative frequency 7 17 (= 7 + 10) 30 (= 17 + 13) 38 (= 30 + 8) 51 (= 38 + 13) 57 (= 51 + 3) 60 (= 57 + 3) 63 (= 60 + 3) 64 (= 63 + 1) 65 (= 64 + 1) ––

Example : From the following table find (a) the less than, and (b) greater than cumulative frequencies and (c) cumulative percentage distributions.

Wages (Rs.) : Frequency :

10–20 21–30 31–40 41–50 51–60 61–70 Total 5 7 12 15 8 3 50

The Class boundaries are 10.5 – 20.5, 20.5 – 30.5 …… etc. The boundary points are 10.5, 20.5, 30.5, ….. etc. There is no frequency below 10.5, so its cumulative frequency (c.f.) is 0, the c.f. below 30.5 is 12 (= 5 + 7) and so on. The detail is shown below :

MATHS

6.23

Cumulative frequencies
STATISTICAL METHODS

Wages (Rs.)

cumulative frequency (c.f.) less than greater than 50 45 (= 50 – 5) 38 (= 45 – 7) 26 (= 38 – 12) 11 (= 26 – 15) 3 (= 11 – 8) 0 (= 3 –3)

cumulative percentage less than 0 5 ´ 100 = 10 50
12 ´ 100 = 24 50 24 ´ 100 = 48 50 39 ´ 100 = 78 50

greater than
50 ´ 100 = 100 50

10.5 20.5 30.5 40.5 50.5 60.5 70.5

0 5 12 (= 5 + 7) 24 (= 12 + 12) 39 (= 24 + 15) 47 (39 + 8) 50 (= 47 + 3)

45 ´ 100 = 90 50
38 ´ 100 = 76 50 26 ´ 100 = 52 50 11 ´ 100 = 22 50

47 ´ 100 = 94 50
50 ´ 100 = 100 50

3 ´ 100 = 6 50
0 ´ 100 = 0 50

Example : Prepare a frequency distribution table with the help of tallymarks for the words in the expression given below taking number of letters in the words as a variable. “Business mathematics and statistics Fundamentals in the Institute of cost and works Accounts of India” [ICWA (F) June 2007] Frequency distribution table with tally marks. Variable 2 3 4 5 6 8 9 10 12 Total tally mark /// /// / // / // / / / – frequency 3 3 1 2 1 2 1 1 1 15

6.24

MATHS

SELF EXAMINATION QUESTIONS : 1. Monthly wages (in Rs.) received by 30 workers in a certain factory are as follows : 310 315 305 320 390 318 325 350 337 354 386 367 370 359 392 335 380 340 300 380 363 397 323 385 331 342 367 375 327 393

Draw a frequency distribution table, classified on the basis of wages, with class-interval of 10. 2. Age at death of 50 persons of a place are as follows : 80 80 73 72 62 75 62 71 71 61 78 67 68 68 78 79 69 70 65 73 66 70 72 63 77 61 71 76 62 79 68 75 78 78 78 72 77 80 79 80 73 69 76 80 63 78 77 75 66 65

(a) arrange the data in a frequency distribution in 10 class-intervals, and (b) obtain the percentage frequency. [Ans. (a) 5, 2, 4, 4, 6, 3, 5, 9, 8 (b) 10, 4, 8, 8, 8, 12, 6, 10, 18, 16] 3. Prepare a frequency distribution table of continuous class-interval of 5 from the following data : 3 4 8 2 7 13 14 18 6 13 0 24 9 7 15 1 5 19 3 20 11 15 7 5 18 21 25 12 4 13 2 6 13 5 7 12 16 8 9 5 22 7 7 8 13 23 17 5 1 10

[Ans. 9, 18, 10, 7, 5, 1] 4. Ages of 100 students are shown below : Age 10 11 12 13 14 15 16 Students 15 20 12 35 4 6 8

Form a cumulative freq. distribution table in the form of ‘less than 11’ ‘less than 12’ and so on. [Ans. 15, 35, 47, 82, 86, 92, 100]

MATHS

6.25

5. The weights in Kilogram of 50 persons are given below :

76 53 64 42 46

64 56 74 52 55

53 65 48 62 65

55 60 59 72 75

66 47 72 43 48

72 55 61 63 59

52 67 43 71 67

63 73 69 64 77

46 44 61 51 64

51 54 58 67 78

Arrange the above data in frequency distribution with class-interval of 5 kg. Construct the frequency polygon on a graph paper with above data.

6. Weekly wages (in Rs.) received by 30 workers in a factory are as follows :

310 370 380 367

320 315 359 363

325 390 305 392

375 350 318

397 327 337

335 323 393

334 360 385

331 386 340

300 342 367

Prepare a frequency distribution table classified on the basis of wages with class interval of 10. Also obtain the percentage frequency in each class interval. [ICWA (F) Dec. ’97]

6.4 PRESENTATION OF DATA GRAPHS AND CHARTS : By classification and tabulation we find systematic presentation of data. Again presentation of data by graphs and charts reveal the true significance of the data. Of course it is true that graphs and charts add nothing to the information already obtained but bring out clearly the relative importance of different figures. The advantages in such presentation of data are attractive to common people. Disadvantage is that charts do not show in detail. Functions : (i) They make complex data simple and easily understandable. (ii) They help to compare the related data, placing in graphic representation. There are various types of graphs, charts, diagrams. A few of them have been shown here.

LINE CHART : We take a rectangular axes. Along the abscissa, we take the independent variable (x or time) and along the ordinate the dependent variable (y or production related to time). After plotting the points, they are joined by a scale, which represents a line chart. The idea will be clear from the following example.
6.26 MATHS

Example : Represent the following data by line chart. The monthly production of motor cars in India during 1959-60

Jan 70

Feb 90

Mar 80

April 120

May 100

June 120

July 110

Aug 125

Sept 130

Oct 150

Nov 100

Dce 120

Graph showing production of motor cars.

Use of false base line : If size of items is big and vertical scale starts from zero, the curve of line chart will be almost at the top of graph paper, as shown above. In such cases, a false base line is usually used.
1.5.1.1.1.1.1.1.1 Generally is vertical scale is broken into two parts and some blank space is left in Generally the vertical scalethe broken into two parts and some blank space is left in between them. between them. The above part starts from upper part starts with starts with a value eq The above part starts from zero and the zero and the upper part a value equal to the minimum value of the

MATHS

6.27

variable. A saw tooth lines are used to break the vertical scale. The idea will be clear from the following graph (refer the previous example) HISTOGRAM (when C.I. are equal)
STATISTICAL METHODS

Let us consider a frequency distribution having a number of class intervals with their respective frequencies. The horizontal axis is marked off to represent the C.I. and on these markings rectangles are drawn by taking the C.I. as breadth and corresponding frequencies as heights. Thus a series of rectangles are obtained whose total area represents the total of the class frequencies. The figure thus obtained is known as histogram. It may be noted here that C.I. must be in continuous form. Even if this is not given, then the discrete C.I. must be transferred to class boundaries and hence to draw the histogram. Example : Draw a histogram of the following frequency distribution showing the number of boys on the register of a school. Age (in years) 2–5 5–8 8–11 11–14 14–17 17–20 C.I. given are in class boundaries. No. of boys (in ’000) 15 20 30 40 25 10

40 35 30

Number of Boys

25 20 15 10 5 0 2 5 8 11 14 Age in Years 17 20

6.28

MATHS

Histogram (when C.I. are uneqal) : If the C.I. are unequal, the frequencies must be adjusted before constructing the histogram. Adjustments are to be made in respect of lowest C.I. For instance if one C.I. is twice as wide as the lowest C.I., then we are to divided the height of the rectangle by two and if again it is three times more, then we are to divide the height of its rectangle by three and so on.

Aliter (with the help of frequency density) : If the width of C.I. are euqal, the heights of rectangles will be proportional to the corresponding class frequencies. But if the widths of C.I. are unequal (i.e. some are equal and others are unequal), then the heights of rectangles will be proportional to the corresponding frequency densities (and not with the class frequencies) Frequency density = Class frequency Width of C. I.

Example : Draw a histogram of the following frequency distribution showing the number of boys on the register of primary school in a certain state :– Age (in years) 2–5 5–11 11–12 12–13 13–14 14–15 15–17 No. of boys (’000) 150 3066 497 477 496 143 162

C.I. (I) 2–5 5–11 11–12 12–13 13–14 14–15 15–17

Width of (II) 3 6 1 1 1 1 2

Frequency (III) 150 3066 497 477 496 143 162

Frequency density (IV) = (III) ÷ (II) 50 511 497 477 496 143 81

1.5.1.1.1.1.1.1.1 Now taking the C.I. on X axis and frequency densities on Y axis histogram can be drawn

Now taking the C.I. on X axis and frequency densities on Y axis histogram can be drawn (left to students for drawing)

MATHS

6.29

(left to students for drawing) HISTOGRAM. (For discontinuous grouped data of equal width).
STATISTICAL METHODS

For discontinuous grouped data first we are to make class-boundaries, then to draw the histogram as usual.

Example : Class-limits 10–19 20–29 30–39 40–49 50–59 60–69 70–79 80–89 Frequency 5 9 14 20 25 15 8 4

In class limits there is gap between upper limit of one class the lower limit of next class. So we are to make class boundaries at first. In making class boundaries we find. Class-boundaries 9.5–19.5 19.5–29.5 29.5–39.5 39.5–49.5 49.5–59.5 59.5–69.5 69.5–79.5 79.5–89.5 Frequency 5 9 14 20 25 15 8 4

Taking the class boundaries on X-axis and corresponding frequencies as heights of rectangles we can draw the diagram (left to the students). HISTOGRAM. (FOR DISCONTINUOUS GROUPED DATA OF UNEQUAL WIDTH). Example : Draw a histogram to present the following frequency distribution and in the graph, make the number of earning members of age 19-32 years. Age (in yrs.)
1.5.1.1.1.1.1.1.1 14–15 63

Earning members

6.30

MATHS

16–17

140 16_17 18–20 21–24 25–29 30–34 35–39

140 150
110 110 100 90

The frequency distribution is in discrete order. So we are to change the class-intervals into corresponding class-boundaries. Again since the widths of the class-intervals are unequal so we are to draw the histogram with the help of frequency density. The frequency distribution after adjustments will be as follows : Class interval (age) 14–15 16–17 18–20 21–24 25–29 30–34 35–39 Classboundaries 13.5–15.5 15.5–17.5 17.5–20.5 20.5–24.5 24.5–29.5 29.5–34.5 34.5–39.5 Width of Class 2 2 3 4 5 5 5 Frequency density 60 140 150 110 110 100 90 60 ÷ 2 = 30 140 ÷ 2 = 70 150 ÷ 3 = 50 110 ÷ 4 = 27.5 110 ÷ 5 = 22 100 ÷ 5 = 20 90 ÷ 5 = 18 Frequency

Now taking class boundaries on x-axis and frequency density on y-axis, we are to draw the histogram taking suitable scale. For convenience white paper is used instead of graph paper.

In the above graph the marked portion indicates the total numbers belonging to the age group 19-32. HISTOGRAM (when only mid-points are given). When only mid-points (of class-intervals) are given, we are to ascertain the upper and lower limits of the various classes and then to construct the histogram. Example : Draw a histogram of the following frequency distribution :

MATHS

6.31

Life of Electric Lamps
STATISTICAL METHODS

(in hrs.) mid values 1010 1030 1050 1070 1090

Firm 10 130 482 360 18

From the mid-values, the class-limits are ascertained as given below : Life of electric lamps 1000–1020 1020–1040 1040–1060 1060–1080 1080–1100 Now the histogram can be draw easily. (ii) Frequency Polygon : The line chart obtained by joining successively the middle-points of the tops (uppermost sides) of the rectangles in histogram by straight lines, is known as Frequency Polygon. It is customary to join the extreme two middle points to the base line at the middle-points of the next class intervals. The area covered by the frequency polygon is nearly the same as by the histogram. The dotted line of the Figure (3) represents the Frequency Polygon The Frequency Polygon can also be drawn without the help of a histogram. Points are plotted by taking the middle-points of the class-interval as abscissa (x-coordinate) and the corresponding frequency as ordinate (y-coordinate). Then the line chart obtained by joining such points by straight lies is known as Frequency Polygon. Draw histogram and frequency polygon of the following data : Wages (Rs.) No. of employees : 50–59 8 60–69 10 70–79 16 80–89 14 90–99 10 100–109 5 110–119 2 Frequency 10 130 482 360 18

[ICWA (F) Dec. 2003] The variates (wages) are is discrete order, so we are to make them in class boundaries at first as follows : Class boundaries : No. of employees : 49.5–59.05 8 59.5–69.5 10 99.5–109.5 5 69.5–79.5 16 109.5–119.5 2 79.5–89.5 14 89.5–99.5 10

6.32

MATHS

20 16
No. of employees

12 8 4 0 49.5 59.5 69.5 79.5 89.5 99.5 109.5 119.5 Wages (Rs)

MATHS

6.33

STATISTICAL METHODS

Note. From the above figure, it is noted that the ogives cut at a point whose ordinate is 85, i.e. half the total frequency corresponding and the abscissa is 51.33 which is the median of the above frequency distribution (see the sum on median in the chapter of Average). Even if one ogive is drawn, the median can be determined by locating the abscissa of the point on the curve, whose cumulative frequency is N/2. Similarly, the abscissa of the points on the less than type corresponding to the cumulative frequencies N/4 and 3N/4 give the Q1 (first quartile) and Q3 (third quartile) respectively, (Q1, Q3 will be discussed after median in the chapter of Average).

Example : Draw an ogive curve to the following data and hence find the value of Q3. [ICWA (F) June 2000] Class-limits 10–19 20–29 30–39 40–49 50–59 60–69 70–79 Frequency 3 8 21 38 15 9 6

Since Q3 is to be estimated from ogive so less than type ogive will be suitable. C.I. Class Boundary 10–19 20–29 30–39 40–49 50–59 60–69 70–79 9.5–19.5 19.5–29.5 29.5–39.5 39.5–49.5 49.5–59.5 59.5–69.5 69.5–79.5 3 8 21 38 15 9 6 f c.f. less than type 3 11 32 70 85 94 100

6.34

MATHS

Ogive (less than type) of given data :

100 90 80 70 60

Frequency

50 40 30 20 Q3 = 53 (approx) 10 0
9.5 19.5 29.5 39.5 49.5 59.5 69.5 79.5

Class Boundaries From the above graph we find that Q3 = 53 (app). ONE DIMENSIONAL DIAGRAM : 1. Simple bar-diagram : Consists of number of bars of uniform width separated by equal intervening spaces. The length of the bars is proportional to the values they represent. The bars may be placed vertically or horizontally. Bar diagram is generally used to represents a time-series. The base-line should be the zero line, when bar-diagrams are used for comparison. Example : The monthly productions of bicycles of a factory are as follows : January February March April May June Represent by simple bar-diagram. Scale : 1 division along Y axis = 10 units. 70 60 90 80 100 110.

MATHS

6.35

STATISTICAL METHODS

Example : Construct a horizontal bar diagram showing expenditure first five year plan of W. Bengal. (Crores of Rs.) On Industries On Irrigation On Agriculture On Transports and Roads On Micellaneous Scale : 1 division along X-axis = 10 crores of Rs. 1.5.1.1.1.1.1.1.1 Expenditure in First Five-Year Plan in West Bengal 110.00 67.50 90.00 42.40 50.00

6.36

MATHS

MATHS

6.37

STATISTICAL METHODS

320 280 240 200 160 120 80 40 0
1990 1991

Overhead Dir. Exp. Dir. Lab. Dir. Mat.

Years

6.38

MATHS

Rs. Cost of material Cost of labour Direct expenses Factory overhead

Rs.
9,600 7,680 2,880 3,840.

Present the above information in a suitable diagram so as to enable comparison among the various components and also in relation to the total. At first we are to express the different costs in percentage as follows :

Items

Amount (Rs.)

In percentage

Cumulative percentage

Cost of material Cost of labour Direct expenses Factory overhead Total

9.680 7,680 2,880 3,840 24,000

9,600/24,000 ´ 100 = 40 32 12 16 100

40 72 84 100

Note. A circular diagram or a pie chart may also be drawn to represent the given data. Subdivided bar diagram on percentage basis. Sub-divided bar-diagram on percentage basis, showing the cost different headings of a manufacturer.

Circular Diagram (or Pie diagram) : It is a pictorial diagram in the form of circles where whole area represents the aggregate and different sectors of the circle, when divided into several parts, represent the different components.

MATHS

6.39

For drawing a circular diagram, different components are first expressed as percentage of the whole.
STATISTICAL METHODS

Now since 100% of the centre of a circle is 360 degrees. 1% corresponds to 3.6 degrees. If p be the percentage of a certain component to the aggregate, then (p ´ 3.6) degrees will be the angle, which the corresponding sector subtends at the centre. Note : A pie diagram is drawn with the help of a compass and a diagonal scale or a protractor. Different sectors of the circle representing different components are to be marked by different shades or signs.
Example : The expenditure during Second Five-year Plan in West Bengal is shown as below : (Rs. in Crores) On Industries ” Irrigation ” Agriculture ” Transports & Roads ” Miscellaneous 127.00 92.50 100.00 92.50 68.00 480.00 – To represent the data by circular diagram. First we express each item as percentage of the aggregate. Industries Irrigation Agriculture Transports & Roads Miscellaneous
= 127.00 ´ 100 = 26.4 480.00

= 19.3 = 20.8 = 19.3 = 14.2

Now 1% corresponds to 3.6 degrees. So the angles at the centre of the corresponding sectros are (in degrees) : Industries Irrigation Agriculture Transp. & Roads Miscellaneous = 26.4 ´ 3.6 = 95.0 = 19.3 ´ 3.6 = 69.5 = 20.8 ´ 3.6 = 74.9 = 19.3 ´ 3.6 = 69.5 = 14.2 ´ 3.6 = 51.1

6.40

MATHS

Agriculture

= 20.8 ´ 3.6 = 74.9 Transp. & Roads Miscellaneous = 19.3 ´ 3.6 = 69.5 = 14.2 ´ 3.6 = 51.1

Now with the help of compass and protractor (or diagonal scale) the diagram is drawn.

