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CONTENTS:UNITI
0140
COLLECTION OF DATA, TABULATION OF STATSTICAL DATA,MEASURE OF CENTRAL TENDENCY, HM, MEDIAN, MODE . UNITII
4188
CHI SQUARE TEST, CORRELATION, REGRESSION, PROBABILITY. UNITIII
89112
TESTS OF SIGNIFICANCE FOR SMALL SAMPLES UNIT – IV
113199
COORDINATE GEOMETRY
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UnitI Collection of Data The findings of an investigation depend on the correctness and completeness of the relevant data. Sources of data are of two kindsprimary source and secondary source. The term source means origin or place from which data comes or is got. According to George Simpson a primary source is one that itself collects the data; a secondary source is the that makes available data which were collected by some other agency. Data are classified under two categoriesPrimary and Secondary data. Primary and Secondary data The data which is collected by actual observation or measurement or count is called Primary data. Either the investigator individually or through his agents or employees collects the data. The data which are complied from the records of others is called Secondary data. Methods of Collection of Primary Data Primary data is collected in any one of the following methods: Direct personal interviews. Indirect oral interviews. Information from correspondence Mailed questionnaire method Schedules sent through enumerators. 1. Direct Personal Interviews: The persons from whom informations are collected are known as informants or respondents. The investigator personally meets them and asks questions to gather the necessary information. Merits: i)
People willingly supply informations because they are approached personally. Hence, more response is noticed in this method than in any other method.
ii)
The collected informations are likely to be uniform and accurate. The investigator is there to clear the double of the informants.
iii) Supplementary informations on informants personal aspects can be noted. Informations on character and environment may help later to interpret some of the results.
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iv) Answers for questions about which the informant is likely to be sensitive can be gathered by this method. Limitations: It is likely to be very costly and time consuming if the number of persons to be interviewed is large and the persons are spread over a wide area. Personal prejudice and bias of the investigator affects the validity of the survey. 2. Indirect Oral Interviews: Under this method, the investigator contacts witness or neighbours or friends or some other third parties who are capable of supplying the necessary information.
This method is
preferred if the required information is on addiction or cause of fire or theft or murder etc. Enquiry committees appointed by Governments generally adopt this method and get peopleâ€™s views and all possible details of facts relating to the enquiry. Merits: For certain surveys, this is the only method available. For almost all the surveys of this kind, the informants live within a closed area. Hence, the time and the cost are less. Limitations: It is not useful in ordinary situations. The informants and the person who conducts a survey easily distort the truth. 3. Informationâ€™s from correspondence: The investigator appoints local agents or correspondents in different places and compiles the information sent by them.
Information to Newspapers and some departments of
Government are collected by this method. The advantage of this method is that it is cheap and appropriate for extensive investigations.
But it may not ensure accurate results because the
correspondence are likely to be negligent, prejudiced and biased. This method is adopted in those cases where information are to be collected from a wide area for a long period at regular intervals of time. Merits: (i)
For certain kinds of primary data collection, this is the only method available.
(ii)
Considering the wide area from which the informations are collected, this method requires less time and less cost.
Limitations: (i)
It is used in a very few surveys.
(ii)
Local agents and correspondents are not likely to be serious and careful.
4. Mailed Questionnaire Method: Under this method, a list of questions is prepared and is sent to all the informants by post. The list of questions is technically called questionnaire. A covering letter accompanying the
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questionnaire explains the purpose of the investigation and the importance of correct information and requests the informants to fill in the blank spaces provided and to return the form within a specified time. Merits: (i)
It is relatively cheap.
(ii)
It is fast if the informants respond duly.
(iii)
It is preferable when the informants are spread over a wide area.
Limitations: (i)
The greatest limitation is that the informants should be literatesable to understand and reply the questions.
(ii)
It is possible that some of the persons who receive the questionnaire do not return them. Their action is known as nonresponse.
(iii)
It is difficult to verify the correctness of the information furnished by the respondents.
5. Schedules sent through Enumerators : Under this method, enumerators or interviewers take the schedules, meet the informants and fill in their replies. Often distinction is made between a schedule and a questionnaire. A schedule is filled by the interviewer in a facetoface situation with the informant. A questionnaire is filled by the informant and he receives and returns it by post. It is suitable for extensive surveys. Merits : (i) It can be adopted even if the informants are illiterate. (ii) Answers for questions of personal and pecuniary nature can be collected. (iii) Nonresponse is almost nil as the enumerators go personally and contact the informants. (iv) The informations collected are reliable. The enumerators can be properly trained for the same. (v) It is most popular. Limitations : (i) It is the costliest method. (ii) Extensive training is to be given to the enumerators for collecting correct and uniform informations. (iii) Interviewing requires experience. Unskilled investigators are likely to fail in the field. Sources of Secondary Data Secondary data can be compiled either from Published sources or from Unpublished Sources. 1. Published Sources : Relevant data can be obtained form the following : Reports and official publications of (a) International organizations such as U.N.O. and its subsidiaries such as International Monetary Fund, International Finance Corporation and United Nations Educational, Scientific and Cultural Organization (UNESCO).
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(b) Central Government organizations such as (i) Office of the Registrar General and Census Commissioner of India, New Delhi, (ii) Central Statistical Organization (C.S.O.), Department of Statistics, Ministry of Planning, (iii) National Sample Survey Organisation (N.S.S.O.), Department of Statistics, Ministry of Planning. (iv) Labour Bureau, Ministry of Labour, (v) Directorate of Economics and Statistics, Ministry of Agriculture and Irrigation. (vi) Directorate General of Commercial Intelligence and Statistics, Ministry of Commerce and (vii) The Indian Army Statistics Organisation (I.A.S.O), Ministry of Defence. (c)SemiGovernment Organizations such as (i) Institute of Foreign Trade, New Delhi, (ii) Institute of Economic Growth, New Delhi, (iii) Gokhale Institute of Politics and Economics, Poona and (iv) Economics and Statistics Department of Reserve Bank of India. (d) Institutions such as (i) Indian Statistical Institute, Calcutta and New Delhi, (ii) Indian Council of Agricultural Research, New Delhi, (iii) Indian Agricultural Statistics Research Institute, New Delhi, (iv) National Council of Applied Economic Research, New Delhi, (v) Institute of Chartered Accountants of India, New Delhi, (vi) Institute of Cost and Works Accountants of India, Calcutta, (vii) Institute of Company Secretaries of India, New Delhi, (viii) Institute of Chartered Financial Analysts of India, Hyderabad. (ix) Stock Exchanges and (x) Indian Cotton Mills Federation. (e) Commissions and Committees appointed by the Government. 2. Unpublished Sources: All statistical data are not published. For example, Colleges and Universities maintain records. They collect the details for administrative purposes. Similarly, details collected internally by private organizations regarding persons, profit, sales, production, etc. become secondary data and are used in certain surveys. Banks collect certain particulars while giving advances. Stock Exchanges get details of the projects of the companies and Government offices gather necessary information during registration, issue of permits, licenses, etc. They can provide the necessary data for others. TABULATION OF STATISTICAL DATA Tabulation is the process of arranging data systematically in rows and columns of a table. Generally, classification and tabulation are done one after the other. Raw data are classified and presented in neat tables consisting of rows and columns. According to Secrist, â€œTables are a means of recording in permanent form the analysis that is made through classification and by placing in juxtaposition things that are similar and should be compared.â€? There are tow methods or modes in which data can be presented. They are i)
Statistical Tables and
ii)
Diagrams or Graphs.
(i) Statistical Tables. 1. Objects : 1. Large and complex data can be presented in a neat and compact form. 2. Nature of the data can be easily understood.
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3. Much of the time which is otherwise necessary to look at the data is saved. 4. The data are so placed in a table that proper comparison is possible and easier. 5. A table facilitates further analysis of data. 6. A table is the convenient form for diagrammatic representation of data. 7. Voluminous data can be presented in a small space. 8. It remains a permanent record and enable ready reference. 9. Sometimes omissions and error s can be detected. These are the advantages of tabulation as well. These show the role of tabulation of data also. 2. Parts of a Table : A good table has the following parts or components : i) Identification Number
ii) Title
iii) Prefatory Note or Head Note iv) Stubs v) Captions
vi) Body of the Table
vii) Footnotes
viii) Source
The positions of these parts are indicated in the specimen table. (i) Identification number : Each table should be numbered for the purpose of identification and reference. Unlike in the specimen table, this may be given just above the title also. (ii) Title: This is a precise description of the contents of the table. This is prominently written in capital letters. This tells what the data is about and where and when the data occurred. (iii) Prefatory Note or Head Note: This is a short explanatory statement and is about the entire and or major portion of it. For example, unit of measurement such as â€˜Crores of Rupeesâ€™ which is not included in any other part of the table can be given as prefatory note or head note. (iv) Stubs: These are the row headings. These constitute the first column and explain what the rows are about. (v) Captions: These are the column headings. These tell what the columns are about. There can be subheadings. (vi) Body of the Table: This contains the classified data. Data are classified according to the stubs and captions and are entered in this part. (vii) FootNotes: Matters of interest which could not be given in other parts are mentioned as footnotes. These may point out certain limitations of the data, if any, such as exgodown price and production during powercut. (viii) Source: This is the source of data and should be clear and complete in all respects. Any one can refer to this for the purpose of authenticity, reliability, more details, etc. 3. Format of a Table: A specimen table indicating various parts is given below.
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TITLE (Head Note) Stub
Caption Heading
Heading Caption Entries
Stub
Body
Entries
Number 4. General Rules of Preparing Tables: No rigid rules are laid down for the preparation of suitable tables. The object of referring the data is to be kept in mind while preparing tables. According to Prof. Bowley, “In collection and tabulation common sense is the chief requisite and experience the chief teacher.” The following points are be borne in mind. (i) Rough Table: A rough sketch of the table is to be prepared and the sufficiency of the spacings for rows and columns is to be ascertained. (ii) Order: Captions and stubs should be in certain order such as alphabetical or chronological or traditional or ascending. The order should make the table easier to grasp. (iii)Units of Measurements: Wherever possible the units of measurements should be given in stubs and captions.
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(iv)Uniformity: All the entries should be upto two decimals or three decimals or in thousand, etc. Uniformly, under footnote, it is to be pointed out how the numbers have been rounded off. Further, rounding off avoids unnecessary details. (v) Loading: Too many information should not be presented in a single table. More table can be used, if necessary. (vi)Important Entries: Any entry which needs prominence in the table should be given distinctly or in a box or a circle. (vii)Lines of Demarcation: Thick lines should be used for major divisions, row totals and column totals. Thin lines should be used for subdivisions. 5. Kinds of Tables: Based on its purpose, a table is called General Purpose Table or Special Purpose Table. Based on the number of characteristics and variables considered, a table is called Simple Table or Complex Table. General Purpose Table: General purpose tables are known also as Reference Tables or Repository Tables. These are designed for storing information. These are tables of reference found as appendices. Tables given in census reports are of this kind. Special Purpose Tables: Special purpose tables are known also as Summary Tables.
These are
generally derived from general purpose tables. These contain less information but highlight any one aspect. These find place in the body of the reports. All the tables in this book are of the kind. Simple Table: Simple tables are the oneway tables based on one criterion of classification. TYPES OF TABLES Statistical tables can be classified into a number of ways. There are many categories depending upon: (1) The basis of coverage which can be further classified into simple table and complex table. A complex table can be classified into twofold, threefold, or manifold table. (2) The basis of objective or purpose. This can be further classified into general purpose table or reference table and special purpose table or summary table. (3) The basis of nature of enquiry, which can further be classified into original or primary table and derived or derivative table. 1. On the basis of converge Simple and Complex In a simple table the data are classified according to only one characteristic. It is termed as one way or single table and it takes form of frequency table. In a complex table two or more characteristics are shown. It is more popular, because it helps appropriate consideration of all related facts.
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Simple table:
Distribution of Marks Class marks
No. of students
20 – 30
10
30 – 40
18
40 – 50
22
Total
50
Twoway table If the caption or stub is classified into two characteristics and if it gives information of two interrelated questions, then such a table is called twoway table; for example, Distribution of Marks (Girls & Boys) Number of students
Class marks Boys
Girls
Total
20 – 30
6
4
10
30 – 40
8
10
18
40  50
10
12
22
Total
24
26
50
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Threeway table In this type of table three characteristics are shown. It gives information regarding three interrelated characteristics of a phenomenon; for example.
Distribution of Population by Age, Sex and Literacy Males
Females
Total
Age group
Literate
Illiterate
Total
Literate
Illiterate
Total
Literate
Illiterate
Total
(years)
0 – 18 18 –15 25 –35 35  45
A large number of interrelated problems or characteristics are represented in the same table; for example, the distribution of students in a college according to faculty, class, sex and residence.
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Manifold or Higher order table Number of students in M.K. University (According to faculty, age, sex and residence) Students
Faculty Boys
age group (year)
Hosteller
Day Scholar
Total Girls
Total
Hosteller
Day Scholar
Total Total
Hosteller
Day Scholar
Total
Commerce 20 – 25 25 – 30 above 30 Arts 20 – 25 25 – 30 above 30 Science 20 – 25 25 – 30 above 30 Law 20 – 25 25 – 30 above 30
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Illustration Draw a blank to show the candidates sexwise, appearing for preUniversity, First year, Second year and Third year Examinations of a University in the faculties of Arts, Science and economics in a certain year. Table showing in the distribution of candidates appearing in various University Examinations Boys Faculty
Year
Preuni
Girls
I
II
Total
III
Year
Preuni
I
II
III
Total Total
Year
Preuni
I
II
III
Total
Art Science Economics Total
Illustration Present the following information in a suitable tabular from, supplying the figures not directly given. In 2005, out of 2,000 workers in a factory, 1,550 were members of a trade union. The number of women workers employed was 250, out of which 200 did not belong to any trade union. In 2006, the number of union workers was 1,725 of which 1,600 were men. The number of nonunion workers was 380, among whom 155 were women. Comparative Study of the Membership of Trade Union in a Factory in 2001 and 2002 Year Trade union Members NonMembers Total
2005
2006
Males
Females
Total
Males
Females
1500*
50*
1,550
1,600
125*
250*
200
450*
225
155
1750
250
2000
1825
280
Total 1,725 380
2105
Complex Table: Complex tables are the twoway and the higher order tables.
MEASURES OF CENTRAL TENDENCY The average lies somewhere within the range of data. Because of this, the average is also described as measure of central value. Measure of Central tendency refers to all those methods of statistical analysis which are used to estimate or calculate the average of a set of data. The most common measures of central tendency are mean, median and mode.
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1. Mathematical Average:
12
Average represented mathematically is termed as mathematical
average. It is calculated by taking into account the values of all the items of the series. Mean is the mathematical average. There are the main types of means or mathematical averages: (a) Arithmetic mean (b) Geometric mean (c) Harmonic mean. 1. Arithmetic Mean(A.M.) The common average of many individual values of observations or items is referred to as the arithmetic mean. It is the number obtained by dividing the sum of values of all the items in a series by the total number of items in that series.
X=
X1 + X 2 + X3 + ...X n ∑ X = n n
X (read as ‘Xbar’) denotes arithmetic mean and
X1 , X 2 , X3 ... etc are different values of variable X. n is the number of observations of variable X. Symbol
Σ is Greek letter sigma. It denotes sum i.e. Σ X is the sum of all values of X.
Types of Arithmetic Mean 1. Simple Arithmetic Mean:
In calculating simple average, all items of a series are given equal
importance. 2. Weighted Arithmetic Mean:
In weighted arithmetic mean, the average reflects the relative
importance of different items of a series. Calculation of Simple Arithmetic Mean Different formulae are used for calculating arithmetic mean of ungrouped or raw data and of grouped data. 1. Arithmetic Mean of Ungrouped or Raw Data We know that ungrouped data consist of individual observations.
This type of arithmetic mean
on the average is calculataed by summing up all the individual observations or measurements of a sample and dividing the total by the number of items, observations or measurements.
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A. Direct Method: As discussed above, the simple Arithmetic Mean can be calculated by using the following formula:
X=
X 1 + X 2 + X 3 + ... X n ∑ X = n n
Arithmetic mean =
Sum of observations No. of observations
The individual observations are represented by
X 1 , X 2 , X 3 ... X n The sum of individual observations is represented by
∑X
The number of observations by n and Their mean by Here, symbol
X (say ‘X bar’)
Σ means summarization.
Example Find the arithmetic mean of the marks obtained by 10 students of a class in mathematics in a certain examination. The marks obtained are: 30,21,55,47,10,15,17,45,35,25 Solution:
Let X be the average marks obtained. Sum of all the observations
∑X
=30+21+55+47+10+15+17+45+35+25=300
Number of students n=10
Arithmetic mean
X=
ΣX 300 = = 30 n 10
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B.
14
Shortcut Method: Short cut method for calculating arithmetic mean is used when the number of items in the series
is very large. The formula used as follows:
d Formula = X = A + ∑ n Here
X is the actual arithmetic mean A is the assumed arithmetic mean d is the deviation of items from the assumed mean, i.e. d=(XA)
∑d
is the sum of deviations from the assumed mean
n is the total number of observations. Steps in Calculation: Step 1: Assumed mean (A) of the series is calculated by dividing the total of maximum and the minimum values of the items of a series with two. Step 2:
Deviation (d) of different values from the assumed mean is calculated by subtracting the
assumed mean from the actual value (i.e. X1A,X2A,….)
Step 3: sum of all these deviations
∑d
is calculated by addition.
Step 4: All these values are placed in the above formula:
X = A+
∑d n
Example The table given below shows the number of colonies of Microbes grown on ten agar plates. Calculate the arithmetic mean by using short cut method.
Plate no 1.
1
2
3
4
5
6
7
8
9
10
No. of Colonies
60
70
80
95
100
110
130
140
140
160
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Solution: Calculation of Arithmetic Mean using shortCut Method: Serial No of plates
No. of Microbial colonies per plate (X)
1
60
2
70
3
80
4
95
5
100
6
110
7
115
8
130
9
140
10
160
Assumed Mean
Deviation from Assumed Mean D = (X110) 60110=50 70110=40 80110=30 95110=15
60 + 160 220 = 2 2 = 110
100110=10 110110=0 115110=5 130110=20 140110=30 160110=50
∑ d = −40 Step 1. Assumed Mean Step 2.
A=
60 + 160 220 = = 110 2 2
∑ d = −40
Step 3. n=10
Step 4. Putting the values in the formula
110 +
=X =A+
−40 40 = 110 − = 106 10 10
∑d n Ans
1. Arithmetic Mean of Grouped Data (Discrete Series): In the case of discrete series, arithmetic mean of group data is calculated by the use of following formula:
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X=
Here N=
∑f .
∑ fX or ∑f
1 ∑ fX N
It is sum of all the frequencies.
Steps in Calculation: When a particular value occurs more than once, the arithmetic mean of group data is calculated: 1. By multiplying each value (X) with the corresponding frequency of its occurrence (obtaining fX of different observations:
f X + f X + f X + .........f n X n ). 1 1 2 2 3 3
2. Adding all the multiplication products to obtain 3. Divide this total value
∑ fx by total number of observations or total frequencies:
∑ fx ∑f Example
( ∑ fx ) .
or
∑ fx N
Find the mean form the following data:
Marks(X) :
5
10
15
20
25
30
35
40
No. of students(f):
5
7
9
10
8
6
3
2
Table 1 marks obtained by different number of student
Marks (X)
No.of Students (f )
fX
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5
5
25
10
7
70
15
9
135
20
10
200
25
8
200
30
6
180
35
3
105
40
2
80
∑ f = 50
∑ fX = 995
X= =
∑ fX ∑f 995 = 19.9 Ans. 50
2. Arithmetic Mean of Group Data (Continuous Series): When there is a continuous class distribution from
x 0 − x1 , x1 − x 2 , x 2 − x 3 ,....... with their corresponding frequencies as f1 , f 2 ,f 3 ..., the arithmetic
mean is calculated by using the formula: Formula
X=
∑ fm ∑f
Here, m = the mid value of various classes f = the frequency of the calss
∑f
= the total frequency
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Steps in Calculation: 1. Find the mid value of each class. 2. Multiply each mid value by respective frequency (f) to obtain value of fm. 3. Add all these fm values to obtain 4. Divide this value or
∑ fm
∑ fm
.
by the sum total of all frequencies, i.e., total number of
observations. Example
The Weight of 50 fishes of a species of fish are given in a frequency table. Calculate the
mean weight. Weight
110
1120
2130
3140
4150
No. of Fishes
3
11
7
4
15
Class
Mid value
Frequency
Interval
m
f
110
1120
2130
10 + 1 = 5.5 2 20 + 11 = 15.5 2
Multiplication of frequency and mid point f.m. 5.5X3=16.5
3 15.5X11=170.5 11
30 + 21 = 25.5 2
25.5X7=178.5 7
3140
4150
40 + 31 = 35.5 2 50 + 41 = 45.5 2
4
35.5X4=142.0
60 + 51 = 55.5 2
15
45.5X15=682.5
5160
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70 + 61 = 65.5 2
6170
0
0
7
65.5X7=458.5
3
75.5X3=226.5
∑ f = 50
∑ fm = 1875
80 + 71 = 75.5 2
7180
Fundamentals of Biostatistics
Calculation:
X=
∑ fm = 1875 = 37.5 ∑ f 50
Ans
1. Calculation of Mean by Assumed Mean Method Mean can be calculated by using another method in which an arbitrary reference point is taken. This arbitrary reference point is termed as Assumed Mean or Provisional Mean. This is the shortcut method which is applied when the frequencies and the values of variables are very large and it becomes very difficult to compute the arithmetic mean. The arithmetic mean is given by the formula: (a) In Case of Ungrouped Data:
X =A+
∑d N
where,
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X = Mean of observations A = Assumed Mean d = Deviation from the arbitrary reference point N = ∑ f = Total number of observations Example Find the mean by shortcut method using the following data: 59
65
71
67
61
63
69
73
Solution: Let us take 65 as assumed mean and prepare the following table: X 59
d =X65 5965=6
65
6565=0
71
7165=+6
67
6765=+2
61
6165=4
63
6365=2
69
6965=+4
73
7365=+8
N=8
∑d = 8
X =A+
∑ d = 65 + 8 = 66cm N
8
Ans
(b) In Case of Grouped Data:
X=A+
∑ fd N
where, f = frequency
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fd = product of the frequency and the corresponding deviation. (c) Data with Equal Class Intervals: In case of frequency table with equal class intervals, the arithmetic mean is given by the formula:
X=A+
∑ fd i N
where i = width of the class interval.
Example Find the mean from the following data using assumed mean method. Marks No. of Students
010
1020
2030
3040
4050
5060
42
44
58
35
26
15
Solution: Mid value Marks
d=
M − 35 10
No. of students (frequency)
fd
3
42
3X42=126
010
M 5
1020
15
2
44
2X44=88
2030
25
1
58
0X35=0
3040
35
0
35
1X26=26
4050
45
1
26
2X15=30
5060
55
2
15
Class interval=10
N=20
∑ fd = −216
Put the values in the following formula:
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X =A+
22
∑ fd i = 35 + −216 X10
N = 35 − 9.8 = 25.2
220 Ans.
2. GEOMETRIC MEAN (G.M) The geometric mean of a set of data for n observations is the xth root of their product. If,
x ,x ,x x 1 2 3,.............. n
are the given n observations then geometric mean is:
G.M. = n x .x .x x 1 2 3......... n 1/n = x x , x ,......x n 1, 2 3
(
)
If n=2,i.e., the number of observations are two only, the GM can be computed by taking the square root of their product. Only the geometric mean can be computed by taking the square root of their product. Say,
G.M. = 4 = 1.41 and of
64 = 8 th
But if n=more than 2, then the computation of the n root is difficult. In that case the calculations are made by making use of the logarithms:
1 log G.M = log(x x , x ,......x n ) 1, 2 3 n 1 = (log x ,log x , log x ...........log x n ) 1 2 3 n 1 = log ∑ f log n Merits of Geometric Mean: It has following merits: 1. It is based on all observations.
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2. Arithmetic mean has a bias for higher values whereas Geometric Mean has bias for smaller observations. 3. It is not affected much by fluctuations of sampling. 4. It is useful in averaging ratios, percentage rate of increase and decrease between two persons. Demerits of Geometric Mean: It has following demerits: 1. Geometric Mean is a mathematical character. It is not easy to understand or to calculate for nonmathematical persons. 2. If any observation
( x , x ) , is zero, Geometric Mean would be zero and if any one of the 1
2
observations is negative, the Geometric Mean becomes imaginary. Example: Find the average rate of increase in population which in the first decade had increased by 20% in the second decade by 30%
and in the third by 40%.
Solution. Here we have to determine the rate of increase in population. The appropriate average to be computed is G.M. and not the arithmetic mean. Hence, the average percentage rate of increase in the population per decade over the entire period=129.7100=29.7. Rate of Increase in
Population at the end of
Population
decade
1.
20%
120
2.0792
2.
30%
130
2.1139
3.
