MEASURES OF CENTRAL TENDENCY

Transcription

1 INTRODUCTION: MEASURES OF CENTRAL TENDENCY Often, one encounters e term Measures of central tendency in a book on statistics (or an examination). One may also find mention of e term in a description of summary statistics. What is a measure of central tendency? Simply put, it is a value at tries to summarize a given dataset. It does so by providing a central (or middle) value at (sort of) represents e entire dataset. Imagine e marks scored by 10 students in an examination were as follows: If I asked you about e performance of e students, you could say at student 1 scored 0 marks; student scored 0 marks; student 10 scored 0 marks. The problem wi is strategy is encountered when we have to deal wi large datasets, of say, 100 values or more. It would take considerable time and effort to describe e entire dataset. Therefore, people tried to find a way by which ey could convey e essence of e dataset. The measures of central tendency do just at. It is like describing a person in one word- smart / creative, etc. In fact, we summarize ings on a routine basis, often wiout realizing it, wheer it is movies, books, classes, or just about anying. Coming back to e example, one could simply say e students scored 0 marks on average. This average value represents e entire dataset, and gives us an idea about e performance of e students. The most common measures of central tendency are e Mean, Median and Mode. MEANING AND DEFINITION OF AVERAGE: (MEASURES OF CENTRAL TENDENCY) A single value at describes e characteristics of e entire series is called e Central Value or an Average. It is e most representative value of e series. DEFINATIONS: According to CROXTON and COWDE an Average value is a single value wiin e range of e data at is used to represent all of e value in e series. Since average is somewhere wiin e range of data, it is also called e measures of central tendency. According to E.A. Waugh, an average is a single value from a group of values to represent em in some way a value which is supposed to stand for whole group of which it is a part, a typical of all e values in e group. According to Prof. Clark, Average is an attempt to find one single figure to describe whole figures. Thus an average represents e whole series and it s value lies in between maximum and minimum values. It is generally located in e centre or middle of e distribution. ARITHMETIC MEAN (A.M.) It is e most popular and widely used measures of central tendency. 1

2 DEFINITION: An Arimetic mean is defined as e value obtained by dividing e sum of all e observations by e number of s. It is also known a e Arimetic Mean. It is denoted by x Arimetic mean can be calculated by e formula. Sum of all values of observations Mean = Total number of observations Symbolically, x = Where, x = Arimetic mean x = Sum of all observations n = Number of s n x CALCULATION OF ARITHMETIC MEAN: (A) Individual Data (Distribution): To calculate mean, add all e values of given observation and divide it by e number of s i.e. Sum of all values of s A.M. = Total number of s x x1 + x + x3 + + xn Symbolically, x = Or = n n Where, x = Arimetic mean x = Sum of all s n = number of s Example: (1) The following are e daily wages of 5 workers in a factory. Calculate mean wage. Daily Wage (Rs.): x x = = = = 700 n 5 5 Mean Wage = Rs. 700/- () Marks of 8 students in an Economics test is given below. Find e mean mark of e students. Marks: 50, 60, 80, 90, 75, 45, 55, 65 x x = = = = 65 n 8 8 Mean Marks = 65 (3) The following are e monly income of 10 families in a village. Calculate mean average income per family. Income (Rs.): 550, 650, 700, 400, 550, 650, 800, 900, 450, 350. x x = = n = = Average Income = (Rs.) 600/- (4) Calculate Arimetic Mean of e following marks obtained by 10 students. Marks: 80, 75, 65, 45, 85, 60, 50, 70, 40, 90

3 x x = Arimetic Mean n Average Marks = = 10 = = 66 (5) Calculate Arimetic Mean form e following data: Data: 3, 6, 5, 4, 7, 9, 9, 5, 7, 9, 8, 6, 5, 3, 8, 6 x x = = = = 6. 5 n x = 6.5 (B) CALCULATION OF ARITHMETIC MEAN: (DISCRETE SERIES DISCRETE DISTRIBUTION or UN GROUPED DATA): In Discrete series, Arimetic Mean can be calculated in any of e two ways: (1) Direct Meods () Short Cut Meod (1) DIRECT METHOD: It is used when values are small and few. Formula x fx = n x = Arimetic mean, F = frequency of an, X = Value of, n = f Steps: (1) Multiply e frequency of each wi respective variable and add em to get Eƒx () Divide ƒx by e total frequency Income & Expenditure, ( ƒ) to get x. Q.1. Marks obtained by 50 students in a class is given below. Calculate Arimetic Mean by Direct Meod Marks No. of Students Marks (x) No. of Students ƒx fx 1500 x = = = 30 N 500 Average Marks = N= ƒ=50 ƒx=1500 Q.. Calculate Arimetic mean from e following Marks No. of Students Marks (x) No. of Students ƒx 3

