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1 0 INTRODUCTION, According to Simpson and Kafka a measure of central tendency is typical. value around which other figures aggregate, According to Croxton and Cowden An average is a single value within the. range of the data that is used to represent all the values in the series Since an. average is somewhere within the range of data it is sometimes called a measure of. central value,1 1 OBJECTIVES, The main aim of this unit is to study the frequency distribution After going through this unit you. should be able to,describe measures of central tendency. calculate mean mode median G M H M, find out partition values like quartiles deciles percentiles etc.

know about measures of dispersion like range semi inter quartile range mean deviation. standard deviation, calculate moments Karls Pearsion s and coefficients skewness kurtosis. 1 2 MEASUIRES OF CENTRAL TENDENCY, The following are the five measures of average or central tendency that are in. common use, i Arithmetic average or arithmetic mean or simple mean. iv Geometric mean,v Harmonic mean, Arithmetic mean Geometric mean and Harmonic means are usually called. Mathematical averages while Mode and Median are called Positional averages. 1 2 1 ARITHMETIC MEAN, To find the arithmetic mean add the values of all terms and them divide sum by the.

number of terms the quotient is the arithmetic mean There are three methods to find. i Direct method In individual series of observations x1 x2 xn the arithmetic mean is. obtained by following formula,x x x x xn 1 xn,A M 1 2 3 4. ii Short cut method This method is used to make the calculations simpler. Let A be any assumed mean or any assumed number d the deviation of the. arithmetic mean then we have, iii Step deviation method If in a frequency table the class intervals have equal width. say i than it is convenient to use the following formula. where u x A i and i is length of the interval A is the assumed mean. Example 1 Compute the arithmetic mean of the following by direct and short cut methods. Class 20 30 30 40 40 50 50 60 60 70,Freqyebcy 8 26 30 20 16. Class Mid Value f fx d x A fd,20 30 25 8 200 20 160. 30 40 35 26 910 10 260,40 50 45 30 1350 0 0,50 60 55 20 1100 10 200.

6070 65 16 1040 20 320,Total N 100 fx 4600 f d 100. By direct method,M fx N 4600 100 46,By short cut method. Let assumed mean A 45,M A fd N 45 100 100 46, Example 2 Compute the mean of the following frequency distribution using step deviation. Class 0 11 11 22 22 33 33 44 44 55 55 66,Frequency 9 17 28 26 15 8. Class Mid Value f d x A u x A i fu,A 38 5 i 11,0 11 5 5 9 33 3 27.

11 22 16 5 17 22 2 34,22 33 27 5 28 11 1 28,33 44 38 5 26 0 0 0. 44 55 49 5 15 11 1 15,55 66 60 5 8 22 2 16,Total N 103 fu 58. Let the assumed mean A 38 5 then,M A i fu N 38 5 11 58 103. 38 5 638 103 38 5 6 194 32 306,PROPERTIES OF ARITHMETIC MEAN. Property 1 The algebraic sum of the deviations of all the variates from their arithmetic. mean is zero, Proof Let X1 X2 Xn be the values of the variates and their corresponding frequencies be.

f1 f2 fn respectively, Let xi be the deviation of the variate Xi from the mean M where i 1 2 n Then. Xi Xi M i 1 2 n,fixi f X M,Exercise 1 a, Q 1 Marks obtained by 9 students in statistics are given below. 52 75 40 70 43 65 40 35 48,calculate the arithmetic mean. Q 2 Calculate the arithmetic mean of the following distribution. Variate 6 7 8 9 10 11 12,Frequency 20 43 57 61 72 45 39. Q 3 Find the mean of the following distribution,Variate 0 10 10 20 20 30 30 40 40 50.

Frequency 31 44 39 58 12,1 2 2 MEDIAN, The median is defined as the measure of the central term when the given terms i e. values of the variate are arranged in the ascending or descending order of magnitudes In. other words the median is value of the variate for which total of the frequencies above this. value is equal to the total of the frequencies below this value. Due to Corner The median is the value of the variable which divides the group into two. equal parts one part comprising all values greater and the other all values less then the. For example The marks obtained by seven students in a paper of Statistics are 15 20 23. 32 34 39 48 the maximum marks being 50 then the median is 32 since it is the value of the. 4th term which is situated such that the marks of 1st 2nd and 3rd students are less than this. value and those of 5th 6th and 7th students are greater then this value. COMPUTATION OF MEDIAN,a Median in individual series. Let n be the number of values of a variate i e total of all frequencies First of all. we write the values of the variate i e the terms in ascending or descending order of. magnitudes,Here two cases arise, Case 1 If n is odd then value of n 1 2th term gives the median. Case2 If n is even then there are two central terms i e n 2 and The mean of. these two values gives the median, b Median in continuous series or grouped series In this case the median Md is. computed by the following formula,Where Md median,l lower limit of median class.