40%
140
2.1461
Decade
Log X
logX=6.3392
G.M.=Antilog
1 log X n
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=Antilog
3. HARMOMIC MEAN
24
6.3392 = ( 2.1131) = 129.7 3
(H.M or H)
Harmonic mean is the reciprocal of the arithmetic mean of the given observations:
(a)In Case of Ungrouped Data: If
x1, x 2 , x 3 ,......x n is a given set of n observations then their harmonic
mean is:
HM =
n 1
∑X
(b) In Case of Grouped Data: In case of frequency distribution the harmonic mean is given by the formula:
HM =
n f
∑X
where, f = total frequency x = value of variable Merits of Harmonic Mean: It has following merits: 1. It is based on all observations. 2. It is not affected much by fluctuations of sampling. 3. As reciprocal values are involved, it gives greater weightage to smaller observations. Demerits of Harmonic Mean: It has following demerits: 1. It is difficult to understand and calculate for biologists. 2. Its value cannot be obtained if any one of the observations is zero.
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Example The following table gives the weight of 31 persons in a sample enquiry. Calculate the mean weight using Geometric and Harmonic Means. Weight (X) No of persons (f):
130
135
140
145
146
148
149
150
157
3
4
6
6
3
5
2
1
1
Solution:
Weight (x)
No of persons
(f)log X
f log X
1/X
f/X
(f)
130
3
2.1139
6.3417
0.00769
0.02308
135
4
2.1303
8.5212
0.00741
0.02963
140
6
2.1461
12.8766
0.00714
0.04286
145
6
2.1614
12.9684
0.00690
0.04138
146
3
2.1644
6.4932
0.00685
0.02055
148
5
2.1703
10.8515
0.00676
0.03378
149
2
2.1732
4.3464
0.00671
0.01342
150
1
2.1761
2.1761
0.00667
0.00667
157
1
2.1959
2.1959
0.00637
0.00637
F=N=31
Log X=39.021
FlogX=66.7710
1/X=0.6250
âˆ‘ X = 0.2177
f
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1 1 f log X = Anti log X66.7710 n 31 = Anti log 2.15390
G.M. = Anti log
therefore, G.M = Anti log 2.15390 = 142.5
H.M. =
Ans
N 31 = = 142.39 âˆ‘ f / X 0.217
HereN = 31 and âˆ‘ f / X = 0.2177 Answer:
(1) Mean weight according to G.M. method=142.5 (2) Mean weight according to H.M. method=142.39
II. MEDIAN (MD) If the values of a variable are arranged in ascending or descending order of magnitude, the median is that value which divides the whole data into two equal parts, one part having all values smaller than the median value and other part having all the values greater than the median value. In other words, 50% of the observations will be smaller than the median while the other 50% will be larger than the median. It means the value of the middle observation or the mean value of two middle observations is called median.
As a matter of fact median is called the positional average.
Median is calculated
differently for ungrouped and grouped data. 1. Median of Ungrouped Data To calculate median of ungrouped data, the values of a variable (i.e., al the observations about the variable) are arranged in the order of magnitude either in ascending or descending order. The middlemost value in this arrangement represents the median. Example To calculate the median of following seven observations: 100
97
110
200
75
120
150
Solution: Step 1. The observations of the raw data are arranged in ascending order of magnitude in the sequence:
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75
97
100
110
27
120
150
200
Step 2. Total number of observations is found out. It is 7 in this case. Step 3. The middle most value in the above series is 110. This is the median. Therefore, the median of above series is 110. Calculation of Median: Median can be calculated in the following ways: 1. Calculation of Median when Number of Observations is Odd: When number of observations (n) is uneven the median is calculated by using the following formula: Median
Number of observations + 1 n + 1 = = 2 2
th
observations
In the above example: (a) the number of observations is 7
(b) Their median will be
7 +1 = 4 = 4 th position 2
(c) In the series fourth position is occupied by 110. (d) Therefore, median of the above series is 110. 2. Calculation of Median when Number of Observations (n) is Even: When number of observations is even, there is no unique median. The median in such cases is located half way between the two th
th
middle items. Therefore, the median is taken as the arithmetic mean of n and (n+1) observations. Example To calculate median of following 6
97
10
120
150
175
observations:75
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Step 1. Arithmetic mean of 6 observations is
=
n 6 = =3 2 2
= third observation Step 2. Arithmetic mean of 6 observations +1 =
n 6 +1 = +1 = 4 2 2
= Fourth observation rd
th
Step 3. Arithmetic mean of 3 +4 observations=Median of 6 observations rd
(a) 3 observation =100 th
(b) 4 observation =120
Arithmetic mean
Answer.
=
100 + 120 220 = = 110 = Median 2 2
Median=110
Example The number of patients that visited a doctor for consultation for 10 consecutive days is arranged in an increasing order in the following table. Find out the median number of the patients that visited the doctor per day. 8
10
12
14
16
18
19
20
22
25
Solution: Step 1. The number of observations (i.e. the number of days) n=10 Step 2. Arithmetic mean of 10 observations
10 =5 2 = 5th observation =
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Step 3. Arithmetic mean of 10 observations+1
10 +1 = 6 2 = 6th observation =
Step 4. Median for above 10 observations=Arithmetic mean of 5th value and of 6th value. (a) 5th observations=16 th
(b) 6 observations=18
16 + 18 34 = = 17 = Median. 2 2 Median = 17 Ans. A.M. =
3. Calculation of Median for Simple Frequency Distribution: For calculation of median for frequency distribution data, cumulative frequency corresponding each value of variable is calculated.The value of the variable corresponding to the cumulative frequency of
N +1 is called median, where N is the total 2
frequency. Example Calculate the median value of the following variables on the basis of following simple frequency distribution. Varaible (X) Frequency (f)
1
2
3
4
5
6
7
1
4
12
9
2
1
1
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Solution: Variable (X1)
Frequency (f)
Cummulative Frequency (Cf)
1
1
1=(0+1)
2
4
5=(4+1)
3
12
17=(12+5)
4
9
26=(9+17)
5
2
28=(2+26)
6
1
29=(1+28)
7
1
30=(1+29)
where total number of frequencies (N)=30
Median value of variable
=
N + 1 30 + 1 = = 15.5 2 2
2. Median of Grouped Data The median of grouped data is calculated by the following formula:
∑f 2 − F Median = L + i
1
fm
Here: L1 = The lower limit of that class interval where median falls. N or
∑f
= Total number of frequencies
f = Frequency
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F = The cumulative frequency just above that class interval where median fall fm = The frequency of that class interval where median falls. i = The width of the class interval. Median has wider application in behavioural biology and environmental pollution, etc Example Compute the median from the data given below: Cumulative Class Interval
Class Boundary
Frequency Frequency
59
4.59.5
2
2
1014
9.514.5
11
2+11=13
1519
14.519.5
26
13+26=39
2024
19.524.5
17
39+17=56
2529
24.529.5
8
56+8=64
3034
29.534.5
6
64+5=70
3539
34.539.5
3
70+3=73
4044
39.544.5
2
73+2=75
4549
44.549.5
1
75+1=76 Cumulative
Total number of frequencies =76
Frequency =76
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Solution:
Here Median =
∑ f = = 76 = 38 2
2
th
The class interval in which 38 (in cumulative frequency ) falls is 1519. see table given above. The exact limits or class boundary of this interval are 14.5 – 19.5. the lower limit of this class interval (L1): L1=14.5 f =13, fm = 13 × 2=26, and I =5
∑f 2 − F ×i Median = L +
1
fm 38 − 13 25 = 14.5 + × 5 = 14.5 + × 5 = 19.31 26 26
The median of the above data is 19.31 Ans. Merits of Median 1. It is rigidly defined. 2. Median is easy to understand and easy to calculate. 3. Median is not affected by extreme observataions and is very useful in the case of skewed distribution. 4. Median can be computed while dealing with a distribution with open end class. 5. Median is best for qualitative data. 6. Median can be calculated even when the extreme values of a variable are not known. Demerits of Median The demerits of median are: 1.Median cannot be determined in the case of even number of observations. We merely estimate it as the arthimetic mean of the two middle terms. 2. Median is relatively less stable than means, particularly for small samples since it is affected more by fluctations of sampling as compared with arithmetic mean.
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3. Median is a positional average. It cannot be accepted for each and every observation. 4. It cannot be subjected to algebraic treatement.
Table Formulae of calculation of median at a glance Types of Series 1. Ungrouped data (individual Series)
Method of Calculation of Median 1. Median (in case of odd numbers of
N + 1 observations) 2 th
=
2. Median ( in case = Average of the even number
of
values
observations
dof
on
two
either
items
lying
idea
of
N +1 observations 2 2. Grouped data
1. Discrete Series
N + 1 M= Size of observations 2 th
Here N=
∑f
(sum of total frequencies)
2.Continuous Series.
∑f − F × i 2 M = L1 + fm III. MODE Mode is the most frequently occurring value in a data. It means for a given data, mode may or may not exist. For example, let’s observe mode for the following 3 sets of data: (a)
10,10,9,8,5,4,12,10 : One mode, i.e., 10
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(b)
10,10,9,9,12,15,5
: Two modes i.e., 10 and 9
(c)
4,6,7,15,12,12,10
: No mode
We note that first set of data (a) has single mode 10, the second set of data (b) has two modes 10 and 9 and the third set of data (c) has no mode. Set (b) has 10 and 9 as modes because they both occur 2 times and they occur more often than other values. DEFINITION OF MODE AND MODAL CLASS According to A.U. Tuttle, â€˜Mode is that value in a sample or data which has the greatest or largest frequency density in the frequency tableâ€™. The class having greatest frequency is called modal class. The modal class can be determined by inspection but the actual value of mode will lie in the class interval and may not be at the mid point of the class. CALCULTION OF MODE A. Calculation of Mode of Dividual Series or Ungrouped Data It can be computed either by inspection or by frequency distribution. 1. Calculation of Mode by Inspection: In this method the data is arranged in increasing order. It is then observed that how many times each values in the data is repeated. The item or value which occurs most frequently represents the mode of that data. Example Calculate the mode for the following data: Variable X
32
22
29
25
17
25
40
Solution: Step: 1 Arrange the data in increasing order, i.e. Variable X: 17 22 25 25 29 32 40
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Step: 2 Value 25 of X in this series has occurred twice while all others are represented just once. Therefore, mode of this data is 25. 2. Calculation of Mode by Frequency Distribution: When the number of items in a series is very large, individual items are converted into frequency distribution. The mode is then calculated as the value corresponding to the highest frequency. Example Calculate mode on the basis of simple frequency distribution of a variable: Variable (X) 1
2
3
4
5
6
7
1
4
12
9
2
1
1
(Number of flowers) Frequency (f)(Plants)
Solution: 1.
In this example, the number of flowers on 30 plants varies from 1 to 7. This is called variable.
2.
Plants bearing 3 flowers are maximum, i.e. 12 in number.
This represents their
frequency. 3.
3 is the value with highest frequency of 12. Therefore, Mode = 3 flowers per plant.
3. Calculation of Mode of Discrete Series: In discrete series, mode can be determined by inspection method or by grouping method. The inspection method is similar to the one discussed in ungrouped data.
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4. Calculation of Mode of Discrete Series by Grouping Method This method is used in cases where there is regularly; and homogeneity in the series. Inspection method is usually not reliable and grouping method is applied. It involves preparation of grouping table as follows: Example From the given data, calculate the value of mode by grouping method Marks (X)
1
2
3
4
5
6
7
8
9
10
11
12
4
7
8
12
16
14
9
7
17
5
3
2
No.of Students (f) 1.
In the column I of the grouping table, the values of variable are arranged in the ascending order
2.
In column 2, corresponding frequencies are written.
3.
The frequencies are grouped in twoâ€™s beginning with the first value.
4.
In column 3, the frequencies are grouped in twoâ€™s beginning with second value of the series.
5.
In column 4, the frequencies are grouped by three values (i.e. 1,2 and 3) starting with the first value. Write them as shown in the table.
6.
In column 5, the frequencies are grouped into three values beginning with the third value.
7.
Underline the maximum grouped frequency in each column. Grouping table is prepared from the above data as shown in the grouping table. Grouping Table
Size of
Frequncies
(f)
3
4
item (Marks)
1
2
5
6
X
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1
4 7
2
3
4
5
6
8 12 16 14 9 7
17 5
7
8
3 2
4 + 7 = 11 8 + 12 = 20 16 + 14 = 30 9 + 7 = 16 17 + 5 = 22 3 + 2 = 5
37
7 + 8 = 15 12 + 16 = 28 4 + 7 + 8 = 19 12 + 16 + 14 = 42 7 + 8 + 12 = 27 14 + 9 = 23 8+12+16= 7 + 17 = 24 9 + 7 + 17 = 35 16 + 14 + 9 = 39 36 5 + 3 = 8 5 + 3 + 2 = 10 7 + 17 + 5 = 29 14+9+7=30
9
10 17+5+3=25
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11
12
8. Prepare the analysis table showing the items corresponding to the maximum 9 + 7 = 1 frequencies in different columns of the grouping table. In analysis table column numbers 1 to 6 showing frequencies are placed on the left hand side and sizes of the items (1to 12) on the top. The maximum frequencies are marketed by √ mark. Analysis Table Column
Size of items showing Maximum fre quency
No. 1
2
3
4
5
6
7
8
10
11
12



√
1 √
2
√
3
√
√
4
√
√
√
√
√
√
3
1
5 6 Total
9


√
√
√
1
3
5

1
9.From the analysis table we find out the variable which has maximum frequency of distribution. Result: From the analysis table it is evident that value 5 or item 5 has maximum frequency.
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3.Calculation of Mode in a Continuous Series Determination of mode in case of continuous frequency distribution is complicated and involves following two steps: 1.
First of all, the modal class is ascertained either by inspection method or by grouping method.
2.
After determining the modal class, the exact value of mode is calculated by using the following formula:
fm − f1 Mode (Z) = L1 + × c 2fm − (f1 − f 2 ) Where L1= Lower limit (boundary) of modal class fm = Frequency of modal class or the maximum frequency f1 = frequency of class just preceding the modal class f2 = frequency of class just succeeding the modal class C= class interval or class width of the classes. Example In a class following is the distribution of marks of 85 students. Calculate the modal class and mode of the following data: Marks (Grouped
2025
2530
3035
3540
4045
4550
5055
5560
5
7
8
18
25
12
7
5
data) No. of Students(f) Solution: Modal class by inspection is 4045
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∴ fm = 25
40
f 2 = 12 Lower lim it of mod al class L = 40
f1 = 18 Class int erval
c=5
Putting these values in the formula fm − f1 Mode (Z) = L1 + ×c 2fm − (f1 − f 2 ) = 40 +
25 − 18 ×5 (2 × 25) − (18 − 12)
= 40 +
7 ×5 50 − 30
= 40 +
7 7 160 + 7 × 5 = 40 + = 20 4 4
=
167 = 41.75 4
Modal class has value = 41.75 Ans Merits of Mode Mode has following merits: 1. Mode is easy to calculate and understand 2. It is not affected by extreme observations and as such is preferred to arithmetic mean value dealing with extreme observations. 3. Mode can be calculated from a grouped frequency distribution with openend classes. Demerits of Mode: The demerits of mode are: 1. Mode is not rigidly defined. It is ill defined if the maximum frequency is repeated or occurs in the very beginning or at the end of the distribution. 2. As compared to mean, mode is affected to a great extent by the fluctations of sampling. 3. It is not suitable for algebraic treatment.
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UNITII CHI SQUARE TEST The
χ2
test is one of the simplest and most widely used nonparametric tests in statistical work.
The symbol
χ2
is the Greek letter chi. The
χ2
quantity
χ 2 test
was first used by Karl Pearson in the year 1890. The
describes the magnitude of the discrepancy between theory and observation. It is defined as,
( Oi − Ei )2 χ = ∑ with (n1) d.f E i =1 i n
2
where, Oi refers to the observed frequencies and Ei refers to the expected frequencies. While comparing the calculated value of
χ2
with the table value we have to determine the
degrees of freedom. By degrees of freedom we mean the number of classes to which the values can be assigned arbitrarily. Note1 The sum of the Observed and Expected frequencies is always zero. Note2
χ2
distribution is a limiting approximation of the multinominal distribution.
Conditions for applying test 1. The total frequency should be sufficiently greater than 50 2. The constraints must be linear. 3. The sample observations should be independent, in the sense no individual item should be included twice or more in the same sample. An increasing use of nonparametric tests in economic and business is an account of the three reasons. 1. The nonParametric tests are distributionfree. 2. They are computationally easier to handle and understand than parametric tests. 3. They can use with the types of measurement that prohibit the use of parametric tests. Uses of
χ 2 test With the help of
χ2
test we can find out whether two or more attributes are associated or not.
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Example The number of parts for a particular spare part in a factory was found to vary from day to day. In a sample study the following information was obtained. Test the hypothesis that the number of parts demanded does not depend on the day of the week. Days
Mon
Tue
Wed
Thur
Fri
Sat
No.of Parts
1124
1125
1110
1120
1126
1115
demanded H0: Let us take the null hypothesis that the number of parts demanded does not depend on the day o the week. Oi
Ei
(OiEi)
(OiEi)
2
( Oi − Ei )
2
Ei
1124
1120
4
16
0.014
1125
1120
5
25
0.022
1110
1120
10
100
0.089
1120
1120
0
0
0
1126
1120
6
36
0.032
1115
1120
5
25
0.022
0.179 Number of parts demanded for six days=6720
∴ parts demanded for 1 day =
6720 = 1120 6
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( Oi − Ei ) 2 = 0.179 Ei
χ2 = ∑ d.f. =n1=61=5 The table value of i.e.
2 χ 0.05
χ2
for d.f. at 5 level of significance in 11.07.
for 5 d.f. =11.07
Calculated value of
χ2
is less than the table value. It is not significant. Hence, we accept the null
hypothesis H0. Conclusion We conclude that the number of parts demanded does not depend on the day of the week Example: A survey of 800 families with 4 children each revealed the following distribution: No. of boys
0
1
2
3
4
No. of girls
4
3
2
1
0
No. of families
32
178
290
236
64
In this result consistent with the hypothesis that the male and female births are equally probable? H0: We set up the null hypothesis that the male and female births are equally probable
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No. of boys
44
Oi
Ei
(OiEi)
2
( Oi − Ei )
2
Ei 0
32
1
178
800 4C0 × (1/ 2 )
0
800 4C1 × (1/ 2 )
1
2
290
3
236
4
64
800 4C2 × (1/ 2 )
× (1/ 2 )
2
800 4C3 × (1/ 2 )
3
800 4C4 × (1/ 2 )
× (1/ 2 )
4
4 −0
4 −1
× (1/ 2 )
× (1/ 2 )
= 200
4−2
4 −3
× (1/ 2 )
= 50
4− 4
= 300
324
6.480
484
2.420
100
.333
1296
6.480
196
3.920
= 200 = 50 19.633
Since Male and Female are equally probable,
p = 1/ 2; q = 1/ 2 Q p + q = 1 P(r success)=
nCr p r q n − r
Calculate Value of
χ 2 =19.633
d.f.=51=4 Table value of
χ2
for 4 d.f. at 5% level of significance=9.488
Since calculated value of
χ2
is greater than the table values, it is significant. Hence, we reject
the Null hypothesis.
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45
Conclusion: We conclude that the male and female births are not equally probable.
Test of Independence of Attributes An information is said to be attribute when it is quantitative. Te chisquare helps to find out whether two or more attributes are associated or not. Example: Examples of attributes are beauty, colour, failure etc The expected frequency for an attribute for each of the cell in contingency table is,
Determined as,
coresponding rowtotal × Column total n
Contingency table A table having ‘r’ rows and ‘c’ colums is known as contingency table. Each row corresponds to a level of one variable, each column to a level of another variable. Entries in the cell are frequencies with which variable combination occur. A table having 2 rows at 2 colums in called 2 × 2 contingency table. Example: The following results were obtained when two sets of items were subjected to two different treatments X and Y, to enhance their tensile strength. Treatment X was applied on 400 items and 80 were found to have gained in strength. Treatment Y was applied to 400 items and 20 were found to have gained in strength. Is treatment Y superior to treatment X? The following results were obtained when two sets of items were subjected to two different treatments X and Y, to enhance their tensile strength. Treatment X was applied on 400 items and 80 were found to have gained in strength. Treatment Y was applied to 400 items and 20 were found to have gained in strength. Is treatment Y superior to treatment X? H0: We set up the null hypothesis that there is no difference in the two types of treatments X and Y.
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Treatment
Gained
Not gained
Total
X
80
320
400
Y
20
380
400
100
700
800
2
2
Oi
Ei
(OiEi)
80
50
900
18.000
20
50
900
18.000
320
350
900
2.571
380
350
900
2.571
(OiEi) /Ei
χ 2 =41.42 400 ×100 = 50 800 400 ×100 E (20) = = 50 800 400 × 700 E (320) = = 350 800 400 × 700 E ( 380 ) = = 350 800 E ( 80 ) =
( Oi − Ei )2 Calculated value of χ = Σ = 41.42 Ei 2
d . f . = ( r − 1)( C − 1) = ( 2 − 1)( 2 − 1) = 1 Table value of
χ2
is greater than the table value. It is significant. Hence, we reject the Null
hypothesis H0.
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Conclusion: We conclude that there is difference in treatment X and Y.
CORRELATION Definition ‘Correlation’ is the tendency of simultaneous variation between two variables’. According to Connor ‘if two or more quantities vary in sympathy and movements in one tend to be accompanied by corresponding movements in the other, then these two quantities are said to be correlated. Explanation 1. Correlation indicates the degree of relationship between two variables. 2. The movements in one variable are accompanied by corresponding movements in the other variable. 3. According to Tuttle, correlation is an analysis of covariation between two or more variables. Significance of Correlation The study of correlation is of great significance in practical life, because of the following reasons: 1. The study of correlation enables us to know the nature, direction and degree of relationship between two or more variables. 2. Correlation studies help us to estimate the changes in the value of one variable as a result of change in the value of related variable. This is called regression analysis. 3. Correlation analysis helps us in understanding the behavior of certain events under specific circumstances. For example, we can identify the factors for rainfall in a given area and how these factors influence paddy production. 4. Correlation facilitates the decision making in the business world. It reduces element of uncertainty in decisionmaking. 5. It helps in making predictions.
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Bivariate and Multivariate Distributions 1. Bivariate Distribution: The distribution in which each unit of the series assumes two values is called bivariate distribution. In a bivariate distribution two variables are said to be correlated. The change in o0ne variable results in corresponding change in the other variable. 2. Multivariate Distribution: If more than two variables of each unit of a distribution are measured, it is called a multivariate distribution. Two variables can be measured in the same individual, e.g., length and weight, oxygen consumption and body weight, etc., or same variable can be measured in two or more related groups such as intelligence quotient in siblings, height in parents and offsprings. Types of Correlation Depending on it s extent and direction the correlation between two variables may be of following types: 1. Positive and Negative Correlations The positive and negative correlations are based on the direction of change in the value of two variables: 1. Perfect Positive Correlation: When two variables move proportionately in the same direction. i.e., the increase in the values of one variable leads to corresponding increase in the values of other variable, the correlation between them is called perfect positive. For example, increase in biody weight with the increase in height presents positive correlation. It is also called direct correlation. Examples of perfect positive correlation are very rare in nature but some examples are: 1. Correlation in day length and temperature, and 2. Correlation in rain and humidity, etc. Following data presents perfect positive correlation:
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Table1: Data showing perfect positive correlation between two variables X and Y Series X
Y
10
20
15
30
20
40
25
50
30
60
Series X
Y
80
40
75
35
70
30
65
25
60
20
2. Moderately Positive Correlation: when two variables are partially positively correlated, the correlation is termed moderately positive correlation, e.g., tallness of plants and the quantity of manure used, nutrition and death rate in pregnancy, and the morality rate and overcrowding, etc. 3. Perfect Negative Correlation: The two variable show negative correlation when one variable increases with a constant interval and another decreases with constant interval. Thus, variables deviate in opposite directions. This is also called inverse correlation. Examples of perfect negative correlation are very rare In nature but some approaching to that extent are temperature and lipid content of the body, number of red blood corpuscles, Hb percentage, etc.
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Plotting of Scatter Diagram Example 1: The following data provides in inches the height of fathers and their eledest sons.
Height of fathers Height of their sons
1
2
3
4
5
6
7
8
65
63
67
64
68
70
71
69
67
66
68
65
69
68
70
68
[ Construct a scatter diagram and indicate the nature of correlation. Solution: For scatter diagram, a graph is plotted on a graph paper. Step 1: Height of fathers is plotted on Xaxis and height of their sons is plotted on the Yaxis. Step2: various values of heights are plotted by dots as shown in the figure below. Conclusion: The scatter diagram shows that there is positive correlation between heights of fathers and sons, i.e., with increase in the height of fathers, the height of sons also increases. Merits of Scatter Diagram The merits of scatter diagram are: 1. Simple Method: this is a simple and nonmathematical method for studying correlation 2. Easy to Understand: It is easy to understand and easy to interpret. It provides quick idea of correlation between two variables just by a glance on the diagram. 3. First Step: It can be regarded as the firs step in studying relation between two variables. 4. Unifluenced: The results are not influenced by the size of extreme observations. Demerits of Scatter diagram The demerits of scatter diagram are: 1. Scatter diagram gives just an idea about the direction of correlation. It does not establish the exact degree of correlation between two variables. 2. It is just a qualitative method of showing relationship between two variables.