4 fx 354 x = = = 7.8 N 50 Mean Marks = N= ƒ=50 ƒx=354 Q. 3. Daily wages of 5 workers in a sugar factory is given below. Calculate mean wages of workers. Wages (Rs.) No. of Workers Wages (x) (Rs.) No. of Workers (ƒ) ƒx fx 1780 x = = = 71.0 N 5 Mean Marks = N= ƒ=5 ƒx=1780 () Short-cut Meod: It is used to avoid mistakes in calculation when e values are large and high. Formula: fd x = A + N Where, A = Assumed mean (Any figure form 1 st figure from 1 st column can be taken as assumed mean) N= total frequency i.e. ƒ, d = deviation i.e.(x-a) Steps: (1) Select Assumed mean () Find out deviation of from assumed mean i.e. d=x-a (3) Multiply ese deviations wi e respective frequency and take total i.e. ƒd. (4) Apply e formula: fd x = A + N Q.1. Height of 50 students in a class is given below. Find mena height of student (use Short cut meod) Height (cm) No. of Students Height (x) No. of Students(ƒ) D=X-A ƒd

5 130 A N= ƒ=50 ƒx=40 x = A+ fd 40 = 130+ N = = Mean Height of a Student = Q.. Marks obtained by 100 students in a class I given below. Calculate Arimetic Mean using short cut meod. Height (cm) No. of Students Height (x) No. of Students(ƒ) D=X-A ƒd A N = ƒ = 100 ƒd = -150 x = A+ fd 150 = 130+ N =130 + (-1.5) = Mean Mark = 8.50 (C) (i) CALCULATION OF ARITHMETIC MEAN: CONTINUOUS SERIES: (GROUPED DATA) In continuous series, Exact value of variables are not known. Only e range is given. So, e midpoints of e various classes are taken as e representative of at particular class. Arimetic mean can be calculated using any of e following meod: Direct Meod: Formula: x fx = N Where, x = Arimetic mean, X = Midpoint of e classes, ƒ = frequency of each class, N = Total frequency i.e. ƒ Upper limit + Lower limit Midpoint = Steps: (1) Obtain midpoint of each class and denote it by X () Multiply midpoint by respective frequency of class and added em up to get ƒx (3) Divide ƒx by e sum of frequencies i.e. N or ƒ. Q. 1. Calculate Arimetic mean from e following Data: (use Direct Meod) No. of Product Sold No. of Salesman No. of Product Sold No. of Salesmen (Mid Value X ) (ƒ) (ƒx) 5

6 N= ƒ=90 ƒx = 880 x fx 880 = = =3 N 90 Q.. Calculate Arimetic Mean from e following data: (use Direct Meod) Marks: No. of Students: Marks (Mid X) No. of Students (ƒ) (ƒx) (ii) N = ƒ = 100 ƒx = 500 x fx 500 = = =5 N 100 Short Cut Meod: Formula: fd x = A + N Where, x = Arimetic Mean; A = Assumed Mean d = Deviation of mid point from Assumed mean A (i.e. X A) N = Total frequency (i.e. ƒ) ƒ = Frequency of e class. Steps: (1) Take assumed mean (A) from e mid points. () Obtain deviation of midpoint from e Assumed Mean i.e. d = X-A (3) Multiply deviation by respective frequencies and sum up it to get ƒd. (4) Find total frequency N = ƒ (5) Apply e formula: fd x = A + N Example Q.1. Calculate Arimetic mean from e following data (use Short Cut Meod). Wages Per Day (Rs.) No. of Workers Wages per day No. of Workers Deviation Mid x (ƒ) D= X-A ƒd

7 A N= ƒ=150 ƒd = -110 fd x = A + = = 35+(-0.73) = 34.7 N 150 Q.. Marks obtained by 100 students is given below. Calculate Arimetic mean for e data (Use short cut meod.) Marks No. of Students Marks No. of Deviation Mid x Students D= X-A ƒd A fd x = A + = N (iii) = N= ƒ=100 ƒd = 0 Step Deviation Meod: It is e shortest meod for calculation of Arimetic Mean. in is meod, each deviation (d=x-a) is divided by e common factor (i) Formula: fd1 x = A + xi N Where, x = Arimetic Mean. A = Assumed Mean N = Total frequency. d = X-A d 1 = X-A/i. X = Mid Point i = Common factor (wid of class interval) ƒd 1 = Product of deviation multiplied by eir respective frequencies. Steps: (1) Obtain Mid Point. () Take Assumed Mean. (3) Calculate deviation from Assumed mean. (4) Divide deviation by common factor. (5) Multiply deviation by respective frequencies. (6) Sum up e product of frequencies and step deviation. (7) Apply e formula: fd1 x = A + xi N Q.1. Calculate Arimetic Mean from e following data by Step Deviation Meod. Class Frequency