IND. 26.4% IRR 19.3%

Transports Roads 19.3%

AGR 20.8%

Note: Additions of all percentages of the items should be equal to 100 and also the addition of all the angles should be equal to 360° (app.). It two aggregates with their components are to be compared, then two circles are required to be drawn having areas proportionate to the ratio of the two aggregates. Deductions from a Pie diagram : Example : In an Institution there are 800 students. Students use different modes of transport for going to the Institution and return. The given pie-diagram represents the requisite data. Read the diagram carefully and answer the following questions. (i) How many students travel by public bus? (ii) How many students go to Institute on foot? Institute Bus

216º 54º

ot fo On

18º

Cycle

lic Pub

s Bu

MATHS

6.41

(iii)How many students do not use Institute bus?
STATISTICAL METHODS

(iv) Find the ratio of students that go to Institution by public bus and Institute bus. (v) Find the percentage of students going to Institution by cycle. For finding number of observation, we shall use the formula given degree ´ total number of observation divided by total degree at the centre (i) Number of students =
54 0 ´ 800 = 120 3600 72 ´ 800 = 160 360

(ii) Number of students travel of foot =

degree at centre (or foot) = 360° – (216° + 54° + 18°) = 72° (iii) Number of students that use Institute bus =
216 ´ 800 = 480 360

\ Number of students not using Institute bus = 800 – 480 = 320 (iv) reqd. ratio = number of students using public bus 120 1 = = number of students using Institute bus 480 4
180 ´ 100 = 5% 1600

(v) Percentage of students going to Institution by cycle =

Example : The cost of manufacturing an article was Rs. 150. A pie diagram was drawn to show the cost. If she labour charges are represented by a sector of 114°, find the sum spent for other expenses. [ICWA (F) ] Total degree at the centre = 360° given degree at centre Amount of other expenses = ´ total amount total degree at centre 360 - 114 246 = ´ 150 = ´ 150 = Rs. 102.50. 360 360 SELF EXAMINATION QUESTIONS : THEORETICAL : 1. Explain with illustrations : (a) (b) (c) (d) (e) (f) Line graph Bar chart Pie chart Multiple chart Component Bar chart Ogive.

2. What is a false base line? Under what circumstances it is used? PROBLEMS :

6.42

MATHS

3. Represent the following sets of data by Line chart. (i) Months : The monthly production of motor-cars in India :–– Jan Feb. March April May June July 125 100 120 130 85 Aug. 95 Sept. 115 Oct 80 Nov. 140 Dec. 125

Production 110 105

(ii) Draw a line diagram of the given data of number of students in a class of a college. Year : 1990 20 ’91 25 ’92 24 ’93 30 ’94 35 ’95 38 (ICWA (F) June, 2005] (ii) Represent the data of production of steel factory by a line diagram. Year : Steel (’000 tons) 1980 9.1 ’81 7.9 ’82 10.3 ’83 11.3 ’84 8.7 ’85 13.6 ’86 10.

[ICWA (F) Dec. 2006] 4. Draw a Bar chart to present the following number of students of college : B. Com. 1 st yr. class –– 600 ,, ,, 2 nd ,, ,, 3 rd ,, ,, –– 540 –– 325

5. The monthly productions of Maruti Udyog Limited for the first six months of the year 1885 are given below : Months : Jan. Feb. 300 March 340 April 320 May 270 June 240

Production: 250

Represent by a bar chart. 6. Draw a simple bar chart of the given productions of a bicycle factory Year : 1995 ’96 7200 ’97 10000 ’98 12000 [ICWA (F) Dec. 2005]

Production : 8400

6. Prepare a Bar Chart from the following data : Indian foreign debt as on 31-3-1967 Source of borrowing Amount of loan in crores of (Rs)

MATHS

6.43

Number of students appeared and passed in B.Com. examination :
STATISTICAL METHODS

College A B C D

Appeared 700 612 507 310

Passed 490 402 390 250

8. The following shows the results Secondary Examination in a school in two consecutive years : Year No. of candidates appeared Passed in all ,, in 1 st div. ,, in 2 nd div. 1980 400 350 80 150 1981 460 390 90 160

You are required to represent the number of candidates passed in 1 st, 2 nd, 3 rd divisions by a suitable chart. [Hint. Use component bar chart] 9. Draw a histogram to represent graphically the following frequency distribution : (i) Weight (in lbs) 80-90 90-100 100-110 110-120 120-130 130-140 Total (ii) Wages/hour(Rs.) No. of workers 5-10 10 10-15 25 15-20 30 20-25 20 25-30 15 [ICWA (F) June, 2006] (iii) Wages/(Rs.) No. of employees 50-59 8 60-69 10 70-79 16 80-89 12 90-99 7 [ICWA (F) June, 2007] (iv) Output (per workers) 500-509
6.44

No. of students 85 300 215 150 50 200 1000

No. of workers 8
MATHS

510-519

18 520-529 530-539 540-549 550-559 560-569 570-579 23 37 47 25 16 5

[Hins. Change to class boundaries and hence draw] 10. Draw a histogram to represent the following frequency distribution :

(i)

Weight (in kg) 30-35 35-40 40-45 45-55 55-65 65-70

Persons 10 12 20 18 10 12

(ii) Wages/hour (Rs.) No. of workers 4-6 6 6-8 12 8-10 18 10-12 10 12-14 4 [ICWA (F) June, 2005] 11. Represent the following table by histogram and frequency polygon : Incomes (in Rs.) No. of Persons 100-149 150-199 200-249 250-299 300-349 350-399 400-449 450-499 [Hint : Change to class boundaries and hence draw.] 21 32 52 105 62 43 18 9

[Hins: Change to class boundaries and hence draw.]

MATHS

6.45

12. Draw ogive or cumulative frequency curve (less than type) form the following frequency
STATISTICAL METHODS

distribution : (a) Weight (kg.) Frequency 8 12 15 20 14 8 4

40–44 45–49 50–54 55–59 60–64 65–69 70–74

(b)

C.I 10–19 20–29 30–39 40–49 50–59 60–69 70–79 and hence for the graph find the value of Q3 (c) daily wages (Rs.) : No. of workers : 0–30 20 30–60 50 60–90 60

Frequency 3 8 21 38 15 9 6 [ICWA(F) 2003] [Ans. 53(app.)]

90–120 40

120–150 30 [ICWA (F) Dec., 2005]

13. Draw a Pie chart to represent the following data on the proposed outlay during a Five–Year plan of a Government : (Rs. in crores) Agriculture Industries and Minerals 12,000 9,000

6.46

MATHS

Irrigation and Power Education Communications

6,000 8,000 5,000

14. Draw a circular diagram from the following data : Revenue of Central Government : (Crores) Customs Excise Income tax Corporation Tax Other sources 160 500 330 110 100

15. Draw a Pie chart to represent the following data relating to the production cost of a manufacture : Rs. Cost of material Cost of labour Direct expenses Overhead 16. Draw a Pie-diagram to represent the following : (i) Items : Expenses (in Rs.) wages 125 materials 110 taxes 180 profit 65 administration 20 [ICWA (F) Dec. 2004] (ii) males Females Girls Boys Total 2,000 1,800 4,200 2,000 10,000 18,360 16,524 3,672 7,344

17. The production cost of manufacturer are as follows : Rs. Direct material Direct labour Direct expenses Overhead 700 800 200 300

Present the above in a suitable diagram so as to enable as to enable among the various components 1.5.1.1.1.1.1.1.1 Present the above in a suitable diagram so comparison comparison among the various components and in relation to and in relation to the total. the total.
MATHS 6.47

6.5 MEASURES OF CENTRAL TENDERCY OR AVERACE :
STATISTICAL METHODS

INTRODUCTION : A given raw statistical data can be condensed to a large extent by the methods of classification and tabulation. But this is not enough for interpreting a given data we are to depend on some mathematical measures. Such a type of measure is the measure of Central Tendency. By the term of ‘Central Tendency of a given statistical data’ we mean that central value of the data about which the observations are concentrated . A central value which ‘enable us to comprehend in a single effort the significance of the whole is known as Statistical Average or simply average. The three common measures of Central Tendency are : (i) (ii) (iii) Mean Median Mode

The most common and useful measure is the mean. As we proceed, we shall discuss the methods of computation of the various measures. In all such discussions, we need some very useful notations, which we propose to explain before proceeding any further. (i) Index or Subscript Notation : Let X be a variable assuming n values x1, x2, …..x3, We use the symbol x j (read “x sub j”) to denote any of the above mentioned n numbers. The letter j, which can stand for any of the numbers x1 , x2 , …..x n is called a subscript notation of index. Obviously, any letter other than j, as I, k, p, q and s could be used.’ (ii) Summation Notation :
n

The symbol
n

åX
j =1

j

is used to denote the sum xj’s from j = 1 to j = n. By definition.

åX
j =1

j

= x1 + x 2 + ... + x n
n

Example 1 :

å j =1

Xj Yj = X1 Y1 + X2 Y2 + …. + X n Yn

Some important result :
1.5.1.1.1.1.1.1.1 (i)

å (x j + y j )= å x j + å y1
j =1 j =1 j =1

n

n

n

6.48

MATHS

(ii)

+ .... å A = A +[nAtimes+ A = nA (A is cons tan t) ] j =1
n

n

(iii)

å Ax j = Ax1 + Ax 2 + ... + Ax n j =1

MEAN : There are there of mean : (i) Arithmetic Mean (A.M.) (ii) Geometric (G. M.) (iii) Harmonic Mean (H.M.) Of these the Arithmetic mean is the most commonly used. In fact, if no specific mention be made by mean we shall always refer to arithmetic Mean (AM) and calculate accordingly. 1. Arithmetic Mean : (i) Simple Arithmetic mean : (Calculating mean from ungrouped data) The simple arithmetic mean x of a given series of values, say, x1, x2,…….. x n is defined as the sum of these values divided by their total number : thus
n

()

x1 + x 2 + .... + x n j =1 x (x bar ) = = n n

åx j åx =
n

Note. Often we do not write xj , x means summation over all the observations. Example : Find the arithmetic mean of 3,6,24 and 48. Required A.M.= 3 + 6 + 24 + 48 = 81 = 20.25 4 4 (ii) Weighted Arithmetic Mean : (Calculating the mean from grouped data) If the number x1, x2 , ……. x n occur f1, f2…….f n times respectively (i.e. occur with frequencies f1, f2 ……..f n) the arithmetic mean is
n

x=

åx j fj j =1
n

f x + f 2 x2 ..... + f n xn = 11 =
f1 + f 2 +....+ f n

å j =1

fj

åf x=åf x f åx N

Where N =

åf

is the total frequency, i.e., total number of cases. This mean x is called the

weighted Arithmetic mean, with weights f1, f2 …….fn respectively. In particular, when the weights (or frequencies) f1, f2……f Arithmetic Mean. Examples 3 : If 5, 8, 6 and 1 occur with frequency 3, 2, 4 and 1 respectively, find the Arithmetic mean.
n

are all equal. We get the simple

MATHS

6.49

STATISTICAL METHODS

Arithmetic mean =

(3 ´ 5)+ (2 ´ 8)+ (4 ´ 6)+ (1 ´ 2)
3 + 2 + 4 +1

57 = 15 + 16 + 24 + 2 = = 5.7 \ x = 5.7 10 10

Calculation of Arithmetic Mean (or simply Mean) from a grouped frequency distribution –– Continuous Series. (i) Ordinary method (or Direct Method) In this method the mid-values of the class-intervals are multiplied by the corresponding classfrequencies. The sum of products thus obtained is divided by the total frequency to get the Mean. The mean x is given by x=

å fx
N

, where x = mid-value of a class and N = total frequency

Example 4 : Calculate the mean of daily-wages of the following table :

Wages 4–6 6–8 8–10 10–12 12–14

No. of workers 6 12 17 10 5

Table 2.1 Calculation of Mean Daily Wages Class Interval 4–6 6–8 8–10 10–12 12–14 Total \ Mean Daily Wages = Mid values (Rs.) x 5 7 9 11 13 –– Frequency f 6 12 17 10 5 50 = N 30 84 153 110 65 442 = fx fx

å fx
N

= 442 = Rs. 8.84 50

1.5.1.1.1.1.1.1.1 (ii) Shortcut Method (Method of assumed Mean)

6.50

MATHS

(ii) Shortcut Method (Method of assumed Mean)
In this method, the mid-value of one class interval (preferably corresponding to the maximum frequency lying near the middle of the distribution) is taken as the assumed mean (or the arbitrary origin) A and the deviation from A are calculated. The mean is given by the formula : x=A+

å fd
N

where, d = x – A = (mid value) – (Assumed Mean).

Step deviation method : å fd ¢ x=A+ ´ i, where d ¢ = x- A i = scale (= width of C.I.) i

å

f

Example 5 : Compute the Arithmetic Mean of the following frequency distribution : Marks 20–29 30 –39 40– 49 50 –59 60 –69 70–79 No. of student 5 11 18 22 16 8

Table 2.2 Calculation of A.M. Class Interval Mid values x Deviation from 54.5 d = x – 54.5 20–29 30–39 40 –49 50–59 60–69 70 – 79 Total 24.5 34.5 44.5 54.5 (=A) 64.5 74.5 –– – 30 – 20 –10 .0 10 20 –– 5 11 18 22 16 8 80 = N – 150 – 220 –180 0 160 160 – 550 +320 = – 230
\ Arithmetic Mean fd \ Arithmetic Mean =A+ = 54.5 + - 230 N 80 fd - 230 = = 54.5 – 2.875 =.5 + A+ = 54 51.625 = 51.6 (approx). N 80
MATHS 6.51

frequency f

fd

=

å fd

å å

= 54.5 – 2.875 = 51.625 = 51.6 (approx).
STATISTICAL METHODS

(iii) Method of Assumed mean

(by using step deviations) x-A i –3 –2 –1 0 1 2 –– fd¢ – 15 – 22 –18 0 16 16 – 23 =

Class 19.5–29.5 29.5–39.5 39.5 –49.5 49.5–59.5 59.5–69.5 69.5 – 79.5 Total

Mid - po int s 24.5 34.5 44.5 54.5=A 64.5 74.5 ––

d´=

f 5 11 18 22 16 8 80 = N

å fd ´

A.M. = A +

å fd´ ´ i = 54.5 – 23 ´10 = 54.5 - 2.88 = 51.6 (app.)
N 80

CALCULATION OF A. M. FROM GROUPED REQUENCY DISTRIBUTION WITH OPEN ENDS

If in a grouped frequency distribution, the lower limit of the first class of the upper limit of the last class are not known, it is difficult to find the A.M. When the closed classes (other than the first and last class) are of equal widths, we may assume the widths of the open classes equal to the common width of closed class and hence determine the AM. But we can find Median or Mode without assumption. Properties of Arithmetic Mean : 1. The sum total of the values fx is equal to the product of the number of values of their A.M. e.g. N x =

å fx.

2. The algebraic sum of the deviations of the values from their AM is zero. [ICWA (F) 2004] If x1, x2……x n are the n values of the variable x and x their AM then x1 – x , x2 – x , ….. x n – x are called the deviation of x1 , x2 ………..xn respectively from from x Algebraic sum of the deviations =

å1(x j - x) j=

n

= (x1 – x ) + ( x 2 – x ) +….. + ( x n – x ) = (x1 + x2 +….+ x n) – n x = n x – n x =0 Similarly, the result for a weighted AM can be deduced. 3. If group of n1 values has AM. x and another group of n2 values has AM x 2 , then A.M. ( x ) of the composite group (i.e. the two groups combined) of n1 + n2 values is given by :

6.52

MATHS

n x + n2 x 2 x= 1 1 In general, for a group the AM ( x ) is given by n1 + n2
r

n1 x j n1 x1 + n2 x 2 + n3 x 3 + ... + nr x r j =1 x= = r n1 + n2 + ... + nr nj

å

å1 j=

Example 6 : The means of two samples of sizes 50 and 100 respectively are 54.1 and 50.3. Obtain the mean of the sample of size 150 obtained by combining the two sample. Here, n1 = 50, n 2 = 100, x 1 = 54.1, x 2 = 50.3 n x + n2 x 2 50 ´ 54.1 + 100 ´ 50.3 \Mean ( x ) = 1 1 = n1 + n2 50 + 100 = 2705 + 5030 = 7735 = 51.57 (approx.) 150 150

FINDING OF MISSING FREQUENCY : In a frequency distribution if one (or more) frequency be missing (i.e. not known) then we can find the missing frequency provided the average of the distribution is known. The idea will be clear from the following example : For one missing frequency : Example 7 : The AM of the following frequency distribution is 67.45. Find the value of f3, Let A = 67. Now using the formula. Height (inch) 61 64 67 70 73 Frequency 15 54 f3 81 24

Calculation of missing frequency Height (x) Height (x) 61
61 64 64 67 67 70
1.5.1.1.1.1.1.1.1 73 70

Calculation of missing frequency Frequency(f) d=x A d´ = d / 3 Frequency(f) d = x6 A – d´ = d2/ 3 15 15 –6 –21 54 3 54 f3 f3 81
81 24 3 1

fd´ fd´ 30 – 30 54
– 540 0 81

–3 0 03

–10 01

6 81

2

48

Total
MATHS

174 +f3
6.53

Let A = 67. Now using the formula
STATISTICAL METHODS

x = A+

å fd´ åf

×i, we get 67.45 = 67 +

45 ´ 3 174 + f 3

or, 0.45 =

135 or, 78.30 + 0.45 f = 135 3 174 + f 3

or, 0.45f3 = 56.70 or, f3 = 56.70 = 126 0.45 For two missing frequencies : Example 8 : The A.M. of the following frequency distribution is 1.46 No. of accidents 0 1 2 3 4 5 Total Find f1 and f2 [ICWA (F) June, 2000] x 0 1 2 3 4 5 Total f 46 f1 f2 25 10 5 200 d=x–2 –2 –1 0 1 2 3 –– fd –92 –f1 0 25 20 15 – 32 –f1 No. of days 46 f1 f2 25 10 5 200

AM = A =

å fd åf
-32 - f1 - (32 - f1 ) or, – 0.54 = 200 200

or, 1.46 = 2 +

or, 108 = 32 + f1 , or, f1 = 76,

f2 = 200 – (46 + 76 + 25+ 110 + 5) = 38

6.54

MATHS

Example : Arithmetic mean of the following frequency distribution is 8.8. Find the missing frequencies : Wages (Rs.) No. workers : 4–6 6 6–8 –– 8–10 16 10 –12 –– 12–14 5 Total 50

[ICWA (F) Dec. 2004]

wages 4–6 6–8 8–10 10–12 12–14 Total

x 5 7 9 11 13 27+f1+f2

f 6 f1 16 f2 5

fx 30 7f1 96 11f2 65 191+7f1+11f2

x=

å fx or, 8.8 = 191 + 7f1 + 11f 2 , 8.8 = 191 + 7f1 + 11(23 - f1 ) 27 + f1 + f 2 27 + 23 åf
or, f1 = 1 i.e. f2 = 23 –1 = 22

or, 8.8 × 50 = 191 + 253 – 4f1 or, 4 f1 = 444 – 440= 4 WRONG OBSERVATION :

After calculating A.M. x of n observations if it is detected that one or more observations have been taken wrongly (or omitted), then corrected calculation of A.M. will be as follows : Let wrong observations x1, y1 being taken instead of correct values x, y then corrected

()

å x = given

å x – (x
Wrong

1

– y1) + (x + y), in this case total no. of observations will be same.