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3. This method is suitable for small number of observations. When number of observations is very large, the method becomes tedious and complicated. 2. Karl Pearson’s Correlation Coefficient A scattered diagram, like a histogram, is a convenient way of displaying existence of correlation and direction of correlation, but it does not give any correlation value. To measure correlation, the coefficient of correlation is worked out by Karl Pearson’s Coefficient of Correlation. Coefficient of correlation is the degree to which two variables are interrelated. It is a mathematical method for measuring the tendency of linear relationship between two variables. This was introduced by Karl Pearson(18671963). This measure of correlation is also known as Pearsonian Correlation Coefficient. If two variables are denoted by X and Y, the coefficient of correlation between them is represented by rxy or r. Pearson’s coefficient correlation method can be used to measure correlation for individual series as well as for grouped data. Properties of Coefficient of Correlation 1. Coefficient of correlation is a measure of closeness between two variables. 2. The correlation may be positive or negative. 3. The range of correlation coefficient is from 1 to +1. 4. If r=+1, the correlation between two variables is perfect and positive. 5. If r=1, the correlation is perfect and negative. 6. If there is a strong positive linear relationship between two variables, the value of r will be close to +1. 7. If there is strong negative linear relationship between the variables, the value of r will be close to 1. 8. If r=0, there is no correlation between two variables. It means variables are independent. The value of r is calculated by using following formula:
r=
__ __ Σ x − x y − y __ 2
__ 2
Σ x − x Σ y − y
__
or
Σx − x Σy 2 __ 2 __ Σx − x Σx Σy − y Σy
Computation of Coefficient of Correlation from Ungrouped Data Coefficient of correlation for ungrouped data is calculated by the following formula:
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r=
__ __ Σ x − x y − y 2
__ __ Σ x − x Σ y − y
2
When coefficient of correlation’s value is +1, it means there is perfect and positive correlation, i.e., the two variables, i.e., temperature and pulse rate change in the same direction. 1. Direct Method of Computation of Coefficient of Correlation __
If x and y are two variates and their arithmetic means are
__
x and y respectively, then
Σdx ×dy Σd 2x Σd 2y __ Here,dx = x  x x = value of X variate __ __ dy = y  y x = arithmetic means of x variate 2 __ 2 dx = x  x y = value of Y variate
r or P(x, y) =
__ __ dy2 = y  y y = arithmetic mean of Y variate
The above formula can also be written as:
r(x, y) =
Sdx.dy nsx×sy
n is the number of observations in X and Y series.
σx,σy
are standard deviations of series X and series Y respectively
Working Rule: Coefficient of correlation can be calculated in the following steps: Step1: The two series are represented by X and Y.
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__
__
Step 2: Arithmetic mean of X and y series, i.e
x and y is calculated. __
Step 3: Deviation of observation (dx) in X series is calculated by the formula dx = x − x
__
Similarly, deviation of observation in Y series is calculated by the formula
dy = y − y . Place
these values in their respective columns as shown in the table:
DerivationDeviation square Series S.No
From
Series
mean
__ dx = x − x 2
X __
Deviation
Deviation
from Mean
Square
2
Y
Product Deviation
__ 2
dy = y − y
dy 2 = y − y
dx.dy
Σdy
Σdy2
Σdx.dy
__
dx = x − x 1 2 2
Σdx
total
Σdx 2
2
Step 4: The values dx and dy are squared and written in respective columns headed as dx and 2
dy . Step 5: In the next column the value obtained by the multiplication of 2
dx × dy is entered.
2
Step 6: Totals of all dx, dx , dy, dy and dxdy are worked out and denoted below their respective columns. These are represented as
Σdx, Σdx 2 , Σdy, Σdy 2 and Σdx.dy.
Step 7: These values are placed in the formula and coefficient of correlation r or rxy is calculated as follows:
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Example: Calculate Karl Pearson’s correlation between X and y series from data given below: X series
12
9
8
10
11
13
7
Y series
14
8
6
9
11
12
3
[ Solution:
Step1: Mean of X series=
12 + 9 + 8 + 10 + 11 + 13 + 7 = 10 7
Step2: Mean of y series =
14 + 8 + 6 + 9 + 11 + 12 + 3 =9 7
Calculation of coefficient of correlation Derivation from Mean S.No
Xseries
Derivation Square
__ dx = x − x
dx 2
Yseries
Derivation
Derivation
Production
from Mean
Square
Deviation
dy = y − y
dy 2
dx.dy
__
× 5=10
1
12
1210=+2
4
14
149=+5
25
2
2
9
910=1
1
8
89=1
1
11
3
8
810=2
4
6
69=3
9
2v3=6
4
10
1010=0
0
9
99=0
0
0
5
11
1110=+1
1
11
119=0
4
1
× 2=2
6
13
1310=+3
9
12
129=3
9
3
× 3=9
7
7
710=3
9
3
39=6
36
3v6=18
Σx = 70
0
Σdx 2 = 28
Σy = 63
0
Σdy 2 = 84
Σdx.dy = 46
× 1=1
Mean=10
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From the above Table:
1.N = 7,
__ ΣX 70 2.MeanX X = = =10 N 7 __ ΣY 63 MeanY Y = = =9 N 7 2 2 3. Σdx = 28, Σdy =84 4. Σdx × Σdy = 46 By putting these values in the following formula:
r=
Σdx.dy 46 46 = = 28 × 84 48.5 Σdx 2 ×Σdy2
r = +0.948 Computation of Rank Correlation Coefficient by Spearman’s Method The value of rank correlation coefficient ® ranges between +1 and 1. It can be calculated in three different situations: 1. When actual ranks are given 2. When actual ranks are not given. 3. When ranks are equal. 1. When Actual ranks are Given When data are given along with the ranks of different items in the two series, the computation of rank correlation is simple and involves following steps:
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Step 1: Prepare a table to show coefficient of rank correlation in the following manner:
st
Rank in 1 series
Rank in 2
R1
nd
series
Difference between
Square of Rank
Ranks
Difference
R1R2=D
D
R2
2
Step 2: Enter ranks R1 and R2 of every item in the respective columns in thetable. Step 3: Calculate the difference of two ranks (i.e. R1R2) and denoted it by d and enter it in column 3 in the table. Step 4: Compute the square of D and finally total them to get ΣD . 2
Step 5: Apply the formula discussed above to get the correlation coefficient.
r = 1−
Example
6ΣD 2 n n2 −1
(
)
Ten competitors in a dance competition were ranked by two judges in the following
order : st
Judges
1
Rank by First Judge
1 4
Rank by Second Judge
nd
rd
2
th
3
4
4
8
9
8
7
5
5
th
th
th
6
7
6
10
7
9
6
10
8
th
th
9
10
3
2
5
2
3
1
Calculate coefficient of Rank Correlation. Solution : Prepare the table for Rank Correlation : Difference in Ranks
Rank given by First
Rank give n by second
Square of Rank
Judge R1
Judge R2
1
4
3
9
4
8
4
16
8
7
1
1
9
5
4
16
2
D = R1 – R2
Difference D
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6
9
3
9
10
6
4
16
7
10
3
9
3
2
1
1
2
3
1
1
5
1
4
16
N = 10
N = 10
∑D = 0
∑D
2
= 94
[[ 1. Number of Pairs N = 10 2.
∑D
2
= 94
Putting these values in the formula :
r = 1−
6∑ D 2
= 1−
6 × 94 103 − 10
n(n − 1) 564 = 1− = 0.43 1000 − 10 r = 0.43 Ans. 2
From the above results, it can be concluded that the judgments of two judges are of similar type because there is positive coefficient of rank correlation between two judgments. 2. When Ranks are not Given Spearman’s Rank Correlation method can be used for calculating coefficient of correlation of quantitative data when ranks are not given. In such cases actual data is converted into ranks. It means ranks are assigned by arranging the data in descending order or in ascending order. In descending order the highest value is given rank 1 and in ascending order the lowest value is ranked I. After determining 2
the ranks in both series of data, D and D are calculated and values are placed in the equation used in first case.
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Example Calculate coefficient of ranks correlation R from the following data showing marks obtained by students in mathematics and physics : Mathematics (X)
8
36
98
25
75
82
92
62
65
35
Physics (Y)
84
51
91
60
68
62
86
58
35
49
Solution : Table for Calculation of Rank Correlation 2
X
Y
x1 = Rank in X
Y1=Rank in Y
D=x1y1
D
8
84
10
3
7
49
36
51
7
8
1
1
98
91
1
1
0
0
25
60
9
6
3
9
75
68
4
4
0
0
82
62
3
5
2
4
92
86
2
2
0
0
62
58
6
7
1
1
65
35
5
10
5
25
35
49
8
9
1
1
N=10
N=10
∑ D =0
∑D
2
[[ From the above table 2
2
n = 10 and n = 10 = 100
∑D
2
= 90
Putting the values in the formula :
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6∑ D 2 6 × 90 = 1− 2 n(n − 1) 10(100 − 1) 540 54 99 − 54 45 = 1− = 1− = = = 0.455 990 99 99 99 r = 0.455 Ans.
r = 1−
3. When Ranks are Equal Sometimes two or more items in one series of data may have equal values. In such a situation, the items with equal value cannot occupy different positions in the series and also they cannot have the same rank. Therefore, a common rank is given to the items having same value. This common rank is the average of the ranks which these items would have occupied in the series, had they differed slightly from each other. For example, two items are ranked equal at fifth rank because of having equal value. Had th
th
they differed slightly, they would have been ranked as 5 and 6 . Therefore, these two items are given a common number which is an average of the two different ranks they were expected to enjoy. Their common rank will be
5+6 = 5.5 . 2
Similarly if three items have equal values, they will have the same common rank. It is calculated by adding their expected ranks and drawing out the average, i.e.
adjustment is done by adding
1 (m3 − m) 12
to the value of
∑D
5 + 6 + 7 18 = = 6. 3 3 2
To simplify, the
, where m stands for the number of
items with the same rank. This value is added as many times as is the number of items having equal values. For example, if two items have same value, the adjustment
1 (m3 − m) 12
is added two times to
6∑ D 2 . The formula for calculation of ranks correlation coefficient in a series where two items have same value becomes :
6 ∑ D2 +
r = 1−
1 1 (m3 − m) + (m3 − m) 12 12 2 n(n − 1)
Example
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Calculate the coefficient of correlation from the following data by Spearman’s Rank Correlation method.
Series X
20
11
24
18
20
22
Series Y
24
9
20
22
9
11
[ Note : In the above data two items in series X have same value 20 and in Y series two items have same value 9. These are represented in box. Step 1. When data in X series is arranged in descending order 24, 22, 20, 20, 18, 11 the two rd
th
items with value 20 come to lie at 3 and 4 places, i.e., their common rank will be
3+ 4 = 3.5 . 2
Step 2. When data in Y series is arranged in descending order 24, 22, 20, 11, 9, 9, the two items th
th
with value 9 come to lie at 5 and 6 ranks. Therefore, their common rank will be
5+6 = 5.5 . 2
Step 3. Total number of items in X and Y series is 6 each, i.e., n = 6. Step 4. Computation of rank correlation :
Series
Rank
Series
Rank
X
R1
Y
R2
Difference in
Square of
Ranks
Rank Difference =
R1 – R2 = D
D2.
20
3.5
24
1
3.51=2.5
6.25
11
6
9
5.5
65.5=0.5
.25
24
1
20
4
14=3.0
9.00
18
5
22
2
52=3.0
9.00
20
3.5
9
5.5
3.55.5=2.0
4.00
22
2
2
3
23=1.0
1.00
n=6
∑D
2
= 29.50
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Note : Items with same rank are represented in box. Step 5. Putting the above values in the formula :
r = 1−
6 ∑ D2 +
1 1 (m3 − m) + (m3 − m) 12 12 2 n(n − 1)
= 1−
1 3 1 (2 − 2) + (23 − 2) 12 12 6(6 × 6 − 1) 1 1 6 29.50 + (6) + (6) 12 12 = 1− 216 − 6 1 1 6 29.50 + + 6 × 30.50 2 2 = 1− = 1− 210 210 183 210 − 183 27 = 1− = = 210 210 210 = 0.128 r = 0.128 Ans. 6 29.50 +
Thus, rank correlation is positive and equal to 0.128. Merits of Rank Correlation Rank correlation method of computation of correlation coefficient has following advantages : 1. Rank correlation method is simple to understand and easy to apply. 2. Information given in the form of ranks is conveniently followed than in the form of quantitative data. 3. Rank correlation method is ideal for presenting qualitative data and ascertaining correlation. 4. Rank correlation method can also be used at times in cases of actual quantitative data. Demerits of Rank Correlation The main demerits of rank correlation are :
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1. Rank correlation method cannot be used for computing correlation in case of grouped frequency distribution. 2. This method can be used only for small number of observations. When number of observations is large, the computation becomes tedious and time consuming. REGRESSION In analyzing data, we find that it is frequently desirable to learn something about the relationship between two variables. For example, we may be interested in studying the relationship between blood pressure and age, height and weight, the concentration of an injected drug and heart rate, the consumption level of some nutrient and weight gain, the intensity of a stimulus and reaction time or total family income and medical care expenditures. The nature of relationship between variables such as these may be examined by regression analysis. OBJECTIVES OF REGRESSION ANALYSIS 1. The regression analysis is used to predict the value of one character or variable from the value of the other character or variable. According to this, the variables may be : (a) Dependent Variable : The variable whose value is influenced or is to be predicted, is called a dependent variable. (b) Independent variable : The variable which influences the values is called an independent variable. 2. The regression analysis is used to find out the measures of error present during the sue of regression line for prediction. For this, standard error of estimate is calculated. 3. From the value of coefficient of correlation, one can find the degree of association between two variables, but from the regression analysis one can predict how a change in one variable is expected to affect the other. Types of Regression Analysis The regression analysis can be of two types : simple and multiple. 1. Simple Regression: The regression analysis confined to the study of only two variables at a time is termed as simple regression. 2. Multiple Regression: The regression analysis for studying more than two variables at a time is known as multiple regression.
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Regression Equation The equation that describes position of any line on a graph is called regression equation. For a linear regression, the equation for a dependent variable XY against independent variable x can be given as follows : y=a+bX Here, values of ‘a’ and ‘b’ are constant and are fixed for a particular line. If the values of ‘a’ and ‘b’ are known, y can be obtained for any corresponding value of x. the values of ‘a’ and ‘b’ are calculated by the following equation :
∑ (X − X)(y − y) = r (SD of y) (SD of x) ∑ (X − X) ∑ xy − X∑ y or b = ∑ X − X∑ X b=
2
or = r
σx σy
2
After obtaining the value of ‘b’ the value of ‘a’ can be calculated. The constant ‘a’ is known as Intercept, and denotes the value of y when the value of x is zero. The constant ‘b’ measures the stops of the line and is called “regression coefficient”. The constant ‘b’ gives an idea of that how changes occurs in variables y when the variable X varies by 1 unit. For instance, if the value of ‘b’ is 5.8, then change in X by one unit will bring out a change in y by 5.8 units. The positive value of ‘b’ indicates the increase in the value of y. It is associated with the increase of X while a negative value will tell the decrease in y with an increase in X. Procedure 1.
Plot a graph between two variables taking independent variable on Xaxis and dependent variable on Yaxis. Find out the values of a and b using the equations given earlier. For drawing the line of best fit (regression line), find out any two values of y associated with corresponding values of x by using the equation : y = a + bx
2. Plot these two values on the graph on which all the points of original values have been put. 3. Make a straight line intersecting through these two points to get the fittest regression line.
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Example Find out regression equation from the following data for 7 fishes of a species : X
13.4
15.1
15.3
16.8
17.5
19.2
21.2
Y
2.1
2.3
2.3
2.6
2.7
3.0
3.3
Solution : The following table is prepared first : 2
X
Y
X.Y
X
13.4
2.1
28.14
179.56
15.1
2.3
24.73
228.01
15.3
2.3
35.19
234.09
16.8
2.6
43.68
282.24
17.5
2.7
47.25
306.25
19.2
3.0
57.6
368.64
21.2
3.3
69.96
449.44
∑ X = 118.5
∑ Y = 18.3
∑ X.Y = 316.55
∑ y = ∑ x.a + ∑ x.b
or y = a + bx
Equation (i)
Equation (ii)
∑ xy = ∑ x.a + ∑ x .b or xy = x.a + b.x 2
∑X
2
= 2048.23
2
Putting the values in the above formula :
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18.3 = a + 118.5b
→ (i)
∑ Y = 18.3 ∑ X = 118.5 316.55 = 118.5a + 2048.23b
→ (ii)
∑ XY = 316.55 ∑ X = 118.5 ∑ X = 2048.23 2
Multiply equation (i) with 118.5
= 18.3 ×118.5 = 118.5a + 118.5 ×118.5b 2168.55 = 118.5a + 14042.25b Subtracting equation (ii) from (iii) (2168.55316.55)=(118.5118.5)a+(14042.252048.23)b (2168.55316.55)=0+(14042.252048.23)b
∴b =
2168.55 − 316.55 1852 = 14042.25 − 2048.23 11994.02
b = 0.16 Now put the values of b in equation (i)
18.3 = 7a + 0.16 ×118.5 18.3 = 7a + 18.96 7a = 18.3 − 18.96 = −0.66 a=
−0.66 = −0.0943 7
Therefore, the Regression equation y = a + bx
y = 0.0943 + 0.16 x Ans.
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Example Determine the equation for the line of best fit from the data given below and calculate the values of variable : 1.
x
8
4
5
1
2.
y
2
0
2
6
Solution : The value of correlation coefficient is calculated and the results are shown in a tabular form as given : x
y
x.y
x
2
y
8
2
16
64
4
4
0
0
16
0
5
2
10
25
4
1
6
6
1
36
∑ x = 16
∑y = 6
∑ xy = −12
∑x
2
2
∑y
= 106
2
= 44
1. Using the formula: correlation coefficient
r=
n ( ∑ XY ) − ∑ X.∑ Y n
( ∑ X ) − ( ∑ X ) × ∑X
r=
2
2
2
(∑ X) − n
n
( ∑ Y ) − ( ∑ Y )
×
∑Y
2
2
2
4(−12) − (16)(6) 4(106) − (16)2 × 4(44) − (6) 2
2
(∑ Y) −
2
n
= 0.939
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Here
67
∑ X = 16 ∑ Y = 6 ∑ xy = −12 ∑x
2
= 106
∑y a=
2
n ( ∑ xy ) − ∑ x.∑ y n
a=
= 44 n = 4
(∑ x ) − (∑ x ) 2
(4)(−12) − (16)(6) (4)(106) − (16)
2
2
=
−12 − 24 = −0.857 106 − 64
( ∑ y ) . ( ∑ x ) − ( ∑ x ) . ( ∑ xy ) b= n (∑ x ) − (∑ x ) 2
2
b=
2
(6)(106) − (16)(−12) = 4.929 4(106) − (16) 2
Therefore, line of best fit will be
y = −0.857x + 4.929 Basic Concept of Probability Probability theory provides us with the ways and means to attain the formal and precise expressions for uncertainties involved in different situations. Uncertainty is a part of human life. Weather stock market prices, product quality are but some of the areas, where, commenting on the future wit certainly, becomes impossible. Decisionmaking in such areas is facilitated through formal and precise expressions for the uncertainities involved. The definition of probability is achieved through the study of certain basic concepts in probability theory, such as experiment. Sample space and event. Experiment The term experiment is used in probability theory in a much broader sense than in Physics or Chemistry. Any action, whether it is tossing of a coin, or measurement of a product’s dimension to ascertain quality constitute an experiment in probability theory
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Definition: Sample space The set of all possible outcomes of an experiment is defined the sample space.
Example: In launching a new product success, and failure defines the sample space.
Event An even is a subset of the sample space, it constitutes one or more possible outcomes of an experiment. Independent Events If the occurance of an event does not affect the occurance of other, such event is said to be independent event.
Example: In tossing a die repeatedly, the event of getting 5 in the 2
nd
throw is independent of getting 5 in
th
the 4 throw.
Dependent event Event is said to be dependent, if the occurance of an event does affect the occurance of other. Example: If we draw a single ball from a packet containing 10 different size balls, the probability, of getting a ball is 1/10. If we do not replace this ball, the probability of taking a next ball is affected and its probability in 1/9. Hence the event is dependent. Mutually Exclusive Events
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Two or more events are said to mutually exclusive if the occurance of any one of them excludes the occurance of all others. Example: In throwing a die, the outcomes are mutually exclusive. The occurance of one face excludes the remaining 5 face. Equally Likely Events The outcomes are said to be equally likely if none of them is expected to occur in preference to other. Example: In tossing a coin the outcomes Head and Tail are equally likely [if the coin is unbiased] Exhaustive Events: The total number of possible outcomes of a random experiment is called exhaustive events Example: When two coins are tossed the exhaustive cases are 4. Simple event: Events are said to be simple when the probability of happening or not happening of single events Example: Probability of drawing taking a pencil from a packet containing 5 pencils, 4 erasers. Compound Events: The joint occurance of two or more events are termed as compound events. Example: In a packet containing 5 pencils & 4 erasers taking 2 pencils & 2 erasers in two successive cases. Complementary Events: If A and B are mutually exclusive and exhaustive events, then, A is called the complementary event of B. Similarly, B is called the complementary event of A. Classical: The probability of an event is defined as
Definition of Probability: The probability of an event is defined
as=
Total number offavourable cases Total Number of equally likely cases
Example: If a fair coin is tossed then there are two equally likely cases, Head and Tail, and hence the probability of head is Â˝.
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Axioms of Probability: Consider a random experiment whose sample space is S, A be the event Then, Axiom1: O ≤ P ( A) ≤ 1; Axuin2 P(S)=1 Let
A1 , A2 ,.... An be disjoint mutually exclusive events, then, ∞ ∞ P U An = ∑ P ( An ) n=1 n =1
Addition Theorem Independent Event: The probability of occurring either event A or event B which are mutually exclusive events, is the sum of the individual probability of A and B.
P ( Aor B ) = P ( A ) + P ( B ) Proof If an event A can happen in n1 ways and B in n2 ways, then the number of ways in which either event can happen is,
n1 + n2 . If the total number of possibilities is n, then by definition of probability of
either A or B event happening is
the number of outcomes favourable to the event total number of comes n +n n n = 1 2 = 1+ 2 n n n where, n P ( A) = 1 n n2 P ( B) = n Hence, P ( Aor B ) = P ( A ) + P ( B ) =
Note: If A,B&C are three mutually exclusive events, then
P ( Aor B or C ) = P ( A ) + P ( B ) + P ( C )
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Addition theorem Dependent events When A and B are not mutually exclusive,
P ( A or B ) = P ( A ) + P ( B ) − P ( A & B )
Events
In the case of three events,
P ( A or B or C )
= P ( A ) + P ( B ) + P ( C ) − P ( AB ) − P ( AC ) − P ( BC ) + P ( ABC ) Example
A bag contains 5 white, 4 Red and 11 black balls. Two balls are drawn at random. Find the probability that they will both be white. Total number of balls in the bag = 5 + 4 + 11 = 20 balls. From 20 balls, drawing 2 balls in
=
20C 2 ways,
20 ×19 = 190 days 1× 2
From 5 white balls, drawing 2 white balls
=
5C2 ways, 5× 4 = 10 ways. 1× 2
∴ Probability for the two balls drawn to be white =
10 1 = 190 19
Example A bag containing 7 Green and 5 Red balls, 5 balls are drawn at fandom. What is the probability that 2 of them are green and 3 of them are red? Total number of balls = 7 + 5 = 12 5 balls are dawn in
12C5 ways = 12C5
=
12 × 11× 10 × 9 × 8 = 792 1× 2 × 3 × 4 × 5
Drawing 2 green from 7 Green balls in
7C2 ways =
7×6 = 21 1× 2
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Drawing 3 red from 5 red balls in
5C3 ways =
5× 4× 3 = 10 1× 2 × 3
∴ The probability of drawing 2 Green and 3 red from the total is, =
=
7C 2 × 5C3 12C5
35 132
Example A bag contains 5 White and 8 red balls Two drawings of 3 balls are made, such that : a ) the balls are replaced before the second trial, b) the balls are not replaced before the second trail. Find the probability that the first drawing will give 3 white and the second 3 red balls in each case. a) Total number of balls = 5 + 8 = 13 3 balls are drawn from 13 in
13C3 ways,
3 white balls drawn from 5 white balls in 3 red balls drawn from 8 red balls in
5C3 ways,
8C3 ways.
∴ The probability of drawing 3 white in the trial is, =
5C3 5 × 4 × 3 ×1× 2 × 3 5 = = 13C3 1× 2 × 3 × 13 ×12 ×11 143
Probability of 3 red balls in second trail [with replacement] is,
=
8C3 28 = 13C3 143
∴ The probability of the compound event that is, when the balls are replaced before the second trial is,
=
5 28 × = 0.007 143 143
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b. In the first trail probability of drawing 3 white balls is,
i.e.,
5C3 5 = 13C3 143
In the 2nd trail probability of drawing 3 red balls,
[ Without replaced ] =
8C3 7 = 10C3 15
∴ The probability of the compound event is,
=
5 7 7 × = = 0.016 143 15 429
Example : The probability of solving a problem by Asha is 2/3 and that by the probability of solving the problem by both Asha & Sophia is
14 4 . The probability of solving by either Asha or Sophaia is . What is 25 5
probability of solving the problem by Sophia? Let us take probability of solving the problem by Asha is A and by Sophia is B.