8 Deviation X A Class Frequency d D= X-A 1 = Product (Mid X) (ƒ) i ƒd (A=35) 1 (i=10) A N= ƒ=0 ƒd 1 =01 fd1 1 x = A + xi = 35 + N x10 = = Q.. Calculate Mean from e following Data (Use Step Deviation Meod) Class Frequency Deviation X A Class Frequency d D= X-A 1 = Product (Mid X) (ƒ) i ƒd (A=35) 1 (i=10) A fd1 x = A + xi N N= ƒ= = 55 + x110 = 55+1 = ƒd=11 (D) (E) MERITS OF ARITHMETIC MEAN: Simple: Definition (meaning) of Arimetic Mean is simple, precise or exact. Easy: It is easy to calculate and understand. Rigidly DEFINED: Arimetic Mean is rigidly defined so, e result obtained will remain same, whatever meod may be used. Arrangement of Data not Necessary: It is not necessary to arrange data in ascending or descending order, to calculate Arimetic Mean. Based on all Items: Arimetic Mean is based each and every of e series. Stability: It is reliable and sufficiently stable and is not much affected by e fluctuation in samples. All Details not Required: Only e data is required to calculate e mean. Oer details are not required. Algebraic Treatment: Arimetic Mean is capable of furer Algebraic manipulation. Centre of Gravity: It balances e values on eier side of it. So, it is e centre of Gravity. DEMERITS OF ARITHMETIC MEAN: Distribution of s: Arimetic means do not equate e entire distribution of s. Affected by Extreme s: Value of Arimetic mean is affected by extreme s (i.e., biggest or smallest). It reduces it s utility as a representative value. For example: If e age of 8

9 Grandfaer is 85 years and e age of 3 grandsons are 10, 8 and 7 respectively, en average age of em is 7.5 years. Not Workable: If data is incomplete or if any single is missed or ignored in e series, mean cannot be calculate. It may loose it s accuracy. Absurd Results: Sometimes, it gives absurd result. For example: if ere are children in a family and 3 in anoer, en, average number of child in ese families is.5. It is absurd. It cannot be in fraction. Cannot be found by inspection. It is difficult to locate mean b inspection. Value may not be in series: Exact value of mean may not be in series. Example: Mean of 10, 0, 30 and 40 is 5. Open-end classes: A.M. is based on assumptions. It cannot be calculated effectively for open-end classes. Not fruitful: It cannot be used for qualitative and deceptive study. COMBINE MEAN: If and are e arimetic mean of two samples of sizes n 1 and n respectively en, e arimetic mean of e distribution combining e two can be calculated as This formula can be extended for still more groups or samples. Justification: = total of e observations of e first sample Similarly = total of e observations of e first sample The combined mean of e two samples = = Example The average marks of ree batches of students having 70, 50 and 30 students respectively are 50, 55 and 45. Find e average marks of all e 150 students, taken togeer. Solution : Let x be e average marks of all 150 students taken togeer. Batch - I Batch - II Batch - III A. marks : = 50 = 55 = 45 No. of students n 1 = 70 n = 50 n 3 = 30 9

10 Example The mean of a certain number of observations is 40. If two or more s wi values 50 and 64 are added to is data, e mean rises to 4. Find e number of s in e original data. Let 'n' be e number of observations whose mean = 40. total of n values. Two more s of values 50 and 64 are added erefore, total of (n + ) values : Now new mean is 4. New 4n + 84 = 40n n = 30 n = 15 Therefore, e number of s in e original data = 15. Example The sum of deviations of a certain numbers of observations measured from 4 is 7 and e sum of deviations of observations measured from 7 is -3. Find e number of observations and eir mean. Let 'n' be e required number of observations, erefore,...note and erefore, Subtracting e two equations we get, (-) (+) (+) 3n=75 n = 5 Putting n = 5 in, we get 10

11 Now Mean is given by What is e 'Geometric Mean': The geometric mean is e average of a set of products, e calculation of which is commonly used to determine e performance results of an investment or portfolio. Technically defined as "e 'n' root product of 'n' numbers", e formula for calculating geometric mean is most easily written as: Where 'n' represents e number of returns in e series. The geometric mean must be used when working wi percentages (which are derived from values), whereas e standard arimetic mean will work wi e values emselves. DEFINITION of 'Harmonic Average' The mean of a set of positive variables. Calculated by dividing e number of observations by e reciprocal of each number in e series. MODE: Mode is at value which occurs most frequently in a set of observations. It is at which repeats maximum number of s. It is e value having maximum, (highest) frequency. It is also known as Modal Value or Norm of e series. (A) CALCULATION OF MODE: INDIVIDUAL SERIES (DATA): Steps: (1) Arrange e data in ascending or descending order. () Use tally mark and make frequency table. (3) Mode is e value repeated maximum number of times. Example: (1) Find mode from e following data: 5, 10, 1, 10, 11, 1, 10, 13, 11, 10, 1, 30 Arrange Data in ascending order and make frequency table Observation Tally Mark Frequency 05 I IIII II 0 1 II 0 13 I 01 1 I I 01 Total 1 In is observation 10 has repeated maximums (i.e. 4) times. Mode = 10 Q.. Find mode from e following data: 65, 40, 48, 5,54, 56, 5, 48, 49, 5, 40, 4, 5, 60, 61, 5, 70, 71, 5, 48 Observation Tally Mark Frequency 40 I 0 4 II III I 01 11