Example 9 : The mean of 20 observations is found to be 40. Later on, it was discovered that a marks 53 was misread as 83. Find the correct marks. [ICWA (F) Dec, 2000]

å x = 20 × 40 = 800,

Correct

å x = 800 – 83 + 53 = 770

\ Correct x = 770 = 38.5 20 Example 10 : A.M. of 5 observations is 6. After calculation it has been noted that observations 4 and 8 have been taken in place of observations 5 and 9 respectively. Find the correct A.M.
x=

åx
n

or, 6 =

åx
5

or,

å x = 30 ,

corrected

å x = 30 –(4+8) + (5+9) = 32

Corrected A.M. = 32 = 6.4 5

MATHS

6.55

STATISTICAL METHODS

Frequency distribution Rs. 0–4 4–8 8–12 f 2 4(=6–2) 7(=13–6) Rs. 0–4 4–8 8–12 12–16 16 –20 Total Let A = 10 x 2 6 10 14 18 –– f 2 4 7 5 2 20

Calculation of A.M. d –8 –4 0 4 8 –– fd –16 –16 0 20 16 4

12–16 5 (= 18–13) 16–20 2 (= 20 –18)

x= A +

å fd = 10 + 4 = 10 + 2 = Rs. 12 2 åf
Calculation of A.M. x 2.5 7.5 12.5 17.5 –– f 2 3 4 1 10 d –5 0 +5 +10 –– d´ –1 0 1 2 –– fd´ –2 0 4 2 4 f 2 (=10–8) 3 (= 8 –5) 4 (=5–1) 1(= 1–0) Marks. 0–5 5–10 10 –15 15–20 Total

(ii) Frequency distribution Marks 0–5 5–10 10–15 15–20

Let A = 7.5

6.56

MATHS

A.M. = A +

å fd´ × i = 7.5 + 4 × 5 = 7.5 + 2 = 9.5 marks. 10 åf

Advantages of Arithmetic Mean (i) It is easy to calculate and simple to understand. (ii) For counting mean, all the data are utilised. It can be determined even when only the number of items and their aggregate are known. (iii) It is capable of further mathematical treatment. (iv) It provides a good basis to compare two or more frequency distributions. (v) Mean does not necessitate the arrangement of data. Disadvantages of Arithmetic Mean (i) It may give considerable weight to extreme items. Mean of 2, 6, 301 is 103 and more of the values is adequately represented by the mean 103. (ii) In some cases, arithmetic mean may give misleading impressions. For example, average number of patients admitted in a hospital is 10.7 per day, Here mean is a useful information but does not represent the actual item. (iii) It can hardly be located by inspection. MORE SOLVED EXAMPLES Example 1: Fifty students appeared in an examination. The results of passed students are given below : Marks 40 50 60 70 80 90 No. of students 6 14 7 5 4 4

The average marks for all the students is 52. Find out the average marks of students who failed in the examination. [ICWA (F) June, 2000, Dec, 2006]

MATHS

6.57

STATISTICAL METHODS

Marks (x) 40 50 60 70 80 90 Total 6 14 7 5 4 4 40

f

fx 240 700 420 350 320 360 2390

x=

å fx = 2390 = 59.75, n = 40 å f 40
1

Let average marks of failed students = x 2, n2 = 10
n x1 + n 2 x 2 x= 1 n1 + n 2

or, 52 =

40 ´ 59.75 + 10x 2 50 \reqd. average marks = 21.

or, x 2 = 21 (on reduction) Class : frequency :

Example 2 : From the following frequency table, find the value of x if mean is 23.5 50–59 x–4 40–49 x–2 30–39 x+3 20 –29 x+5 10–19 x + 10 0–9 x –2

[ICWA (F) June, 2005]

Class

mid. pt. x

f

d = x– 34.5



fd

50–59 40 –49 30 – 39 20 – 29 10 – 19 0–9 Total

54.5 44.5 34.5 24.5 14.5 4.5

x–4 x –2 x+3 x+5 x + 10 x–2 6x + 10

20 10 0 –10 –20 – 30

2 1 0 –1 –2 –3

2x – 8 x–2 0 – x –5 – 2x – 20 –3x + 6 – 3x –29

x = A+

å fd´ × i, åf

or 23.5 = 34.5 + -3x - 29 × 10 6x + 10

or, x = 5 (on reduction).

6.58

MATHS

Example 3 : The mean salary of all employees of a company is Rs. 28,500.The mean salaries of male and female employees are Rs. 30,000 and Rs. 25,000 respectively. Find the percentage of males and females employed by the company. [ICWA (F) June, ’98]

n x +n x Let number of male employees be n1 and that of female be n2. We know x = 1 1 2 2 n1 + n2 or, 28500 =
30000n1 + 25000n2 n1 + n2

n or, 7n2 = 3n1 (on reduction) or , 1 = 7 n2 3 Percentage of n1 (male) = 7 ´ 100 = 70% 10 ,, ,, n2 (female) = 7 ´ 100 = 30% 10

SELF EXAMINATION QUESTIONS : 1. The weight of 6 persons are as follows (in kg.) 70, 42, 85, 75, 68, 55. Find the mean weight. [Ans. 65.83 kg.] 2. Find A.M. of the following numbers : (i) 1, 2, 3, …… upto 10 th term (ii) The first 10 even numbers (iii) The first 10 odd numbers 3. Find A.M. of the following numbers : (i) 77, 73, 75, 70, 72, 76, 75, 71, 74, 78 (ii) 4, 5, 6, 7, 5, 4, 8, 6, 2, 5, 3 4. Find A.M. of the given frequency distribution : (i) Weight (kg.) 50 55 60 65 70 Total Persons 15 20 25 30 30 100 [Ans. 60 kg.] [Ans. 79.68] [Ans. 5] [Ans. 5.5] [Ans. 11] [Ans. 10]

MATHS

6.59

STATISTICAL METHODS

(ii)

Weight (kg.) 20 21 22 23 24 25 26 27 28

Workers 8 10 11 16 20 25 15 9 6 [Ans. Rs. 24.05]

5.

Find the weekly average wage from the given frequency : Wages (Rs.) 30–40 40–50 50–60 60–70 70–80 80–90 No. of Workers 80 20 40 18 10 4 [Ans. Rs. 56.40]

6.

Find A.M. from the following table : Wages (Rs.) 20–25 25–30 30–35 35–40 40–45 45–50 No. of Workers 200 700 900 800 600 400 [Ans. Rs. 34.40]

6.60

MATHS

7.

Compute A.M. of the following distribution : Class Interval 1–4 4–9 9–16 16–27 Frequency 6 12 26 20 [Ans. 13.52]

8.

A.M. of the following distribution is 124 lb. Weight (lb) 100 110 120 135 x+5 Total No. of Persons 1 2 3 2 2 10 [Ans. 140 ]

Find the value of x. [Hint. Use direct method 9. A.M. of the following frequency distribution is 5.4. Find the missing frequency. x 2 4 6 8 10 f 4 2 –– 3 2 [Ans. 6 ] 10. A.M. of the distribution is Rs. 56.47. Find missing frequencies. Daily Wages (Rs.) 45 50 55 60 65 70 75 Frequency 5 48 f3 30 f5 8 6

Total

150

[ICWA (F) June, 2004] [Ans.41, 12 ]

MATHS

6.61

GEOMETRIC MEAN (G. M.)
STATISTICAL METHODS

Definition. : The geometric mean (G) of the n positive values x1, x2, x3 ………….xn is the nth roof of the product of the values i.e. G = n x1 . x2 ....., xn It means, G = (x1. x2,……… xn )1/ n Now taking logarithms on both sides, we find
log G = 1 log ( x1. x 2 ........., x n ) = 1 (log x1 + ..... + log x n ) = 1 n n n

å log x .....(1)

\G = antilog é 1 ên ë

å log x ù ú û

Thus, from formula (1) we find that the logarithm of the G. M. of x1, x2 ….., xn = A.M. of logarithms of x1, x2 , …..., x n . Properties : 1. The product of n values of a variate is equal to the nth power of their G. M. i.e., x1 , x2 , ……, xn = Gn (it is clear from the definition)] 2. The logarithm of G. M. of n observations is equal to the A.M. of logarithms of n observations. [Formula (1) states it] 3. The product of the ratios of each of the n observations to G. M. is always unity. Taking G as geometric mean of n observations x1 , x2 , ……., xn the ratios of each observation to the geometric mean are x1 x 2 x , …… n G G G By definition, G = n x1 , x 2 ,....., x n or, Gn = (x1 , x2, ……, x n). Now the product of the ratios.
n x1 x 2 x x1.x 2 ....x n . …… n = = G =1 G G G G.G.....to n times G n

4. If G1 , G2……, are the geometric means of different groups having observations n1, n2………respectively, then the G. M. (G) of composite group is given by G=
N

G1n1 . G 2 n 2 .... where N = n1 + n2 + …..i.e., log G =

1 [n log G + n log G + .....] 1 2 2 N 1

Example 1 : Find the G. M. of the number 4, 12, 18, 26. Solution : G = 4 4 .12 .18.18 ; here n = 4 Taking logarithm of both sides, Log G = 1 (log 4 + log 12 + log 18 + log 26) 4 = 1 (0.6021 + 1.2553 + 1.4150) = 1 (4, 3516) = 1.0879) 4 4 \ G = antilog 1.0879 = 12.25.

6.62

MATHS

WEIGHTED GEOMETRIC MEAN : If f1, f2 , f3……f n are the respective frequencies of n variates x1 , x2 , x3 ,…….x n, then the weighted G. M. will be G = x1f1 ´ x 2 f 2 ´ x 3f3 ´ ..... ´ x n f n Now taking logarithm. Log G = 1 (f1 log x 1 + f2 log x2 +f3 log x3 +…..+ f n log x n) N = 1 N

(

1 )/ n where N = f + f + ……+ f = å f
1 2 n

å f log x.

G = anti log æ 1 ç èN

÷ å f log x ö ø

Steps to calculate G. M. 1. Take logarithm of all the values of variate x. 2. Multiply the values obtained by corresponding frequency. 3. Find

åf

log x and divide it by

å f , i.e., calculate å f

log x/

åf.

4. Now antilog of the quotient thus obtained is the required G. M. The idea given above will be clear from the following example. Example 2 : Find (weigh) G. M. of the table given below : –– x 4 12 18 26 f 2 4 3 1

Calculations of G.M x 4 12 18 26 Total \ log G = f 2 4 3 1 10 = 10.7019 = 1.07019 10 log x 0.6021 1.0792 1.2553 1.2553 ––– f log x 1.2032 4.3168 3.7659 1.4150 10.7019

å f log x åf

\ G = antilog 1.07019 = 11.75

MATHS

6.63

Advantages Geometric Mean
STATISTICAL METHODS

(i) It is not influenced by the extreme items to the same extent as mean. (ii) It is rigidly defined and its value is a precise figure. (iii) It is based on all observations and capable of further algebraic treatment. (iv) It is useful in calculating index numbers.

Disadvantages of Geometric Mean : (i) It is neither easy to calculate nor it is simple to understand. (ii) If any value of a set of observations is zero, the geometric mean would be zero, and it cannot be determined. (iii) If any value is negative, G. M. becomes imaginary. [Use. It is used to find average of rates of changes.]

SELF EXAMINATION QUESTIONS 1. Find G.M of the following numbers : (i) 3, 9, 27 (ii) 3, 6, 24, 48 2. Weekly wages of 6 workers are 70, 42, 85, 75, 68, 53 (in Rs.). Find the G. M. 3. Calculate G. M. (upto 2 decimal places) : (i) 90, 25,m 81, 125 3 (ii) 125, 700 , 450, 87 3 4. Compute G. M. :] (i) 4, 16, 64, 256 (ii) 1, 2, 4, 8, 16 (iii) 2, 79; 0.375, 1000 5. Monthly expenditure of 5 students are as follows : Rs. 125, 130, 75, 10, 45, find G.M.
1.5.1.1.1.1.1.1.1 of 2574, 475, 75, 5, 0.8, 0.005,, 0.0009. 6. Calculate G.M.6. Calculate G.M. of 2574, 475, 75, 5, 0.8, 0.005,, 0.0009.

[Ans.9] [Ans. 12]

[Ans. 64.209]

[Ans. 69.08] [Ans. 183.90]

[Ans. 32] [Ans. 4] [Ans. 10.877]

[Ans. Rs. 55.95]
[Ans. 1.841] [Ans. 1.841]

7. Find G.M. of the table given : x 44.5 7.05 91.72 f 2 3 4 [Ans. 71.38]
6.64 MATHS

8. Find G.M. of 111, 171, 191, 212, having weight by 3, 2, 4 and 5 respectively.

[Ans. 173.4]

9. Increase of productions for the first three years are respectively 3%, 4%and 5%. Find average production of the three years. [Hint : Use G.M] [Ans. 3.9%]

HARMONIC MEAN (H. M.) : Definition. The Harmonic Mean (H) for n observations, x1, x2,…….x n is the total number divided by the sum of the reciprocals * of the numbers.
i.e. H = n = 1 + 1 + .... + 1 x2 x2 xn

å

n .1 x

Again , 1 = H

å. 1 x
n

(i.e. reciprocal of H. M = A. M. of reciprocals of the numbers).

* For ab = 1. a = 1 , i.e. a reciprocal of b. And for 1 = b, b is reciprocal of a. Reciprocal of 2 is. 1 . b a 2 Example 1 : Find the H. M. of 3, 6, 12 and 15. H.M. = 4 4 = 1+1+ 1 + 1 20 + 10 + 5 + 4 3 6 12 15 60

Example 2 : Find the H.M. of 1, 1 , 1 ,……. 1 2 3 n H.M. =
n n = = 2n 1 + 2 + 3 + ... + n n (2 + n - 1) n (n + 1) 2

[Note. The denominator is in A..P. use S = n { a + ( n - 1)d} 2 2 Example 3 : A motor car covered distance of 50 miles four times. The first time at 50 m. p. h, the second at 20 m. p. h., the third at 40 m. p. h, and the fourth at 25 m.p.h Calculate the average speed and explain the choice of the average. Average Speed (H.M) = 4
1 + 1 + 1 + 1 50 20 40 25

= =

4
20 + 50 + 25 + 40 1000

= 4 ´ 1000 = 29.63 135

= 30 (app.) m. p. h.

MATHS

6.65

For the statement x units per hour, when the different values of x (i.e. distances) are given, to find
STATISTICAL METHODS

average, use H.M. If again hours (i.e., time of journey) are given, to find average, we are to use A.M. In the above example, miles (distances) are given, so we have used H.M. Weighted H.M. The formula to be used is as follows : H .M . = N f1 f 2 + + ..... + x1 x2 fn xn ,

åf

=N

Example 4 : (a) A person travelled 20 k.m. at 5 k.m.p.h. and again 24 k.m. at 4 k.m.p.; to find average speed. (b) A person travelled 20 hours at 5 k.m.p.h. and again 24 hours at 4.m.p.h.; to find average speed.

(a) We are to apply H.M. (weight) in this case, since, distances are given. Average speed (H.M.) = 20 + 24 = 44 = 44 = 4.4 k.m.p.h. 20 + 24 4 + 6 10 5 4 (b) We are to apply A.M. (weighted), since times of journey are given. Average speed (A.M.) = = 20 ´ 5 + 24 ´ 4 = 100 + 96 = 196 = 4.45 k.m.p.h (app.) 20 + 24 44 44 Example 5 : Find the harmonic mean of the following numbers : 1, 1 , 1 , 1 2 3 4 H.M.= 4 4 = = 4 =2 1 + 1 + 1 + 1 1 + 2 + 3 + 4 10 5 1 12 13 14 a square and sides of which measure 100 kms. Each. The

Example 6 : An aeroplane flies around

aeroplane cover at a speed of 10 Kms per hour the first side, at200 kms per hour the second side, at 300 kms per hour the third side and at 400 kms per hour the fourth side. Use the correct mean to find the average speed round the square. Here H.M. is the appropriate mean. Let the required average speed be H kms per hours then H =
4 4 = 4 ´ 1200 = 4 × 48 = 192 kms/hr. = 25 1 + 1 + 1 + 1 12 + 6 + 4 + 3 100 200 300 400 1200

[ICWA (F) Dec, 2000]

ADVANTANGES OF HARMONIC MEAN : (i) Like A.M. and G. M. it is also based on all observations. (ii) Capable of further algebraic treatment. (iii) It is extremely useful while averaging certain types of rates and rations. DISADVANTAGES OF HARMONIC MEAN : (i) It is not readily understood nor can it be calculated with ease.
6.66 MATHS

(ii) It is usually a value which may not be a member of the given set of numbers. (iii) It cannot be calculated when there are both negative and positive values in a series or one of more values in zero.