Given,
2 4 P ( A ) = , P ( A or B ) = 3 5
P ( AB ) =
14 15
By addition theorem,
P ( A or B ) = P ( A ) + P ( B ) − P ( AB ) 4 2 14 = + P ( B) − 5 3 15 ∴ P ( B) =
4 2 14 − + 5 3 15
The probability of solving the problem by Sophia is
=
4 9
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Example Pappu is know to hit the target in 3 out of 4 shots, whereas Chattu is known to hit the target in 2 out of 3 shots. Find the probability of the target hit when they both try. Given, A and B be the event of hitting the target by Pappu and Charru.
Probability when Pappu hits the target
=
3 i.e., 4
3 2 P ( A ) = ; P ( B) = 4 3 A & B are not mutually exclusive events. By addition theorem,
3 2 3 2 P ( A or B ) = P ( A ) + P ( B ) − P ( AB ) = + − × 4 3 4 3 Probability of hitting the target when they both try = 0.917 Example A salesman has 65 percent chance of making a sale to each customer. The behavior A salesman customer is independent. If three customers Bharathi, Milton & Murali enter, what is the probability that the salesman will make a sale to A or B or C? Let A be the event making sale to Bharathi Let B be the event making sale to Miltan Let C be the events making sale to Murali Given,
P ( A ) = 0.65; P ( B ) = 0.65; P ( C ) = 0.65
P ( A or B or C ) = P ( A ) + P ( B ) + P ( C ) − P ( AB ) − P ( BC ) − P ( AC ) + P ( ABC ) = 0.65 + 0.65 + 0.65 − ( 0.65 × 0.65 ) − ( 0.65 × 0.65 ) −
( 0.65 )( 0.65) + ( 0.65 × 0.65 × 0.65 )
P ( A or B or C ) = 0.9571
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Example : A bag contains 10 white and 4 red balls. Two balls are drawn in succession at random what is the probability that one of them is white and another are in red. Total number of balls
= 14
Drawing 2 balls from 14 balls
=
14C 2
Drawing 1 white from 10 white balls
=
10C1
Similarly, Drawing 1 Red from 4 Red balls
=
4C1
Drawing 1 white and 1 Red, number of favourable cases
= 10C1 × 4C1 Probability of getting 1 white and 1 red
=
10 × 4 ×1× 2 40 = 14 × 13 91
Example : The M.B.A. class consists of 60 students. In that, 20 of them are rich, 10 of them are rich, 5 of them are fair complexioned. What is the probability of selecting a fair complexioned rich girl ? Let A,B & C be the three events of selecting a girl, rich and fair complexioned.
Given, P(A)
P (A) =
5 1 = 60 12
20 1 = 60 3 10 1 P ( C) = = 60 6 P ( B) =
∴ P ( A and Band C ) =
1 1 1 1 × × = 12 3 6 216
Conditional Events Multiplication for dependent events A conditional probability indicates that the probability that an event will occur is subject to the condition that another event has already occurred.
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B P ( AB ) i.e., P = A P ( A) A P ( AB ) P = B P ( B) By the rule of multiplication,
B P ( A and B ) = P ( A ) + P A A P ( A and B ) = P ( B ) P B Baye’s Theorem It states that, if
A1 and A 2 are two mutually exclusive and collective exhaustive events and B is
another event defined in the context of the same experiment, then given the values of
P ( A1 ) ,
B B P ( A 2 ) , P and P , the conditional probability A1 A2 B P ( A1 ) P A A1 P 1 = B B B P P ( A1 ) + P P ( A2 ) A1 A2 B P ( A2 ) P A2 A2 P = B B B P P ( A1 ) + P P ( A2 ) A1 A2 In the case of n events,
A1 , A 2 ,...........A n which are mutually exclusive and collectively
exhaustive
B P ( A1 ) P A A1 P 1 = n B B P ( A1 ) P ∑ i =1 A2 FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +919999554621
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Proof : [This theorem is based on conditional probability] Let A1 and A2 are events which are mutually exclusive and exhaustive and B the event which is the intersection of A1 and A2 By the definition of conditional probability,
A P ( A1 and B ) P 1 = P ( B) B
A P ( A 2 and B ) P 2 = P ( B) B The part of B within
→ (1) → ( 2)
A1 is A1 and B, similarly the part of B within A2 is A2 and B.
∴P ( B ) = P ( A1 and B ) + P ( A 2 and B ) From 1 & 2,
B B P ( B ) = P ( A1 ) P + P ( A 2 ) P A1 A2 2 B = ∑ P ( Ai ) P i =1 Ai
In general, if there are n mutually exclusive and exhaustive events, then,
P ( B ) = ∑ PA i P B A i i =1 n
where,
A i.e., P 1 = B
P ( Ai ) P B Ai n P ( A i ) P B ∑ Ai i =1
A i.e., P k B
P ( A k ) P B Ak n P ( Ai ) P B ∑ Ai i =1
=
A k is the event between A1 to A n .
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Posterior Probability Probability before revision by Baye’s rule is called prior Probability, A probability which has undergone revision via rule is called posterior probability because it represents a probability computed after the sample information is taken into account. Posterior probability is always conditional probability, the condition event being the sample information. Exercise: A factory has two machines, past records show that machine 1 produces 20 % of the items of output and machine 2 produces 80% of the items, 5% of the items produced by machine 1 were defective and only 1% produced by machine 2 were defective. If a defective item is drawn at random, what is the probability that the defective item was produced by machine by 1 or machine 27. Let
A1 , A 2 be the events of drawing an item produced by machine 1 and 2 respectively. Let B be
the event of drawing a defective item produced by machine 1 or 2. Given,
P ( A1 ) = 20% = 0.2 P ( A 2 ) = 80% = 0.8 B P = 5% = 0.01 A1 B P = 1% = 0.01 A2 P ( B ) = P ( A1 and B ) + P ( A 2 and B )
B B = P ( A1 ) P + P ( A 2 ) P A1 A2 = 0.2 × 0.05 + 0.8 × 0.01
P ( B ) = 0.01 + 0.08 = 0.09 ∴ Probability that a defective item was produced by machine 1 is,
A P ( A1 and B ) 0.01 P 1 = = = 0.11 P ( B) 0.09 B ∴ Probability that a defective item was produced by machine 2 is, FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +919999554621
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A P ( A 2 and B ) 0.08 P 2 = = = 0.88 P ( B) 0.09 B ∴ Probability that the defective item from machine 1 or machine 2 is = 0.111 + 0.888 = 0.999 The values can be derived in the following tabular form. Event
Prior
Conditional
Joint Probability
Posterior Probability
Probability
Probability
A1
0.2 = P(A1)
0.05 = P B A1
P ( A1 & B ) = 0.01
A 0.01 P 1 = = 0.111 B 0.09
A2
0.8 = P ( A 2 )
0.01 = P B A2
P ( A 2 & B ) = 0.08
A 0.08 P 2 = = 0.888 B 0.09
1.00
0.999
Binomial Distribution The probability of success or failure remains constant from trial to trial. The success of an event is p and its failure by q. since the binomial distribution is a set of dichotomous alternatives p + q = 1. By expanding the binomial terms, we obtain probability distribution which is called the binomial probability distribution. By expanding the terms of the binomial
(q + p) where
n
(q + p)
n
we get,
= q n + nC1q n −1P + nC2 q n − 2 + P 2 + ...... + P n
1, nC1 , nC 2 ,.... are called binomial coefficients. The general form of binomial distribution is
P ( r ) = nC r , p r q n − r , Note The terms of this distribution are symmetrical ascending upto the Middle of the series, and then descending. Constants of Binomial Distribution Mean = nP
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S.D. = npq first movement
µ1 =0
µ 2 = npq
µ3 = npq ( q − p ) µ 4 = 3n 2 p 2q 2 + npq (1 − 6pq )
β1
(q − p) =
2
npq 1 − 6pq β2 = 3 + npq Example The probability that a part time student will get the degree is 0.4. Determine the probability that out of 5 students, (a) one; (b) at lest one will be graduate Given, probability of getting a degree is 0.4 i.e., P = 0.4 p+q=1 q = 0.6
P ( one will be a graduate ) = 5C1 , ( 0.4 )( 0.6 ) = 0.2590 4
P ( at least one will be graduate ) = 1 − P ( none will be graduate )
P ( none will be a graduate ) = 5Co ( 0.4 ) ( 0.5 ) o
5
= 1×1× ( 0.0777 ) = 0.7777 ∴P ( atleast one graduate ) = 1 − 0.08 = 0.92
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Example : Bring out the fallacy, if any, in the following statements : a) The mean of a binomial distribution is 12, and its standard deviation is 4. Given Mean = 12 i.e., np = 12 S.D. = 4
i.e., npq = 4. (variance) npq = 16
16 4 = >1 12 3 Qp + q =1 q >/ 1 q=
Hence the statement is wrong. Poisson Distribution This distribution describes the behaviour of rare events and ahs been known as the Law of Impropable events, It was found by French mathematician Simeon D. Poisson, Poisson distribution can study when we kow the mean value of occurrences of an event without knowing the sample space. Mathematically, the Poisson distribution is limiting form of binomial distribution as n tends to infinity and P approaches zero such that np = m remains constant. It defined as,
P (r) =
e− m mr ; r = 0,1, 2.... Lr
e = 2.71823 Constants in Poisson Distribution
µ1 = 0 µ2 = m µ3 = m FOR MORE DETAILS VISIT US ON WWW.IMTSINSTITUTE.COM OR CALL ON +919999554621
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µ 4 = m + 3m 2 β1 =
1 m
β2 = 3 +
1 m
Example : The mean of the Poisson distribution is 2. Find the other constants of the distribution. Given m = np = 2
σ = m = 2 = 1.414 µ1 = 0 µ2 = 2 µ3 = 2 µ 4 = 2 + 3 22 = 14 1 1 = = 0.5 m 2 1 β2 = 3 + m 1 β2 = 3 + = 3.5 2 β1 =
Example : It is given that 3% of electric bulbs manufactured by a Company are defective. Using Poisson approximation, find the probability that a sample of 100 bulbs will contain (i) no defective (ii) exactly are defective. Given number of defective bulbs in a sample 100 is 3. i.e., m = 3
e−m mo ∴P ( no defective ) = = e − m = e −3 = 0.05 0! −3 e − m m 2 e ( 3) P ( exactly 2 defective ) = = = 0.05 × 4.5 = 0.225 2! 2 2
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Example : In a street 2 accidents took place in 10 days. Assuming that the number of accidents per day follows Poisson distribution, find the probability that there will be three or more accidents in a day.
Given, the average number of accidents per day =
2 = 0.2 10
∴ P ( 3or more acci de nts ) = 1 − P ( 2 or less acci de nts ) = 1 − P ( 0 ) + P (1) + P ( 2 ) 2 e −0.2 × ( 0.2 ) = 1 − e−0.2 + e −0.2 × 0.2 + 2 = 1 − 0.8187 ×1.22
= 1 − 0.999 = 0.001
Normal Distribution The normal distribution is the cornerstone of modern statistics. The binomial and the Poisson distributions are most useful theoretical distributions for the discrete variable. The normal distribution happens to be most useful theoretical distribution for continuous variables. The normal distribution is an approximation to binomial distribution whether or not p is equal to q, the binomial distribution tends to the form of the continuous curve and when n becomes large at least for the material part of the range. The general form of normal distribution is
Px =
1 −(x − µ) 2 e 2σ 2 σ 2π
X   Values of the continuous random variable
µ
  Mean of the normal random variable
e = 2.7183
π
= 3.14.16
The normal distribution with mean
µ
and standard deviation
σ
is generally denoted by N(m,
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The normal distribution is symmetric about the mean. The area on each side of the mean is 0.5. Though the range of the variable is specified from  ∞ to
∞ , 99.7% of the values of the random variable
fall with in ±3σ . ie,
P [ µ − 3σ ≤ x ≤ µ + 3σ ] = .997
Because of the symmetry and the points of inflexion at ±1σ distance, the curve has a bell shape. Example: Heights, Weights and dimensions of a product are some of the continuous random variable which are found to be normally distributed Constants of The mean of the normal distribution is
x
The Normal distribution S .D = σ
µ2 = σ 2 µ3 = 0 µ4 = 3σ 2
Example: In a distribution exactly normal 7% of the items are under 35 and 89% are under 63. what is the mean and standard deviation of the distribution.
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Since 7% of items are under 35 The Area between
x and 35 is 43%
Similarly, 89% of item are under 63. The Area between
x and 63 is 39%
The standard Normal Variate corresponding to 43% is 1.48
Z=
1.48 =
Similarly,
−1.48 σ + x = 35
x−x
σ
35 − x
σ
→ (1)
The standard Normal Variate Corresponding to 39% is 1.23
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63 − x
σ
= 1.23
1.23σ + x = 63 Solving
1&2
We get,
x = 50.3
→ (2)
The mean of distribution is 50.3 and S.D=10.33 Example A sample of 100 dry battery cells tested to find the length of life produced the following results:
x = 12 σ = 3 hrs. Assuming the data to be normally distributed, what percentage of battery cells are expected to have life, i.
More than 15 hours,
ii.
Less than 6 hours and
iii.
Between 10 and 14 hours?
Let x denote the length of life of dry battery cells (i) x=15
x−x σ 15 − 12 Z= =1 3
Z=
Standard normal variate: P(x>15)=P(z>1)
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P(X>60) = P(Z>0.75) =Area to the right of Z= 0.75 = 0.5 – 0.2734 = 0.2266 i.e. The accounts have a balance in excess of Rs.60 is 22.66% (ii) P(40<X<60)
X=40
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Z=
40 − 48 −8 = = 0.5 16 16
X=60
Z=
60 − 48 12 = = 0.75 16 16
P(40<<60) = P(0.5<Z<0.75) = Area between0.5 to 0 + Area between 0 to o.75 = 0.1915+0.2734=0.4649 The accounts have an average balance between Rs.40% & Rs.60% is 46.99%
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UNITIII TESTS OF SIGNIFICANCE FOR SMALL SAMPLES Student’s t Distribution Theoretical work on tdistribution was done by W.S. Gosset (18761937) in the early 1900. Gosset was employed by the Guinness & Son, a Dublin bravery, Ireland, which did not permit employees to publish research findings under their own names. So Gosset adopted the pen name “student” and published his findings under this name. Thereafter; the tdistribution is commonly called student’s tdistribution or simply Student’s distribution. The tdistribution is used when sample size is 30 or less and the population standard deviation is unknown. The “tstatistic” defined as: __
t=
x− µ × n S
__ ∑ X − X where S = n −1
2
The tTable. The ttable given at the end is the probability integral of tdistribution. It gives, over a range of values of v, the probabilities of exceeding by chance value of t at different levels of significance. The tdistribution has a different value for each degree of freedom and when degrees of freedom are infinitely large, the tdistribution is equivalent to normal distribution and the probabilities shown in the normal distribution tables are applicable.
Application of the tDistribution The following are some of the examples to illustrate the way in which the ‘student’ distribution is generally used to test the significance of the various results obtained from small samples. Type – I To test the significance of the mean of a random sample In determining whether the mean of a sample drawn from a normal population deviates significantly from a stated value (the hypothetical value of the population mean), when variance of the population is unknown we calculate the statistic:
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__ X− µ n t= S where __
X = the mean of the sample µ = the actual or hypothetical mean of the population N= the sample size __
S = the standard deviation of the sample 2
__ X − X or S = S= n −1
__ d − n ∑ d n −1
2
2
2 d) ( 1 ∑ 2 = ∑ d − n n −1
where
d = deviation from the assumed mean. __
If the calculated value of
t exceeds t0.05, we say that the difference between X and µ is significant at
5% level, if it exceeds t0.01 the difference is said to be significant at % level. __
If
t < t 0.05 , we conclude that the difference between X and µ is not significant and hence the
sample might have been drawn from a population with mean= µ . Fiducial Limits of Population Mean. Assuming that the sample is a random sample from a normal population of unknown mean the 95% fiducial limits of the population mean ( µ ) are: __
X±
S t 0.05 n
and 99% limits are
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__
X±
S t 0.05 n
The following examples will illustrate this test: Illustration: The manufacturer of a certain make of electric bulbs claims that his bulbs have a mean life of 25 months with a standard deviation of 5 months. A random sample of 6 such bulbs gave the following values. Life in months 24,36,30,20,20,18. Can you regard the producer’s claim to be valid at 1% level of significance? (Given that the table values of the appropriate test statistic at the said level are 4.032, 3.707 and 3.499 for 5,6 and 7 degrees of freedom respectively). Solution. Let us take the hypothesis that there is no significant difference in the mean life of bulbs in the sample and that of the population. Applying ttest __
t=
x−µ × n S __
CALCULATION OF
X AND S 2
X
__ X − X
X
24
+1
1
26
+3
9
30
+7
49
20
7
9
20
3
9
18
5
25
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∑ x = 138 __
X=
∑x
2
= 102
∑ X = 138 = 23 n
6
∑x
2
102 = 20.4 = 4.517 n −1 5 23 − 25 2 × 2.449 = = 1.084 6= 4.517 4.517 v = n − 1 = 6 − 1 = 5. For v = 5.t 0.01 = 4.032
S=
=
The calculated value of t is less than the table value. The hypothesis is accepted. Hence the producer’s claim is valid at 1% level of significance. Illustration: A random sample of size 16 has 53 as mean. The sum of the squares of the deviations taken from mean is 135. Can this sample be regarded as taken from the population having 56 as mean ? Obtain 95% and 99% confidence limits of the mean of the population. (for v=15, t0.05=2.13 for v=15 t0.01=2.95) Solution. Let us take the hypothesis that there is no significant difference between the sample mean and hypothetical population mean. Applying t test:
F=
L arg er estimate of var iance smaller estimate of var iance
The calculated value of t is more than the table value. The hypothesis is rejected. Hence the sample has not come from a population having 56 as mean. 95% confidence limits of the population mean __
X±
S t 0.05 n 3 × 2.13 16 = 53 ± 1.6 = 51.4 to 54.6 = 53 ±
99% confidence limits of the population mean
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X±
93
S t 0.01 n = 53 ±
3 × 2.95 16
3 = 53 ± 2.95 = 53 ± 2.212 = 50.788 to 55.212 4 Illustration: The life time of electric bulbs for a random sample of 10 from a large consignment gave the following data: Item
1
2
3
4
5
6
7
8
9
10
Life in
4.2
4.6
3.9
4.1
5.2
3.8
3.9
4.3
4.4
5.6
000 hours Can we accept the hypothesis that the average life time of bulbs is 4,000 hours. Solution. Let us take the hypothesis that there is no significant difference in the sample mean and the hypothetical population mean. Applying the ttest. __
t=
x−µ n S __
CALCULATION
X AND S
X
__ X − X
4.2
0.2
0.04
4.6
+0.2
0.04
3.9
0.5
0.25
4.1
0.3
0.09
__ X − X
2
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5.2
+0.8
0.64
3.8
0.6
0.36
3.9
0.5
0.25
4.3
0.1
0.01
4.4
0
0
5.6
+1.2
1.44
2
__ x − ∑ x =3.12
2
__ ∑ X − X 3.12 S= = = 0.589 n −1 9 4.4 − 4 0.4 × 3.162 t= 10 = = 2.148 0.589 0.589 v = n − 1 = 10 − 1 = 9. For v = 9, t 0.05 = 2.262
The calculated value of t is less than the table value. The hypothesis is accepted. The average life time of the bulbs could be 4000 hours. Testing Difference between Means of Two Samples (Independent Samples) ___
Given two independent random samples of size n1 and n2 with means
___
X 1 and X 2 and standard
deviations S1 and S2 we may be interested in testing the hypothesis that the samples come from the same normal population. To carry out the test, we calculate the statistic as follows: ___
___
X −X2 n1n 2 t= 1 × S n1 + n 2 where ___
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n1 = number of observations in the first sample n2 = number of observations in the second sample S = combined standard deviation. The value of S is calculated by the following formula: 2
__ __ X − X + X − X ∑ 1 1 ∑ 2 2 S= n1 + n 2 − 2
2
Illustration 3. Two types of drugs were used on 5 and 7 patients for reducing their weight. Drug A was imported and drug B indigenous. The decrease in the weight after using the drugs for six months was as follows: Drug A
10
12
13
11
14
Drug B
8
9
12
14
15
10
9
Is there a significant difference in the efficacy of the two drugs? If not, which drug should you buy. (for v=10, t0.05=2.223)
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Solution. Let us take the hypothesis that there is no significant difference in the efficacy of the two drugs. Applying ttest: __
__
x −x t= 1 2 S X1
___ X − X 1 1
10
2
12
n1 n 2 n1 + n 2 X2
__ X − X 2 2
4
8
3
9
0
0
9
2
4
13
+1
1
12
+1
1
11
1
1
14
+3
9
14
+2
4
15
+4
16
10
1
1
9
2
4
∑X
1
= 60
___ X1 − X1
2
2
___ X − X ∑ 1 1 = 10
∑X
2
= 77
__ X2 − X2
2
2
__ X − X ∑ 2 2 = 44
However, it is advisable to take account of bias.
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X1 =
∑X
1
n1
=
97
__
60 12; 5
X2 = 2
∑X n2
2
=
77 = 11 7 2
__ __ − + − X X X X ∑ 1 1 ∑ 2 2 10 + 44 54 = = = 2.324 S= n1 + n 2 − 2 5+7−2 10 __
__
X − X2 t= 1 S
n1 n 2 n1 + n 2
12 − 11 5 × 7 1.708 = = 0.735 2.324 5 + 7 2.324 v = n1 + n 2 − 2 = 5 + 7 − 2 = 10 For v = 10, t 0.05 = 2.228. The calculated value of t is less than the table value, the hypothesis is accepted. Hence there is no significance in the efficacy of two drugs. Since drug B is indigenous and there is no difference in the efficacy of imported and indigenous drug, we should buy indigenous drug in B. Illustration: For a random sample of 10 persons, fed on diet A, the increase to weight in pounds in a certain period were: 10,6,16,17,13,12,8,14,15,9 For another random sample of 12 persons, fed on diet B, the increase in the same period were: 7,13,22,15,12,14,18,8,21,23,10,17 Test whether the diets A and B differ significantly as regards their effect on increase in weight. Given the following: Degrees of freedom value of t at 5% level 19
20
21
22
23
2.09
2.09
2.08
2.07
2.07
Solution: Let us take the null hypothesis that diets A and B do not differ significantly wieth regard to their effect on increase in weight. Applying ttest.
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98
__
x −x t= 1 2 S
n 1n 2 n1 + n 2 2
__ __ X − X + X − X ∑ 1 1 ∑ 2 2 S= n1 + n 2 − 2
2
Calculating the required values: Persons fed on diet A Increas
Deviations from
e
mean
in
weight
12
Persons fed on diet B
__ X1 − X1
2
Increase in
Deviations from
weight X2
actual mean 15
__ X − X 1 1
X1
__ X2 − X2
__ X − X 2 2
10
2
4
7
8
64
6
6
36
13
2
4
16
+4
16
22
+7
49
17
+5
25
15
0
0
13
+1
1
12
3
9
12
0
0
14
1
1
8
4
16
18
+3
9
14
+2
4
8
7
49
15
+3
9
21
+6
36
9
3
9
23
+8
64
10
5
25
17
+2
4
∑X
1
∑ X
1
=120
__ − X1 =
2
2
__ X − X ∑ 1 1 =1
∑X
2
0
=18
∑ X
2
__ − X2 =
2
__ X − X ∑ 2 2 =3
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99
20
0
14
Mean increase in weight of 10 persons fed on diet A ___
X1 =
∑X
1
n1
=
120 = 12 Pounds 10
Mean increase in weight of 12 persons fed on diet B ___
X2 =
∑X n2
2
=
180 = 15 Pounds 12 2
2
__ __ X − X + X − X ∑ 1 1 ∑ 2 2 120 + 314 S= = = 4.66 n1 + n 2 − 2 10 + 12 − 2
__
__
X1 = 12, X 2 = 15, n1 = 10, n 2 = 12, S = 4.66 Substituting the values in the above formula: 12 − 15 10 × 12 2 × = × 2.34 = 1.51 4.66 10 + 12 4.66 v = n1 + n 2 − 2 = 10 + 12 − 2 = 20.
t=
For v = 20, the table value of t at 5 per cent level is 2.09. The calculated value is less than the table value and hence the experiment provides no evidence against the hypothesis. We, therefore, conclude that diets A and B do not differ significantly as regards their effect on increase in weight is concerned. The value of S is calculated as follows: 2
__ d − ∑ d S= or n −1
__ d − n ∑ d n −1
2
2
It should be noted that t is based on n1 degrees of freedom. The following examples will illustrate the application of difference test: Illustration. To verify whether a course in accounting improved performance, a similar test was given to 12 participants both before and after the course. The original marks recorded in alphabetical order of the
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participants were 44,40,61,52,32,44,70,41,67,72,53, and 72. After the course, the marks were in the same order, 53,38,69,57,46,39,73,74,60 and 78. was the course useful?
