N 1
ASSIGNMENT SOLUTIONS GUIDE (015-016) E.C.O.-7 Element of Statistics Disclaimer/Special Note: These are just the sample of the Answers/Solutions to some of the Questions given in the Assignments. These Sample Answers/Solutions are prepared by Private Teacher/Tutors/Authors for the help and guidance of the student to get an idea of how he/she can answer the Questions given the Assignments. We do not claim 100% accuracy of these sample answers as these are based on the knowledge and capability of Private Teacher/Tutor. Sample answers may be seen as the Guide/Help for the reference to prepare the answers of the Questions given in the assignment. As these solutions and answers are prepared by the private teacher/tutor so the chances of error or mistake cannot be denied. Any Omission or Error is highly regretted though every care has been taken while preparing these Sample Answers/ Solutions. Please consult your own Teacher/Tutor before you prepare a Particular Answer and for up-to-date and exact information, data and solution. Student should must read and refer the official study material provided by the university. N Attempts all the questions. Q. 1. Define classification. Explain the purpose and methods of classification of data giving suitable examples Ans. Meaning of Classification: Classification is the process of arranging data in groups or classes according to resemblances, similarities and dissimilarities. Homogeneous data should be included in a particular class. It should possess the characteristics of diversification. OBJECTIVES OF CLASSIFICATION The main objectives of the classification are: 1. The presentation of data must be simple, brief and easily understandable.. It should identify clearly the points of similarities and dissimilarities in the statistical data to facilitate comparison. 3. It must present the raw data i.e., the information collected through censuses and surveys in original form in a systematic manner to draw inferences. 4. To present the data scientifically, it provides the basis for tabulation. 5. It helps us to recognise the possible characteristic of data. METHODS OF CLASSIFICATION The statistical data is broadly classified on the basis of attributes and on the basis of variables. Classification According to Attributes Attributes describes to what extent a character possesses natural, in born characteristics common to all characters. The attributes are not capable of being measured quantitatively i.e., cannot be expressed numerically. The examples of attributes are religion, literacy, unemployment, occupation, sex, etc. When the data is classified on the basis of qualities, it is called classification on the basis of attributes. There are two types of classification based on attributes. 1. Simple Classification: When the data is classified on the basis of only one attribute i.e., one characteristic, it is called simple classification. For example the number of people working in an office is classified according to sex.. Manifold Classification: When the data is classified on the basis of more than one attribute i.e. two or more characteristics, it is called manifold classification. For example, the number of people working in an office is classified according to sex and their education. Classification According to Variables Any character which is capable of taking several numerical values for what is being observed is called a variable. Variables are capable of being measured quantitatively i.e., can be expressed numerically. The examples of variables are height, weight, marks obtained, age, salary etc.
When the data is classified on the basis of variables, they are presented in the form of frequency distribution. Frequency distribution is the tabular arrangement of the data showing the frequency with which each successive value of the variable occurs and get organised into classes. Frequency distribution can be categorised into: 1. Uni-variate Frequency Distribution: The frequency distribution with one variable is called a univariate frequency distribution.. Bi-variate Frequency Distribution: The frequency distribution having two variables is called a bi-variate frequency distribution. 