It is useful in averaging speed, if the distance travelled is equal. When it is used to give target weight to smallest item, this average is used. RELATIONS BETWEEN A.M., G.M. AND H.M. : 1. The Arithmetic Mean is never less than the Geometric Mean, again Geometric Mean is never less than the Harmonic Mean. i.e. A.M. ³ G. M. ³ H. M. For the observations x1 and x2, we know

( x1 -

x2

)2 …0 or, x + x
1

2

– 2 x1 x2 …0,

x + x2 or, x1 + x2 ³ 2 x1x2 or, 1 ³ x1x2 2 Again for 1 and 1 (observation) x1 x2
æ 1 ç ç x è 1 1 x2 ö ÷ ³ 0 or, 1 + 1 ÷ x1 x2 ø
2

or, A.M. ³ G. M.

2 ³0 x1x2 1 2 + 1 or, G. M. ³ H.M. x2

or, 1 + 1 ³ x1 x2

2 or, x1x2

x1x2 ³

x1

A.M. ³ G. M. ³ H. M Uses of H.M. : Harmonic mean is useful in finding averages involving rate, time, price and ratio. Example 7 : For the numbers 2, 4, 6, 8, 10, find GM & HM and show that AM > GM > HM. G. M. =
5

2 × 4 × 6 × 8 ×10 = (2 × 4 × 6 × 8 × 10 )

15

Log GM = 1 (log 2 + log4 + log6 + log8 + log 10) 5 = 1 (0.3010 + 0.6021 + 0.7782 + 0.9031 + 1.0000) = 1 × 3.5844 = 0.7169 5 5 \ G. M. = antilog 0.7169 = 5.211 H.M. =
5 5 = 1+1+1+1+ 1 1 (60 + 30 + 20 + 25 + 12) 2 4 6 8 10 120

= 5 × 120 = 600 = 4.379 137 137 Again A.M. = 1 (2+4+6+8+10) = 1 × 30 = 6 5 5

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We get A.M. =6, G. M. = 5.211, H.M. = 4.379 i.e. A.M. ³ G. M. ³ H. M
STATISTICAL METHODS

Note : In only one case the above relation is not true. When all the variates are equal, we will find that AM = GM = HM Example 8 : A.M. and G.M. of two observations are respectively 30 and 18. Find the observations. Also find H.M. x+ y Now = 30 or, x + y = 60 …….(1) again xy = 18 2 Or, xy = 324 or, (60 – y). y = 324, from (1) Or, y2 – 60 y + 324 = 0 or, (y –54) (y –6) = 0, y = 54, 6 \y = 54 , x = 6 or, y = 6, x = 54. \ Required observations are 6, 54. 2 \H.M. = = 2 = 2 ´ 54 = 10.80. 10 1 +1 9 +1 6 54 54 A. M . G. M . 2. For a pair of observations only. = or (G. M)2 = A. M. × H.M. G.M H .M Let the pair observations be x1 and x2. Now. 22 x +x x + x 2x x 2 x1 x 2 R.H.S. = 1 2 . = 1 2 . 1 2 = x1x2 = x1x2 2 1 x1 + 1 x2 2 2 = (G.M.)2 = L.H.S

(

( )

SELF EXAMINATION QESTIONS : 1. Find H.M. of the numbers : (i) 3, 6, 24, 48 [Ans. 7.1] (ii) 2, 4, 6, 8 [Ans. 3.84] 2. Calculate H.M. of the following numbers (i) 1, 1 , 1 , ……, 1 [Ans. 0.18] 10 2 3 (ii) 1, 1 , 1 ,….. 1 [Ans. 2 ] 2 3 n n +1 1 , 1 ,….., 1 (iii) 1, [Ans. 1 ] 3 5 2n - 1 n 3. Places A, B and C are equidistant from each other. A person walks from A to B at 5 km.p.h. : from B to C at 5 km.p.h. and from C to A at 4 km.p.h.. Determine his average speed for the entire trip. [Ans. 4 37 km. p.h. ] 37 4. A person covered a distance from X to Y at 20 km.p.h.. His average speed is 22 km.p.h.. Is the statement correct? [Ans No.] [Hints. Use H.M.]

MEDIAN : Definition : If a set of observation are arranged in order of magnitude (ascending or descending), then the middlemost or central value gives the median. Median divides the observations into two equal pats, in such a way that the number of observations smaller than median is equal to the number greater than it. It is not affected by extremely large or small observation. Median is, thus an average of position. In certain sense, it is the real measure of central tendency.

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MATHS

Calculation of Median : (A) For simple data or Series of Individual Observations : Individual observations are those observations (or variates) having no frequencies or frequency is unit every case. At first, the numbers are to arranged in order of magnitude (ascending or descending). Now for n (the total number of items of items) odd. Median = value of n + 1 th item and for n even 2 Median = average value n th item and æ n + 1ö th item. ç ÷ 2 è2 ø or, median = value of n + 1 th item (n = odd or even) 2 [Note : n + 1 th item gives the location of median, but not its magnitude] 2 Steps to calculate Median 1. Arrange the data in ascending or descending order. 2. Find n (odd or even). 3. Apply usual formula and calculate. Example 1 : To find the median of the following marks obtained by 7 students : 4, 12, 7, 9, 14, 17, 16. (i) Arrangement of marks : 4, 7, 9, 12, 14, 16, 17. (ii) n = 7 = an odd number (iii) Median = value of n + 1 th item = value of 7 + 1 th item = value of 4 th item = 12 (from the 2 2 arranged data \median is 12 marks. [Note : Unit of the result will be same as given in original variate.] Example 2 : To find the median of marks : 4, 12, 7, 9, 14, 17, 16, 21 (i) Arrangement : 4, 7, 9, 12, 14, 16, 17, 21. (ii) n = 8 = an even number.

(iii) Median = average value of n th item and æ n + 1ö th i.e. ç ÷ 2 è2 ø = average value of 8 th item and the next item 2 = average value of 4th item and the 5th item = average value of 12 and 14 marks = 12 + 14 = 13 marks. 12
1.5.1.1.1.1.1.1.1 Alternative way

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Alternative way
STATISTICAL METHODS

Median = value of n + 1 th item = value of 8 + 1 th item = value 4.5th item = 1 (value of 4th item 2 2 2 and value of 5th item) = 1 (12+14) = 1 × 26 = 13 marks. 2 2 (B) for Direct Series (or simple Frequency Distribution) Cumulative frequency (less than type) is calculated. Now the value of the variable corresponding to the cumulative frequency n + 1 gives the median, when N is the total frequency. 2 Example 3 : To find the median of the following x: y: 1 7 2 12 3 17 4 19 5 21 6 24

x 1 2 3 4 5 6

f 7 12 17 19 21 24

cum . freq. (c.f) 7 19 36 55 76 100 (= N)

Now, median = value of n + 1 th item = value of 100 + 1 th item = value of 50.5th item. 2 2 From the last column, it is found 50.5 is greater than the cumulative frequency 36, but less than the next cum. Freq. 55 corresponding to x = 4. All the 19 items (from 37, to 55) have the same variate 4. And 50.5 item is also one of those 19 item. \Median = 4. (C) For Continuous Series (Grouped Frequency Distribution) We are to determine the particular class in which the value of the median lies. by using. The formula n (and not by N + 1 , as in continuous series N divides the area of the curve into two equal parts). 2 2 2 After locating median, its magnitude is measured by applying the formula interpolation given below:
l -l Median = l1 + 2 1 (m – c), where m = N fm 2 é ù m-c êor median = l1 + f ´ i , where i = l2 - l1 ú m ë û

Where l1 = lower limit of the class in which median lies, l2 = Lower limit of the class in which median lies.
6.70 MATHS

fm = the frequency of the class in which median falls. m = middle item (i.e., item at which median is located or N th item). 2 C = cumulative frequency of the class preceding the median class, [Note : The above formula is based on the assumption that the frequencies of the class-interval in which median lies are uniformly distributed over the entire class-intrerval] Remember : In calculating median for a group frequency distribution, the class-intervals must be in continuous forms. If the class-intervals are given in discrete forms. They are to be converted first into continuous or class-boundaries form and hence to calculate median, apply usual formula. Example 4 : Find the median and median-class of the data given below :–– Class-boundaries 15–25 25–35 35–45 45–55 55–65 65–75 Frequency 4 11 19 14 0 2 [ICWA Jan. 1965] Class-boundaries 15–25 25–35 35–45 45–55 55–65 65–75 Frequency 4 11 19 14 0 2 Cumulative frequency 4 15 34 48 48 50 (= N)

Median = value of N th item = value of 50 th item = value of 25th item, which is greater than cum. 2 2 Freq. 15. So median lies in the class 35–45. Now. Median l -l = l1 + 2 2 (m –c), where l1 = 35, l2 = 45, f = 19, m = 25, c = 15 f = 35 + 45 - 35 (25 - 15) = 35 + 10 ´ 10 = 35 + 5.26 = 40.26 19 19 reqd. median is 40.26 and median-class is (35 – 45).
1.5.1.1.1.1.1.1.1 Example 5 : Calculate the median of the table given below :

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Class interval :
STATISTICAL METHODS

0–10 5

10–20 4

20–30 6

30–40 2

40–50 2

Frequency :

[ICWA(F) June, 2007]

C.I 0–10 10–20 20–30 30–40 40–50

f 5 4 6 3 2

c.f 5 9 15 18 20 (= N)

median = value of N th term = value of 20 (=10) th term, median class is (20–30). 2 2 l -l Median = l1 + 2 1 æ N - c ö ç ÷ f è2 ø = 20 + 30 - 20 (10–9) 6 = 20 + 10 = 20 + 1.67 = 21.67 6 CALCULATION OF MEDIAN FROM DISCRETE GROUPED DISTRIBUTION If the class intervals of grouped frequency distribution are in discrete form, at first they are to be converted into class-boundaries and hence to find median by applying usual formula. The idea will be clear from the following example. Example 6 : Marks obtained by 62 students in English are as follows:–– Marks 10–19 20–29 30–39 40–49 50–59 60–69 Total Compute median class and median. The class intervals are in discrete form. They are to be converted to class boundaries first, which is shown below : No. of students 5 8 14 20 11 4 62

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MATHS

Class boundaries 9.5 – 19.5 19.5–29.5 29.5–39.5 39.5–49.5 49.5–59.5 59.5–69.5

frequecy 5 8 14 20 11 4

Cumulative Frequency 5 13 27 47 58 62 (N)

Median = value of N th term = values of 62 th term or value of 31st term 2 2 \ Median lies in (39.5 – 49.5) l -l Now median = l1 + 2 1 (m – c), here l1 = 39.5, l2 = 49.5, fm = 20, m = 31, c = 27 fm Median = 39.5 + 10 (31 – 27) = 39.5 + 1 × 4 = 39. 5 + 2 = 41.5 marks. 20 2 Calculation of median from cumulative frequency distribution In this case at first cumulative frequency is to be converted into general group frequency distribution. Then applying usual formula median is to be calculated. Examples 7 : Compute median from the table given below : Marks less than 10 less than 20 less than 30 less than 40 less than 50 The general group frequency distribution is as follows : –– Marks 0–10 10–20 20–30 30–40 40–50 Students(f) 3 5(=8-3) 9 (=17-8) 3 (= 20-17) 2 (=22–20) c.f 3 8 17 20 22 (N) No. of students(f) 3 8 17 20 22

median = value of N th term = value of 22 th term = value of 11th term 2 2 \median class is ( 20 – 30)
l -l 1.5.1.1.1.1.1.1.1 \median = l1 + 2 1 (m – c), fm

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STATISTICAL METHODS

= 20 + 10 (11 – 8), here l1 = 20 l2 = 30, fm = 9, m = 11, c = 8 9 = 20 + 10 = 20 + 3.33 = 23.33 marks. 9 Note : If the cumulative frequency distribution is given in ‘more than type’ form then also the same procedure is to be followed. Example 8 : Calculate the median of the frequency distributions Marks : No. of students : 1–20 3 21–40 5 41–60 9 61–80 3 81–100 2

The class intervals: are in discrete forms, so they are to be made in class boundaries at first Class boundaries 0.5–20.5 20.5–40.5 40.5–60.5 60.5–80.5 80.5–100.5 f 3 5 9 3 2 c.f 3 8 17 20 22 (=N)

Median = value of 22 th term = values of 11th term \ Median class is (40.5 – 60.5) 2 l -l 60.5 - 40.5 Median = l1 + 2 1 æ N - c ö = 40.5 + (11 - 8) ç ÷ fm è 2 ø 9 = 40.5 + 10 ´ 3 = 40.5 + 3.33 = 43.83 marks. 9 Calculation of median from open ends class intervals : Since the first and last class intervals are not required in computing median, so in case of open end class-intervals median is calculated by usual process. For example, in the above example it the lower-limit of first class interval (i.e.0) and upper limit of last class (i.e. 5) are not given question, there would be no difficulty to compute median. In case of open and class-intervals, median is preferred than A.M. as average Finding of missing frequency The idea of finding missing frequency will be clear from the following example. Examples 9: An incomplete frequency distributions given below :––] Marks 10–20 20–30 30–40 40–50 50–60 No. of students(f) 3 5 –– 3 1

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MATHS

It is given that median of the above distribution is 32.5 marks. Find the missing frequency. Marks 10–20 20–30 30–40 40–50 50–60 f 3 5 f3 3 1 c.f 3 8 8 + f3 11 + f3 12 + f3

Here Median = 32.5 (given), so median class is (30–40). 20 + f 3 Let f3 be the missing frequency, N = = 6 + f3/2 = m , c = 8, l1 = 30 12=40, fm = f3 2 2
l -l From the formula, med. = l1 + 2 1 (m – c) fm f æ ö æf ö We get, 32.5 = 30 + 40 - 30 ç 6 + 3 - 8 ÷ or, 2.5 = 10 ç 3 - 2 ÷ or, 2.5 f3 = 5 f3 – 20 or f3 = 20 = 8. f3 è 2 f3 è 2 2.5 ø ø

(D) Graphic Method : Median can be determined graphically by the following methods : (i) Draw less than (or greater than) type ogive, taking the variation on X-axis and the cumulative frequency on Y-axis. Now corresponding to N/2 on Y-axis draw a horizontal line to meet at ogive, and again from the point of intersection, perpendicular is now drawn on X-axis. The point on X-axis is read off, which gives the median. Example 8 : To find the median graphically from the following Type : Wages (Rs.) 10–20 20–30 30–40 40–50 50–60 60–70 70–80 No. of workers 5 10 12 16 8 5 4

It is given that median of the above distribution is 32.5 marks. Find the missing frequency. Wages (Rs.) (less than) 20 30 40 5 15 27 No. of workers Wages (Rs.) (greater than) 10 20 30 No. of workers 60 55 45

MATHS

6.75

50
STATISTICAL METHODS

50 48

48 40

40 33

33

60 70 80

51 56 60

50 60 70 80

17 9 4 0

We draw less than type ogive, as shown before.

60 50 40 30

Cum. Freq

20 10 0 20 30 40 50 Wages (Rs.) 60 70 80 90

Median = size of N th item = size of 30 th item. Now take 30 on Y-axis, and from 30 draw a 2 horizontal line to meet the ogive. From this point on ogive, draw a perpendicular on X-axis. The point on X-axis is read off. The point is 42, which gives median. So median is Rs. 42. [Note : If we draw greater than types ogive, we should get the same result.] (ii) Draw two ogives. From the point of intersection of the curves, (i.e., ogives), draw a perpendicular to meet the X-axis. The point on the X-axis is read off, which gives the median. Advantages of Median : (i) The median, unlike the mean, is unaffected by the extreme values of the variable. (ii) It is easy to calculate and simple to understand, particularly in a series of individual observations a discrete series. (iii) It is capable of further algebraic treatment. It is used in calculating mean deviation. (iv) It can be located by inspection, after arranging the data in order of magnitude. (v) Median can be calculated even if the items at the extreme are not known, but if we know the central items and the total number of items. (vi) It can be determined graphically. Disadvantage of Median :

(i) For calculation, it calculation, it is necessary to arrange the averages averages do any 1.5.1.1.1.1.1.1.1 (i) For is necessary to arrange the data; other data; otherdo not need not need any such arrangement. such arrangement.
6.76 MATHS

(ii) It is amenable to algebraic treatment in a limited sense, Median cannot be used to calculate the combined median of two or more groups, like mean. (iii) It cannot be computed precisely when it lies between two items. (iv) Process involved to calculate median in case of continuous series is difficult to follow. (v) Median is affected more by sampling fluctuations than the mean.

SELF EXAMINATION QESTIONS : 1. Define median, Mention merits and demerits of median. 2. Find median of the following numbers : (i) 38, 56, 31, 70, 41, 62, 53, 57 (ii) 14, 15, 30, 40, 10, 25, 20, 35 (iii) 25, 1275, 748, 162, 967, 162 3. Of the numbers 78, 82, 36, 38, 50, 72, 68, 70, 64 find median. 4. The heights (in cm) of few students are as follows : 69, 75, 72, 71, 73 , 74, 76, 75, 70 Find second quartile. 5. Find the median of the following numbers : 6, 4, 3, 6, 5, 3, 3, 2, 4, 3, 4, 3, 3, 4, 3, 4, 2, 2, 4, 3, 5, 4, 3, 4, 3, 3, 4, 1, 1, 2, 3. 6. Find the median of the following distribution : x 1 2 3 4 5 6 y 22 31 40 42 24 12 [Ans.3] [Ans. 3] [Ans. Q2 = 72.5 cm] [Ans. 54.5] [Ans. 22.5] [Ans. 455] [Ans. 68]

7. Find the median class and median from the table given : (i) C.I 0–10 10–20 20–30 30–40 40–50 Frequency 5 4 6 3 2 [ICWA (F) June, 2004][Ans. (20–30); 21.67]
MATHS 6.77

(ii)
STATISTICAL METHODS

Score 5–10 10–15 15–20 20–25 25–30 30–35 Total

Frequency 4 7 10 12 8 3 44 [Ans.(20 –25); 42 score]

8. Find the following distribution find median class and median : Score 30–39 40–49 50–59 60–69 70–79 80–89 90–99 Frequency 1 4 14 20 22 12 2 [Ans. (59.5–69.5); 68.75 score]

9. Calculate median of table given : Marks (i) Less than10 Less than 20 Less than 30 Less than 40 Less than 50 Students 5 9 15 18 20 Marks (ii) Less than 45 Less than 40 Less than 35 Less than 30 Less than 25 Students 20 17 12 5 2

[Ans. (i) 21.67 marks, (ii) 33.57 marks] 10. In the following frequency distribution one frequency is missing. It is given that median of the distribution is 53.5, find the missing frequency.