cf total of all frequencies before median class,f frequency of median class. i class width of median class, Example 1 According to the census of 1991 following are the population figure in. thousands of 10 cities,1400 1250 1670 1800 700 650 570 488 2100 1700. Find the median,Solution Arranging the terms in ascending order. 488 570 650 700 1250 1400 1670 1800 2100, Here n 10 therefore the median is the mean of the measure of the 5th and 6th terms.

Here 5th term is 1250 and 6th term is 1400,Median Md 1250 14000 2 Thousands. 1325 Thousands, Examples 2 Find the median for the following distribution. Wages in Rs 0 10 10 20 20 30 30 40 40 50,No of workers 22 38 46 35 20. Solution We shall calculate the cumulative frequencies. Wages in Rs No of Workers f Cumulative Frequencies c f. 0 10 22 22,10 20 38 60,20 30 46 106,30 40 35 141,40 50 20 161. Here N 161 Therefore median is the measure of N 1 2th term i e 81st term Clearly 81st. term is situated in the class 20 30 Thus 20 30 is the median class Consequently. Median M d l 2 i,20 161 60 46 10,20 205 46 20 4 46 24 46.

Example 3 Find the median of the following frequency distribution. Marks No of students Marks No of students,Less than 10 15 Less than 50 106. Less than 20 35 Less than 60 120,Less than 30 60 Less than 70 125. Less than 40 84, Solution The cumulative frequency distribution table. Class Marks Frequency f Cumulative,No of students Frequency C F. 0 10 15 15,10 20 20 35,20 30 25 60,30 40 24 84,40 50 22 106.

50 60 14 120,60 70 5 125,Total N 125,Median measure of term. Clearly 63rd term is situated in the class 30 40,Thus median class 30 40. Median M d l 2 i,30 125 2 60 24 10,30 1 04 31 04,1 2 3 MODE. The word mode is formed from the French word La mode which means in. fashion According to Dr A L Bowle the value of the graded quantity in a statistical. group at which the numbers registered are most numerous is called the mode or the. position of greatest density or the predominant value. According to other statisticians The value of the variable which occurs most. frequently in the distribution is called the mode, The mode of a distribution is the value around the items tends to be most heavily. concentrated It may be regarded at the most typical value of the series. Definition The mode is that value or size of the variate for which the frequency is. maximum or the point of maximum frequency or the point of maximum density In other. words the mode is the maximum ordinate of the ideal curve which gives the closest fit to. the actual distribution,Method to Compute the mode.

a When the values or measures of all the terms or items are given In this case the. mode is the value or size of the term or item which occurs most frequently. Example 1 Find the mode from the following size of shoes. Size of shoes 1 2 3 4 5 6 7 8 9,Frequency 1 1 1 1 2 3 2 1 1. Here maximum frequency is 3 whose term value is 6 Hence the mode is modal size number. b In continuous frequency distribution the computation of mode is done by the following. Mode M 0 l i i,2 f1 f 0 f 2,l lower limit of class. f1 frequency of modal class, f 0 frequency of the class just preceding to the modal class. f 2 frequency of the class just following of the modal class. i class interval, Example 2 Compute the mode of the following distribution. Class 0 7 7 14 14 21 21 28 28 35 35 42 42 49,Frequency 19 25 36 72 51 43 28.

Solution Here maximum frequency 72 lies in the class interval 21 28 Therefore 21 28 is. the modal class,l 21 f1 72 f 0 36 f 2 51 i 7,Mode M 0 l i. 2 f1 f 0 f 2,2 72 36 51, c Method of determining mode by the method of grouping frequencies This method is. usually applied in the cases when there are two maximum frequencies against two different. size of items This method is also applied in the cases when it is possible that the effect of. neighboring frequencies on the size of item of maximum frequency may be greater The. method is as follows, Firstly the items are arranged in ascending or descending order and corresponding. frequencies are written against them The frequencies are then grouped in two and then in. threes and then is fours if necessary In the first stage of grouping they are grouped i e. frequencies are added by taking first and second third and fourth After it the. frequencies are added in threes The frequencies are added in the following two ways. 1 i First and second third and fourth fifth and sixth seventh and eighth. ii Second and third fourth and fifth,2 i First second and third fourth fifth and sixth. ii Second third and fourth fifth sixth and seventh. iii Third fourth and fifth sixth seventh and eighth. Now the items with maximum frequencies are selected and the item which. contains the maximum is called the mode For illustration see following example 1. Example 3 Compute the mode from the following distribution. Size of Item 4 5 6 7 8 9 10 11 12 13,Frequency 2 5 8 9 12 14 14 15 11 13.