Solution. Let us take the hypothesis that there is no difference in the marks obtained before and after the course, i.e., the course has not been useful. Applying ttest (difference formula): __
t=
d n S
Participants
Before(1st Test)
After(2nd Test)
(2nd1st Test) d
d2
A
44
53
+9
81
B
40
38
2
4
C
61
69
+8
64
D
52
57
+5
25
E
32
46
+14
196
F
44
39
5
25
G
70
73
+3
9
H
41
48
+7
49
I
67
73
+6
36
J
72
74
+2
4
K
53
60
+7
49
L
72
78
+6
36
∑ d = 60
∑d
2
= 578
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d=
101
∑ d = 60 = 5 n
12
2
2 __ d n − ∑ d 578 − 12 ( 52 ) S= = = 5.03 n −1 12 − 1
5 × 12 5 × 3.464 = = 3.443 5.03 5.03 v = n − 1 = 12 − 1 = 11; For v = 11, t 0.05 = 2.201
t=
The calculated value of t is greater than the table value. The hypothesis is rejected. Hence the course has been useful. Illustration 36. A drug is given to 10 patients, and the increments in their blood pressure were recorded to be 3,6,2,4,3,4,6,0,0,2. Is it reasonable to believe that the drug has no effect on change of blood pressure? (5% value of t for 9 d.f.=2.26) Solution. Let us take the hypothesis that the drug has no effect on change of blood pressure. Applying the difference test: __
t=
d n S 2
d
d
3
9
6
36
2
4
4
16
3
9
4
16
6
36
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0
0
0
0
2
4
∑ d =20
__
d=
∑d
2
=130
∑ d = 20 = 2 n
10
2
__ ∑ d − n d 130 − 10 ( 2)2 S= = = 3.162 n −1 10 − 1 2
2 10 2 × 3.162 = =2 3.162 3.162 v = n − 1 = 10 − 1 = 9 For v = 9, t 0.05 = 2.26 t=
The calculated value of t is less than the table value. The hypothesis is accepted. Hence it is reasonable to believe that the drug has no effect on change of blood pressure. Testing the significance of an observed correlation Coefficient Given a random sample from a bivariate normal population. If we are to test the hypothesis that the correlation coefficient of the population is zero, i.e., the variables in the population are uncorrelated, we have to apply the following test:
t=
r 1− r2
× n−2
Here t is based on (n2) degrees of freedom. If the calculated value of t exceeds t0.05 for (n2) d.f., we say that the value of r is significant at 5% level. If t<t0.05 the data are consistent with the hypothesis of an uncorrelated population. The following examples will illustrate the test: Illustration. A study of the hights of 18 pairs of husbands and their wives in a factory shows that the coefficient correlation is 0.52. Apply ttest to fin whether correlation is significant.
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(For 16 degrees of freedom at 5% level of significance, the table value of t=2.12) Solution. Let us take the hypothesis that there is no significant difference in the sample correlation and correlation in the population. Applying ttest:
r
t=
× n−2 1 − r2 r = 0.52 n = 18 52 .52 × 4 t= × 18 − 2 = = 2.44 2 0.854 1 − ( 52 ) v = ( n − 2 ) = (18 − 2 ) = 16 v = 16, t 0.05 = 2.12 The calculated value of t is greater than the table value. The hypothesis is rejected. The given value of r is significant. Illustration: How many pairs of observations must be included in a sample in order that an observed correlation coefficient of value 0.42 shall have a calculated value of t greater than 2.72
t=
r 1− r2
× n−2
We are given the value of t and r and we have to find out n.
∴ Or Or
0.42 1 − ( 0.42 )
2
× n − 2 > 2.72
0.42 × n − 2 > 2.72 0.908 0.42 × n − 2 > 2.72 × 0.908 2.72 × 0.908 = 5.88 0.42
Or
n−2 >
Or
n − 2 > ( 5.88 ) = 34.57
Or
2
n = 34.57 + 2 = 36.57 or 37
Hence we should include 37 observations. Illustration:
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The following table gives the ages in years of 10 husbands and their wives at marriage. Compute the correlation coefficient and test for its significance. Husbands’s age
3
7
8
9
0
1
3
5
6
9
8
2
3
4
5
6
8
9
0
2
Wife’s age
Solution. We set up the hypothesis that there is no correlation in the population, Applying ttest
t =
r 1− r
2
× n−2 =
0.939 1 − ( 0.939 )
2
× 10 − 2
0.939 0.939 × 2.8284 = × 2.8284 = 7.72 0.344 1 − 0.882 v = n − 2 = (10 − 2 ) = 8 =
For v = 8, t 0.05 = 2.31 The calculated value being much higher than the table value, the correlation is significant. Illustration: Is a correlation coefficient of 0.5 significant if obtained from a random sample of 11 pairs of values from a normal population? Use ttest. Solution. Let the null hypothesis be that there is no correlation or the given correlation coefficient is not significant.
t=
r
× n−2 1 − r2 r = 0.5, n = 11 t=
0.5 1 − ( 0.5 )
2
× 11 − 2 =
0.5 0.5 × 9= × 3 = 1.732 0.866 1 − 0.25
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The calculated value of t is less than the table value and hence the given correlation coefficient is not significant. Ftest The object of the Ftest is to find out whether the two samples may be regarded as drawn from the normal populations having the same variance. Since Ftest is based on the ratio of two variances, it is known as variance ratio test.
F=
L arg er estimate of var iance smaller estimate of var iance
With (n11,n21) d.f. The calculated value of F is compared with the table value for (n11,n21) d.f. at 5% or 1% level of significance. Analysis of variance: It is a statistical technique specially designed to test whether the means of more than two quantitative populations are equal. It consists of classifying and crossclassifying statistical results and testing whether the means of a specified classification differ significantly. The analysis of variance is studied by: a)
One –way classification and
b)
Twoway classification
Oneway classification In oneway classification the data are classified according to only one criterian i.e. different levels of a single factor is controlled in the experiment. The experiment. The null hypothesis is,
H 0 : µ1 = µ 2 = ... = µ k H1 : µ1 ≠ µ 2 ≠ ... ≠ µ k i.e., the arithmetic means of populations from which the K sample were randomly drawn are equal to one another. The variance ration that µ
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F=
Between Column var iance Within Column var iance
Analysis of Variance Table (ANOVA): Oneway classification:
Source of
Sum of squares
variation Between
Degrees of
Mean Square
Variance ration
freedom
of F
SSC
c1
SSC/c1
SSE
nc
SSE/nc
SST
n1
MSC MSE
samples Within samples
SST – Total sum of squares of variations SSC – Sum of squares between samples (columns) SSE – Sum of squares between samples (rows) MSC – Mean Sum of squares between samples MSC – Mean Sum of squares within samples. 1. To assess the significance of possible variation in performance in a certain test between the grammar schools of a city, a common test was given to a number of students take at random from the senior fifth class of each of the four schools concerned. The result are given below. Make an analysis of variance of data Schools
A
B
C
D
8
12
18
13
10
11
12
9
12
9
16
12
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8
14
6
16
7
4
8
15
Null hypothesis: We set up the null hypothesis that there is no significant difference in the means of the samples. Let the samples, I, II, III and IV be
Sample I
x1 , x 2 and x 3 and x 4 respectively.
Sample II
Sample III
Sample IV
x1
x12
x2
x 22
x3
x 32
x4
x 24
8
64
12
144
18
324
13
169
10
100
11
121
12
144
9
81
12
144
9
81
16
36
12
144
8
64
14
196
6
36
16
256
7
49
4
16
8
64
15
225
45
421
50
558
60
824
65
875
Sum of all the items of the samples 9T)
∑x +∑x +∑x +∑x 1
Correction factor =
2
(T) N
3
2
( 220 ) = 20
4
= 45 + 50 + 60 + 65 = 220
2
= 2420
Total sum of squares
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= ∑ x12 + ∑ x 22 + ∑ x 32 + ∑ x 24 − C.F = 421 + 558 + 824 + 875 − 2420 = 258 Sum of squares between the samples Two ways ANOVA. To study the performance of three detergents and three different water temperature, the following “whiteness” readings were obtained with specially designed equipment. Water temp.
Detergent A
Detergent B
Detergent C
Cold water
57
55
67
Warm water
49
52
68
Hot water
54
46
58
Perform a twoway Analysis of Variance using 5% level of significance. Let us take 3 different detergents to be
Cold
x1 , x 2 , x 3 respectively.
x1
x2
x3
Total
x12
x 22
x 32
57
55
67
179
3249
3025
4489
49
52
68
169
2401
2701
4624
54
46
58
158
2916
2116
3364
160
153
193
506
8566
7845
12477
water Warm water H ot water Total
Null hypothesis
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We set up the null hypothesis that there is no significant difference in the three varieties of detergents and water temperature.
(T) Corrector factor= N
2
( 506 ) =
2
9
= 28448.44
(160 ) Sum of the squares between detergents= 3
2
(153) +
2
3
(193) + 3
2
− C.F
= 8533.33 + 7803 + 12416.33 − 28448.44 = 304.22
Sum of squares between water temperature =
(179 ) 3
2
(169 ) + 3
2
(158) + 3
2
− C.F
= 10680.33 + 9520.33 + 8321.33 − 28448.44 = 73.55 Total sum of squares = 8566 + 7845 + 12477 = 439.50 Analysis variance table Source of Variation
Sum of squares
d.f.
Mean square
Between Columns
SSC = 304.25
C1=2
152.110
Between rows
SSR = 73.55
r1=2
36.775
residual
SSE = 61.79
(C1) (r1)=4
15.445
total
439.56
(detergent)
F=
8
152.11 = 9.85 15.445
d.f. (2.4) The table value of F for (2,4) d.f. at 5% level of significance is 6.94 Calculated value of F is greater than the table value. It is significant. Hence we reject the null hypothesis H 0 .
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Conclusion: We conclude that there is significant difference in the three varieties of detergents.
(water temp)
F=
36.775 = 2.38 15.445
d.f. =(2.4) The table value for (2,4) d.f. at 5% level of significance in 6.94. Since Calculated value of F is less than the table value. It is not significant hence we accept the null hypothesis H 0 . Conclusion: We conclude that water temperature do not make a significant difference. Example: The following table gives the number of units of production per day turned out by four different types of machines. Employee
Types of machines
M1
M2
M3
M4
E1
40
36
45
30
E2
38
42
50
41
E3
36
30
48
35
E4
46
47
52
44
Using Analysis of Variance (i) test the hypothesis that the ean production is the same for the four machines, and (ii) test the hypothesis that the employees do not differ with respect to mean productivity.
H 0 : We set up the null hypothesis that the mean production is the same for four machines, the employees do not differ with respect to mean productivity.
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In order to simplify the calculations subtract 40 from the given values, Let the four different machines be
x1 , x 2 , x 3 & x 4 .
x1
x2
x3
x4
Total
x12
x 22
x 32
x 24
0
4
5
10
9
0
16
25
100
2
2
10
1
11
4
4
100
1
4
10
8
5
11
16
100
64
25
6
7
12
4
29
36
49
144
16
0
5
35
10
20
56
169
333
142
Total = ∑ x12 + ∑ x 22 + ∑ x 32 + ∑ x 42 i.e., T=20
( 20 ) C.F.= C.F. = 16
2
= 25
Sum of Shares between machines
( 0) =
2
4
( −5 ) +
2
4
( 35) +
2
4
( −10 ) + 4
2
− C.F = 312.5
Sum of squares between employees
( −9 ) = 4
2
(11) +
Total sum of squares
4
2
(11) + 4
2
( 29 ) + 4
2
− C.F = 266
= ∑ x12 + ∑ x 22 + ∑ x 32 + ∑ x 24 − C.F = 56+169+333+14225=675
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ANOVA Table
Source of
Sum of squares
D.F.
Mean square
F
312.5
3
104.17
104.17 = 9.72 10.72
Between Rows
266.0
3
88.67
88.67 = 8.27 10.72
Residual
96.5
9
10.72
1
Total
675.00
15
Variation Between Columns
Calculated value of F [Types of machines] = 9.72 For (3,9) d.f. table value of F=3.86 at 5% level of significance. Since calculated value of F greater than the table value of F.
∴ It is significant. We reject the null hypothesis. Conclusion: We conclude that the mena production is not the same for all the four machines. Calculated value of F [Employees mean productivity] = 8.27 d.f. = (3,9) The table value of F for (3,9) d.f. at 5% level of significane is 3.86. Since calculated value of F is greater than the table value of F It is significant
∴ We reject the null hypothesis H 0 . Conclusion: We conclude that the employees differ w.r.t mean productivity.
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UNIT  IV COORDINATE GEOMETRY Coordinates: Let
XOX ' and YOY ' two perpendicular lines intersecting at, O. These two lines are called axes.
XOX ' is called the Xaxis, YOY ' is called the Y=axis and O is called the origion. These axes help us to locate any point P.
Let P be any point on the plane XOY. Draw PL and PM perpendicular to X and y axes. The length PM=x is called the x coordinate of the point P and the length PL=y is called the y coordinate of P. PM and PL are called the coordinates of P. this ordered pair is denoted by (x,y). Note that OL=MP=x and OM=PL=y. The x coordinates of the point P is also called the abscissa and the y coordinate of P is called the ordinate; the coordinates of the origin O are (0,0) OX is called the positive direction of Xaxis and
OX ' is called the negative direction of the X'
axis. OY is called the positive direction of the Yaxis and OY is called the negative direction of the Y'
axis. The lines XOX and YOY
'
divide the plane into four quadrants. The four quadrants are
represented in the following figure.
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The sign of coordinates in the four quadrants is given below: Quadrant
Coordinate x
y
I
+
+
II

+
III


IV
+

Example Show that the points (4,2), (7,5) and (9,7) are collinear.
Solution: If the area of the triangle formed by these points is zero then the points are collinear.
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1 x1 ( y 2 − y3 ) + x 2 (y3 − y1 ) + x 3 (y1 − y 2 ) 2 1 = [ 4(5 − 7) + 7(7 − 2) + 9(2 − 5)] 2 1 = [−8 + 35 − 27] = 0 2
∆=
∴ The given points are collinear. Example: Find the locus of the point which is equidistant from the two given points (2,3) and (4,1) Solution Let P(x,y) be a point such that PA=PB where A and B are the points (2,3) and (4,1).
PA 2 = PB2 (x − 2)2 + (y − 3)2 = (x + 4) 2 + (y − 1) 2 x 2 + 4 − 4x + y 2 + 9 − 6y = x 2 + 16 + 8x + y 2 + 1 − 2y (ie) 12x + 4y = 4 or 3x + y = 1 Example Find the locus of the point which moves from the point (4,3) keeping a constant distance of 5 units from it. Solution Let C(4,3) be the given point and P(x,y) be any point such that CP=5.
∴ CP 2 = 25 ( x − 4) 2 + ( y − 3) 2 = 25 x 2 + y 2 + 8 x − 6 y = 0 . This is the required locus. Straight Line: (i) Equation of Straight line paralled to an axis
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Let P(x,y) be any point on the straight line parallel to xaxis and is at a distance of k units from the xaxis. The oridinate of any point on this line is k. for every point on this line y=k. hence the equation of the line parallel to xaxis and is at a distance of k units from it is x=k. Similarly the equation of the line paralled to yaxis and is at a distance of h units from it is x=h. Note: The equation of the yaxis is x=0 The equation of the xaxis is y=0 (ii) Slope of a straight line
If a straight line makes an angle
θ
with the positive direction of xaxis, tan
θ
is called the slope
of the straight line or the gradient of the straight line. It is denoted by the symbol m. (ie) m=tan
θ
Slope in terms of coordinates
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Let a straight line make an angle
117
θ
with positive direction of the xaxis. Let
P( x1 , y1 ) and
Q( x2 , y2 ) be the two given points on the line. Draw PL,QM perpendicular to OX and QR perpendicular to LP respectively. Then
QR = ML = OL − OM = x1 − x2 RP = LP − LR = LP − MQ = y1 − y 2 Since QR is parallel to xaxis
∠RQP = ∠XOP = θ In
∆PQR , tan θ =
∴ m=
RP y1 − y2 = QR x1 − x2
y1 − y2 x1 − x2
(iii) Equation of straight line which makes an angle
θ
with the positive direction of xaxis and cuts off an
intercept c on the yaxis
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Let P(x,y) be any point on the straight line which makes an angle
118
θ
with xaxis.
∠OAP = θ , OB = c = y  intercept Draw PL perpendicular to xaxis and BN perpendicular to LP Then NBP = θ BN = OL = x
∴ NP=LPLN=LPOB=YC In ∆NBP ,
(ie)
m=
tan θ =
NP BN
y−c x
yc = mx or y = mx+c This equation is true for all positions of P on the straight line. Hence this is the equation of the required line. (iv) Equation of the straight line which cuts off intercepts a and b on the x and y axes.
Let P(x,y) be any point on the straight line which meets x and y axes at A and B respectively, Let OA = a and OB = b ON=x, NP = y NA=OAON=a0x Triangles PNA and BOA are similar.
∴
PN NA y a−x (ie) = = OB OA b a
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y x x y = 1 − or + = 1 b a a b This result is true for all positions of P on the straight line and hence this is the equations of the required line. (V) Equation of the straight line with given shape m and passing through the given point
( x1 , y1 )
Theon of the straight line with a given slope m is
y = mx+c → (1) Here c is unknown. This straight line passes through the point
( x1 , y1 ) . This point has to satisfy
the equation y=mx+c. Substituting the value of c in (1) we get the equation of the line as
y = mx+y1 mx1 (ie)
yy1 = m ( xx1 )
(vi) The equation of the straight line passing through two given points
( x1 , y1 ) and ( x 2 , y 2 )
yy1 =m ( xx1 ) → (1) where m is unknown. The slope of the straight line passing thro the points
m=
y1 − y 2 x1 − x 2
( x1 , y1 ) and ( x1 , y1 ) and ( x 2 , y 2 )
is
→ (2)
Substituting (2) in (1), the equation of the required straight line is
yy1 = (ie)
y1 − y 2 ( xx1 ) x1 − x 2
yy1 y1 − y 2 = xx1 x1 − x 2
(vii) The equation of a straight line in terms of the perpendicular P from the origin to the line and the angle that perpendicular makes with x axis.
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Draw OL perpendicular to AB. Let OL=p Let AOL=α ∴ ∠OBA=α
OA = sec α OL OA = OL sec α = p.sec α OB = sec α ∴ OB = OL cos ec α = P cos ec α OL ∴ The equation of the straight line AB is
x y + =1 p.sec α P cos ec α
(ie) x cos α + y sin α = p (viii) Point of intersection of two Lines: Let
( x1 , y1 ) be
a 2 x + b 2 y + c2 = 0 . Since
the
point
( x1 , y1 )
of
intersection
of
the
two
lines
a1x + b1 y + c1 = 0 and
lies on two lines ordinates of the point have to satisfy the two
equations.
a1x + b1 y + c1 = 0 a 2 x + b 2 y + c2 = 0 Solving these two equations we get
x1 y1 1 = = b1c 2 − b 2 c1 c1a 2 − c 2 a1 a1b 2 − a 2 b1
∴ The point of intersection is,
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b1c2 − b 2c1 a1b 2 − a 2 b1
121
c1a 2 − c 2a1 a1b 2 − a 2 b1
(IX) Concurrent Lines: If the lines
a1x + b1 y + c1 = 0 , a 2 x + b 2 y + c2 = 0 and a 3 x + b3 y + c3 = 0 , are concurrent then
the point of intersection of first two lines should also lie on third line
a 3 x + b 3 y + c3 = 0
Substituting the point
b 2 c2 − b 2c1 a1b 2 − a 2 b1 We get
c1a 2 − c 2a1 rd in the 3 equation. a1b 2 − a 2 b1
a 3 ( b1c 2 − b 2 c1 ) b3 ( c1a 2 − c2 a1 ) + + c3 = 0 a1b 2 − a 2 b1 a1b 2 − a 2 b1 ∴ a 3 ( b1c2 − b 2 c1 ) + b3 ( c1a 2 − c 2 a1 ) + c3 ( a1b 2 − a 2 b1 ) = 0
(X) Angle between two lines:
Let the two lines by
y = m1x + c1 and y = m 2 x + c2 . Let these lines make angles θ1 and θ2
respectively with OX. Then
m1 = tan θ1 m 2 = tan θ2 Let θ be the angle between the lines Then
θ1 = θ2 + θ or θ = θ1 − θ2
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tan θ =
tan θ1 − tan θ2 1 + tan θ1 tan θ2
tan θ =
m1 − m 2 → (1) 1 + m1m 2
122
m − m2 θ = tan −1 1 1 + m1m 2 The acute angle between the lines is given by
θ = tan −1
m1 − m 2 1 + m1m 2
→ (2)
Conditions for parallel and perpendicular lines (a) If these lines are parallel then θ = 0
∴ tan θ = 0 ∴ From (1), m1 − m 2 =0 ∴ m1 = m 2 Hence if two lines are parallel their slopes are equal. (b) If the lines are perpendicular θ = 90° and
tan θ = 90°
∴ From (1), 1 + m1m 2 = 0 or m1m 2 = −1 ∴ The condition for two lines to be perpendicular is that the product of their slopes is equal to 1 Example: Find the angle between the lines
a1x + b1 y + c1 = 0 and a 2 x + b 2 y + c2 = 0 and deduce the
condition for the lines to be (1) parallel (2) perpendicular, The slope of the two lines are
m1 = −
a1 , b1
m2 = −
a2 b2
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a1 a 2 + m1 − m 2 b1 b 2 tan θ = 1 + m1m 2 1 + a1 . a 2 b1 b 2 −
=
a1b2 − a2b1 a1a2 + b1b2
θ = tan −1
a1b 2 − a 2 b1 a1a 2 + b1b 2
If the lines are parallel than
θ =0
∴ a1b2 − a2b1 = 0 or
a1 b1 = a2 b2
If the lines are perpendicular the
θ = 90°
∴ a1a2 + b1b2 = 0 (xi) perpendicular distance from the origin on the line
ax + by + c = 0
Draw OM perpendicular to AB. Let OM=p ax+by=e
x y + =1 −c a −c a OA =
−c −c , OB = a b
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( Area
of ∆AOB ) = ( OM.OB ) 2
p p
124
2
2
c 2 c2 c2 c2 + = + a 2 b 2 a 2 b 2 c a 2 + b2 c4 = 2 2 a 2 + b 2 a b p2 =
c2 a 2 + b2
(xii) This perpendicular distance from
Though
∴p =
c a + b2 2
( x1 , y1 ) on the line ax+by+c=0
P ( x1 , y1 ) draw lines PX and PY parallel to ox and oy. Let Q be any point (x,y) with
reference to axes ox, oy and (x,y) with reference to axes PX and PY. Then
x = X + x1 y = Y = y1 ∴ The equation of the straight line AB with reference to axes PX, PY is
a ( X + x1 ) + b ( Y + y1 ) + c = 0
(ie.) aX + bY + ( ax1 + by1 + c ) = 0
The length of the perpendicular from P(x,y) on the line is
p=
a1x1 + by1 + c a 2 + b2
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Important results: 1. Slope of a straight line is m = tan θ . 2. slope of a straight line passing through
( x1 , y1 )
and
( x2 , y2 ) is m =
y1 − y2 x1 − x2
3. the various forms of equation of a straight line are: a.
y = mx + c ; m is slope and c is the intercept
b.
x y + = 1 ; a and b are intercepts on the axes. a b
c.
y1 − y2 = m ( x1 − x2 ) ; m is the slope and ( x1 , y1 ) is a point on the line.
d.
yy1 y1 − y2 = xx1 x1 − x2
e.
x cos α + y sin α = p, p is the r length from the origin on the line and α is the
( x1 , y1 )
and
( x2 , y2 ) are two points on the line.
inclination of the perpendicular to xaxis. 4. Angle between two lines
θ = tan −1
If the lines are parallel
m1 − m 2 1 + m1m 2
m1 = m 2
If the lines are perpendicular, 5. the perpendicular distance from
m1 , m 2 = −1
( x1 , y1 ) , on the line ax + by + c = 0 p=
is
ax1 + by1 + c a 2 + b2
Example 1: Find the slope of the line joining the points (2,3) and (4,5) Solution:
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The slope of the line joining two given points
m=
126
( x1 , y1 ) and ( x 2 , y 2 )
is
y 2 − y1 x 2 − x1
∴ The slope of the line joining the given points is m=
−5 − 3 −8 = = −4 −4 − 2 2
Example 2: Find the slope of the line 2x − 3y + 7 = 0 Solution: The equation of the line is 2x − 3y + 7 = 0 (ie.)
3y = 2x + 7
y=
2 7 x+ 3 3
∴ The slope of the line =
2 3
Example 3: Find the equation of the straight line making an angle of
135° with the positive direction of xaxis
and cutting an intercept 5 on the yaxis. Solution: The slope of the straight line is m = tan θ
= tan135° = tan(180° − 135° ) = − tan 45° = −1 y intercept = c = 5 The equation of the straight line is y = mx + c
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(ie) y=x+5 or x+y=5 Example 4: Find the equation of the straight line cutting off intercepts 2 and 5 on the axes. Solution
The equation of the straight line is
x y + =1 a b
Here a=2 and b=5
∴ the equation of the straight line is x y − = 1 or 5x2y=10 2 5 Example 5: Find the equation of the straight line passing through the points (7,3) and cutting off equal intercepts on the axes. Solution:
The equation of the straight line is
x y + =1 a b
(ie.) x+y=a This straight line passes through the point (7,3)
∴ 73=a(ie.) a=4 ∴ The equation of the straight line is x+y=4.
Example 6: Find the equation of the straight line, the portion of which between the axes is bisected at the point (2,5)
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Solution:
Let the equation of the straight line be
x y + =1 a b
Let the line meet the x and y axes at A and B respectively. Then the coordinates of A and B are (a,0) and (0,b). The mid point AB is
a b , 2 2
But the midpoint is given as (2,5)
a b ∴ = 2 and = −5 2 2 ∴ a=4 and b=10 Hence the equation of the straight line is
x y − =1 4 10
(ie.) 5x2y=20 Circle A circle is the locus of a point in a plane such that its distance from a fixed point in the plane is a constant. The fixed point is called the centre of the circle and the constant distance is called radius of the circle. 1. Find the equation of the circle whose centre is (h,k) and radius r. Let C(h,k) be the centre of the circle and p(x,y) be any point on the circle. CP=r is the radius of the circle.