12 5 IIIII I I I I I I I 01 Total = 0 In is, observation 5 has repeated maximum (i.e. 6) times Mode = 5. (B) CALCULATION OF MODE; DISCRETE SERIES: (UNGROUPED DATA): In case of Discrete series, mode can be located by Inspection. It is e value in e series, having maximum frequency. Example: (1) Calculate mode from e following Data: Size of Shirt No. of Persons Size of Shirt No. of Persons In is, e observation 50 has maximum frequency (i.e. 8). Mode is e value having maximum frequency. Mode = 50 () Calculate mode form e following frequency distribution: Daily Wages (in Rs.) No. of Workers Daily Wages (in Rs.) No. of Workers In is, e observation 73 has e highest frequency (i.e. 160). Mode is e value having maximum frequency. Mode = 73 (C) CALCULATION OF MODES: CONTITUOUS SERIES (GROUPED DATA) Steps: (1) Find e class interval having maximum frequency (model class) () Use interpolation formula. (L + L1)(f1 f0) Mode = L1 + find exact value of mode f + f f 1 0 Where, L 1 = Lower limit of e class L = Upper limit of e class ƒ 1 = Frequency of modal class (highest frequency) ƒ 0 = Frequency of e class preceding modal class. ƒ = Frequency of e class succeeding modal class 1

13 Examples: Q. 1. Calculate mode from e following data: Sales (in Rs. crores) No. of Firms Sales (in Rs. Crores) No. of Firms ƒ ƒ ƒ In is, 1 is e highest frequency. Corresponding modal class = 8-1 (L + L1)(f1 f0) Mode = L1 + f + f f = (1 8)(1 6) 8 + (x1) x6 4 = 8 + = = = Mode = Q.. Find mode from e following data: Daily Wages No. of Workers Daily Wages (Rs.) No. of Workers ƒ ƒ ƒ In is, 5 is e highest frequency. Corresponding modal class = (L + L1)(f1 f0) Mode = L1 + f + f f = 1 (40 30)x(5 0) 30 + (5) x5 50 = 30 + = = Mode =

14 (D) (a) (E) MERITS AND DEMERITS OF MODE: Merits of Mode: Easy and Simple: Mode is simple to understand and easy to calculate. Not Influenced by Extreme Items: Extreme s has no effect on mode. All Detail no required: To calculate mode, all details of s are not required. Only e norm or most frequent is to be known. Value Exist in Series: Mostly, Modal Value exists in series. Most Representative Average: Mode is e most descriptive average in use. Example: Model height, Modal wage, Average number of Railway accidents in an year. It is also useful for qualitative data. Graphical Location: Mode can be determined graphically rough Histograms. Useful for Open end Classes: It s value can be determined in open-end classes. Inspection: Mode can be determined by inspection. It is e most frequent observation in e series. DEMERITS OF MODE: Ambiguous Definition: Mode is Illustration-defined and indefinite. It is not properly defined. There can be more an one Mode in a series. Example: (Bimodal tri-modal). If e values do not repeat, en Mode is absent. Not Based on all Items: Mode is not based on each and every of e series. It consider only most frequent. Not Algebraic Properties: Mode is not capable of furer Algebraic Manipulations. Not a Good Measure: Mode is not a good measure of central tendency as it consider only e modal class and ignore oer variables in e series. Limited Scope: In small sample scope of Mode is limited. When non of e variable repeats, Mode is absent. Unstable Average: Mode is e most unstable average. So it is difficult to determine it s true value. MEDIAN: Median is e middle most value in e series arranged in ascending or descending order. It divides e series into equal parts. It is a positional average. (A) CALCULATION OF MEDIAN: INDIVIDUAL DATA: (a) If e number of s in e distribution is odd (uneven) Steps: (1) Arrange data in ascending or descending order. ( n + 1 ) () Apply e formula: M = Size of Where, M = Median n = Numbers of s Q. 1. Marks obtained by 7 students in a class test is given below. Find Mark. Marks 7, 10, 11, 5, 8, 1, 15 Step: Arrange Data in ascending order. Marks 5, 7, 8, 10, 11, 1, 15 ( n + 1 ) Median = Size of ( ) = Size of = Size of 4 Corresponding 4 Value in e series is 10. Median = 10 14

15 Q.. Find Median from e following data: 35, 38, 40, 39, 35, 36, 37 Arrange Data in ascending order. 35, 35, 36, 37, 38, 39, 40 ( n + 1 ) Median = Size of ( ) = Size of = Size of 4 Corresponding 4 Value in e series is 37. Median = 37 (b) If e number of s in e distribution is even: (1) Arrange Data in Ascending or Descending order. ( n + 1 ) () Apply e formula: M = Value of It will be in traction. So, add values before and after e fractional value and divided total by to get Median value. Q. 1. Suppose 8 Students secured following marks in Economics: Find Median Marks. 83, 47, 59, 5, 36, 30, 9, 74 Arrange Data in ascending order. 5, 30, 36, 47, 59, 74, 83, 9 ( n + 1 ) Median = Size of ( ) = Size of = Size of 4.5 Size of 4 + size of 5 = = = = 53 Median Marks = 53 Q.. Find Median from e following data: 14, 13, 1, 11, 15, 16, 18, 17, 19, 0 Arrange Data in ascending order. 11, 1, 13, 14, 15, 16, 17, 18, 19, 0 ( n + 1 ) Median = Size of ( ) = Size of = Size of 5.5 Size of 5 + size of 6 = = = = 15.5 Median =