3. Multi-variate Frequency Distribution: The frequency distribution with more than two variables is called a multi-variate frequency distribution. Q.. Statistical methods are most dangerous tools in the hands of the inexpert. Explain the significance of this statement. Ans. DISTRUST OF STATISTICS: Lack of reliability, confidence and trust in statistical methods and data create distrust in statistics. While applying statistical tool one should understand its proper use and its real purpose. If used in inappropriate way, wrong conclusions can be drawn, which has resulted in lot of distrust. Some of its important reasons are as follows: 1. Numerical figures or data are capable of being easily manipulated in any desired manner.. Results based on inadequate Nsample can mislead (i.e., if the person reach the conclusion on the basis of very small sample of study). 3. If a person concludes without understanding the whole circumstances and conditions, then even if the correct figures are used, wrong decisions can be taken. 4. Statistical data do not carry the label of their quality; therefore unintentionally faulty conclusions may be drawn. 5. By applying wrong statistical tool to even correct and complete data, one can reach far away from the correct result. In spite of great use of statistics in every discipline, there is some amount of misgiving, in the minds of a few people with regards to its reliability. Lack of reliability, confidence and trust in statistical methods and data create distrust in statistics. Some of its important reasons are as follows: 1. Results based on inadequate sample can mislead (i.e., if the person reach the conclusion on the basis of very small sample of study).. Numerical statements, figures or data are capable of being easily manipulated in any desired manner. 3. Statistical data, which has been selected, do not carry the label of their quality; therefore unintentionally faulty conclusions may be drawn. 4. If we apply wrong statistical tool to even correct and complete data, one can reach far away from the correct result. 5. If anyone concludes without understanding the whole circumstances and conditions, then even if the correct figures are used, wrong decisions can be taken. 6. A slight alteration in the definition of any term used in statistics provides a base for misguidance. 7. Inappropriate method of selecting the sample for any statistical enquiry can become a cause for wrong interpretation. The usefulness of statistics depends to great extent upon its user. While applying statistical tool one should understand its proper use and its real purpose. If used properly by an efficient and unbiased statician, it will prove to be an efficient tool. If used in inappropriate way, wrong conclusions can be drawn, which may result in lot of distrust. It should be kept in mind that statistics neither proves anything nor disproves anything. It is only a method of approach (statistical tool) which should be used with caution by experts. Q. 3. Find out missing frequencies in the following incomplete distribution : Measurement 0-10 10-0 0-30 30-40 40-50 Frequency 3? 0 1? The values of Median and Mode are 7 and 6 respectively. 3
Ans. Mode 3Median Mean Mean 3Median Mode Mean 7.5. Measurement Frequency Cummulative Frequency 0-10 3 3 10-0 x 1 3 + x 0-30 0 3 + x 1 30-40 1 40-50 x 35 + x 1 35 + x 1 + x Now Mean 7.5 N Given median 7, which lies in class 0-30 Median 7 7.5...(i) 3 x 1 14...(ii) Also 3 + x 1 + 0 + 1 + x n 35 + x 1 + x n...(iii) Comparing (1) and (3) n 55 Substituting n 55 in (ii), we get 3 x 1 14 7.5 x 1 17 x 1 7.5 17 x 1 10.5 and x 9.5 Thus, required frequencies are 10.5 and 9.5. Q. 4. The arithmetic mean and the standard deviation of a series of 0 items are 0 cms and 5 cms. resectively. But while calculating them, an item 13 misread as 30. Find the correct mean and correct standard deviation. Sol. Uncorrected ΣX 0 0 400 Correct ΣX 400 30 + 13 383 Correct mean 19.15 Now, uncorrected S.D. 5 4
Standard Deviation (σ) 5 ΣX ( 0 ) 0 500 ΣX 8000 ΣX 8,500, which is uncorrected value. Correct ΣX 8,500 (30) + (13) 7769 Hence, corrected Standard Deviation (σ) X X Σ n 1.73 Σ3 X + 1 X d 18.75 Q Q 5 M 5 nq 8 Q 3 1 N 7769 (19.15) 0 ( 388.45 366.75) 4.67. Q. 5. Calculate Karl Pearson s coefficient of skewness from the following data : Class 40-60 30-40 0-30 15-0 10-15 5-10 3-5 0-3 Frequency 5 15 1 8 6 4 3 Ans. Since the class intervals are unequal, so skewness can be measured by Bowley s method Class interval Frequency Cummulative Frequency 0-3 3-5 3 5 5-10 4 9 10-15 6 15 15-0 8 3 0-30 1 35 30-40 15 50 40-60 5 75 S kb Now Q has n/4 18.75 observation below if hence it lies in the class 15 0. Substituting the values in the formula: Q 1 L 15 + tn c + 4 i t 1,, 3 15 +.34 17.34 Q 3 has 3n/4 56.5 observations below it. Hence it lies in the class 40-60. f 5
Now, S kb Q 3 40 + 40 + 5 45 45 + 17.34 ( 31.66) 45 17.34 0.0354. It means the distribution Nis negatively skewed. 6