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CP 2 = r 2 (ie.)
(x − h) + ( y − k) 2
2
= r2
This is the equation of the required circle. Note: If the centre of the circle is at the origin then the equation of the circle is 2. Show that the equation
x 2 + y2 = r 2 .
x 2 + y 2 + 2gx + 2fy + c = 0 represents a circle. Find its centre and radius.
x 2 + y 2 + 2gx + 2fy + c = 0 x 2 + y 2 + 2gx + 2fy = −c Adding
g 2 + f 2 to both sides, x 2 + y 2 + 2gx + 2fy + g 2 + f 2 = g 2 + f 2 − c
(x + g) + ( y + f ) 2
2
=
(
g2 + f 2 − c
)
2
→ (1)
This equation is of the form
(x − h) + ( y − k) 2
2
= r 2 which is a circle with centre (h,k) and radius r.
∴ Equation (1) represents a circle whose centre is (g,f) and radius
g2 + f 2 − c
Note 1: 2
A second degree equation in x and y will represent a circle if the coefficients of x and
y 2 are
equal and xy term is absent. Note 2: If
g 2 + f 2 − c is positive then the equation represents a real circle. If g 2 + f 2 − c is zero then the
equation represents a point. If
g 2 + f 2 − c is negative then the equation represents an imaginary circle.
3. Find the length of the tangent from the point
P ( x1 , y1 ) to the circle
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x 2 + y 2 + 2gx + 2fy + c = 0
The centre of the circle is
C ( −g, −f ) and
g2 + f 2 − c
radius r =
Let PT be the tangent from P to the circle. The
PT 2 = PC2 − r 2 = ( x1 + g ) + ( y1 + f ) − ( g 2 + f 2 − c ) 2
2
= x12 + 2gx1 + g 2 + y12 + 2fy1 + f 2 − g 2 − f 2 + c = x12 + y12 + 2gx1 + 2fy1 + c ∴ PT = x12 + y12 + 2gx1 + 2fy1 + c Note: (i) If PT > 0 then the point 2
(ii) If
( x1 , y1 ) lies outside the circle
PT 2 = 0 then the point ( x1 , y1 ) lies on the circle
(iii) If PT < 0 then the point lies 2
(4) Find the equation of the tangent at The centre of the circle is
( x1 , y1 ) inside the circle
( x1 , y1 ) to the circle
x 2 + y 2 + 2gx + 2fy + c = 0
( −g, −f )
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The slope of the radius CP= CP =
y1 + f x1 + g
∴ The equation of the tangent at ( x1 , y1 ) is y − y1 = m ( x − x1 )
y − y1 = −
(ie.)
( x1 + g ) y1 + f
( x − x1 )
(ie.)
( y − y1 )( y1 + f ) = − ( x1 + g )( x − x1 )
(ie.)
yy1 − y12 + fy − fy1 = − xx1 − gx + x12 + gx1
(ie.)
xx1 + yy1 + gx + fy = x12 + y12 + gx1 + fy1
Adding
gx1 + fy1 + c to both the sides, xx1 + yy1 + g ( x + x1 ) + f ( y + y1 ) + c = x12 + y12 + 2gx1 + 2fy1 + c = 0 since the point
The equation of the tangent at
( x1 , y1 ) lies on the circle
( x1 , y1 ) is
xx1 + yy1 + g ( x + x1 ) + f ( y + y1 ) + c = 0 5. Find the equation of the circle on the line joining the points A ( x1 , y1 ) and B ( x1 , y1 ) as the ends of a diameter. A ( x1 , y1 ) and B
( x1 , y1 )
are the ends of a diameter. Let
P ( x, y ) be any point on the
circumference of the circle.
Then
∠APB = 90° i.e. AP ⊥ r to PB
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The slope of AP is
y − y1 = m1 x − x1
The slope of BP is
y − y2 = m2 x − x2
Since AP ⊥ r PB ,
m1 , m 2 = −1
132
i.e.,
( y − y1 )( y − y2 ) = − ( x − x1 )( x − x 2 ) = 0
i.e.,
( x − x1 )( x − x 2 ) + ( y − y1 )( y − y 2 ) = 0
This is the equation of the required circle. 6. Find the condition that the straight line y=mx+c is a tangent to the circle The centre of the circle is (0,0). The radius of the circle is a. If
x 2 + y2 = a 2
y=mx+c is a tangent to the circle
the ⊥ r distance from the centre on the straight line y=mx+c = radius of the circle.
∴
c 1 + m2
= ±a
c 2 = a 2 (1 + m 2 )
This is the required condition. Example 1: Find the equation of the circle whose centre is (3,2) and radius 3 units. Solution: The equation of the circle is
i.e.,
( x − 3) + ( y + 2 ) 2
2
(x − h) + ( y − k) 2
2
= r2
= 32
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i.e., x 2 − 6x + 9 + y 2 + 4y + 4 = 9 x 2 + y 2 − 6x + 4y + 4 = 0 Example 2: Find the equation of the circle whose centre is (a,a) and radius ‘a’. Solution: The centre of the circle is (a,a) The radius of the circle is a. The equation of the circle is
(x − a) + (y + a) 2
2
= a2
x 2 − 2ax + a 2 + y 2 + 2ay + a 2 = a 2 i.e., x 2 + y 2 − 2ax + 2ay + a 2 = 0 Example 3: Find the centre and radius of the circle
(1) x 2 + y 2 − 14x + 6y + 9 = 0 (2)5x 2 + 5y 2 + 4x − 8y − 16 = 0 Solution:
(1) x 2 + y 2 − 14x + 6y + 9 = 0 Centre is (7,3) Radius = =
g2 + f 2 − c 49 + 9 − 9 = 49 = 7 units
(2)5x 2 + 5y 2 + 4x − 8y − 16 = 0 Dividing by 5,
4 8 16 x 2 + y2 + x − y − = 0 5 5 5
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Centre is
Radius=
134
−2 4 , 5 5 4 16 16 100 + + = = 2 units 25 25 5 25
Example 4: Find the equation of the circle whose centre is (2,2) and which passes through the centre of the circle
x 2 + y 2 − 6x − 8y − 5 = 0 Solution: Centre of the required circle is (2,2). The centre of the circle
x 2 + y 2 − 6x − 8y − 5 = 0 is (3,4) The radius of the required circle is given by
r 2 = ( 2 − 3) + ( −2 − 4 ) 2
2
= 1 + 36 = 37
∴ The equation of the required circle is
( x − 2) + ( y + 2) 2
i.e.,
2
= 37
x 2 + y 2 − 4x + 4y − 29 = 0
Example 5: Show that the line 4x − y = 17 is a diameter of the circle
x 2 + y 2 − 8x + 2y = 0 .
Solution: The centre of the circle
x 2 + y 2 − 8x + 2y = 0 is (4,1)
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Put x = 4 and y = 1 in the equation
135
4x − y = 17 .
∴ 16+1=17 which is true. ∴ The line 4x − y = 17 passes through the centre of the given circle. Hence the given line is a diameter of the circle. Example 6: Prove that the centres of the circles
x 2 + y 2 + 4y + 13 = 0 , x 2 + y 2 + 6x + 8y − 17 = 0 and
x 2 + y 2 − 30x − 16y − 42 = 0 are collinear. Solution: The centres of the 3 given circles are A(0,2), B(3,4) and C(15,8)
The slope of AB is
−2 + 4 2 = 0+3 3
The slope of BC is
−4 − 8 −12 2 = = −3 − 15 −18 3
Since the slopes of AB and BC are equal and B in a common point the points A,B,C are collinear.
PARABOLA If a point moves in a plane such that its distance from a fixed point bears a constant ration to its perpendicular distance from a fixed straight line then the path described by the moving point is called a conic. In other words, if S is the fixed point, l is a fixed straight line and P is a moving point and PM is the perpendicular distance from P on l, such that
SP = a constant, then the locus of P is called a conic. This PM
constant is called the eccentricity of the conic and is denoted by e. If e=1, the conic is called a parabola. If e<1, the conic is called an ellipse. If e>1, the conic is called a hyperbola.
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The fixed point S is called the focus of the conic. The fixed straight line l is called the directrix of the conic. The property
SP = e is called the focusdirectrix property of the conic. PM
1. Equation of a Parabola Let S be the focus and the line l be the directrix. We have to find the locus of a point P such that its distance from the focus S is equal to its distance from the fixed line l.
i.e.,
SP =1 where PM is the perpendicular to the directrix. PM
Draw SX perpendicular to the directrix and bisect SX. Let A be the point of bisection. Let SA=AX=a.
Then the point A is a point on the parabola since
SA =1 AX
Take AS as the xasix and AY perpendicular to AS as the yaxis. Then the coordinates of S are (a,0). Let (x,y) be the coordintes of the point P. draw PN perpendicular to xaxis. PM=NX=NA+AX=x+a
Since
SP = 1, PM
(x − a)
2
SP 2 = PM 2
+ y2 = ( x + a )
2
i.e., y 2 = ( x + a ) − ( x − a ) 2
2
y 2 = 4ax
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This being the locus of the point P, is the equation of the parabola this equation is the simplest possible equation to a parabola and is callec the standard equation of the parabola. Note: 1. The line As (xaxis) is called the axis of the parabola. 2. the point A is called the vertex of the parabola. 3. AY (yaxis) is called the tangent at the vertex. 4. the perpendicular through the focus is called the latusrectum. 5. the double ordinate though the focus is called the length of the latus rectum. 6. the equation of the directrix is x+a=0. 7. the equation of the latus rectum is xa=0. 2. To find the length of the latus rectum Draw
LM ' perpendicular to the directrix
Then
SL =1 LM '
To trace the curve
∴ SL = LM ' = SX = 2a
y 2 = 4ax
1. If x<0, y is imaginary ∴ the curve does not pass through the left side of yaxis. 2. when y=0, we get x=0. ∴ the curve meets the yaxis at only one point i.e., (0,0). 3. when x=0,
y 2 = 0 , i.e., y=0, 0. Hence the yaxis meets the curve at two coincident points 0,0.
hence yaxis is a tangent to the curve at (0,0). 4. if (x,y) is a point on the parabola
y 2 = 4ax, (x, − y) is also a point. Therefore the curve is
symmetrical about the xaxis. 5. as x increases indefinitely, the values of y also increase indefinitely. Therefore the parts of the curve lying on the opposite sides of xaxis extend to infinity towards the positive side of xaxis.
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Different forms of parabola 1. If the focus is taken at the point (a,0) with the vertex at the origin and its axis as xaxis, then its equation is
y 2 = −4ax .
2. if the axis of the parabola is the yaxis, vertex at the origin and focus at (0,a), the equation of the parabola
x 2 = 4ay
If the focus is at (0,a), vertex (0,0) and axis as yaxis, then the equation of the parabola is
x 2 = −4ay
Equation of parabola in parametric form
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The point
139
( at , 2at ) satisfies the equation 2
y 2 = 4ax for all values of t. Hence ( at 2 , 2at ) is a
point on the parabola for all values of t. here the coordinates of a point are expressed in terms of a single
y 2 = 4ax are
variable t. t is called a parameter and the parametric equation of the parabola
x = at 2 , y = 2at . The point ( at 2 , 2at ) is denoted by ‘t’. 3. Condition for the line y = mx + c to be a tangent to the parabola
y 2 = 4ax .
Solution: The equation of the parabola is
y 2 = 4ax
→ ( 2)
y = mx + c
The equation of the line is
→ (1)
Solving equations (1) and (2) we get their points of intersection. The x coordinates of the points of intersection are given by
( mx + c ) = 4ax m 2 x 2 + 2 ( mc − 2a ) x + c2 = 0 2
i.e.,
→ ( 3)
If y = mx + c is a tangent to the parabola then the roots of this equation are equal. The condition for that is the discriminate=0
∴ 4 ( mc + 2a ) = 4m 2 c2 2
i.e., m 2c 2 + 4a 2 − 4mca = m 2c2 4a 2 = 4mca
i.e.,
c=
or
a m
Hence the condition for the line y = mx + c to be a tangent to the parabola is
Substituting
c=
c=
a . m
a in (1) we get m
m 2 x 2 − 2ax +
a2 =0 m2
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140
2
a a mx − = 0 ∴ x = 2 is the x coordinate of the point of contact. m m Also
y = mx + = m.
a m
a a 2a + = 2 m m m
a 2a ∴ The point of contact is 2 , . m m Note:
Any tangent to the parabola
y 2 = 4ax is y = mx +
a . m
4. Show that always two tangents can be drawn from a given point to the parabola. Solution: Let the equation of the parabola y Let
2
= 4ax .
( x1 , y1 ) be a given point.
Any tangent through
( x1 , y1 ) is of the form
y = mx +
a m
Since this tangent passes through ( x1 , y1 ) . We have
a m 2 my1 = m x1 + a y1 = mx1 +
i.e.,
or m 2 x1 − my1 + a = 0 This is a quadratic equation in m.
∴ There are two values for m. for each value of m, there is a tangent and there are two tangents to the parabola from a given point. Note:
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If
141
m1 and m 2 are the slopes of the two tangents, then m1 + m 2 =
y1 a and m1m 2 = . x1 x1
5. Show that the locus of the point of intersection of tangents. The equation of any tangent in the parabola is
y = mx +
a . m
If this passes through
( x1 , y1 ) then
y1 = mx1 +
a . m
or m 2 x1 − my1 + a = 0 If
m1 and m 2 are the slopes of the two tangents from ( x1 , y1 ) then m1m 2 =
a . x1
Since the tangents are perpendicular.
m1m 2 = −1 ∴
a = −1 or x1 + a = 0 x1
∴ The locus ( x1 , y1 ) is x+a=0 which is the directrix. 6. Find the locus of the point of intersection of two tangents to the parabola which make complementary angles with the axis of the parabola. Solution:
Let
( x1 , y1 ) be the point of intersection of tangents. Any tangent to the parabola is
y = mx +
a m
This tangent passes through ( x1 , y1 ) .
∴m 2 x1 − my1 + a = 0 If
m1 , m 2 are the slopes of the two tangents, m1 + m 2 =
y1 x1
Since the tangents make complementary angles with the xaxis their slopes are and m 2
m1 = tan θ
= tan ( 90 − 0 ) . °
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∴ m1.m 2 = tan θ.tan ( 90 − 0 ) ∴
142
°
y1 = 1 or x1 − y1 = 0 x1
( x1 , y1 ) is x − y = 0 .
The locus of
7. Find the equation of the tangent at
( x1 , y1 ) on the parabola y2 = 4ax .
Solution: Let
P ( x1 , y1 ) and Q ( x 2 , y 2 ) be two points on the parabola y 2 = 4ax .
∴ y12 = 4ax1
→ (1)
y = 4ax 2
→ ( 2)
2 2
The equation of the chord joining the points
y − y1 y1 − y 2 = x − x1 x1 − x 2 From (1) & (2)
( x1 , y1 ) and ( x 2 , y 2 )
is
→ ( 3)
y12 − y 22 = 4a ( x1 − x 2 ) y − y1 4a = x − x1 y1 + y 2
i.e., y − y1 =
When the point
4a ( x − x1 ) y1 + y 2
Q ( x 2 , y 2 ) tends to coincide with P ( x1 , y1 ) , the chord PQ becomes the tangent
at P.
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∴ the equation of the tangent at ( x1 , y1 ) is
y − y1 =
2a ( x − x1 ) y1
yy1 − y12 = 2ax − 2ax1 yy1 = 2ax − 2ax1 + 4ax1
i.e., yy1 = 2a ( x + x1 )
Aliter: The equation of the curve is
y 2 = 4ax
Differentiating with respect to x.
2y
∴
dy = 4a dx
dy 2a ( x1 , y1 ) = = Slope of the tangent at ( x1 , y1 ) dx y1
The equation of the tangent at
y − y1 =
( x1 , y1 ) is
2a ( x − x1 ) y1
yy1 − y12 = 2ax − 2ax1
yy1 = 2ax − 2ax1 + 4ax1 (Q y12 = 4ax1 )
i.e., yy1 = 2a ( x + x1 ) 8. Find the equation of normal at
( x1 , y1 ) to the parabola
y 2 = 4ax .
Solution:
The slope of the tangent at
( x1 , y1 ) is
2a y1
∴ the slope of the normal at ( x1 , y1 ) is −
y1 2a
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The equation of the normal at
y − y1 = −
144
( x1 , y1 ) is
y1 ( x − x1 ) 2a
i.e., 2ay − 2ay1 = − xy1 + x1 y 1 i.e., xy1 + 2ay = 2ay1 + x1y1 9.Find the equation of the chord joining the points
' t1 ' and ' t 2 ' on the parabola y 2 = 4ax and deduce
the equation of tangent at ' t ' . Solution: The two points are
( at , 2at ) and ( at , 2at ) . 2 1
2 2
1
2
The equation of the chord is
y − y1 y1 − y 2 = x − x1 x1 − x 2 y − 2at1 2at1 − 2at 2 = x − at12 at12 − at 22 i.e.,
y − 2at1 2 = 2 x − at1 t1 + t 2
y ( t1 + t 2 ) − 2at1 ( t1 + t 2 ) = 2x − 2at12
y ( t1 + t 2 ) = 2x + 2at1t 2
This chord becomes tangent at t if t1
= t2 = t .
∴ the equation of the tangent at ‘t’ is y ( 2t ) = 2x + 2at 2 i.e., yt = x + at 2 10. Find the equations of the tangent and normal at‘t’ on the parabola
y 2 = 4ax .
Solution: The equation of the parabola is
y 2 = 4ax
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Differentiating with respect to x.
dy = 4a dx dy 2 4a 1 at , 2at ) = = ( dx 2.2at t 2y
= the slope of the tangent at ‘t’. The equation of the tangent at‘t’ is
1 x − at 2 ) ( t 2 yt − 2at = x − at 2 y − 2at =
i.e.,
yt = x + at 2
The equation of the normal at ‘t’ is
y − 2at = − t ( x − at 2 ) y + xt = 2at + at 3 11. Find the equation of the chord of contact of tangents from
( x1 , y1 ) on the parabola
y 2 = 4ax .
Solution:
Let QR be the chord of contact of tangents from Let Q and R be the points
( x 2 , y2 )
and
P ( x1 , y1 ) .
( x 3 , y3 )
respectively.
Then the equations of the tangent at Q and R are
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yy3 = 2a ( x + x 3 ) These two tangents pass through
( x1 , y1 ) .
∴ y1y 2 = 2a ( x1 + x 2 ) y 1 y3 = 2a ( x1 + x 3 ) These two conditions show that the points
( x 2 , y 2 ) and ( x 3 , y3 ) lie
on the straight line
yy1 = 2a ( x + x1 ) . ∴ the equation of QR, the chord of contact, is yy1 = 2a ( x + x1 ) . 12. The coordinates of the ends of a focal chord of the parabola Prove that
y 2 = 4ax are ( x1 , y1 ) and ( x 2 , y 2 ) .
x1x 2 = a 2 , y1y 2 = −4a 2 .
Solution: Let the ends of the chord be
( at , 2at ) 2 1
1
and
( at , 2at ) . 2 2
2
The equation of the chord is
y ( t1 + t 2 ) = 2x + 2at1t 2 . This passes through the focus ( a, 0 ) . ∴ 0 = 2a + 2at1t 2 i.e., t1t 2 = −1
x1x 2 = at12 .at 22 = a 2 ( t1t 2 ) = −1 2
y1 y 2 = 2at1.2at 2 = 4a 2 t1t 2 = −4a 2 13. The normal at the point
t 2 = − t1 −
' t1 ' on the parabola y 2 = 4ax cuts the parabola again at ' t 2 ' . Show that
2 . t1
Solution:
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The equation of the normal at
147
y + xt1 = 2at1 + at13
The equation of the chord joining
→ (1)
t1 & t 2
y ( t1 + t 2 ) = 2x + 2at1t 2 → ( 2 ) Equations (1) and (2) represent the same line.
∴ Identifying we get
t1 + t 2 2 2at1t 2 =− = 1 t1 2at1 + at13 ∴ t1 + t 2 = −
2 t1
t 2 = − t1 −
or
2 t1
14. If the normal at two points of the parabola ordinates of the points is
y 2 = 4ax meet on the curve, prove that the product of the
8a 2 .
Solution: Let the two points on the curve be Let these normal at
' t1 ' and ' t 2 '
t1 and t 2 meet at t 3
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∴
t 3 = − t1 −
2 t1
t3 = −t 2 −
2 t2
∴ t1 +
2 2 = t2 + t1 t2
148
( from (1) and ( 2 ) )
1 1 t1 − t 2 = 2 − t 2 t1 t −t t1 − t 2 = 2 1 2 t1 t 2 Since
t1 ≠ t 2 , t1 t 2 = 2
∴ Product of the ordinates = 2at1.2at 2 = 4a 2 t1t 2 = 8a 2 Note: If the normal at
t1 and t 2 meet at t 3 then t1 + t 2 + t 3 = 0 . (add (1) and (2) and use t1t 2 = 2 )
HYPERBOLA 1. Standard Equation
Let S be the focus and the line l be the directrix. Draw SX perpendicular to the directrix. Divide SX internally and externally in the ratio e :1 ( e > 1) . Let A and
A ' be the points of division.
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Since
149
SA ' SA = e and ' = e , the points A and A ' lie on the curve. AX AX
Let AA = 2a and C be its middle point. '
SA = e.AX
→ (1)
SA ' = e.A ' X
→ ( 2)
SA + SA ' = e ( AX + A ' X )
Adding
( CS − CA ) + ( CS + CA ) = e.AA ' 2CS = e.2a CS = ae
Subtracting
SA − SA ' = e ( AX − A ' X ) AA ' = e ( AC + CX ) − ( CA − CX ) a 2a = e.2CX ∴ CX = e
Take CS as the xaxis and CY perpendicular to CX as the yaxis. Then the coordinates of S are (ae,0). Let P(x,y) be any point on the curve. Draw PM perpendicular to the directrix and PN perpendicular to xaxis.
From the focus directrix property of hyperbola,
SP =e PM
SP 2 = e 2 PM 2
( x − ae )
2
+ y 2 = e2 .NX 2 = e2 ( CN − CX )
2
2
a = e2 x − e 2 2 2 2 2 2 x − 2aex + a e + y = e x − 2aex + a 2 x 2 ( e2 − 1) − y 2 = a 2 ( e2 − 1)
Dividing by
a 2 ( e 2 − 1)
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x2 y2 − =1 a 2 a 2 ( e2 − 1) i.e.,
x 2 y2 − 2 = 1 where b 2 = a 2 ( e2 − 1) 2 a b
This is called the standard equation of hyperbola. Note: 1. The curve meets the xaxis at the points 2. When
( a,0 ) and ( −a, 0 ) .
x = 0, y 2 = −a 2 ∴The curve meets the yaxis only at imaginary points i.e., there are
no real points of intersection of the curve and yaxis. 3. If
( x, y )
is a point on the curve,
( x, − y )
and
( −x, y ) are also points on the curve. This
shows that the curve is symmetrical about both the axes. 4. For any value of y, there are two values of x. As y increases, x increases and when y → ∞ , x also → ∞ . The curve consists of two symmetrical branches, each extending to infinity in both the directions. 5.
AA ' is called the transverse axis and its length is 2a.
6.
BB' is called the conjugate axis and its length is 2b.
7. A hyperbola in which a=b is called a rectangular hyperbola. Its equation is x eccentricity is
e=
2
− y 2 = a 2 . Its
a 2 + b2 = 2. a2
8. The double ordinate through the focus S is called latus rectum and its length is
2b 2 . a
'
9. There is a second focus S and a second directrix r to the hyperbola. 2. An important property of hyperbola The difference of the focal distances of any point on the hyperbola is equal to the length of transverse axis.
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SP = ePM S'P = ePM '
S' P − SP = E ( PM ' − PM )
= eMM ' = e.XX ' a = e ( CX + CX ' ) = e 2. = 2a e Equation of hyperbola in parametric form
( a sec θ, b tan θ ) is
a point on the hyperbola
x 2 y2 − = 1 for all values of θ , θ is called a a 2 b2
parameter and is denoted by ‘ θ ’. The parametric equations of hyperbola are x = a sec θ , y = b tan θ .
3. Find the equation of the tangent and normal at
( x1 , y1 ) on the hyperbola
x 2 y2 − =1 . a 2 b2
Solution:
The equation of the hyperbola
x 2 y2 − =1 a 2 b2
Differentiating with respect to x.
2x 2y dy − =0 a 2 b 2 dx dy b 2 x ∴ = dx a 2 y
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b2 x dy = 2 1 = Slope of the tangent at ( x1 , y1 ) dx ( x1 ,y1 ) a y1 The equation of the tangent at
( x1 , y1 ) is
a 2 yy1 − a 2 y12 = b 2 xx 1 − b 2 x12 b 2 xx1 − a 2 yy1 = b 2 x12 − a 2 y12 2
2
Dividing by a b ,
xx1 yy1 x12 y12 − 2 = 2− 2 a2 b a b (Since
( x1 , y1 ) is a point on the curve).