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MATHS

Variate

Frequency 20–30 30–40 40–50 50–60 60–70 70–80 8 5 f3 20 10 4 [Ans. 12]

[Note. That the class interval are unequal.]

11. The expenditure of 1000 families are as follows : Expenditure (Rs.) 40–59 60–79 80–99 100–119 120–139 No. of families 50 –– 500 –– 50

In the above table median is Rs. 87.50. Find the missing frequency. 12. Estimate median graphically from the table : (i) Class boundary 2–4 4–6 6–8 8–10 Frequency 3 4 2 1 [Ans. 5]

(ii)

Marks 0–10 10–20 20–30 30–40 40–50

Student 10 20 35 25 10 [Ans. 25.7 marks]

MODE
1.5.1.1.1.1.1.1.1 Definition :

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MODE
STATISTICAL METHODS

Definition :
Mode is the value of the variate which occurs most frequency. It represents the most frequent value of a series. When one speak of the ‘average student’, we generally mean the modal wage, the modal student. If we say that the modal wages obtained by workers in a factory are Rs. 70, we mean that the largest number of workers get the same amount. As high as Rs. 100 and as low as Rs. 50 as wages are much less frequented and they are non-modal. Calculation. Mode cannot be determined a series of individual observations unless it is converted to a discrete series (or continuous series). In a discrete series the value of the variate having the maximum frequency is the modal class. However, the exact location of mode is done by interpolation formula like median. Location of modal value in case of discrete series is possible if there is concentration of items at on point. If again there are two or more values having same maximum frequencies, (i.e. more concentration), it becomes difficult to determine mode. Such items are known as bimodal, tri-modal or multi-modal accordingly as the items concentrate at 2, 3 or more values. (A) For. Individual Observations The individual observations are to be first converted to discrete series (if possible). Then the variate having the maximum will be the mode. Example 1 : Calculate mode from the data (given) : (Marks) : 10, 14, 24, 27, 24, 12, 11, 17.

Marks 10 11 12 14 17 24 27 (Individual observation are converted into a discrete

Frequency 1 1 1 1 1 2 1 series)

Here marks 24 occurs maximum number of times, i.e. 2. Hence the modal marks is 24, or mode = 24 marks. Alternatively : Arranging the numbers : 10, 11, 12, 14, 17, (24, 24) 27.
1.5.1.1.1.1.1.1.1 Now 24 occurs maximum number of times, i.e. 2. \ Mode = 24 marks.

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MATHS

Now 24 occurs maximum number of times, i.e. 2. \ Mode = 24 marks.
[Note. When there are two or more values having the same maximum frequency, then mode is illdefined. Such a sense is known as bimodal or multi-modal as the case may be.]

Example 2 : Compute mode from the following data. Marks obtained : 24, 14, 20, 17, 20, 14. Marks 14 17 20 24 [Here 14 occurs 2 times (max.) and 20 occurs 2 times (max.) \ mode is ill-defined.] (B) For Simple Frequency Distribution Discrete Series. To Find the mode from the following Table : Height (in inches) 57 59 61 62 63 64 65 66 67 69 No. of Persons 3 5 7 10 20 22 24 5 2 2 Frequency 2 1 2 1

1.5.1.1.1.1.1.1.1 Frequenciescolumn (1) are grouped (1) two's in column (2) andcolumn (2) and (3) and Frequencies given below, in given below, in column by are grouped by two’s in (3) and then by three's then by three’s in(5), and (6). The maximum frequency infrequency in each columnby marked by Bold in columns (4), columns (4), (5), and (6). The maximum each column is marked is Bold Type. Type. We do not find any fixed point having maximum frequency but changes with the change of We do not find following table, the sizes of maximum frequency changes with the change of grouping. grouping. In the any fixed point having maximum frequency but in respect of different columns are In the following table, the sizes of maximum frequency in respect of different columns are arranged. arranged.

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Grouping Table
STATISTICAL METHODS

Inches

(1)

Grouping Table Height Frequency (2) (3) (4)

(5) (6)

Analysis Table Colume 1 2 3 4 5 6 No. of items 1 3 62 63 63 63 64 64 64 64 64 5 65 65 4 66 1 65 Sizes of items having maximum frequency 65

From the above table, we find 64 is the size of the item which is most frequented. The mode is, therefore, located at 64. [Note. At glance from column (1) one might think that 65 is the mode since it contains maximum frequency. This impression is corrected by the process of grouping . So it is not advisable to locate the mode merely by inspection.] (C) For continuous Series.
1.5.1.1.1.1.1.1.1 or byinspections or by preparing Grouping TableTable, ascertain the modal class. modal By inspections By preparing Grouping Table and Analysis and Analysis Table, ascertain the class. Then tothe exact value of mode, apply thethe following formula. Then to find find the exact value of mode, apply following formula.

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MATHS

Mode = l +

f1 - f 0 ´ i. 2 f1 - f 0 - f 2

Where, 1 = lower class-boundary of modal class f1 = frequency of modal class. f0 = frequency of the class preceding modal class. f2 = frequency of the class succeeding the modal class. i = size of the class- interval of modal class. Note : the above formula may also be expressed as follows : Mode = l +

( f1 - f 0 )+ ( f1 - f 2 )´ i

f1 - f 0

=l+

D1 ´ i . Where D1 = f1 – f0.; D2 = f1 – f2. D1 + D2

Example 3 : Compute mode of the following distribution. Marks 10–20 20–30 30–40 40–50 50–60 60–70 No. of students 5 8 12 16 10 8

Marks 10–20 20–30 30–40 40–50 50–60 60–70

No. of students 5 8 12® f0 16® f1 10® f2 8

From the table it is clear that the maximum frequency is 16th : modal class is (40–50) Here l = 40, f0 = 12, f1 = 16, f2 = 10 (marked in table), i = 10 (= 50 –40) Mode = l +
f1 - f 0 ´i 2 f1 - f 0 - f 2

16 - 12 16 - 12 1.5.1.1.1.1.1.1.1 = 40 + ´ 10 = 40 + ´ 10 2´ 2 ´ 16 - 12 - 1016 - 12 - 10

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= 40 +
STATISTICAL METHODS

4 ´ 10 32 - 22

= 40 + 4 ´ 10 = 40 + 4 = 44 marks. 10 Alternatively, D1 = f1–f0 = 16 –12 = 4, D2 = f1 – f2 = 16–10 = 6, i = 10, 1 = 40 Mode = 40 + 4 ´ 10 = 40 + 4 ´ 10 4+6 10

= 40 + 4 = 44 marks. Calculation of Mode From discrete group frequency distribution. In such cases at first class boundaries are to be formed for applying formula.] Example 4 : Compute mode from the following frequency distribution : Marks 50–59 60–69 70–79 80–89 90–99 100–109 No. of students 5 20 40 50 30 6

The class intervals which are in discrete form are first converted into class boundaries. Calculation of mode Class boundaries 49.5–59.5 59.5–69.5 69.5–79.5 79.5–89.5 89.5–99.5 99.5–109.5 Frequency 5 20 40 50 30 6

Now modal class is (79.5 – 89.5), since this class has the highest frequency. Here l = 79.5, f0 = 50, f1 = 50, f2 = 30, i = 10 Mode = l +
f1 - f 0 50 - 40 ´ 10 ´ i = 79.5 + 100 - 40 - 30 2 f1 - f 0 - f 2

= 79.5 + 10 ´ 10 = 79.5 + 10 = 79.5 + 3.33 = Rs. 82.83. 30 3

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MATHS

Calculation of mode from cumulative frequency distribution : Example 5 : From the following cumulative frequency distribution of marks of 22 students in Accountancy, calculate mode : Marks No. of students below 20 3 below 40 8 below 60 17 below 80 20 below 100 22 [ICWA(F) June, 2005] At first we are to transfer the above cumulative frequency distribution into a equal group frequency distribution and hence to calculate mode. Marks 0–20 20–40 40–60 60–80 80–100 students(f) 3 5 (= 8 –3) 9(=17–8) 3 (=20–17) 2 (=22–20)

Modal class is (40–60), as this class has highest frequency. Here l = 40, f0 = 5, f1 = 9, f2 = 3, I = 20 Mode l + f1 - f0 9-5 ´ 20 = 40 + 4 ´ 20 ´ i 40 + 2´9 - 5 - 3 10 2 f1 - f0 - f 2 = 40 + 8 = 48 marks.

Calculation of missing frequency : Example 6. Mode of the given distribution is 44, find the missing frequency Marks No. of students 10–20 5 20–30 8 30–40 12 40–50 –– 50–60 10 60–70 8

Since mode is 44, so modal class is 40–50.

Marks 10–20 20–30 30–40 40–50 50–60 60–70 let the missing frequency be f1

Frequency(f) 5 8 12 –– 10 8

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Now mode = l +
STATISTICAL METHODS

f1 - 12 ´ 10 2 f1 - 12 - 10

or, 44 = 40 + or, 4 =

f1 - 12 ´ 10 2 f1 - 22 or, f1 = 16 (on reduction)

f1 - 12 ´ 10 2 f1 - 22

Location of Mode graphically In case of the Frequency Distribution, Mode can be located graphically. Draw a histogram of the data given. In the inside of the modal class-bar, draw two lines diagonally starting from each upper corner of the adjacent bar (as show in the next figure). Now draw a perpendicular from the point of intersection of the diagonal lines to X-axis. The point on the X-axis is read off, which gives the modal value. Example 7 : The monthly profits in rupees of 100 shops are distributed as follows:

Profits per shop 000–100 100–200 200–300 300–400 400–500 500–600

No. of shops 12 18 27 20 17 6

Draw the histogram to the data and hence find modal value, Cheek this value by calculation. From the graph, Mode is found to be Rs. 256 (app) Now for direct calculation, we find modal class as (200–300) since the class has got the highest frequency. Again, l = 200, f0 = 18, f1 = 27, f2 = 20, i
30 25 20

No. of Shops

15 10 5 0 100 200 300 Profits 400 500 600

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MATHS

\ Mode = l +

f1 - f 0 27 - 18 ´ 100 ´ i = 200 + 54 - 18 - 20 2 f1 - f 0 - f 2

= 200 + 9 ´ 100 = 200 + 56.25 = Rs. 256.25 16 Calculation of Mode when class-intervals are unequal. If the class-intervals are unequal, then we are to make them equal, having frequencies adjusted. Then, the formula for computing the value of mode is to be applied. Miscellaneous examples : 1. If two variates x and y are related by 2x = 3y – 1, and mean of y be 9 ; find the mean of x. 2x = 3y –1 or, 2x = 3y - 1 or, 2x = 3y - 1 or, 2x = 3 ´ 9 - 1 = 26 or, x = 13 2. If 2u = 5x is the relation between two variables x and u and harmonic mean of x is 0.4, find the harmonic mean of u.
u= 5x 5 or, u = ´ 0.4 = 1.0 2 2

[ICWA (F) June 2005] \ reqd. H.M is 1.0

3. The relation between two variables x and y is 3y – 2x + 5 = 0 and median of y is 40, find the median of x. From 3y – 2x + 5 = 0 we get, x =
3 5 . 40 + = 62.5 2 2 3 5 y + . As the median is located by position, so median of x is 2 2

4. Mode of the following frequency distribution is 24 and total frequency is 100. Find the values of f1 and f2. C.I : Frequency : 0 –10 14 10–20 f1 20–30 27 30–40 f2 40–50 15 27 - f 1 ´ 10 54 - f1 - f 2

Mode is 24, so modal class is (20–30). From the formula of mode we find. 24 = 20 + 270 - 10f1 54 - (f1 + f 2 ) 270 - 10f1 270 - 10f1 = 54 - 44 10 14 + f1 + 27 + f2 + 15 = 100

or, 4 = or, 4 =

or, f1 + f2 = 100 – 56 = 44 …….….(1) or, 40 = 270 – 10f1
1.5.1.1.1.1.1.1.1 or, 10f1 = 230 or, f1 = 23. From (1) , f2 = 44 – 23 = 11 \ f1 = 23, f2 = 11

MATHS

9.87

5. The following are the monthly salaries in rupees of 20 employees of a firm :
STATISTICAL METHODS

130 145

125 151

110 65

100 71

80 118

76 140

98 116

103 85

122 95

66 151

The firm gives bonuses of Rs. 10, 15, 20, 25 and 30 for individuals in the respective salary group : exceeding Rs. 60 but not excedding Rs. 80, exceeding Rs. 80 but not exceeding Rs. 100 and so on up to exceeding Rs. 140 but not exceeding Rs. 160. Find the average bonus paid per employee. From the monthly salaries of the employees, we find the number of employees lying in the salary groups mentioned as follows : Calculation of average bonus

Salary (Rs.) f Exceeding 60 but not exceeding 80 100 120 140 Total
A.M. (x ) =

bonus x 10 15 20 25 30 fx 50 60 80 100 90 380

80 100 120 140 160

5 4 4 4 3 20

å fx = 380 = Rs. 19. å f 20

6. Marks obtained by 30 students in History of a Test Examination 2004 of some school are as follows : 34 48 40 36 22 51 10 39 35 21 26 52 31 34 41 32 39 32 22 10 30 43 17 35 48 47 53 36 38 23

construct a frequency table with class intervals 10–19. 10–29 etc. Calculate the median and mode from the frequency distribution. Construction of frequency table and hence calculation of median and mode.

Marks 10–19 20–29 30 – 39 40 – 49
1.5.1.1.1.1.1.1.1

tally mark /// //// //// //// /// //// /
50 – 59 /// 3

f 3 5 13 6
30

cf 3 8 21 27
49.5 – 59.5

class boundaries 9.5 – 19.5 19.5 – 29.5 29.5 – 39.5 39.5 –49.5

6.88

MATHS

Median = value of

N 30 th i.e., i.e. 15th firm 2 2

So median class is (29.5 – 39.5) \ median = 29.5 +
39.5 - 29.5 (15 - 8) = 29.5 + 10 ´ 7 13 13

= 29.5 + 5.38 = 34.88 marks Highest frequency is 13 (= f1), f0 = 5, f2 = 6 \ Mode = l + f1 - f 0 13 - 5 ´ 10 ´ i = 29.5 + 2f1 - f 0 - f 2 2 ´ 13 - 5 - 6
8 ´ 10 = 29.5 + 5.33 = 34.83 marks. 15

= 29.5 +

Advantages of mode : (i) It can often be located by inspection. (ii) It is not affected by extreme values. It is often a really typical value. (iii) It is simple and precise. It is an actual item of the series except in a continuous series. (iv) Mode can be determined graphically unlike Mean. Disadvantages of mode : (i) It is unsuitable for algebraic treatment. (ii) When the number of observations is small, the Mode may not exist, while the Mean and Median can be calculated. (iii) The value of Mode is not based on each and every item of series. (iv) It does not lead to the aggregate, if the Mode and the total number of items are given.

EMPIRICAL RELATIONSHIP BETWEEN MEAN, MEDIAN AND MODE A distribution in which the values of Mean, Median and Mode coincide, is known symmetrical and if the above values are not equal, then the distribution is said asymmetrical or skewed. In a moderately skewed distribution, there is a relation amongst Mean, Median and Mode which is as follows : Mean – Mode = 3 (Mean – Median) If any two values are known, we can find the other. Example 6 : In a moderately asymmetrical distribution the mode and mean are 32.1 and 35.4 respectively. Calculate the Median. From the relation, we find 3 Median = 2 Mean + Mode or 3 Median = 2 × 35.4 + 32.1 = 70.8 + 32.1 = 102.9 \ Median = 34.3

MATHS

6.89

SELF EXAMINATION QUESTIONS :
STATISTICAL METHODS

1. Define mode. Mention the advantages and disadvantages of mode. 2. Calculate the mode of the following numbers : (i) 25, 1275, 748, 169, 876, 169 (ii) 4, 3, 2, 5, 3, 4, 5, 1, 7, 3, 2, 1 (iii) 69, 75, 57, 70, 71, 75, 76 (iv) 1, 3, 4, 7, 9, 10, 11, 13, 14, 16 3. Find the mode of the numbers : 7, 4, 3, 5, 6, 3, 3, 2, 4, 3, 3, 4, 4, 2, 3 4. Find the mode of the following frequency distribution :x 0 1 2 3 4 5 6 7 8 5. Compute mode from the following frequency distribution : f 5 22 31 43 51 40 35 15 3 [Ans. 4] [Ans. 3] [Ans. 169] [Ans. 3] [Ans. 75] [Ans. 11]

Marks 0–10 10–20 20–30 30–40 40–50 (ii) Score 25–30 30–35 35–40 40–45
1.5.1.1.1.1.1.1.1 45–50 9

Students 3 7 10 6 2 [Ans. 25 marks]

Frequency 3 5 6 10
[Ans. Rs. 44]

6.90

MATHS

6. Calculate mode of the distribution given below : (i) Marks less than 10 less than 20 less than 30 less than 40 less than 50 (ii) Wages No. of students 5 9 15 18 20 No. of students 0 and above 20 and above 40 and above 60 and above 80 and above 100 and above 50 45 34 16 6 0 [Ans. (i) 24 marks, (ii) Rs. 49.33]

7. Daily wages of 100 worker are given in the table :

Daily Wages (Rs.) 2–3 4–5 6–7 8–9 10–11 Compute the modal value. OBJECTIVE QUESTIONS :

No. of workers 5 8 12 10 7

1. What is the sum of deviations of a variates from their A.M.? 2. Write the relation of AM. G.M. and H.M. 3. For a pair of variates, write the relation of AM. G.M and H.M 4. Write the emperical relation of mean, median and mode.

[Ans. zero] [Ans. AM ³ G.M ³ H.M] [Ans. (GM)2 = AM × H.M]

[Ans. Mean-mode = 3 (Mean – median)]

1.5.1.1.1.1.1.1.1 5. In case of open end class intervals frequency distribution to calculate average, which is most appropriate average? [Ans. medain]

MATHS

6.91

6. Find A.M and mode of : 7, 4, 10, 15, 7, 3, 5, 2, 9, 12
STATISTICAL METHODS

[Ans. 7.4 ; 7] [Ans. 12]

7. Find G.M. of 3, 12, 48

8. For a symmetry distribution mode and A.M. are respectively Rs. 12.30 and Rs. 18-48 ; find median of the distribution. [Ans. 16.42]

9. The mean marks of 100 students was found to be 40. Later on it was discovered that marks 53 was misread as 83. Find the corrected mean marks 10. A.M. of 7, x – 2, 10, x + 3 is 9 find x [Ans. 39.70] [Ans. 9]

11. Find G.M. of 8 observations : 2 occuring 4 times, 4 occuring twice 8 and 32 occuring once each.[Ans. 4] 12. Find H.M. of the observations
1 1 1 1 1 and , , , 2 4 5 6 8

[ICWA (F) June 2007] [Ans.