16 (B) CALCULATION OF MEDIAN: DISCRETE SERIES (UNGROUPED DATA): Steps: (1) Arrange Data in Ascending order (if not arranged) and write corresponding frequency. () Find cumulative frequency (f) by successively adding e frequencies of each in e series. ( N + 1 ) (3) Apply e formula: Value of. Where, N = ƒ. ( N + 1 ) (4) Look C.ƒ. and find e value equal to or greater an (5) Value corresponding to at C.ƒ. will be Median. Q. 1 Find Median Marks obtained by e students: Marks: No. of Students: Marks No. of Students (ƒ) C.ƒ ( N + 1 ) Median = Size of ( ) = Size of N= ƒ=48 = Size of 4.5 (It lies in Cƒ 31) The value corresponding to e Cƒ 31 is : 0 Median = 0 Hence, Median Marks = 0 Q. Find Median from e following data: Marks: No. of Students: Marks No. of Students (ƒ) C.ƒ ( N + 1 ) Median = Size of ( ) = Size of N= ƒ=49 = Size of 5 (It lies in Cƒ 33). 16

17 The value corresponding to e Cƒ 33 is : 13 Median = 13 Q. 3 Find Median from e following data: Income (Rs.) No. of Persons Income (Rs.) No. of Persons (ƒ) C.ƒ. 1, , , , , , ( N + 1 ) Median = Size of ( ) = Size of N= ƒ=90 = Size of 45.5 (It lies in Cƒ 70) The value corresponding to e Cƒ 70 is : 4,000 Median Income = Rs. 4,000 Q. 4 Find Median from e following data: Size (X) Frequency (ƒ) Size (X) Frequency (ƒ) C.ƒ ( N + 1 ) Median = Size of ( ) = Size of = Size of 47 (It lies in Cƒ 79) The value corresponding to e Cƒ 79 is : 9 Median = 9 N= ƒ=493 17

18 (C) CALULATION OF MEDIAN: CONTINUOUS SERIES (GROUPED DATA) Steps: (1) Arrange data in ascending or descending order (if not) and write frequencies of respective class. () Calculate cumulative frequencies (C.ƒ.) (3) Determine e class in which Median Class lies: (using e formula) N Rank of Median = Size of, where N = ƒ (4) To get Exact value of e median, use e interpolation, formula: N N C.f. C.f. M L = 1 + X ( L L1 ) OR M = L 1 + Xf f f Where, M = Median L 1 = Lower limit of e Median Class L = Upper limit of e Median Class ƒ = Frequency of e Median Class C.ƒ. = Cumulative frequency of e class preceding e Median Class i = (L L 1 ) Example: Q. 1. Calculate median from e following data: Income (Rs.) No. of Persons Income (Rs.) No. of Persons (ƒ) C.ƒ Rank of Median = Size of N N= ƒ=40 40 = Size of = Size of 0 (It lies in Cƒ 7) Corresponding median Class = N 40 L 1 = 30, L = 40, ƒ 1 = 10, Cƒ. = 17, i=10, = = 0 N C.f. M L 0 17 = 1 + Xi = = 30 + X10 f 03 =30+ X10=30+03=33 10 Median Income = Rs. 33/- 18

19 Q.. Calculate median from e following data: Working Hours (per week) No. of Workers Income (Rs.) No. of Persons (ƒ) C.ƒ N= ƒ=50 N 50 Rank of Median = Size of = Size of = Size of 5 (It lies in Cƒ 6) Corresponding median Class = 15-0 N 50 L 1 = 15, L = 0, ƒ = 15, Cƒ. = 11, i=5, = = 5 N C.f. M L 5 11 = 1 + Xi = = 15 + X5 f =15+ X5 15 = =19.67 Q. 3. Calculate median from e following data: Wages per hours (Rs.) No. of Workers Wages per hours (Rs.) No. of Workers (ƒ) C.ƒ N= ƒ=50 N 50 Rank of Median = Size of = Size of = Size of 5 (It lies in Cƒ 7) Corresponding median Class = 0-30 N 50 L 1 = 0, L = 30, ƒ = 1, Cƒ. = 15, i=10, = = 5 19

20 N C.f. L 5 15 = + Xi = = 0 + X5 f 1 M 1 10 =0+ X10 1 =0+8.33=8.33 Q. 4. Calculate median from e following data: Marks No. of Students Marks No. of Students (ƒ) C.ƒ N= ƒ=60 N 60 Rank of Median = Size of = Size of = Size of 30 (It lies in Cƒ 36) Corresponding median Class = 15-0 N C.f. M L 30 0 = 1 + Xf = = 15 + X5 f =15+ X5 16 = =18.15 or Q. 5. Calculate median from e following data: Income (Rs.) No. of Persons Income (Rs.) No. of persons (ƒ) C.ƒ N= ƒ=40 N Rank of Median = Size of = Size of = Size of 0 (It lies in Cƒ -8) Corresponding median Class =