∴ The equation of the tangent is
xx1 yy1 − 2 = 1. a2 b The slope of the normal at
( x1 , y1 ) is
The equation of the normal at
y − y1 = −
−
a 2 y1 b 2 x1
( x1 , y1 ) is
a 2 y1 ( x − x1 ) b 2 x1
b 2 yx1 − b 2 x1 y1 = −a 2 y1x + a 2 x1 y1 a 2 y1x + b 2 x1y = ( a 2 + b 2 ) x1 y1
Dividing by
x1 y1 , a 2 x b2 y + = a 2 + b2 x1 y1
4. Find the equation of tangent and normal at the point
( a sec θ, b tan θ )
on the hyperbola
x 2 y2 − =1 . a 2 b2
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Solution: Differentiating with respect to x.
2x 2y dy − =0 a 2 b 2 dx dy b 2 x ∴ = dx a 2 y b 2 .a sec θ dy = 2 dx ( a sec θ,b tan θ) a .b tan θ
=
b sec θ = Slope of the tangent at θ . a tan θ
The equation of tangent at θ is
y − b tan θ =
b sec θ ( x − a tan θ ) a tan θ
Simplifying we get,
x y sec θ − tan θ = 1 a b The slope of the normal at θ is
−
a tan θ b sec θ
The equation of the normal is
y − b tan θ = −
a tan θ ( x − a sec θ ) b sec θ
Simplifying we get,
ax by + = a 2 + b2 sec θ tan θ
x 2 y2 5. Find the condition that the straight line y = mx + c is a tangent to the hyperbola 2 − 2 = 1 . a b
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Solution:
The equation of the hyperbola is
The equation of the line is
x 2 y2 − =1 a 2 b2
y = mx + c
→ (1)
→ ( 2)
The xcoordinates of the points of intersections are given by
x 2 ( mx + c ) − =1 a2 b2 2
b 2 x 2 − a 2 ( mx + c ) = a 2 b 2 2
x 2 ( b 2 − a 2 m 2 ) − 2a 2 mcx − a 2 ( b 2 + c 2 ) = 0
→ ( 3)
If y = mx + c is a tangent to the hyperbola then the roots of the equation are equal. The condition for that is, The condition for that is,
4a 4 m 2 c 2 + 4 ( b 2 − a 2 m 2 ) a 2 ( b 2 + c 2 ) = 0 a 2 m 2 c2 + b4 + b2 c2 − a 2 m 2 − a 2c 2 m2 = 0 ∴ c2 = a 2 m 2 − b2 This is the required condition. Note: Any tangent to the hyperbola is
y = mx + a 2 m 2 − b 2 .
6. Try: (i) Show that two tangents can always be drawn from a given point to a hyperbola and the locus of the point of intersection of perpendicular tangents is a circle.
(ii) show that the chord of contact of tangents from
( x1 , y1 )
to the hyperbola
x 2 y2 − = 1 is a 2 b2
xx1 yy1 − 2 = 1. a2 b
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Asymptotes: An asymptote to a curve is a straight line which touches the curve at infinity but does not lie altogether to infinity. FUNCTIONS Meaning By the term function, we mean the relationship between any two variable like, supply and price, time and distance, volume and freight etc. which are so related with each other that for any value of one of them, there corresponds a definite value for the other, and thus the second variable is said to be the function of the first one. Definitions “A function is relationship between two real values, x and y which are so related that corresponding to every value of x we get a finite value of y, whereby y is said to be a function of x”. Characteristics From the above definitions, the essential characteristics of a function may be analysed as under: 1. it is a logical relation between two variables or sets in which each element of the first set is related to only one element of the second set. For example, if the two sets A and B are, and
A = {1, 2,3, 4}
B = {3, 6,9,12,15,18} , and we construct another set ‘f’ there from basing on the rule, a is treble of
b, then the resulting set
'f ' = {(1,3) , ( 2, 6 ) , ( 3,9 ) , ( 4,12 )} . This can be diagrammatically represented as
under where we find that ‘f’ is a correspondence in which every element of A is associated with a unique element of B:
(i) Domain The domain of a function consists of the set of all the first elements (coordinates) of the pairs of a functions(s). thus, if a function
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'f ' = {(1, 2 ) , ( 2, 4 ) , ( 3, 6 ) , ( 4,8 )} Then the domain of the function would be,
d ( f ) = {1, 2,3, 4} However, the domain of the algebraic function would be determined as under:
(i) Where,
f (x) =
1 , d ( f ) = set of all real numbers excluding zero. x
(ii) Where,
f (x) =
1 , d ( f ) = set of all real numbers excluding 1. 1− x
(iii) Where,
f ( x ) = x and f ( x ) ≤ 0, d ( f ) = 0 ≤ x ≤ ∞
(iv) Where,
f ( x ) = 4 − x , f ( x ) ≥ 0, d ( f ) = −∞ < x < 4.
(ii) Codomain The co – domain of a function is the set of all elements of the second set B some or all of which may
have
relation
with
the
elements
of
the
Domain
of
the
function.
Thus,
if
A = {1, 2,3, 4} , B = {2, 4, 6,8,10} and f : A → B = {(1, 2 ) , ( 2, 4 ) , ( 3, 6 ) , ( 4,8 )} , then the codomain of the function would be set B i.e.,
{2, 4, 6,8,10} .
(iii) Range The range of a function refers to the set of all second elements (coordinates) of the pairs of a function. Thus, if a function, be,
f : A → B = {(1,3) , ( 2, 6 ) , ( 3,9 ) , ( 4,12 )} , the range of the function would
r(f ) = {3, 6,9,12} . When all the elements of a codomain have relation with some elements of the
Domain, the codomain itself is the range of the function. In case of algebraic functions, however, the range would be determined as under:
(i) where
f (x) =
1 , r ( f ) = set of all real number excluding zero. x
(ii) where
f (x) =
1 , r ( f ) = set of all real numbers excluding zero. 1− x
(iii) where,
f ( x ) = x and f ( x ) ≤ 0, r ( f ) = −∞ ≤ f (x) ≤ 0
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(iv) where
157
f ( x ) = 4 − x , f ( x ) ≥ 0, r ( f ) = 0 < f (x) < +∞.
(iv) Images The term ‘image’ refers to the value of the dependent variable y of the pairs (x,y) of a function: Thus, if the
f = {(1, 2 ) , ( 2, 4 ) , ( 5, 6 )} , then the respective images are, 2, 4 and 6.
(v) PreImage The term ‘preimage’ refers to the value of the independent variable, x of the pair (x,y) of a function. Thus, in the above example, the respective preimages are 1,3, and 5. The images and preimages illustrated above can be shown visàvis as under: Preimages
Images
1
2
3
4
5
6
8. A function of algebraic form may consists of a constant that retains the same value throughout a set of mathematical operations. Conventionally, the initial alphabets like, a,b,c, etc. are used as symbols for constants. There are two types of constants used in a function, viz (i) absolute constant, Illustration: Given,
A = {1, 2,3, 4,5} , B = {1, 2,3, 4,5, 6, 7,8,9,10,..........,15} and f (x) = 3x − 1 , find (i)
the elements in the function f : A → B
(ii)
the domain of the function,
(iii)
the codomain of the function,
(iv)
the range of the function,
(v)
the preimages and images of the function.
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Solution: Given
A = {1, 2,3, 4,5}
B = {1, 2,3, 4,5, 6, 7,8,9,10,..........,15} and the rule, f (x) = 3x − 1 thus, we have,
f (1) = 3(1) − 1 = 2 f (2) = 3(2) − 1 = 5 f (3) = 3(3) − 1 = 8 f (4) = 3(4) − 1 = 11 f (5) = 3(5) − 1 = 14 Now, the various required results will be as under: (i) the elements in the function, f : A → B or (ii) the domain of the function, or
d(f ) = {1, 2,3, 4,5}
(iii) the codomain of the function or (iv) the range of the function or
f {(1, 2 ) , ( 2,5 ) , ( 3,8 ) , ( 4,11) , ( 5,14 )}
cd(f ) = {1, 2,3, 4,5,6, 7,8,9,10,..........,15}
r(f ) = {2,5,8,11,14}
(v) the preimages and the images of the function are: Preimages
Images
1
2
2
5
3
8
4
11
5
14
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Illustration: Given,
A = {1,3,5, 7} and B = {0,1, 2,3, 4,5, 6, 7,8,9,10,..........,30} find, (i)
the function
f (x) = 2x + 5
(ii)
the function
f (x) = x 2 − 1
(iii)
The domain and range of each of the above two functions.
Solution: Given
A = {1,3,5, 7} and B = {0,1, 2,3, 4,5, 6, 7,8,9,10,..........,30}
(i) Under the rule
f (1) = 2(1) + 5 = 5 f (3) = 2(3) + 5 = 11 f (5) = 2(5) + 5 = 15 f (7) = 2(7) + 5 = 19 Hence, the required function or
f = {(1, 7 ) , ( 3,11)( 5,15 ) , ( 7,19 )} ; the required domain or
d(f ) = {1,3,5, 7} and the required range or r(f ) = {7,11,15,19} (ii) Under the rule
f (x) = x 2 − 1 we have,
f (1) = (1) 2 − 1 = 0 f (3) = (3)2 − 1 = 8 f (5) = (5)2 − 1 = 24 f (7) = (7)2 − 1 = 48 Hence the required function or
f = {(1, 0 ) , ( 3,8 )( 5, 24 ) , ( 7, 48 )} the required domain or
d(f ) = {1,3,5, 7} and the required range or r(f ) = {0,8, 24, 48} DIFFERENT TYPES OF FUNCTION There are different types of functions on two different basis viz: (i) theory of sets and (ii) theory of number system. There was briefly identified here as under: (i) On the basis of theory of sets
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I. Into function A function is said to be an into function when there remains at least one element of the codomain which has no relation with any element of the domain. In other words, a function is considered s an into function when its range is a proper subset of its codomain i.e. r(f ) ⊂ cd(f ) .
Examples:
(i) when (ii)
A = {1, 2,3} , B = {1, 2,3, 4} and f : A → B = {(1, 2 ) , ( 2,3) , ( 3, 4 )}
when
A = {a, b, c, d} , B = {a, e,i, o, u}
and
f : A → B = {( a, a ) , ( b, e ) , ( c,i ) , ( d, o )} The following arrow diagram gives the picture of such a function:
2. onto function A function is said to be an onto function when all the elements of the codomain of the function are related to some elements of the domain of the function. In other words, a function is considered an onto function when its range is equal to its codomain i.e.,
r(f ) = cd(f ) . Such functions are also
otherwise called Surjective functions. Examples:
(i)
when
A = {1, 2,3} , B = {a, b, c} and f = {(1, a ) , ( 2, b ) , ( 3, b )}
(ii)
when
X = {1, 2,3, 4,5} , Y = {a, b, c, d} and
f = {(1, a ) , ( 2, b ) , ( 3, c ) , ( 4, d ) , ( 5, d )} The following arrow diagram exhibits the picture of such a function.
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3. One one (11) function A function is said to be oneone function when no element of its codomain is related with more than one element of its domain. Such functions are also called injective functions. Examples: (i)
A = {1, 2,3} , B = {3, 6,9,12} and f = {(1,3) , ( 2, 6 ) , ( 3,9 )}
(ii)
X = {1,3,5, 7} , Y = {1,9, 25, 49} and f = {(1,1) , ( 3,9 ) , ( 5, 25 ) , ( 7, 49 )}
The following arrow diagram exhibits the shape of such a function:
4 Manyone function A function is said to be a manyone function when two or more elements of its domain are related to one element of its codomain. In other words, a function is considered Manyone function where two or more preimages of the domain have the same image in its codomain. Examples: (i) where
A = {a, e,i, o, u} , B = {p, q, r,s} and f : A â†’ B = {( a, p ) , ( e, q ) , ( i, r ) , ( o,s ) , ( u,s )}
(ii) where,
P = {1,3,5} , Q = {14,15,16} and
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The following arrow diagram gives the picture of such a function.
5. oneone into function A function is said to be a oneone into function when a satisfies the requirements of both the oneone and into functions as under: (i) each element of the domain is associated with a separate element of the codomain. (ii) there remains at least one element in the codomain which is not associated with any element of the domain. Examples
(i)
where,
A = {a, b, c} , B = {d, e, f , g} and f : A â†’ B = {( a, d ) , ( b, e ) , ( c, f )}
(ii)
where
X = {1, 2,3} , Q = {1, 4,9,10, 25} and f = {(1,1) , ( 2, 4 ) , ( 3,9 )}
The following arrow diagram bears a testimony of such a function:
6. oneone onto function A function that satisfies the characteristics of both the 11 and onto function as under is called oneone onto function or a Bijective function: (i) each element of the domain is associated with a separate element of the codomain. (ii) no element of the codomain remains unrelated with an element in the domain. Examples: (i) where
A = {c, h, a, r, m} , B = {m, a, r, c, h} and
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f : A → B = {( c, m ) , ( h, a ) , ( a, r ) , ( r, c ) , ( m, h )} (ii) where,
A = {1,3,5, 7} , Q = {1, 27,125,343} and
f : A → B = {(1,1) , ( 3, 27 ) , ( 5,125 ) , ( 7,343)} The following arrow diagram bears a testimony of such a function:
7. Manyone into function: A function that satisfies the characteristics of both manyone and into functions as under is called manyone into function: (i) two or more elements the domain are related to one element of the codomain. (ii) there remains at least one element in the codomain which is not associated with any element of the domain. Examples (i)
P = {h, a, r, m} , Q = {f , a, r, m, e, r} and f : P → Q = {( c, m ) , ( h, a ) , ( a, r ) , ( r, c ) , ( m, h )}
(ii) where,
S = {1, 2,3, 4} , T = {1, 4,9,10,12} and
f : S → T = {(1,1) , ( 2, 4 ) , ( 3,9 ) , ( 4,9 )} The following arrow diagram bears a testimony of such a function:
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8. Manyone onto function A function is said to be a manyone onto function, when it satisfies the characteristics of both the Manyone and onto function as under: (i) two or more elements of the domain are related to one element of the codomain. (ii) no element of the codomain is left unrelated with any element of the domain. Examples: (i) Where
A = {1, 2,3, 4} , B = {3, 6,9} and
f : A â†’ B = {(1,3) , ( 2,3) , ( 3, 6 ) , ( 4,9 )} (ii) Where,
C = {c, r, e, d,i, t} , D = {c, a,s, h} and
f : C â†’ D = {( c, c ) , ( r, a ) , ( e,s ) , ( d, h ) , ( i, h ) , ( t, h )} The following arrow diagram gives the picture of such a function:
9. Constant function: A function is said to be a constant one when all the elements of its domain are associated with a single element of its codomain. Example: (i) where,
A = {1, 2,3, 4} , B = {4,5, 6, 7,8} and f = {(1, 6 ) , ( 2, 6 ) , ( 3, 6 )}
The following arrow diagram bears a testimony of such a functions:
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10. Inverse function A function that is obtained by interchanging the ordered pairs of a oneone onto function is called an
inverse
function.
It
is
denoted
by
f −1 : B → A,
or
by
g(y) .
Thus,
where
A = {1, 2,3} , B = {a, b, c} and f : A → B = {(1, a ) , ( b, 2 ) , ( c,3) , ( 2, b ) , ( 3,c )} , the inverse function or f −1 : B → A = {( a,1) , ( b, 2 ) , ( c,3)} . The above phenomena of a function and its inverse can be diagrammatically represented as under:
Where f : A → B =
Where
f −1 : B → A =
From the above description, it must be noted that to obtain an inverse function from a given function the following conditions must be fulfilled: (i) The number of elements in the domain of the function must be equal to those of its codomain. (ii) The given function must be of the type of 11 and onto. If an y of the above two conditions is not satisfied, then there can not exist an inverse of the given function. The examples of such a function can be taken as under:
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Function (f)
If
y = x2
If y = x − 5
166
Inverse function
f −1
x=± y
Then
Then x = y + t
If
y=
5x + 3 2x + 9
Then
x=
3 − 9y 2y − 5
If
y=
ax + b cx + d
Then
x=
dy − b a − ay
If
y = sin −1 x
If
Then x = sin y
x=3y
Then
y = x3
On the basis of theory of number system 1. Real value function When a function is a set of some real numbers it is called a real value function. A real function is of three types, viz: (i) Single value function (ii) Double value functions, and (iii) Multivalue functions. When a function has only one value corresponding to each value of the independent variable (i.e., of domain). It is called a single or one value function, e.g.,
y = x 2 . When a function has two values
corresponding to each value of the independent variable, it is called a two or double value function, e.g.,
y = x , i.e.,
x and − x . When a function has several values corresponding to each value of the
independent variable, it is called a Multi value or Many valued function. E.g.
y = x 2 + 3x − 7
2. Absolute value function A function is said to be an absolute value function or a Modulus function when it is defined as follows:
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y = f (x) = x, for x > 0 = 0, for x = 0 = − x, for x < 0 Example: When
x = 5, y = 5 = 5 x = 0, y = 0 = 0 x = −5, y = −5 = −5 3. Explicit function A function that is expressed directly in terms of the dependent variable is called an Explicit function. For example,
y = x 2 + 7x − 4
4. Implicit function A function that is not expressed directly in terms of the dependent variable and in which either of the two variables determines the other, is called an implicit function. For example,
x=
4x − 5y = 0 where
5 4 y and y = x 4 5
5. Continuous function A function is said to be continuous at a point x=a, if f(x) possesses a finite and definite limit as x tends to the value, a from either lower or upper side of ‘a’ say viz. : 2.9, 2.99 etc. or 31, 301 etc. respectively, and each of these limits is equal to f(a). 6. Discontinuous function If any of the essential conditions of a continuous function is not satisfied then the function is called a Discontinuous function. 7. Even function A function that satisfies
f ( − x ) = f ( x ) is called as Even function. The examples of such a
function are:
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8. Odd function A function that satisfies are:
f ( x ) = −f ( x ) is called an odd function. The example of such a function
f ( x ) = sin x, f ( x ) = x 3 , f ( x ) = 5x + 7x 3 etc.
9. Periodic function A function in which the range can be separated into equal subintervals in such a manner that the graph of the function remains the same in each of the part intervals, is called a periodic function. 10. Composite function A function is said to be a composite function or a function of functions, when it becomes a function of another function. Thus, if
y = g ( u ) , u = g ( x ) , and y = ( g ( x ) ) then y is a function of
function or a composite function. For example, the volume of a tank is the function of its area while the area itself is the function of its radius (i.e.
A = πr 2 ), and so the volume is the function of the radius of the
area. The phenomenon of a composite function can be better understood from the following diagram:
Illustration: Find the composite function (i)
and
i) f ( g ( x ) ) , and ( ii ) g ( f ( x ) ) from the following two functions:
(a)
f ( x ) = x 2 + 3, x ∈ R
(b)
g ( x ) = x + 3,
x∈R
also comment on the result. Solution
i)
f ( g ( x ) ) = f ( x + 3) = ( x + 3) + 3 2
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(ii)
g ( f ( x ) ) = g ( x 2 + 3) = x2 + 3 + 3 = x2 + 6 Comment From the above result it is observed that
f (g ( x )) ≠ g (f ( x )) .
11. Algebraic function An algebraic function is one, the variables x, y etc. of which are finite in number and are affected by the operations of addition, subtraction, multiplication, division, power, and roots, the examples of such functions are:
(i) y = 3x 2 + 2x − 7 (ii) y =
1 − x , etc x3
12. Transcendental function A function that is not algebraic in nature is called a transcendental or nonalgebraic function. Such functions include exponential functions, logarithmic functions, trigonometric functions, and inverse trigonometric functions etc. which are explained in the paragraphs to follows: 13. Exponential function An exponential function is a nonalgebraic function in which the independent variable, x is used as an index. For example,
y = e x , y = a 2x y = ka x , y = b x etc. where K ≠ 0 , a and e>0 but ≠ 1 .
Illustration 8. Find the elements of the exponential function,
y = 3x up to first 5 numbers
Solution: Given
y = 3x (up to first five numbers)
Thus,
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f (1) = 31 = 3 f ( 2 ) = 32 = 9 f ( 3) = 33 = 27 f ( 4 ) = 34 = 81 f ( 5 ) = 35 = 243 Thus,
f ( 3x ) = {(1, 3) , ( 2,9 ) , ( 3, 27 ) , ( 4,81) , ( 5, 243)}
14. Logarithmic function A nonalgebraic function in which logarithmic terms are used is called a logarithmic function. The examples of such function are:
y = log x, f ( x ) = log ( 5x − 7 ) , y = log.3x 2 − 2x + 5etc 15. Trigonometric function A nonalgebraic function in which the trigonometric ratios (i.e., the relations between any two of the three sides of a triangle) are used is called a Trigonometric function. The examples of such a function are:
y = sin −1 x, f ( x ) = cos x, y = cos ec2 ( x + 2 ) , y = tan x And the inverse trigonometric functions like,
y = sin −1 x, f ( x ) = cos −1 x etc. 16. Polynomial function A function of the form
f ( x ) = ax n + a1x n −1 + a 2 x n − 2 + ....... + a n −1x + a n is called a polynomial
function, where x is a positive integer, and
a 0 , a1 , a 2 etc. are all constants, Such functions are of various
types such as (i) Constant functions of degree 0. (ii) Linear function of degree 1, (iii) Quadratic function of degree 2 and, (iv) Cublic function of degree 3.
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These are briefly identified as under: 17. Rational function An algebraic function that can be expressed as a ration of two polynomials is called a Rational function. The form of such a function is
r (x) =
p(x) , where p ( x ) and q ( x ) are polynomials q (x)
and q ( x ) ≠ 0 . The examples of such a function are:
(i)
r (x) =
x 2 − 5x + 2 x2 + 5 , (ii) etc. x2 + 3 3x 2 + 7
18. Irrational function A function in which the independent variable, x has root extraction of terms is called an irrational function. The examples of such a function are:
(i ) f ( x ) =
x , ( ii ) y = x 2 + 5x + x − 3 + 7x − 5
19. Menotonic function A function in which the dependent variable changes with a change in the value of the independent variable is called Monotonic function. When the dependent variable increases with an increase in the independent variable the function is called a monotonically increasing function. For example, the supply as a function of price. But when the dependent variable decreases with an increase in the independent variable, it is called a monotonically decreasing function. For example, the demand as a function of price. Thus the function, f : x → 2x, x ∈ R is a monotonic increasing function, and
f :x →
1 is a monotonic decreasing function. Where x>0 x
20. Identity function A function in which each element of the domain corresponds to itself is called an identity functions. The expression of such a function runs as under:
f ( x ) = {(1,1) , ( 2, 2 ) , ( 3,3)}
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21. Signum function A function is called a signum function when it is defined as follows:
+1, when x > 0 sign. x = 0, when x = 0 −1, when x < 0 Such a function is denoted by
sign. x =
x x
, where x ≠ 0
22. Greatest Integer function A function is called the Greatest integer function when it is expressed as
f ( x ) = x for the
greatest integer less than or equal to x. The examples, of such a function are:
f ( 2.9 ) = 2, f ( 0.7 ) = 0, f (−2.9) = −3 The elements of the function with the above element of the domain will be as under:
f=
{( ( 2.9 ) , 2) , ( ( 0.7 ) , 0 ) , ( ( −2,9 ) , −3)}
Such functions are, also, otherwise called as step functions. 23. Equal function A function is said to be an equal to another functions, if and only if its domain and range are equal to the corresponding domain and range of that another function. Thus,
f ( x ) = g ( x ) , where, d ( f ) = d ( g ) and r ( f ) = r ( g )
24. Parametric function A function is said to be a parametric one when both of its independent variable, x and dependent variable, y are expressed in terms of another third variable. Thus, if
y = f ( t ) , where x = f ( t ) and y = t 2 + 1 where x=2t are the examples of parametric
function.
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Illustration: If
f ( 0 ) , f (1) , f ( 2 ) , f ( −2 ) ,f ( a ) and f ( m ) as well as the ordered pairs of the function.