1 ] 5

13. If the means of two groups of m and n observations are 40 and 50 respectively and the combined group mean is 42, find the ratio m : n. 14. Find mean and mode of the 9 observations 9, 2, 5, 3, 5, 7, 5, 1, 8 [ICWA (F) June 2007] [Ans. 4 :1] [ICWA (F) Dec. 2006] [Ans. 5, 5] 15. If two groups have number of observations 10 and 5 and means 50 and 20 respectively, find the grouped mean. 16. Two variables x and y are related by y = 45, find the mean of y. 17. Find H.M. of [ICWA (F) Dec. 2006] [Ans. 40]
x -5 and each of them has 5 observation. If mean of x is 10

[ICWA (F) Dec. 2006] [Ans. 4]

1 2 3 n , , , ....., , occuring with frequencies 1, 2, 3, …., n respectively 2 3 4 n +1 [ICWA (F) June 2006] [Ans. 1] n (n + 1)/ 2 n (n + 1)/ 2 = = 1] 2 + 3 + 4 + .... + n + 1 n (n + 1)/ 2

[Hints : H.M. =

1 + 2 + 3 + .... + n n +1 1.2 + 2. 3 + 3. 4 + ..... + n. 2 3 n

=

18. If the relation between two variables x and y be 2x + 5y = 24 and mode of y is 4, find mode of x. [Ans. 2] 19. Find median of the 10 observations 9, 4, 6, 2, 3, 4, 4, 6, 8, 7 [Ans. 5]

20. The mean of 10 observations was found to be 20. Later on one observation 24 was wrongly noted as 34. Find the corrected mean. 21. Prove that for two numbers 2 and 4, AM × HM = (G.M.) . 22. If the relation between two variables x and y is 2x + 3y = 7, and median of y is 2, find the median of x. [ICWA (F) Dec. 2005] [Ans.
1 ] 2
2

[ICWA (F) June 2006] [Ans. 19]

1.5.1.1.1.1.1.1.1 23. If two groups of 50 and 100 observations have means 4 and 2 respectively, find the 23. If two groups of 50 and 100 observations have means 4 and 2 respectively, find the mean 2 meanof the combinedgroup. of the combined group. [ICWA (F) Dec. 2005] [Ans. 3 ] 3

6.92

MATHS

24. If a variable x takes 10 values 1, 2, 3, …., 10 with frequency as its values in each case, then find the arithmetic mean of x. [Hints : A.M. = 12 + 2 2 + 32 + ...... + 10 2 & etc. ] 1 + 2 + 3 + ..... + 10 [ICWA (F) June 2005] [Ans. 7]

25. If first of two groups has 100 items and mean 45 and combined group has 250 items and mean 5/, find the mean of second group. 26. Find the median of the following distribution Weight (kg) No. of students : : 65 5 66 15 67 17 68 4 [Ans. 12] [Ans. 15] [Ans. 67] [ICWA (F) June 2005] [Ans. 55]

27. Find G.M. of 3, 6, 24, 48 28. A.M. of two numbers is 25 and their H.M. is 9, find their G.M.

29. The means of samples of sizes 50 and 75 are 60 and x respectively. If the mean of the combined group is 54, find x. [ICWA (F) Dec. 2004] [Ans. 50]

30. Find the median of the given distribution : Value (x) Frequency (f) : : 1 7 2 12 3 18 4 4 [Ans. 3]

31. If each of 3, 48 and 96 occurs once and 6 occurs twice verify that G.M. is greater than H.M. 32. Find G M. of 1, 2, 3,
1 1 , . What will be G.M. if ‘0’ is added to above set of velues? 2 3

[ICWA (F) June 2003] [Ans. 1 ; 0] 33. The G.M. of a, 4, 6 is 6, find a 34. A.M. of a variable x is 100, find the mean of the variable 2x – 50. [Ans. 9] [Ans. 150]

35. The variable x and y are given by y = 2x + 11. If the median of x is 3, find the median of y. [Ans. 17]

6.6 MEASURES OF DISPERSION :
DISPERSION : A measure of dispersion is designed to state the extent to which individual observations (or items) vary from their average. Here we shall account only to the amount of variation (or its degree) and not the direction (which will be discussed later on in connection with skewness). Usually, when the deviation of the observations form their average (mean, median or mode) are found out then the average of these deviations is taken to represent a dispersion of a series. This is why measure of dispersion are known as Average of second order. We have seen earlier that mean, median and mode, etc. are all averages of the first order. Measures of dispersion are mainly of two types– (A) Absolute measures are as follows :
MATHS 6.93

(i) Range, (ii) Mean deviation (or Average deviation), (iii) Standard deviation
STATISTICAL METHODS

(A) Among the Relative measures we find the following types : (i) Coefficient of dispersion. (ii) Coefficient of variation. Absolute and Relative measures : If we calculate dispersion of a series, say, marks obtained by students in absolute figures, then dispersion will be also in the same unit (i.e. marks). This is absolute dispersion. If again dispersion is calculated as a ratio (or percentage) of the average, then it is relative dispersion. RANGE : For a set observations, range is the difference between the extremes, i.e. Range = Maximum value – Minimum value Example 1 : The marks obtained by 6 students were 24, 12, 16, 11, 40, 42. Find the Range. If the highest mark is omitted, find the percentage change in the range. Here maximum mark = 42, minimum mark = 11. \ Range = 42 – 11 = 31 marks If again the highest mark 42 is omitted, then amongst the remaining. Maximum mark is 40. So, range (revised) = 40 – 11 = 29 marks. Change in range = 31 – 29 = 2 marks. \ Reqd. percentage change = 2 ÷ 31 × 100 = 6.45% Note : Range and other obsolute measures of dispersion are to be expressed in the same unit in which observations are expressed. For grouped frequency distribution : In this case range is calculated by subtracting the lower limit of the lowest class interval from the upper limit of the highest. Example 2. For the following data calculate range : Marks 10–15 15–20 20–25 25–30 Here upper limit of the highest class interval = 30 And lower limit of the first class interval = 10 \ Range = 30 – 10 = 20 marks Note : Alternative method is to subtract midpoint of the lowest class from that of the highest. In the above case, range = 27.5 – 12.5 = 15 marks.
1.5.1.1.1.1.1.1.1 In practice both the methods are used.

Frequency 2 3 4 1

6.94

MATHS

Coefficient of Range : The formulae of this relative measure is difference of extreme value i.e. range sum of extreme vales In the above example, Coefficient of range =
30 - 10 20 1 = = = 0 .5 30 + 10 40 2

Advantages of Range : Range is easy to understand and is simple to compute. Disadvantages of Range : It is very much affected by the extreme values. It does not depend on all the observations, but only on the extreme values. Range cannot be computed in case of open-end distribution. Uses of Range : It is popularly used in the field of quality control. In stock-market fluctuations range is used. MEAN DEVIATION (or Average Deviation) : Mean deviation and standard deviation, however, are computed by taking into account all the observations of the series, unlike range. Definition : Mean deviation of a series is the arithmetic average of the deviations of the various items from the median or mean of that series. Median is preferred since the sum of the deviations from the median is less than from the mean. So the values of mean deviation calculated from median is usually less than that calculated from mean. Mode is not considered, as its value is indeterminate. Mean deviation is known as First Moment of dispersion. Computation of Mean Deviation : (a) For individual Observation (or Simple Variates) The formula is Mean Deviation (M.D.) =

å |d|
n

Where | d | within two vertical lines denotes deviations from mean (or median), ignoring algebraic signs (i.e., + and –). Steps to find M. D. (1) Find mean or median (2) Take deviation ignoring ± signs (3) Get total of deviations (4) Divide the total by the number of items. Example 3 : To find the mean deviation of the following data about mean and median :

(Rs.) 1, 6, 11, 14, 16, 1, 6,23. 14, 16, 19, 23. 1.5.1.1.1.1.1.1.1 (Rs.) 19, 11,

MATHS

6.95

Computation of Mean Deviation.
STATISTICAL METHODS

About Mean Serial No. (Rs.) x Dev. From A.M. ignoring ± signs |d| 1 2 3 4 5 6 7 Total 2 6 11 14 16 19 23 –– 11 7 2 1 3 6 10 40 1 2 3 4 5 6 7 Total 2 6 11 14 16 19 23 –– Serial No. (Rs.) x

About Median Dev. From Med. ignoring ± signs |d| 12 8 3 0 2 5 9 39

A.M. =

1 (2 + 6 + 11 + 14 + 16 + 19 + 23)= 1 ´ 91 = Rs. 13 7 7 7 +1 th iterm = size of 4th item = Rs. 14 2

Median = size of

Mean deviation (about mean) =

å | d | = 40 = Rs. 5.71
n 7

Mean deviation (about median) = Note : The sum of deviation

å | d | = 39 = Rs. 5.57
n 7

( | d |)about median is 39, less than | d | about mean (= 40). Also M.D. å
M.D. 5.71 = = 0.44 (app.) Mean 13 M.D. 5.57 = 0.40 (app.) = Median 14

about median.(i.e.5.57) is less than that about mean, (i.e., 5.71) Coefficient of Mean Deviation : About mean, Coefficient of M.D =

About median, Coefficient of M. D. =

(b) For Discrete Series (or Simple Frequency Distribution) The formula for computing M.D. is
M.D. =

åf x - x åf
i i i

or

åf x - x = åf | d | åf åf
MATHS

Where | d | = deviations from mean (or median) ignoring ± signs.
6.96

Steps of find M.D. (i) Find weighed A.M. or median. (ii) Find deviations ignoring ± signs. i.e., | d | (iii) Get (iv)

åf | d | ; Divide å f | d | by å f
x (marks) 5 10 15 20 25 Total f (student) 6 7 8 11 8 40

About Mean Example 4 : To calculate mean deviation of the following series :

Find also the coefficient of dispersion. Computation of Mean Deviation (About Mean) Marks Deviation From Mean (16) X (1) 5 10 15 20 25 Total f (2) 6 7 8 11 8 40 (d= x – 15) (3) –10 –5 0 5 10 – d¢ = d/5 (4) –2 –1 0 1 2 – fd¢ (5) = 2 ´ (4) –12 –7 0 11 16 8 x-x (6) 11 6 1 4 9 – f x-x (7) = (2) × (6) 66 42 8 44 72 232 from actual

A.M. = A +

å fd¢ ´ i = 15 + 8 ´ 5 = 15 + 1 = 16 marks 40 åf
= 232 = 5.8 marks. 40
M.D. 5.8 = = 0.363 Mean 16
6.97

M.D. =

åf x - x åf

Coefficient of dispersion (about mean) =

MATHS

STATISTICAL METHODS

M.D. =

åf x - x åf

=

232 = 5.8 marks. 40 M.D. 5.8 = = 0.363 Mean 16

Coefficient of dispersion (about mean) = About Median : Example 5 : The same example as given above.

Computation of Mean deviation (About median)

Marks X f

Cum. Freq. c.f.

Dev. From median (15) |d| f|d| 60 35 0 55 80 230

5 10 15 20 25 Total Median = value of the

6 7 8 11 8 40
40 + 1 th item 2

6 13 21 32 40 (= N) –

10 5 0 5 10 –

= value of 20.5 th item = 15 marks. M.D. =

å f | d | = 230 = 5.75 marks å f 40
M.D. 5.75 = 0.383 = Median 15

Coefficient of dispersion (about median) =

(c) For Class Intervals (or Group Distribution) Steps to compute (M.D.) (i) Find mid-value of the class intervals (ii) Compute weighted A.M. or median (iii) Find | d | and f | d | (iv) Divide

å f | d | by å f
Daily wages (Rs.) 3.50–5.50 5.50–7.50 7.50–9.50 No. of workers 6 14 16
MATHS

Example 6 : Find M.D. about A.M. of the following frequency distribution :

6.98

9.50–11.50

10 11.50–13.50 4

Calculate also M.D. about median and hence find coefficient of mean dispersion. Computation of M.D. about A.M.

Wages (Rs.) 3.50–5.50 5.50–7.50 7.50–9.50 9.50–11.50 11.50–13.50 Total

Mid-value x 4.50 6.50 8.50 10.50 12.50 – f 6 14 16 10 4 50

d¢ =

x - 8.50 2

fd¢

|d| = x-x f|d| 22.08 23.52 5.12 23.20 17.28 91.20

–2 –1 0 1 2 –

–12 –14 0 10 8 –8

3.68 1.68 0.32 2.32 4.32 –

x (A.M.) = A +

å fd¢ ´ i = 8.50 + (- 8)´ 2 = 8.50 – 0.32 = 8.18 50 åf
= 1.824 = Rs. 1.82 Calculation of M.D. about median

M.D. =

å f | d | = 91.20 50 åf

Wages (Rs.) 3.50–5.50 5.50–7.50 7.50–9.50 9.50–11.50 11.50–13.50 Total

f

c.f.

mid-value

|d|

f|d|

6 14 16 10 4 50

6 20 36 46 50 (N) –

4.50 6.50 8.50 10.50 12.50 –

3.63 1.63 0.37 2.37 4.37 –

21.78 22.82 5.92 23.70 17.48 91.70

Median = value of N/2th item = value of 50/2, i.e., 25 th item. So median class is (7.50 – 9.50) \ Median = l1 + l2 - l1 (m - c)= 7.50 + 9.50 - 7.50 (25 - 20) f1 16

MATHS

6.99

= 7.50 +
STATISTICAL METHODS

2 ´ 5 = 7.50 + 0.625 = 8.125 = Rs. 8.13 16

M.D. =

å f | d | = 91.70 50 åf

= 1.834 = Rs. 1.83 M.D. 1.824 = 0.223 = 0.22 = A.M. 8.18
M.D. 1.834 = 0.225 = 0.23 = Median 8.13

Coeff. Of dispersion (about A.M.) =

Coeff. Of dispersion (about median) = Advantages of Mean Deviation :

(1) It is based on all the observations. Any change in any item would change the value of mean deviation. (2) It is readily understood. It is the average of the deviation from a measure of central tendency. (3) Mean Deviation is less affected by the extreme items than the standard deviation. (4) It is simple to understand and easy to compute. Disadvantages of Mean Deviation : (1) Mean deviation ignores the algebraic signs of deviations and as such it is not capable of further algebraic treatment. (2) It is not an accurate measure, particularly when it is calculated from mode. (3) It is not popular as standard deviation. Uses of Mean Deviation : Because of simplicity in computation, it has drawn the attention of economists and businessmen. It is useful reports meant for public. STANDARD DEVIATION : In calculating mean deviation we ignored the algebraic signs, which is mathematically illogical. This drawback is removed in calculating standard deviation, usually denoted by ‘ s ’ (read as sigma) Definition : Standard deviation is the square root of the arithmetic average of the squares of all the deviations from the mean. In short, it may be defined as root-mean-square deviation from the mean. If x is the mean of x1, x2, ……., xn, then s is defined by é1 2 2 ù ê n (x1 - x ) + ...... + ( x n - x ) ú = ë û

{

}

1 n

å (x

i

- x)

2

Different formulae for computing s. d. (a) For simple observations or variates. If x be A.M. of x1, x2 …….., xn, then s = 1 n

å (x

i

- x)

2

6.100

MATHS

(b)For simple or group frequency distribution For the variates x1, x2, x3, ……., xn, if corresponding frequencies are f1, f2, f3, ….., fn Then s =

å f (x - x ) åf
i i i

2

…. (2) Where, x = weighted A. M.

Note : If variates are all equal (say K), then s = 0, as x = K and Example 7 : For observations 4, 4, 4, 4, s = 0 as x = 4 and Short cut method for calculating s.d.

å (x - x )= 0

å ( 4 - 4) = 0

If x (A. M.) is not an integer, in case (1), (2) ; then the calculation is lengthy and time consuming. In such case, we shall follow the following formulate for finding s.d.

(c) For simple observations, s =

åd
n

2

æ dö ÷ -ç ç n ÷ è ø

å

2

…. (3)

Where, d = x – A, A is assumed mean. (a) For simple (or group) frequency distribution

s=

å fd åf

2

æ -ç ç è

å fd ö ÷ f ÷ å ø

2

, Where, d = x – A

(e) For group frequency distribution having equal class interval æ fd ¢ ö ÷ ´i …. (5) s= -ç ç f ÷ è ø (This is known as step deviation method) Observation s =

å fd¢ åf

2

å å

2

where, d¢ =

x-A i

å (x

i

- x)

2

n

=

åx
n

2

æ åx ö -ç ç n ÷ ÷ è ø

2

…. (6)

(The proof is not shown at present) Note : Formula (3) may be written as, for step deviation where d¢ =
x-A i

s=

å d¢
n

2

æ d¢ ö ÷ ´i -ç ç n ÷ è ø

å

2

…. (7)

Computation for Standard Deviation : (A) For individual observations computation may be done in two ways : (a) by taking deviations from actual mean. Steps to follow––

MATHS

6.101

(1)Find the actual mean, i.e. the actual mean, i.e. . (1) Find x .
STATISTICAL METHODS

(2) Find the deviations from the mean, i.e., d. (1) Make squares of the deviations, and add up, i.e.

åd . åd
2

2

(4) Divide the addition by total number of items, i.e., find make square root of it. (b) by taking deviations from assumed mean. Steps to follow––

/ n and hence

(1) Find the deviations of the items from an assumed mean and denote it by d find also

å d.

(2) Square the deviations, find

åd .
2

2

(3) Apply the following formula to find standard deviation.
S.D. (s ) =

åd
n

2

æ dö ÷ -ç ç n ÷ è ø

å

Example 8 : Find s.d. of (Rs.) 7, 9, 16, 24, 26. Calculation of s.d. by methods (a) and (b).