21 N C.f. L 0 15 = + Xi = = 30 + X10 f 13 M 1 5 =30+ X10 = = (D) (E) MERITS OF MEDIAN: Easy: Median is simple to understand and easy to calculate. Inspection: In some cases, median can be located by mere inspection. Rigidly defined: Median has well defined formula. All Details Not Required: Details about e values are not required to calculate median. It can be determined even if complete data is not available. Not Influenced by Extreme Items: It is not affected by e value of extreme (end) s in e series. Useful to Study Attributes: Median is an ideal average to study qualitative characteristics (attributes) like intelligence, honesty, heal, etc. Exist in Series: Mostly, e median value exist wiin e series. Graphical Location: Median can also be determined graphically. DEMERITS OF MEDIAN: Array is Essential: It is necessary to arrange data in ascending or descending order, to calculate Median. If e series is lengy, e task of finding e median is difficult. Not Precise: It cannot be precisely determined in a series wi even number of s. In such a case, e value estimated may not be e actual value in e series. Not Based on All Observations: Median is a positional average. So, its value is not based on all e observations in e series. Ignores Extreme ITEMS: Since e median ignore extreme s, it is not useful in ose cases where large weights are to be given to extreme s. Not capable of Algebraic Manipulation: It is not capable of furer algebraic treatment. Not a True Representative: Median is not actual (true) representative value in e series, as it ignore e extreme s. It is not a representative especially when e distribution lacks uniformity. Sampling Fluctuations: Value of median is more affected by sampling fluctuations. QUARTILES Definition: Quartiles are e values (data) which divide e series (distribution) into four equal parts. They are e 3 values at divide e distribution into 4 part, each representing one quarters of e score. These 3 values are called as first quartile (Q 1 ), second quartile (Q ) and ird quartile (Q 3 ). Second quartile is noing but e median. 1,, 3, 4, 5, 6, 7, 8, 9, 10, 11 Q 1 Q Q 3 The figure shows at quartiles are e ree points: Q 1, Q, and Q 3 which divided e series into four parts in such a way at each part contains equal number of s. (A) FIRST AND THIRD QUARTILES FOR INDIVIDUAL DATA: Steps: (1) Let n observations of Individual Data be arranged in ascending order. () Apply e formula s First Quartile (Q 1 ) n + 1 = value of Observation. Second Quartile (Q ) 1

22 n + 1 = value of Third Quartile (Q 3 ) n + 1 = value of 3 Observation. Observation. Q. 1. Find e first and ird quartiles from e following data. 4, 6, 7, 9, 11, 5, 8, 10, 14, 1, 13 After arranging e given data in ascending order we get, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14 n = number of observations n = 11 First quartile (Q 1 ) = value of n + 1 observation = value of observation. = value of 3 rd observation. Ans.: Q 1 = 6 Third quartile (Q 3 ) n + 1 = value of 3 observation = value of 3 observation. = value of (3 x 3) observation = value of 9 observation. Ans.: Q 3 = 1 Q.. Find e first and ird quartiles from e following data. 30, 35, 37, 3. 36, 33, 38, 31, 34, 39, 40 After arranging e given data in ascending order we get, 30, 31, 3, 33, 34, 35, 36, 37, 38, 39, 40 n = number of observations n = 11 First quartile (Q 1 ) = value of n + 1 observation = value of observation. = value of 3 rd observation. Ans.: Q 1 = 3 Third quartile (Q 3 ) n + 1 = value of 3 observation = value of 3 observation. = value of (3 x 3) observation = value of 9 observation.

23 Ans.: Q 3 = 38 Q. 3. Calculate e first quartile, second quartile and ird quartiles from e following data. 1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15 After arranging e given data in ascending order we get, 1,, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15 n = number of observations n = 15 First quartile (Q 1 ) = value of n + 1 observation = value of observation. = value of 4 observation. Ans.: Q 1 = 4 Second quartile (Q ) n + 1 = value of observation = value of observation. = value of 8 observation. Ans.: Q = 8 Third quartile (Q 3 ) n + 1 = value of 3 observation = value of 3 observation. = value of 1 observation. Ans.: Q 3 = 1 Q. 4. Find out e first and ird quartile from e following daily pocket money of eleven students. Rs. 50, 40, 60, 90, 80, 70, 100, 140, 130, 10, 110 After arranging e given data in ascending order we get, 40, 50, 60, 70, 80, 90, 100, 110, 10, 130, 140 n = number of observations n = 11 First quartile (Q 1 ) = value of n + 1 observation = value of observation. = value of 3 rd observation. Ans.: Q 1 = 60 Third quartile (Q 3 ) n + 1 = value of 3 observation. 3

24 = value of 3 observation. = value of (3 x 3) observation = value of 9 observation. Ans.: Q 3 = 10 Q. 5 Find e first and ird quartiles from e following data. 13, 16, 5, 30, 3, 45, 65 After arranging e given data in ascending order we get, 13, 16, 5, 30, 3, 45, 65 First quartile (Q 1 ) = value of n + 1 observation = value of observation. = value of nd observation. Ans.: Q 1 = 16 Third quartile (Q 3 ) n + 1 = value of 3 observation = value of 3 observation. = value of (3 x ) observation = value of 6 observation. Ans.: Q 3 = 45 (B) FIRST AND THIRD QUARTILES FOR DISCRETE DISTRIBUTION (UNGROUPED DATA): Steps: (1) Arrange data in ascending order. () Calculate e cumulative frequency s. (3) Apply e following formula for first and ird quartiles. Q 1 = value of N + 1 Observation N + 1 Q 3 = value of 3 Observation Where, N = ƒ For Example: Q.1: Find e first and ird quartile for e following ungrouped data. X F X ƒ C.ƒ