Solution
f ( x ) = x 3 − 4x + 9
Given, rule of the function, Hence, we have,
f ( 0 ) = 03 − 4 ( 0 ) + 9 = 9 f (1) = 13 − 4 (1) + 9 = 6 f ( 2 ) = 23 − 4 ( 2 ) + 9 = 9 f ( −2 ) = −23 − 4 ( −2 ) + 9 = 9 f ( a ) = a 3 − 4 ( a ) + 9 = a 3 − 4a + 9 f ( m ) = m3 − 4 ( m ) + 9 = m 2 − 4m + 9 And
{
}
f = ( 0,9 ) , (1, 6 ) , ( 2,9 ) , ( −2,9 ) , ( a, a 2 − 4a + 9 ) , ( m, m 2 − 4m + 9 )
Illustration If
( 2 ) , g ( −2) and g ( −1 2 )
g ( x ) = 5x , find g ( 0 ) , g (1) , g ( 2 ) , g 1
Solution: Given,
g ( x ) = 5x
Thus,
g ( 0 ) = 50 = 1 g (1) = 51 = 5 g ( 2 ) = 52 = 25
( 2) = 5
g 1
1
2
= 5
g ( − 2 ) = 5 −2 =
( 2) = 5
g −1
−1
2
1
( 5) =
2
=
1 25
1
=
( 5)
1
2
1 5
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ALGEBRA OF FUNCTIONS By algebra of functions we mean the application of the four basic operations of addition, subtraction, multiplication, and division on any two or more functions. It is performed just in the same manner as it is done in case of simple arithmetic. Thus, if f(x) and g(x) are any two functions, then their four algebraic operations will be performed as under: (i) Addition of f(x) and g(x)=f(x)+g(x) (ii) Subtraction of g(x) from f(x)=f(x)g(x) (iii) Multiplication of f(x) and g(x)=f(x).g(x)
(iv) Division of f(x) by
g (x) =
f (x) g (x)
The following illustration will show how algebra of functions are performed. Illustration Perform the four basic operations on the following two functions: Solution: (i) Addition:
f ( x ) + g ( x ) = ( 2x 2 − 5 ) + ( 3x + 5 )
= 2x 2 − 5 + 3x + 5 = 2x 2 + 3x (ii) Subtraction:
f ( x ) − g ( x ) = ( 2x 2 − 5 ) − ( 3x + 5 ) = 2x 2 − 5 − 3x − 5 = 2x 2 − 3x − 10
(iii) Multiplication:
f ( x ) .g ( x ) = ( 2x 2 − 5 ) . ( 3x + 5 ) = 6x 3 − 15x + 10x 2 − 25 = 6x 3 + 10x 2 − 15x − 25
2 f ( x ) ( 2x − 5 ) 2x 2 − 5 (iv) Division: = = g ( x ) ( 3x + 5 ) 3x + 5
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Illustration
f ( x ) = [ x ] , x ∈ R , where x ∈ R find
(i)f + g, (ii) f − g, (iii) f .g, (iv)
f g
Solution Given
f ( x ) = [ x ] and g ( x ) = x
Thus,
(i) f ( x ) + g ( x ) = x + x = 2x, when x > 0 or = − x + x = 0, when x < 0 (ii) f ( x ) − g ( x ) = x − x = 0, when x > 0 or = − x − x = −2x, when x < 0 or = 0 − x = − x, when x = 0
(iii) f ( x ) .g ( x ) = x.x = x 2 , when x > 0 or = − x.x = − x 2 , when x < 0 or = 0.x = 0, when x = 0 (iv)
f (x) x = =1 g (x) x or = − or =
x = −1, when x < 0 x
0 = −1, when x = 0 x
Illustration If
f ( x ) = 2x 2 + 4, and h ( x ) = 5x + 3
find
( i )( f + h )( 3) , ( ii )( fh )( 7 ) ,
( iii )( f / h )( 2 ) , (iv)f ( h (1) )
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Solution
( i )( f + h )( 3) = f ( 3) + h ( 3) 2 2 = 2 ( 3 ) + 4 + 5 ( 3 ) + 3 = 18 + 4 + 15 + 3 = 40
( ii )( fh )( 7 ) = f ( 7 ) + h ( 7 ) 2 = 2 ( 7 ) + 4 . 5 ( 7 ) + 3 = [98 + 4] .[35 + 3] = 102 × 38 = 3876
( iii )( f / h )( 2 ) =
f ( 2)
h ( 2)
2 ( 2 ) + 4 12 = 5 ( 2 ) + 3 13 2
=
(iv)f ( h (1) ) = f ( 5 ×1× 3) = f ( 8 )
= ( 2 × 82 + 4 ) = 128 + 4 = 132
Note:
(i) c ( f + h ) = cf + ch
(ii) c ( f − h ) = cf − ch (iii) c ( fh ) = cf.ch (iv)c ( f / h ) =
cf ch
VALUE OF A FUNCTION By value of a function we mean the computed value of a given function which we obtain by substituting the given value of x in place of x. thus, f(5) is the value of the function f(x), for x=5; f(a) is the value of the function, f(x), for x=a. The following illustrations will show how the value of a function is determined under the procedure explained above.
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Illustration Determine the value of the function
f ( x ) = 5x + 7 , for x=0, 1 and 1/5
Solution: Given,
f ( x ) = 5x + 7
Thus (i)
the value of
f ( 0 ) = 5(0) + 7 = 7
(ii)
the value of
f (1) = 5(1) + 7 = 12
(iii)
the value of
f 1
( 5 ) = 5 ( 15 ) + 7 = 8
Illustration:
If f and g are real functions defined by value of the function
f (x) =
x+2 3 2 , and g ( x ) = x − 2x + x + 2 , find the x −3
(i) g ( −2 ) , (ii) f ( 9 ) , (iii)f ( g ( 3) ) , (iv)f .g ( 2 )
Solution: Given f ( x ) =
x+2 3 2 , and g ( x ) = x − 2x + x + 2 x −3
Thus (i)
the value of the function
g ( −2 ) = ( −2 ) − 2 ( −2 ) − 2 + −2 2
2
= −8 − 2 ( 4 ) = −16 9 + 2 11 = 9−3 6
(ii)
the value of the function
f (9) =
(iii)
the value of the function
f ( g ( 3) ) = f
{(3) − 2 (3) 3
2
}
−3+ 2
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= f {9 − 18 − 1} = f ( −10 ) = (iv)
−10 + 2 −8 8 = = −10 − 3 −13 13
the value of the function
f .g ( 2 ) = f ( 2 ) .g ( 2 )
x+2 3 2 = ( 2 ) ( x − 2x − x + 2 ) ( 2 ) x−2 2 2+2 3 = 2 − 2 ( 2 ) − 2 + 2 2−3 4 = ( 8 − 8 ) = −4 ( 0 ) = 0 −1 1.5. GRAPHIC REPRESENTATION OF FUNCTIONS When the ordered pairs of values of any two variables given by a function are represented through a graph drawn on a graph paper. It is called graphic representation of a function. The picture of a graph will be different for the function set under different rules. However, for representing any function through a graph we are to go by the following steps in turn. Steps (i)
Tabulate the ordered pairs of values of the two variables x and y according to the rule or definition of the given function.
(ii)
Draw the two lines of x abscissa, and y ordinate on a graph paper marking the point of their intersection as 0.
(iii)
Divide both X line and the Y line into a number of equal parts on the basis of a natural scale so as to accommodate thereon the maximum values of the respective variables.
(iv)
Plot the dots at the coordinated points of each of the pairs of the values of x and y, and draw a curve by joining all these points in a freehand manner. The curve thus obtained is the required graph of the given function.
While drawing such a curve, the points for which the curve is not defined are to be excluded and encircled to indicate that they are not included in the graph. Such an occasion arises only when the function is a discontinuous one. The following illustration will show how various functions can be represented through graphs.
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Illustration: Make a graphic representation of the function given below:
f ( x ) = 2 + 2x Solution: The given function f ( x ) = 2 + 2x is a linear function as the index of the independent variable x is only 1. For representing such a function through a graph we need only two coordinating points to draw the curve of a straight line. Thus, we tabulate the two ordered pairs for the two variables x and y on the basis of
f ( x ) = 2 + 2x as under: X
0
4
Y
2
10
With the above ordered pairs, the graph of the function will be drawn as under: Graphic representation of the linear functions,
f ( x ) = 2 + 2x
Illustration. x
Draw the graph of the function f(x) = 2 . Solution. The given function is an exponential one. The ordered pairs of values for such a function may be tabled as under :
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x
0
1
2
3
4
y
1
2
4
8
16
With the above pairs of values, the required graph will be drawn as under : Graphic representation of the function (fx) = 2
x
Illustration. Represent graphically the following function : y = f(x) = x
2
Solution: The given function is a polynomial function of degree two. The ordered pairs of values for such a function may be tabulated as under : x
5
4
3
2
1
0
1
2
3
4
5
y
25
16
9
4
1
0
1
4
9
16
25
With the above pairs of values of x and y, the required graph will be drawn as under : Graphic representation of the function, f(x) = x
2
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SEQUENCE AND SERIES A set of numbers arranged in a specific order is a sequence. Each number in the sequence is called a term. Generally a sequence is written as
u1 , u 2 , u 3 ,....
u1 , u 2 and u 3 are the first, second and third terms of the sequence respectively. From the first three terms, succeeding terms can be determined when they have specific relation. The following are some of the examples of sequences. (i)
1, 5, 9, …………………………
(ii) 2, 6, 18, ………………………. (iii) 7, 49, 343, ……………….. When the successive terms of a sequence are connected by plus or minus signs, the sequence is called a series. For example,
(i)
u1 + u 2 + u 3 + .............
(ii)
1+5+9+……………………….
(iii)
7+49+343+………………. are series.
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A series is called a finite series if it contains finite number of terms. It is called an infinite series if it contains infinite number of terms. Two important types of series, namely, (i) Arithmetic series of arithmetic Progression (A.P.) and (ii) Geometric series or Geometric Progression (G.P.) are considered hence forth. Harmonic series or Harmonic Progression (H.P.) is mentioned later. ARITHMETIC PROGRESSION (A.P.) If the successive terms increase or decrease by a constant (quantity), the series is called Arithmetic Progression (A.P.). That constant quantity is called the common difference. e.g. 2,7,12,………………. is an A.P. 2 is the first term while 5(=72=127) is the common difference. The alphabets ‘a’ and ‘d’ are used to denote the first term and the common difference re respectively. a,a+d,a+2d,………………is the standard form of an A.P. When d is positive, the series is an increasing A.P. The series is a decreasing A.P. when d is negative.
In this, n th term, denoted by
t n is the general term and t n = a + (n − 1)d.
Having known that a sequence is an A.P., any two of its terms determine others. Example: The fourth and seventh terms of an A.P. are 3 and 36. Find the A.P. and its fifteenth terms. Solution:
th
=3……………… (1)
7 term, a+6d
th
=36……………. (2)
(2)  (1),
3d=33
4 term, a+3d
d=33/3=11 by substituting in (1), a + 3 ×11 = 3 a =333=30
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Hence, the A.P. is – 30, 30+11,
30 + 2 ×11,...............
That is, 30, 19, 8,……………………is the A.P.
t15 = a + 14d
th
The 15 term,
= −30 + 14 ×11 = 124 Example: In an Arithmetic series, the seventh and the ninth terms are respectively 16 and 20. Find the nth term. Solution: Given
t 7 a + 6d = 16.............(1) t 9 a + 8d = 20.............(2) (2) − (1), 2d = 4 ∴ d = 4/2 = 2 By substituting in (1),
(1),a + 6 × 2 = 16
∴ a = 16 − 12 = 4 Hence the nth term,
t n = a + (n −1)d = 4 + (n −1)2
= 4 + 2n − 2 = 2 + 2n Example: The sum of three numbers in Arithmetic Progression is 24 and their product is 440. Find the numbers. Solution: Let the three numbers be ad, a, a+d. Given: (ad)+a+(a+d)=24 That is,
3a=24
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∴ a=24/3=8………..(1) Also given: (ad).a.(a+d).=440] By (1), this becomes, (ad).8.(8+d)=440 That is,
(8d).(8+d)=440/8=55]
That is,
64 d =55
2
d 2 = 64 − 55 = 9
That is,
∴
d = ± 9 = ±3
Case 1: a=8 and d=3. The three numbers are 5, 8 and 11. Case 2: a=8 and d=3. The three number are 11, 8 and 5. Note: If four numbers are to be found, assume them as a3d, ad, a+d and a+3d.
their common
difference is 2d. Example: Find three numbers in A.P. whose sum is 12 and the sum of whose cubes is 408.
Solution: Let the three numbers be ad, a and a+d Given:
(ad)+a+(a+d) That is,
3a
∴ Also given
=12 =12
a = 12 / 3 = 4..........(1)
(a − d)3 + a3 + (a + d)3 = 408
By (1), this becomes
(4 − d)3 + 43 + (4 + d)3 = 408
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That is,
185
43 − 3.42 d + 3.4d 2 − d3 + 64 + 43 + 3.42 d + 3.4d 2 + d3 = 408 2
That is, 192+24 d =408
∴ 24d 2 = 408 − 192 = 216 ∴ d 2 = 216 / 24 = 9 ∴ d = ± 9 = ±3 Case 1: a=4 and d=3. The three numbers are 1,4 and 7 Case 2: a=4 and d=3. The three numbers are 7,4 and 1 Formula for the sum of the first n terms of an A.P.:
Prove
Sn =
n [ 2a + (n − 1)d ] Vn ∈ N where Sn denotes the sum of the first n terms of an A.P. and n 2
is a natural number, that is, for all the values of n=1,2,3,……….. N denotes the set of natural numbers. (B.Com. Bharathiar, A94 ) Derivation:
Sn = a + (a + d) + (a + 2d) + ...................... + (1 − d) + 1 where n term. ∴1 = a + (n − 1)d. In reverse order, th
By adding,
1 denotes the last term. i.e.,
Sn = 1 + (1 − d) + (1 − 2d) + ............. + (a + d) + a
2Sn =(a+1)+(a+1)+(a+1).......... = n(a+1)
n n (a + 1) = [ a + a + (n − 1)d ] 2 2 n = [ 2a + (n − 1)d ] 2
∴ Sn =
Note:
Sn =
n (a + 1) is also useful (when 1 is known). 2
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Example : Find the sum of the following series.
(i) 8+13+18+…………………upto 23 terms (ii) 3
1 1 3 1 + 5 + 7 + ...................... + 23 4 2 4 2
(iii) 40+36+32+…………………… Solution: (i) The given series is an A.P. with common difference, d=138=1813=5 and first term, a=8. N=23. Required sum is given by
n Sn = [ 2a + (n − 1)d ] 2 23 = [ 2 × 8 + (23 − 1)5] 2 = 11.5[16 + 110] = 1449
(ii) The given series is an A.P. with common difference,
first term,
1 a = 3 = 3.25 4
Assuming
1 23 as n th term,. 2
i.e., i.e., i.e., i.e.,
1 1 3 1 1 d = 5 − 3 = 7 − 5 = 2 = 2.25 and 2 4 4 2 4
t n = a + (n − 1)d = 23.5 3.25 + (n − 1)2.25 = 23.5 (n − 1)2.25 = 23.5 − 3.25 = 20.25 (n − 1) = 20.25 / 2.25 = 9 n = 9 + 1 = 10
Required sum is given by
n Sn = (a + 1) 2 =
10 (3.25 + 23.50) = 133.75 2
(iii) The given series is an A.P. with common difference, d=3640=3236=4 and first term, a=40. Assuming 0 as nth term,
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t n = a + (n − 1)d = 0 40 + (n − 1)(−4) = 0 (n − 1)(−4) = 0 − 40 = −40 (n − 1) = −40 / − 4 = 10 n = 10 + 1 = 11
i.e., i.e., i.e., i.e., Required sum is given by
187
n Sn = (a + 1) 2 =
11 (40 + 0) = 220 2
Example: If S1, S2 and S3 be respectively the sum of the first n, 2n and 3n terms of a A.P., prove that S3=3(S2S1)
Solution: By substituting the values of n in
n Sn = [ 2a + (n − 1)d ] , weget 2 3n S = [ 2a + (3n − 1)d ] 3 2 9 3 = 3na + n 2d − nd → (1) 2 2 2n S = [ 2a + (2n − 1)d ] 2 2 n 1 1 S = [ 2a + (n − 1)d ] = na + n 2d − nd → (3) 1 2 2 2 (2) − (3)is multiplied by3and hence, 3 1 9 3 3 (S − S ) = 3 na + n 2d − nd = 3na + n 2d − nd = S by (1) 2 1 3 2 2 2 2
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Example: The first and the last terms of an A.P. are –4 and 146 and the sum of the A.P. is 7171. Find the number of terms in the A.P. and the common difference. Solution: We know that
n ( a + 1) 2 Given a = −4 1 = 146 and sn = 7171 n That is, ( −4 + 146 ) = 7171 2 That is 71n = 7171 sn =
∴
No of terms in the A.P n=7171/71=101
∴
By substituting the known values in tn=a+(n1)d
We get 146=4+100d
∴
100d=150
and common difference, d=150/100=1.5 Example: 2
Sum of the first n terms of a series is 3n +6n. show that it is an A.P. Which term of the series is 105?. Solution: Given
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Sn = 3n 2 + 6n ∴ s n −1 = 3 ( n − 1) + 6 ( n − 1) 2
= 3 ( n 2 − 2n + 1) + 6n − 6 = 3n 2 − 3
t n = Sn − Sn −1
= 3n 2 + 6n − ( 3n 2 − 3) = 6n + 3
By substituting n=1,2 and 3. We get
t1 = 9, t 2 = 15 and t 3 = 21. As t 2 − t1 = t 3 − t 2
the series is an
A.P. Further, a=9 and d=6 Let tn=105
a + ( n − 1) d = 105
9 + ( n − 1) d = 105
6 ( n − 1) = 105 − 9 = 96 n − 1 = 96 / 6 = 16 n = 16 + 1 = 17 th
Hence, 17 term of the given A.P. is 105. Example: The first term of an Arithmetic Series is 5. The number of terms is 15 and their sum is 390. Find the common difference and the middle term. Solution:
Given
a = 5, n = 15 and S15 = 390. To find d and t8 consider
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n 2a + ( n − 1) d = Sn 2 15 [ 2 × 5 + 14d ] = 390 2 2 10 + 14d = 390 × = 52 15 14d = 52 − 10 = 42 ∴ Common difference, d = 42 /14 = 3 The middle term,
t 8 = a + ( n − 1) d
= 5 + ( 7 × 3) = 26
Formula for the Sum of Natural Number 1,2,3….are called natural numbers. 1,2,3,…n are the first n natural numbers. They are in A.P. with a=1, d=1, 1=n and n=n. Hence the sum of the first n natural numbers,
n ( a + 1) 2 n ( n + 1) n (1 + n ) = 2 2 sn =
Example: Find the sum of the first 100 natural numbers. Solution: By numbers=
substituting
n=100
in
the
formula,
the
sum
of
the
first
100
natural
100 ×101 = 5,050 2
Example: Find the sum of all natural numbers between 100 and 1000 which are divisible by 13. Solution:
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Between 100 and 1000, 104 and 988 are the first and last numbers divisible by 13. Hence the required sum is 104+117+130+….+988 Hence, a=104, d=13 and 1=988 To find n, consider
t n = 988
a + ( n − 1) d = 988
104 + ( n − 1)13 = 988
( n − 1)13 = 988 − 104 = 884
n − 1 = 884 /13 = 68 n = 68 + 1 = 69 n ∴ 104 + 117 + 130 + .... + 988 = ( a + 1) 2 69 = (104 + 988 ) 2 = 37,674 First, 100+101+102+…..+1000 is found out as
n 901 (100 + 1000) = 4,95,550 . ( a + 1) = 2 2 Therefore, the sum of numbers between 100 and 1000 which are not divisible by 13.
= 4,95,550 − 37,,674 = 4,57,876 Example: Find the sum of integers from 1 to 100 that are divisible by 2 or 5. Solution: Sum o integers from 1 to 100 that are divisible by 2 is 2+4+6+….+100. As a=5, 1=100 and n=20, their sum
2 + 4 + 6 + .... + 100 =
50 ( 2 + 100 ) = 2550 2
(1)
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5 + 10 + 15 + .... + 100 =
20 ( 5 + 100 ) = 1050 2
(2)
The number which are divisible by both 2 and 5 have been added in both the series. Their sum is to be subtracted once.
10 + 20 + 30 + ..... =
10 (10 + 100 ) = 550 2
(3)
(1)+(2)(3) gives the sum of integers from 1 to 100 that are divisible by 2 or 5 as
2550 + 1050 − 550 = 3050 . ARITHMETIC MEANS (A.M) If a series of n terms is in A.P., the first and the last terms are called the extremes and the intermediate terms are called arithmetic means. When there are only 3 terms, there is only one A.M.It is the mean of the first and the last terms. That is, if a, A,b are in A.P., A is the A.M of a and b and
A=
(A) between two given numbers (a and b), we find their mean
a+b 2
Aa=bA. Hence, to insert one A.M
a+b A = 2
To insert n arithmetic means (x1,x2,..xn) between 2 numbers (a and b), we have to remember that a, x1,x2,….xn , b is an A.P.
∴ The last term which is (n+2)th term is b
b = a + ( n + 2 − 1) d
∴ ( n + 1) d = b − a b−a ∴ d= n +1 In words, common difference,=
Hence, the first A.M.,
Last term − First term Number ofA.M.'s inserted + 1 b−a n +1
x1= a +
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b−a n +1
The second A.M.., x2 = a + 2
b−a n +1
The nth (last) A.M., xn= a + n
Example: Insert one A.M. between 70 and 50. Solution: Given a=70 and b=50
Required to find
A=
a + b 70 + 50 = = 60 2 2
Example: Insert 5 A.M’s. between 23 and 47. Solution: Given n=5, a=23 and b=47
∴ Common difference, d =
b − a 47 − 23 = =4 n +1 5 +1
The five A.M.’s between 23 and 47 are 27, 31, 39 and 43 GEOMETRIC PROGRESSION (G.P) If the successive terms increase or decrease by a constant factor, the series is called Geometric Progression(G.P) That constant factor is called the common ratio and is denoted by x.
e.g. 7, 42,252………. is a G.P 7 is called the first term while
42 252 = =6 7 42
is called the common ratio
(constant factor). The alphabets a and r are used to denote the first term and the common ratio respectively.
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a, ar,ar 2 ,...
194
is the standard form of a G.P
When r>1, the series is an increasing G.P. The series is a aecreasing G.P. When r<1. In this series, nth term, denoted by tn, is the general term and tn=arn1 Having known that a series is a G.P., any two of its terms determine others.
Example: If the third and the seventh terms of a G.P. are 2 and 1/8, find the G.P. and its tenth term. Solution: Given
t ,ar 2 = 2 3
→ (1)
t 7 , ar 6 = 1/ 8
→ (2)
∴ (2) ÷ (1), r 4 = 1/16 = (1/ 2 ) r = ±1/ 2 ∴
4
Case I. By substituting r=1/2 in (1)
a (1/ 2 ) = 2 a =8 2
th
The G.P. is 8,4,2…and 10 term,
t10 = ar 9 = 8 (1/ 2 ) = 1/ 64 9
Case II. By substituting r=1/2 in (10
a ( −1/ 2 ) = 2 a =8 2
th
The G.P. is 8, 4,2….and 10 term.
t10 = ar 9 = 8 ( −1/ 2 ) = −1/ 64 9
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Example: Find the number of terms in the geometric series
0.03 + 0.06 + 0.09 + ..... + 1.92
Solution:
We find
a = 0.03l; r =
0.12 =2 0.06
Let 1.92 be the nth term
t n = 1.92 ar n −1 = 1.92
( 0.03) ( 2n −1 ) = 1.92 2n −1 =
1.92 = 64 = 26 0.03
Equating the powers, n1=6 N=7 Number of terms in the geometric series=7 Example: The sum of 3 numbers in G.P. is 35 and their product is 1000. Find the numbers. Solution:
Let the three numbers be
a ,a and ar. r
a Given + a + ar = 35 r a .a.ar = 1000 r a 3 = 1000 = 103
→ (1) and
a = 10 Substituting in (1),
10 + 10 + 10r = 35 r
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Multiplying by r,
196
10 + 10r + 10r 2 = 35r 2
Transposing 35r, 1025r+10r =0
10 − 20r − 5r + 10r 2 = 0 10 (1 − 2r ) − 5r (1 − 2r ) = 0
(1 − 2r )(10 − 5r ) = 0
1 − 2r = 0or 10 − 5r = 0 r = 1/ 2 or r = 2 Case I: a=10 and r=1/2 The three numbers are 20,10,5 Case II: a=10 and r=2 The three numbers are 5,10,20 Formula for the sum of the first n terms of a G.P. Derive the formula to find the sum of n terms in Geometric Progression. Derivation: Let Sn be the sum
Sn = a + ar + ar 2 + .... + ar n −2 + ar n −1
→ (1)
r.Sn = ar + ar 2 + ar 3 + .... + ar n −1 + ar n
→ (2)
(1) − (2), Sn − rSn = a − ar n
(1 − r ) Sn = a (1 − r n ) Sn =
a (1 − r n ) 1− r
It can also be written as
Note:
Sn =
a (1 − r n ) 1− r
1 Usually the first form is used when r<1 and the second, when r>1.
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n
2. When r<1, r becomes smaller and smaller as n becomes larger and larger. When
n→∞,
rn → 0 ∴ The formula becomes S∞ =
a −1 = a (1 − r ) 1− r
Example: The first three terms of a G.P are x, x+3 and x+9. Find the value of x and the sum of the first eight terms. Solution: As x,x+3 and x+9 are in G.P., 2 ( x + 3) = x ( x + 9 )
i.e, x 2 + 6x + 9 = x 2 + 9x ∴
− 3x = −9 or x = 3
As the first three terms are 3,6 and 12 a=3 and
r=
6 12 = =2 3 6
The sum of the first 8 terms,
S8 = =
a ( r n − 1)
r −1 3 ( 28 − 1)
2 −1 = 765 Example:
A person is entitled to receive an annual payment which for each year is less by one tenth of what it was for the year before. If the first payment is Rs.100, show that he can not receive more than Rs.1000 however long he may live.
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Solution: First payment, a=100
Second payment=
Third Payment
9 ×100 less by one tenth 10
2 99 9 ×100 = 100 10 10 10
Number of Payments is unlimited or
∞
The person is entitled to receive 2
100 +
9 9 (100 ) + 100 + ....∞ 10 10 =
100 a 9 Q S∞ = and a = 100; r = 9 1− r 10 1− 10
100 = 100 ×10 = Rs.1000 1 10 Hence, however long he may live, he can not receive more than Rs.1000
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Published on Jun 19, 2014
IMTS BBA (Business mathematics & statistics)