Method (a) : Calculation from A.M. Variate (Rs.) x 7 9 16 24 26 Total Dev. From A.M. (16.4) d –9.4 –7.4 –0.4 7.6 9.6 –– d2 88.36 54.76 0.16 57.76 92.16 293.20
82 = 16.40 5
2

Method (b) : Calculation from assumed mean Variate (Rs.) x 7 9 16 24 26 – Dev. From A.M. (16) d –9 –7 0 8 10 2 d2 81 49 0 64 100 294

For method (a) : x (A.M.) = s (= s.d ) = 1 n

å (x - x )

=

1 n

åd

2

=

1 ´ 293.20 = 58.64 = Rs. 7.66 5

Here the average or A.M. 16.40 and the variates deviate on an average from the A.M. by Rs. 7.66.
1.5.1.1.1.1.1.1.1 : Let A (assumed mean) = 16 For method (b) For method (b) : Let A (assumed mean) = 16

6.102

MATHS

s (= s.d.) =

åd
n

2

æ dö ÷ , by using formula (3) -ç ç n ÷ è ø 294 æ 2 ö - ç ÷ = 58.8 - (0.4)2 = 58.8 - 0.16 = 58.64 = Rs. 7.66. 5 è5ø
2

å

2

=

Note : If the actual mean is in fraction, then it is better to take deviations from an assumed mean, for avoiding too much calculations. (B) For discrete series (or Simple Frequency Distribution). There are three methods, given below for computing Standard Deviation. (a) Actual Mean, (b) Assumed Mean, (c) Step Deviation. For (a) the following formula are used. This method is used rarely because if the actual mean is in fractions, calculations take much time. s=

å f (x - x ) åf

2

or

å fd åf

2

; d = x -x

(In general, application of this formula is less)

For (b), the following steps are to be used :– (i) Find the deviations (from assumed mean), denote it by d.

å fd. (iii) Find å fd , i.e. (fd × d and then take å , and hence use the formula.
(ii) Obtain
2

fd fd å fd 22 - ææå fd öö ç = -ç å ÷÷ å f çç å ff ÷÷ è å øø è

22

Example 9 : Find the Standard deviation of the following series : x 10 11 12 13 14 Total f 3 12 18 12 8 48

MATHS

6.103

STATISTICAL METHODS

Calculation of Standard Deviation

Devn. From Ass. Mean (12) X (1) f (2) d (3) fd (4) = (2) × (3) 10 11 12 13 14 Total 3 12 18 12 3 48 –2 –1 0 1 2 –6 –12 0 12 6 0 d2 (5) = (3) × (3) 4 1 0 1 4 fd2 (6) (2) × (5) 12 12 0 12 12 48

s=

å fd åf

2

æ -ç ç è

å fd ö ÷ f ÷ å ø

2

=

48 0 = 1 =1 48 48

For (c) the following formula is used. The idea will be clear from the example shown below : Formula is, s =

å fd ¢ åf

2

æ -ç ç è

å fd ¢ åf

2

ö ÷ ´ i where d¢ = step deviation, i = common factor. ÷ ø

Example 10 : Find the standard deviation for the following distribution : x 4.5 14.5 24.5 34.5 44.5 54.5 64.5 f 2 3 5 17 12 7 4

6.104

MATHS

Calculation of Standard Deviation

x 4.5 14.5 24.5 34.5 44.5 54.5 64.5

f 2 3 5 17 12 7 4

d –30 –20 –10 0 10 20 30 –

d¢ =

d 10

fd¢ –6 –6 –5 0 12 14 12

fd¢2 18 12 5 0 12 28 36

–3 –2 –1 0 1 2 3 –

å f = 50
ì ï s= í ï î
=

å fd¢ = 21

å fd¢

2

= 111

å fd¢ - æ å fd¢ ö ç ÷ f ç åf ÷ å è ø

2

ü ï ý ´i = ï þ

2ü ì ï111 æ 21 ö ï - ç ÷ ý ´10 í ï 50 è 50 ø ï î þ

(2.22 - 0.1764) ´ 10

= 1.4295 × 10 = 14.295.

(C) For Continuous Series (or group distribution) : Any method discussed above (for discrete series) can be used in this case. Of course, step deviation method is convenient to use. From the following example, procedure of calculation will be clear. Example 11 : Find the standard deviation from the following frequency distribution.

Weight (kg.) 44–46 46–48 48–50 50–52 52–54 Total

No. of persons 3 24 27 21 5 80

MATHS

6.105

STATISTICAL METHODS

Calculation of s.d.

Weight (kg.)

mid. pt. x

Frequency f 3 24 27 21 5 80

devn.

(d = x - 49 )
–4 –2 0 2 4 –

d¢ =

d 2

fd¢ –6 –24 0 21 10 1

fd¢2 12 24 0 21 20 77

44–46 46–48 48–50 50–52 52–54 Total

45 47 49 51 53 –

–2 –1 0 1 2 –

Let A (assumed mean) \ 49 ì ï s= í ï î

å fd¢ åf

2

æ -ç ç è

å fd¢ ö ÷ f ÷ å ø

2

ü 2 77 æ 1 ö ï -ç ÷ ´2 ´i = ý 80 è 80 ø ï þ

1 ö æ = ç 0.9625 ÷´2 6400 ø è
= 0.9625 - 0.00016 ´ 2 = 0.96234 ´ 2 = 0.9809 × 2 = 1.96 kg.

MATHEMATICAL PROPERTIES OF STANDARD DEVIATION : Combined Standard Deviation. We can also calculate the combined standard deviation for two or more groups, similar to mean of composite group. The required formula is as follows : s12 = where n1s12 + n 2 s 2 2 + n1d12 + n 2 d 2 2 n1 + n 2 s12 = combined standard deviation of two groups. s12 = standard deviation of 1st group. s12 = standard deviation of 2nd group.
1.5.1.1.1.1.1.1.1

d1 = x1 - x12 ; d 2 = x 2 - x12 , where x12 =

n1x1 + n 2 x 2 n1 + n 2

6.106

MATHS

For Three Groups s123 = n1s12 + n 2 s 2 2 + n 3s 33 + n1d12 + n 2 d 2 2 + n 3d 3 2 n1 + n 2 + n 3

Where d1 = x1 - x123 ; d 2 = x 2 - x123 ; d 3 = x 3 - x123 Example 12 : Two samples of sizes 40 and 50 respectively have the same mean 53, but different standard deviations 19 and 8 respectively. Find the Standard Deviations of the combined sample of size 90. Here, n1 = 40, x1 = 53, s1 = 19 ; n 2 = 50, x 2 = 53, s 2 = 8 Now, x12 = Now, n1x1 + n 2 x 2 40 ´ 53 + 50 ´ 53 2120 + 2650 4770 = = 53 = n1 + n 2 40 + 50 90 90 [ICWA (F) Dec. 2003]

d1 = x1 - x12 = 53 - 53 = 0, d 2 = x1 - x12 = 53 - 53 = 0,

ì 19 ï 40 ( ) + 50 (8) + 40 (0) + 50 (0 ) s12 = í 40 + 50 ï î

2

2

2

2

ü ï ý ï þ

17640 æ 14440 + 3200 ö = ç = 196 = 14 ÷= 90 90 è ø 2. Prove that the Standard Deviation does not depend on the choice of origin. For the n observations, x1, x2, ….., xn let d1, d2, …., dn are respective quantities obtained by shifting the origin to any arbitrary constant, say A, so that d1 = x1 – A (for I = 1, 2 ….. n). Now we are to show sx = dd. We know, s x 2 = å (x i - x )2 / n, where x =

å xi
n

Now, di = xI – A so that å d i = å x i - å A (taking å to both sides), Again

å d i = å x i - å A (dividing by n)
n n n

or d = x - A or, x = A + d Now, x i - x = (A + d i )- A + d = d i - d So, s x 2 = å (x i - x )2 / n = å d i - d / n = s d 2 . 3. Prove that the Standard Deviation calculated from two value x1 and x2 of a variable x is equal to half their difference. We know s 2 = [ICWA (F) June 2000] according to definition of s and where

(

)

(

)

2

(x1 - x )2 + (x 2 - x )2
2

MATHS

6.107

STATISTICAL METHODS

x=

1 (x1 + x 2 ), i.e., x is A.M. of x1 and x2. 2

or s 2 =

x + x2 ö æ x + x2 ö 1 éæ êç x1 - 1 ÷ + ç x2 - 1 ÷ 2 êè 2 ø è 2 ø ë
2 2ù

2



ú (Putting the value of x ) ú û

=

1 éæ x1 - x 2 ö æ x1 - x 2 ö êç ÷ +ç ÷ 2 êè 2 ø è 2 ø ë

1 é1 2 2 2 ùù é1 ú = ê (x1 - x 2 ) + (x1 - x 2 ) = ê (x1 - x 2 ) ú ú 2 ë4 4 ë ûû ú û

{

}

\s =

1 (x1 - x 2 ), since s is always positive. 2

4. Prove that the standard deviation is independent of any change of origin, but is dependent on the change of scale. For the n observations x1, x2, …… xn, let the origin be changed to A and the scale to d, then y1 = x1 - A or x1 =A +dy1 which means y1, y2 …. Yn are the deviations of x1, x2 ….. xn from an d

arbitrary constant A, in units of another constant d. Now, x = A + dy i.e., mean of x¢ s = A + d (mean of y¢ s) Again, x i - x = (A + dy i ) - (A + dy ) = d (y i - y )
\ sx =
2

å (x i - x )
n

2

=

d å { (y i - y )} n

2

=

d 2 å (y i - y )2 2 = d 2s y n

\ s x = ds x (A is absent, but d is present). This shows S.D. is unaffected by any change of origin, but depends on scale. VARIANCE : The square of the Standard Deviation is known as Variance. COEFFICIENT OF VARIATION : It is the ratio of the Standard Deviation to the Mean expressed as percentage. This relative measure was first suggested by Professor Kari Pearson. According to him, coefficient is the percentage variation in the Mean, while Standard Deviation is the total variation in the Mean. Symbolically, Coefficient of variation (V) =
s ´ 100 = Coefficient of stand. deviation × 100. x

Note : The coefficient of variation is also known as coefficient at variability.
1.5.1.1.1.1.1.1.1 Example 13 : If Mean and Standard deviation of a series 40 and 10, then the coefficient Example 13 : If Mean and Standard deviation of a series are respectivelyare respectively 40 and 10, then the coefficient of variations would be 10 / 40 × 100 = 25%, which means the standard deviation is 25% of of variations would be 10 / 40 × 100 = 25%, which means the standard deviation is 25% of the mean. the mean.

6.108

MATHS

Example 14 : An analysis of the monthly wages paid to workers in two firms, A and B, belonging to the same industry gives the following results : Firm A No. of wage-earners Average monthly wages Variance of distribution of wages 586 Rs. 52.5 100 Firm B 648 Rs. 47.5 121

(a) Which firm A and B pays out the largest amount as monthly wages? (b) Which firm A and B has greater variability in individual wages? (c) Find the average monthly wages and the standard deviation of the wages of all the workers in two firms A and B together. (a) For firm A : total wages = 586 × 52.5 = Rs. 30,765. For firm B : Total wages = 648 × 47.5 = Rs. 30,780. i.e. Firm B pays largest amount. (b) For firm A : s2 = 100 \ s = Rs. 10 Now, v =
s 10 ´ 100 = ´ 100 = 19.04 Mean 52.5 11 ´ 100 = 23.16 (here s = Rs. 11) 47.5

For firm B : V =

\ Firm B has greater variability, as its coefficient of variation is greater than that of Firm A. (c) Here, n1 = 586, x1 = 52.5, s1 = 10
n 2 = 648, x 2 = 47.5 s 2 = 11

\ x12 = =

n1 x1 + n 2 x 2 586 ´ 52.5 + 648 ´ 47.5 30,765 + 30,780 = = n1 + n 2 586 + 648 1234 61,545 = 49.87 = Rs. 49.9 1,234

Again, d1 = x1 - x 2 = 52.5 - 49.9 = 2.6 ; d2 = 47.5 – 49.9 = – 2.4
2 2 2 2ü ì ï n s + n 2 s 2 + n 1d 1 + n 2 d 2 ï \ s12 = í 1 1 ý n1 + n 2 ï ï î þ

ì 10 11 ï 586 ( ) + 648 ( ) + 586 (2.6 ) + 648 (- 2.4 ) = í 586 + 648 ï î ì 58600 + 78408 + 3962 + 3733 ü = í ý 1234 î þ
1.5.1.1.1.1.1.1.1 =

2

2

2

2

ü ï ý ï þ

144703 = 10.83 (Calculation by log table) 1234

MATHS

6.109

Example 15 : In an examination a candidate scores the following percentage of marks :
STATISTICAL METHODS

English 62

2nd language 74

mathematics 58

Science 61

Economics 44

Find the candidates weighted mean percentage weighted of 3, 4, 4, 5 and 2 respectively are allotted of the subject. Find also the coefficient of variation. [ICWA (F) Dec. 2003] fd2

Marks

f

fx

d = x – 61

fd

62 74 58 61 44 Total

3 4 4 5 2 18

186 296 232 305 88 1107

1 13 –3 0 –17

3 52 –12 0 –34 9

3 676 36 0 578 1293

Weighted mean percentage =

å fx = 1107 å f 18

= 61.5 marks

s.d. (s) =

å fd 2 åf

2 æ å fd ö 1293 æ 9 ö ÷ = -ç - ç ÷ = 71.83 - 0.25 ç åf ÷ 18 è 18 ø è ø

2

= 71.58 =8.46 Coeff. of variation =
s.d 8.46 ´ 100 = ´ 100 = 13.76%. A.M. 61.5

Example 16 : The A.M. of the following frequency distribution is 1.46. Find f1 and f2. No. of accidents : No. of days : 0 46 1 f1 2 f2 3 25 4 10 5 5 total 200 [ICWA (F) June 2003]

Also find coefficient of variation.

For A.M. calculation, see solved examples in Average chapter (example for two missing frequencies) Putting these values of f1 and f2 we find the following distribution :

6.110

MATHS

x 0 1 2 3 4 5 Total
AM (x ) = 2 +

f 46 76 38 25 10 5 200

d –2 –1 0 1 2 3 –

fd –92 –76 0 25 20 15 –108

fd2 184 76 0 25 40 45 370

-108 = 2 - 0.54 = 1.46 200
2

s=

370 æ - 108 ö -ç ÷ 200 è 200 ø

= 1.85 - (0.54 )

2

= 1.85 - 0.2916 = 1.5584 = 1.248 = 1.25 (app.) Now coeff. of variation =
1.25 ´ 100 = 85.62% (app.) 1.46

Example 17 : For the numbers 5, 6, 7, 8, 10, 12 if s1 and s2 be the respective root mean square deviation about the mean and about an arbitrary number 9, show that 17s22 = 20 s12. [ICWA (F) June 2003] d2 9 4 1 0 4 16 34 d2 16 9 4 1 1 9 40

x 5 6 7 8 10 12 48

d=x–8 –3 –2 –1 0 2 4

d=x–9 –4 –3 –2 –1 1 3 –6

s1 =

34 34 , s12 = , 6 6
40 40 , s22 = , 6 6

20s12 = 20 ´

34 680 = 6 6
40 680 = \ 17s 2 2 = 20s12 ; 6 6 x (mean ) = 48 =8 6

s2 =

17s 2 2 = 17 ´

Advantages of Standard Deviation :

MATHS

6.111

Advantages of Standard Deviation :
STATISTICAL METHODS

1. Standard deviation is based on all the observations and is rigidly defined. 2. It is amenable to algebraic treatment and possesses many mathematical properties. 3. It is less affected by fluctuations of sampling than most other measures of dispersion. 4. For comparing variability of two or more series, coefficient of variation is considered as most appropriate and this is based on standard deviation and mean. Disadvantages of Standard Deviation : 1. It is not easy to understand and calculate. 2. It gives more weight to the extremes and less to the items nearer to the mean, since the squares of the deviations of bigger sizes would be proportionately greater than that which are comparatively small. The deviations 2 and 6 are in the ratio of 1 : 3 but their squares 4 and 36 would be in the ratio of 1 : 9. Uses of Standard Deviation : It is best measure of dispersion, and should be used wherever possible. SELF EXAMINATION QUESTIONS (A) Regarding Range. 1. Daily wages in Rs. of 7 workers are as follows : (Rs.) : 12, 8, 9, 10, 7, 14, 15. Calculate range. 2. Find range : Weight (kg) : 40, 51, 47, 39, 60, 48, 64, 61, 57. [Ans. 25 kg] [Ans. Rs. 8]

3. The marks obtained by 6 students are 24, 12, 16, 11, 40, 42. Find range. If now the highest mark is obtained, find the percentage change in range. (B) Regarding MEAN DEVIATION : 4. Find Mean Deviation about mean of the numbers given : (i) 31, 35, 29, 63, 55, 72, 37. (ii) 29, 35, 51, 63, 78, 106, 128 5. Find M.D. about median of : 13, 84, 68, 24, 96, 139, 84, 27. 6. Find M.D. about A.M. of the table given below : x 2 4 6 8 19 Find also coefficient of mean dispersion.
6.112

[Ans. 31 marks, 6.45]

[Ans. 14.9] [Ans. 29.143]

[Ans. 33.88]

f 1 4 6 4 1 [Ans. 1.5, 0.25]
MATHS

7. From the following table find coefficient of mean dispersion about : (i) A.M., (ii) Median. Marks 10 15 20 30 40 50 Frequency 8 12 15 10 3 2 [Ans. (i) 0.363, (ii) 0.36]

8. From the following frequency distribution find M.D. about median : C.I. 2–4 4–6 6–8 8–10 9. Find M.D. about A.M. of the table :– Weight (lb) 95–105 10–115 115–125 125–135 Students 20 26 38 16 [Ans. 8.6 lbs] (C) Regarding STANDARD DEVIATION : 10. Calculate standard deviation of the following numbers : (i) 9, 7, 5, 11, 3 (ii) 1, 2, 3, 4, 5 (iii) 1, 2, 3, 4, ….. 9, 10 (iv) 4, 5, 6, 6, 7, 8 (v) 9, 7, 5, 11, 1, 5, 7, 3. 11. The frequency distribution of heights of 50 persons is shown below : Height (inches) 62 64 66 68
1.5.1.1.1.1.1.1.1 Find s.d. and variance.

f 3 4 2 1 [Ans. 1.4]

[Ans. 2.83] [Ans. 1,414] [Ans. 2.87] [Ans. 1.29] [Ans. 3.072]

No. of persons 8 13 17 12
[Ans. 2.02 inch. 4.50 sq. inch]

MATHS

6.113

12. Find s.d. from the tables :
STATISTICAL METHODS

(i)

Age (yrs.) 30 40 50 60 70

Persons 64 132 153 140 51 Frequency 1 5 12 22 17 9 4 [Ans. 13.25] [Ans. 11.64 yrs.]