25 Q 1 = e size of n + 1 N=59 observation = e size of observation. = e size of 15 observation Ans.: Q 1 = 6 n + 1 Q 3 = e size of 3 observation = e size of 3 observation. = e size of 6 observation Ans.: Q 3 = 1 Q.: Find e first and ird quartile for e following ungrouped data. X F X ƒ C.ƒ Q 1 = e size of n + 1 N=79 observation = e size of observation. = e size of 0 observation Ans.: Q 1 = 5 n + 1 Q 3 = e size of = e size of 3 observation. observation. 5

26 = e size of 60 observation Ans.: Q 3 = 9 (C) FIRST AND THIRD QUARTILES FOR CONTINUOUS FREQUENCY DISTRIBUTION (UNGROUPED DATA): First and ird quartiles for continuous frequency distribution are calculated by following formula. N C.f Q 1 = L X i f Where, L 1 = Lower limit of Q 1 class. i = L L 1. N = Total number of frequencies (i.e. ƒ) C.ƒ = Cumulative frequency of class preceding in Q 1 class. ƒ = frequency of Q 1 class. 3N C.f Q 3 = L X i f Where, L 1 = Lower limit of Q 3 class. i = L L 1. N = Total number of frequencies (i.e. ƒ) C.ƒ = Cumulative frequency of class preceding in Q 3 class. ƒ = frequency of Q 3 class. For Example: Q. 1: Find first and ird quartile from e following data (distribution) C.I Frequency C.I. Frequency C.ƒ N = 150 N = Total number of observation = 150 N 150 = = N = 3 x 37.5 = Hence, is Q 1 class is Q 3 class For Q 1 L 1 = 30, C.ƒ = 5, ƒ=5, i=10 6

27 N C.f Q 1 = L x i f = x i = 30+ x 10 = 30 + = Ans: Q 1 = 35 For Q 3 L 1 = 50, C.ƒ = 90, ƒ=35, i=10 3N C.f Q 3 = L x i = 50 + x = Ans: Q 3 = x 10 = 50 + = Q.. Calculate Q 1 and Q 3 for e following data. C.I ƒ C.I. ƒ C.ƒ N = 45 N = Total number of observation = 45 N 45 = = N = 3 x 11.5 = Hence, 10-0 is Q 1 class is Q 3 class For Q 1 L 1 = 10, C.ƒ = 07, ƒ=10, i=10 N C.f Q 1 = L x i f 7

28 = 30+ x 10 = = Ans: Q 1 = 14.5 For Q 3 : L 1 = 30, C.ƒ = 3, ƒ= 8, i=10 3N C.f Q 3 = L x i = 30 + x = 30+ x = Ans: Q 3 = 3.18 = DECILES: Definition: Deciles divide e series into ten equal parts. Thus, ere must be nine points which will divide e arrayed series in such a way at each part contains an equal numbers of s. The value of ese none points are called deciles. They are denoted as D 1, D, D 3,... D 9. There are nine deciles in a distribution. The 5 decile D 5 being e median. While calculating ese values in individual and discrete values in e series, we add all observations i.e. 1 st to n (total numbers). For calculating D 1 to D 9 we use e following formula: (1) First Decile: D 1 = size of N + 1, N + 1 () Second Decile: D = size of N + 1 (3) Third Decile: D 3 = size of 3 N + 1 (4) Four Decile: D 4 = size of 4 N + 1 (5) Fif Decile: D 5 = size of 5 N + 1 (6) Six Decile: D 6 = size of 6,, N + 1 (7) Seven Decile: D 7 = size of 7 N + 1 (8) Eight Decile: D 8 = size of 8 8

29 N + 1 (9) Nin Decile: D 9 = size of 9 (A) INDIVIDUAL DATA: Steps: (1) Arrange data in ascending order. () Apply e formula for given decile. Q. 1. Find D, D 4 and D 7 for e following data. 11, 15, 1, 16, 13, 17, 14, 18, 0, 19, 5, 1, 4,, 3, 9, 8, 7, 6 Arranging e values in ascending order, we get, 11, 1, 13, 14, 15, 16, 17, 18, 19, 0, 1,, 3, 4, 5, 6, 7, 8, 9 The number of Value N = 19 Second Decile: Hence, Rank of D, = N = = ( X ) = 4 Ans: D = 14 Four Deciles: Hence, Rank of D 4, = 4 N = 4 = (4 X ) = 8 Ans: D 4 = 18 Seven Deciles: Hence, Rank of D 7, = 7 N = 7 = (7 X ) = 14 Ans: D 7 = 4 Q.. Find D 1, D 4 and D 8 for e following data: 61, 6, 63, 64, 65, 66, 67, 68, 69, 70, 71, 7, 73, 74, 75, 76, 77, 78, 79 Arranging e values in ascending order, we get, 61, 6, 63, 64, 65, 66, 67, 68, 69, 70, 71, 7, 73, 74, 75, 76, 77, 78, 79 The number of values N = 19 First Decile: 9