Solution From the given date we observe that size 11 has the maximum. frequency 15 but it is possible that the effect of neighboring frequencies on the size. of the item may be greater Thus it may happen that the frequencies of size 10 or 12. may be greater and 11 may not remain mode We shall apply the method of. Size of I II III IV V VI, We have used brackets against the frequencies which have been grouped Now we. shall find the size of the item containing maximum frequency. Column Size of item having maximum frequency,IV 10 11 12. VI 9 10 11, Here size 8 occurs 1 time 9 occurs 3 times 10 occurs 5 times 11 occurs 4 times 12. occurs 1 time,Since 10 occurs maximum number of times 5 times. Hence the required mode is size 10,1 2 4 EMPIRICAL RELATION BETWEEN MEDAIN AND MODE.

For moderately asymmetrical distribution or for asymmetrical curve the relation. Mean Mode 3 Mean Median, approximately holds In such a case first evaluate mean and median and then mode. is determined by,Mode 3 Median 2 Mean, If in the asymmetrical curve the area on the left of mode is greater than area on the. right then,Mean median mode i e M Md M0,Median Mode. M Md M0 M0 Md M, If in the asymmetrical curve the area on the left of mode is less than the area on the. right then in this case,Mode median mean i e M0 Md M.

Exercise 1 c, Q 1 Find the Mode of the following model size number of shoes. Model size no of shoes 3 4 2 1 7 6 6 7 5 6 8 9 5, Q 2 Compute the Mode of the following distribution. Class 0 7 7 14 14 21 21 28 28 35 35 42 42 49,Frequency 19 25 36 72 51 43 28. 1 2 5 GEOMETRIC MEAN, If x1 x2 xn are n values of the variate x none of which is zero Then their. geometric mean G is defined by,G x1 x2 xn 1 n 1, If f1 f2 fn are the frequencies of x1 x2 xn respectively then geometric mean G.

is given by,G x1f1 x2f2 xnfn 1 N,N f1 f2 fn,Taking log of 1 we get. Log G 1 N f1 log x1 f2 log x2 fn log xn,Log G fi log xi. Log G f log x, Example 2 Compute the geometric mean of the following distribution. Marks 0 10 10 20 20 30 30 40,No of students 5 8 3 4. Solution Here,Class Mid value Frequency Log Product.

Log10 x F log x,0 10 5 5 0 6990 3 4950,10 20 15 8 1 1761 9 4088. 20 30 25 3 1 3979 4 1937,30 40 35 4 1 5441 6 1764,N f 20 flogx. Log G log x N 23 2739 20 1 1637,G anti log 1 1637 12 58 marks. 1 2 6 HARMONIC MEAN, The Harmonic mean of a series of values is the reciprocal of the arithmetic means of. their reciprocals Thus if x1 x2 xn none of them being zero is a series and H is its. harmonic mean then,H N x1 x 2 xn, If f1 f2 fn be the frequencies of x1 x2 xn none of them being zero then harmonic.

mean H is given by, Example 1 Find the harmonic mean of the marks obtained in a class test given below. Marks 11 12 13 14 15,No of students 3 7 8 5 2,Marks Frequency 1 x f 1 x. 11 3 0 0909 0 2727,12 7 0 0833 0 5831,13 8 0 0769 0 6152. 14 5 0 0714 0 3570,15 2 0 0667 0 1334,N f 25 f x 1 9614. Required harmonic mean is given by,250000 19614,12 746 marks.

Property For two observations x1 and x2 we have, Where A arithmetic mean H harmonic mean and G geometric mean. 1 3 PARTITION VALUES, If the values of the variate are arranged in ascending or descending order of. According to Simpson and Kafka a measure of central tendency is typical value around which other figures aggregate According to Croxton and Cowden An average is a single value within the range of the data that is used to represent all the values in the series Since an average is somewhere within the range of data it is sometimes called a measure of central value 1 1 OBJECTIVES The

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