(ii)

Class-limits 4.5 14.5 24.5 34.5 44.5 54.5 64.5

13. Compute s.d. from the following tables : (i) Height (inch) 60–62 62–64 64–66 66–68 68–70 Total (ii) Marks 0–10 10–20 20–30 30–40 40–50 14. Find the coefficient of variation of numbers : 1, 2, 3, 4, 5. Students 34 27 20 13 5 100 Students 5 8 15 16 6 [Ans. 10.77 marks] [Ans. 47.13%] [Ans. 2.41 inch]

6.114

MATHS

15.

Marks 10 20 30 40 50 60

Students 8 12 20 10 7 3 [Ans. 64.81%]

Find coefficient of variation. 16. Run-scores in 10 innings of two cricketers are as follows : A 31 28 47 63 71 39 10 60 96 14 Find which batsman is more consistent in scoring. B 19 31 48 53 67 90 10 62 40 80

[Ans. Batsman B]

17. The A.M.’s of two samples of sizes of 60 are 90are respectively 52 and 48, the s.d. are 9 and 12. Obtain the mean and s.d. of the sample of size 150 obtained by combining the two samples. [Ans. 49.6, 11.1] 18. The first of two samples has 100 items with mean 15 and s.d. 3. If the whole group has 250 items with mean 15.6 and s.d. 13.44 , find the s.d. of the second group. OBJECTIVE QUESTIONS : 1. Find the range of 6, 18, 17, 15, 14 2. Find Mean Deviation (M.D.) of 4, 8, 12 (cm) about A.M. 3. Find M.D. about median of 4, 8, 10 (kg) 4. Find S.D. of (i) 2, 5, 8 (ii) 2, 6 [Ans. 12] [Ans. 2.67 cm] [Ans. 2 kg.] [Ans. (i) 6 (ii) 2] [Ans. 4]

5.Find variance of 2, 5, variance of 2, 5, 8 [Ans. 6] 1.5.1.1.1.1.1.1.1 5. Find 8

[Ans. 6]

MATHS

6.115

6. Find S.D. from the given data : n = 10, Sx = 40, Sx2 = 250
STATISTICAL METHODS

[Ans. 3] [Ans. 5] [Ans. 8%] [Ans. 9] [Ans. 40]

7. If n = 10, Sx = 120, Sx2 = 1690 ; find s.d. 8. If variance = 16, A.M. = 50, find coefficient of variation 9. Find variance of x, if it’s A.M. is 6 and coefficient of variation is 50%. 10. Find mean, if c.v. = 5% and variance = 4 11. Coefficient of variation of a distribution is 25%, it it means what?

[Ans. s.d. is 25% of A.M.] 12. If each term of variates is increased by 2, what will be the effect on (i) A.M. (ii) range and (iii) s.d. [Ans. (i) increased by 2, (ii) & (iii) no change] 13. If each item is doubled what will be effect on (i) A.M. (ii) Range (iii) s.d. [Ans. (i), (ii) & (iii) doubled] [Ans. 8] [Ans. 6] [Ans. 1] [Ans. 1]

14. Two variables x and y are related by y = 4x – 7. If s.d. of x is 2, find s.d. of y 15. Two variates x and y are given by y = 2 – 3x, s.d. of x is 2, find s.d. of y 16. Compute s.d. of 6 numbers 7, 7, 7, 9, 9, 9. 17. Compute M.D. of 6 numbers 4, 4, 4, 6, 6, 6 [ICWA (F) June 2007] 18. Means and S.D. of runs of 10 innings of two players are as follows : First player : mean = 50, s.d. = 4 Second player, mean = 40, s.d. = 5 Find who is more consistent in scoring runs?

[ICWA (F) June 2007] [Ans. First player]

19. If 2xI + 3yI = 5 for I = 1, 2, …., n and mean deviation of x1, x2, ….., xn about their mean is 12, find the mean deviation of y1, y2, …., yn about their mean. [ICWA (F) Dec. 2006] 20. If the means of two groups of 30 and 50 observation are equal and their standard deviation are 8 and 4 respectively, find the grouped variance.
10

[Ans.

19 ] 3

[Ans. 5.83]

21. For 10 values x1, x2, …., x10 of a variable x, variance of x

å x i = 110,
i =1

and

å (x i - 5)
i =1

10

2

= 1000, find

[ICWA (F) June 2006] [Ans. 64]

22. If the relations between two variables x and y be 2x – y + 3 = 0 and range of x be 10, then find the range of y. [Ans. 20]

23. Runs made by two groups G1 and G2 of cricketers have means 50 and 40 and variance 49 and 36 respectively. Find which group is more constant in scoring runs. [Ans. G1]
1.5.1.1.1.1.1.1.1 24. If A.M. and coefficient a variation of a variable 50% respectively, find the variance 24. If A.M. and coefficient of variation of of variable x are 10 andx are 10 and 50% respectively, find the variance of x. [Ans. 25] of x.[Ans. 25]

6.116

MATHS

25. If the relation between two variables x and u is x – 10 = 2u and mean deviation of x about its mean is 10, find the mean deviation of u about its mean. 6.7 SKENESS AND KURTOSIS : 8.2.1. SKEWNESS ––Introduction We may have frequency distributions that differ widely both in nature and composition, but still have same average and dispersion. Let us take the following two frequency distributions A and B. They have the same average (mean = 30) and dispersion (s.d. = 12.04) but of different nature. If they are represented by histograms, we will find different shape and size. Class A 0–10 10–20 20–30 30–40 40–50 50–60 10 30 60 60 30 10 Frequency B 10 40 30 90 20 10 [Ans. 5]

To analyse further we need two other measures skewness and kurtosis to reveal clearly the salient feature of frequency distribution. Skewness helps us to study the shape of distribution i.e. symmetry or asymmetry, while kurtosis indicates the flatness or peakedness of curve. A frequency distribution is said to be ‘symmetrical’ if the frequencies are distributed symmetrically (or evenly) on either side of an average. When plotted on a graph paper, such distribution will show a normal or ideal curve. In a normal curve mean, median or mode coincide. A normal curve is a bell shaped curve, in which the values on either side of an average are symmetircal. If general frequency distributions are not symmetrical, they are slightly or highly asymmetrical. Skewness is oppositie to symmetrical. Measures of skewness will not only show the amount of skewness, but also its direction. A distribution is said to be positively skewed when it has a long tail towards the higher values of the variable and negatively skewed when the longer tail is present towards the lower values of the variable.

(a)
MATHS

M = Me = Mo
6.117

STATISTICAL METHODS

(b)

Mo Me M

(c)

M Me Mo

The figures give us an idea about the shape of symmetrical and asymmetrical curves. Fig. (a) shows the shape of an ideal symmetrical curve. It is a bell-shaped and the values of mean, median and mode would be equal. Fig. (b) indicates a moderately skewed curve. In it the value of the mean would be greater than that of median, which would be also greater than mode. The curve is skewed to the right and is known as positively skewed. Fig (c) the value of mean would be less than that of median which would be again less than mode. It is skewed to the left and is known as negatively skewed. Test of skewness : 1. The values of mean, median and mode would not be the same. 2. When plotted on a graph paper, a skew distribution would not show a bell-shaped curve. [as in fig. (a)] Measures : 1. Pearson’s first formula : Absolute skewness = Mean – Mode. Coefficient of skewness = 2. Pearson’s second formula : Coefficient of skeness = SOLVED EXAMPLES : Example 1 : Comment on the following results or average of any distribution : (i) AM is 10, median is 11.
6.118 MATHS

mean - mod e s.d.

3 (mean - median ) s.d.

(ii)AM is (ii) median 15, median is 12 15, AM is is 12 (ii) Mode is is 11, median is 13 (iii) Mode 11, median is 13

(iv) Median 10, A.M. is 14 (iii) Median is is 10, A.M. is 14
(v) Medain is 12, Mode is 13. (i) AM (10) < median (12), distribution is negatively skewed (ii) AM (15) > median (12), distribution is positively skewed (iii) Med. (13) > mode (11), distribution is positively skewed (iv) AM (14) > median (10), distribution is positively skewed (v) Median (12) < mode (13), distribution is negatively skewed Example 2 : From the following table, calculate absolute skewness and also coeff. of skewness.

Weekly wages (Rs.) 15 20 25 30 35 40 45 Total

No. of earners 3 25 19 16 4 5 6 78

Wages (Rs.) 15 20 25 30 35 40 45 Total

f 3 25 19 16 4 5 6 78

d = x – 30 –15 –10 –5 0 5 10 15

d¢ =

d 5

fd¢ –9 –50 –19 0 4 10 18 –46

fd¢2 27 100 19 0 4 20 54 224

–3 –2 –1 0 1 2 3

x=A+

(-46 ) ´ 5 = 30 - 2.95 = Rs.27.05 Sfd¢ ´ i = 30 + Sf 78
Sfd¢ 2 æ Sfd¢ ö -ç ÷ ´i = Sf è Sf ø
2

s.d.(= s )=

224 æ - 46 ö ´ç ÷ ´5 78 è 78 ø

2

MATHS

6.119

= 2.87 - 0.3481 ´ 5 = 2.5219 ´ 5
STATISTICAL METHODS

= 1.5 88 × 5 = Rs. 7.9 4 Mode = 20. Now, Absolute skewness = Mean – Mode = 27.05 – 20 = Rs. 7.05 Coeff. of skewness =
mean - mod e 7.05 = = 0.89. s.d. 7.94

Example 3 : For a group of 11 items Sx = 65, Sx2 = 493 and mode = 6. Find the pearson’s coefficient of skewness.
x (mean ) = Sx 65 = = 5.91 n 11 Sx 2 æ Sx ö -ç ÷ = n è n ø
2

s.d. (s) =

493 æ 65 ö -ç ÷ 11 è 11 ø

2

= 44.82 - 34.93 = 9.89 = 3.145 \ Coeff. of skewness = =
mean - mod e s.d.

5.91 - 6 -.09 = = - 0.029 3.145 3.145

Note : The distribution is negatively skewed. Example 4 : For a moderately skewed distribution, AM = 112, Mode = 110 and s.d. = 40. Find (i) coefficient of variation, (ii) Pearson’s coefficient of skewness, and (iii) median. (i)
C.V. = s.d. 40 ´ 100 = ´ 100 = 35.71 mean 112 mean - mode 112 - 110 2 = = = 0.05 s.d. 40 40

(ii) Coefficient of skewness =

(iii) Mean – mode = 3 (mean – median) or, 112 – 110 = 3 (112 – median) or, 2 = 3 (112 – med.) = 336 – 3 med. or, 3 med. = 334
\ Median = 334 = 111.33. 3

Example 5 : In a distribution mean = 65, median = 70, coefficient of skewness is –0.6. Find– (i) mode
1.5.1.1.1.1.1.1.1 coefficient variation. (ii) coefficient of of variation.

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(i) We know, mean – mode = 3 (mean – median) or, 65 – mode = 3 (65 – 70) or, 65 – mode = 3 × (–5) = – 15 or, mode = 80 (ii) Coeff. of skewness = or, - 0.6 =
mean - mode s.d.

65 - 80 -15 or, - 0.6 = or, s.d. = 25 s.d. s.d.
s.d. 25 ´ 100 = ´ 100 = 38.46 A.M. 65

\ Coeff. of variation =

Example 6 : Given A.M. = 50, coeff. of variation = 40%, coeff. of skewness = –0.4, find s.d., mode and median. Coeff. of variation =
s.d. s.d. ´ 100 or, 40 = ´ 100 or, s.d. = 20 A.M. 50 AM - mode s.d.

Again, coeff. of skewness =

or, - 0.4 =

50 - mode or, mode = 50 + 20 × 0.4 = 58 20

AM – mode = 3 (AM – median) or, 50 – 58 = 3 (50 – med.) or, –8 = 3 (50 – med.) or, 3 med. = 158 or, median = 52.67. Example 7 : Which group is more symmetircally skewed? (i) AM = 22, median = 24, s.d. = 10 (ii) AM = 22, median = 25, s.d. = 12
3 (AM - med ) 3 (22 - 24) - 6 = = = -0.6 s.d. 10 10 3 (22 - 25) - 9 = = -0.75 12 12

(i) Coeff. of skewness = (ii) Coeff. of skewness =

From the above results it is clear that the first group is more symmetrically skewed. Example 8 : For the following frequency distributions, determine mean, mode, s.d. and coefficient of skewness : Marks : No. of students : Marks x 0–10 10 f 10–20 30 d = x – 25 0–10 10–20 20–30 30–40 Total 5 15 25 35 10 30 40 20 100 –20 –10 0 10 –2 –1 0 1 –20 –30 0 20 –30 40 30 0 20 90 20–30 40 d¢ = d 10 30–40 20 fd¢ fd¢2

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x (= mean) = A +
STATISTICAL METHODS

Sfd¢ -30 ´ i = 25 + ´ 10 = 25 - 3 = 22 marks Sf 100

Modal class is (20 – 30), as it has highest frequency f0 = 30, f1 = 40, f2 = 20, I = 10 \ mode = 20 +
40 - 30 10 ´ 10 = 20 + ´ 10 = 20 + 3.33 = 23.33 . 2 ´ 40 - 30 - 20 30
2

s.d. (= s ) =

90 æ - 30 ö -ç ÷ ´ 10 = 0.9 - 0.09 ´ 10 = 0.81 ´ 10 100 è 100 ø

= 0.9 × 10 = 9 Coeff. of skewness = KURTOSIS : Three measures i.e. average, dispersion and skewness that we have so far studied are not even sufficient for analysis of a frequency distribution completely. In the following diagram all the three different curves A, B and C are symmetrical about mean and hence have same variations too, which verifies the point. So to identify a distribution completely we need one more measure, about which Prof. Peason suggested the convexity of the curve (or its Kurtosis) towards the middle part (or about the mean) of frequency curve. A B C Mode We know skewness suggests us to identify the right or left tails of a frequency curve. Kurtosis helps us to get an idea about the shape and nature of the hump (middle part) of frequency distribution. Thus we can say Kurtosis is referred to flatness or peakedness of the frequency curve. Definition : Kurtosis refers to the degree of peakedness of the hump of the distribution. Description : The curve (type B) is neither flat nor peaked and hence the shape of its hump is accepted as standard one. Such curve is known as normal curve. Normal curves are said to have normal kurtosis and are known as Mesokurtic. The curve (type A) having more peaked than normal curve are known as Lopto kurtic. They are also said to lack kurtosis or to have negative kurtosis. Lepto Kurtic Meso Kurtic Platy Kurtic mean - mode 22 - 23.33 -1.33 = – 0.148 (app.) = = s.d. 9 9

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Lastly, the curve (type C) that are flattered than normal curve are called Platy kurtic. They are also said to possess kurtosis in excess or have positive kurtosis. SELF EXAMINATION QUESTIONS : 1. Comment on the following average values of distribution : (i) Median is 21, mode is 20 (ii) A.M. is 12, median is 14 (iii) A.M. is 10, mode is 9 (iv) Mode is 15, median is 12 (v) Mode is 12, A.M. is 10 (vi) A.M. is 10, median is 10, mode is 10 [Ans. (i) +, (ii) –, (iii) +, (iv) –, (v) –, (vi) symmetry] 2. Using Pearson’s first and second formula, find the coefficients of skewness of the following distribution : (i) x 10 11 12 13 14 Total f 3 12 18 12 8 48 [Ans. 0, 0] (ii) x 10 11 12 13 14 15 Total
1.5.1.1.1.1.1.1.1

f 2 4 10 8 5 1 30
[Ans. (ii) 0.3602 ; 1.0806]

[Ans. (ii) 0.3602 ; 1.0806]

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3. Find the coefficient of skewness using Pearson’s formula regarding mean, mode and s.d. of the
STATISTICAL METHODS

given table :

x 14.5 15.5 16.5 17.5 18.5 19.5 20.5 21.5

f 35 40 48 100 125 87 43 22 [Ans. (–0.186)]

4. For moderately skewed distribution A.M. = 172, median = 167, s.d. = 60. Find the coefficient of skewness and mode. [Ans. 0.25 ; 157]

5. Pearson’s coefficient of skewness of a distribution is 0.32. It s.d. is 6.5 and A.M. is 29.6. Find the mode and median of the distribution. 6. In the above example if mode is 24.8, find s.d. [Ans. 27.52 ; 28.91] [Ans. 15]

7. For a distribution A.M. = 65, median = 70, coefficient of skewness = – 0.6, find (i) mode, (ii) coeff. of variation. [Ans. (i) 80 (ii) 38.46%]

8. You are given : A.M. = 50, C.V. = 40%, coeff. of skewness = – 0.43. Find the s.d., mode and median. [Ans. 20, 58.6, 52.87]

9. For a certain distribution, the following results were obtained : mean = 45 ; median = 48, coefficient of skewness = – 0.4. The person who gave the above results forgot to give the value of s.d. of the same distribution. From the given data find s.d. [Ans. 22.5]

10. The mean, median and the coefficient of variation of 100 variations are found to be 90, 84 and 60. Find the coefficient of skewness of the above system of 100 observations. OBJECTIVE QUESTIONS 1. Which group is more skewed? Group I : AM = 20, mode = 25, s.d. = 8 Group II : AM = 18, mode = 27, s.d. = 9 [Ans. Group II] [Ans. + 025]

2. If the coefficient of skewness, mean and variance of a variable are – 6, 80 and 4, find the mode of that variable. 3. Calculate which of the two distributions is more skewed? (i) mean = 22, mode = 20, s.d. = 2 (ii) mean = 24, mode = 18, s.d. = 3
1.5.1.1.1.1.1.1.1 [Ans. 2nd distribution]

[Ans. 92]

[Ans. 2nd distribution]

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