30 Hence, Rank of D 1 = = Ans: D 1 = 6 Four Deciles: N + 1 = nd N + 1 Hence, Rank of D 4 = = 4 = (4 X ) = 8 Ans: D 4 = 68 Eight Deciles: N + 1 Hence, Rank of D 8 = 8 = (8 X ) = 16 Ans: D 8 = 76 (B) DISCRETE DISTRIBUTION (UNGROUPED FREQUENCY DISTRIBUTION) Steps: (1) Arrange Data in ascending order (if not) () Find C.ƒ. (3) Find rank of required decile. (4) Determine Decile, based on C.ƒ. value Example: Q. 1. Find D 3 and D 4 for following data. Marks No. of students Marks No. of Students Cumulative Frequency N = Rank of D 3 = 3 N D 3 = 3 D 3 = 3(4.6)

31 D 3 = 13.8 Hence, D 3 = 3 Marks Rank of D 4 = 4 N ` + 1 D 4 = 4 D 4 = 4(4.6) D 4 = 18.4 Hence, D 3 = 3 Marks (C) CONTINUOUS (GROUPED) DISTRIBUTION: For calculation of decile in continuous data. We use following formula. Steps: (1) Find C.ƒ. () Find Rank of Decile. (3) Apply e formula. N C.f D = L X i f Where, D = Decile L1 = lower limit of e decile class. i = L1 L c.ƒ = Cumulative frequency of preceding class of decile class. ƒ = frequency of Decile class. Example: Q. 1. Calculate e D, D 7, and D 9 for e following distribution. Marks No. of Students Marks No. of Students C.ƒ N = 80 Decile Rank of D = N 10 X Hence, 0-30 is e class of D L 1 = 0, i = 10, C.ƒ. = 13, ƒ = 14 = = 31

32 D = L 1 + = 0 + N 10 x C.f. x i f = 0 + x = 0 + x = = Ans: D =.14 x 10 7N 7X80 Rank of D 7 = = = Hence, is e class of D 7 L 1 = 40, i = 10, C.ƒ. = 5, ƒ = 18 7N C.f. D 7 = L x i f 56 5 = 40 + x = 40 + x = = Ans: D 7 = 4. N X80 Rank of D = = = Hence, 0-30 is e class of D L 1 = 0, i = 10, C.ƒ. = 13, ƒ = 14 9N C.f. D 9 = L x i f 7 70 = 50 + x 10 6 = 50 + x 10 6 = Ans: D 9 =

33 Q. : Find D 5 for e following distribution. X F X ƒ C.ƒ N = 150 5N 5x150 Rank of D 5 = = = Hence, is e class of D 5 L 1 = 40, C.ƒ. = 74, ƒ = 30 5N C.f. D 5 = L x i f = 40 + x = 40 + x = Ans: D 5 = PERCENTILES Definition: Percentiles divide e series into 100 equal parts. There are 99 percentiles giving ninety none dividing points, e value of which is called percentiles. There are 99 percentiles in a distribution, e 50 percentiles being e median or second quartile or 5 deciles. They are denoted by P 1, P, P 3... P 99 SYMBOLICALLY: 33 First Percentile: P 1 = size of N N + 1 Second Percentile: P = size of 100 N + 1 Third Percentile: P 3 = size of Or Or Or N + 1 Ninety-nin Percentile: P 99 = size of n 100 n 100 3n n Or 100

34 (A) INDIVIDUAL DATA Steps: (1) Find Rank of concerned percentile. () Based on result, count e number and determine e percentile. Example: Calculate 50 percentile for e following data. 10, 0, 30, 40, 50, 60, 70, 80, 90 n + 1 Rank of P 50 = P 50 = P 50 = 50 X 10 P 50 = 5 5 is 50 Hence, P 50 = 50 (B) DISCRETE DISTRIBUTION (UNGROUPED DATA): Steps: (1) Find C.ƒ. () Find e rank of concerned percentile. (3) Based on c.f. and rank determine e percentile. Example: Calculate P 60 for e following Marks No. of Students Marks No. of Students C.ƒ N = 19 n + 1 Rank of P 60 = Size of P 60 = P 60 = 100 P 60 = 1 1 lies in C.ƒ. = 14 Hence, P 60 = 40 (C) CONTINUOUS DISTRIBUTION (GROUPED DATA): Example: From e following frequency distribution find P 40 and P Marks

35 No. of Students Marks No. of Students C.ƒ Here, N = 5 N Rank of P 40 = N = 5 value 5 = 40 value 100 = 40 x ¼ value = 10 value This lies in e class L 1 = 30, C.ƒ.= 9, ƒ = 8, i = 10 n P 40 = L C.ƒ x i f n = 30 + x 10 f 10 9 = 30 + x 10 8 = x 10 = P 40 = 31.5 N Rank of P 80 = = value value 1 = 80 x value 4 = 0 value This lies in e class L 1 = 40, C.ƒ.= 17, ƒ = 4, i = 10 n P 80 = L C.ƒ x i f 0 17 = 40 + x 10 35

36 3 30 = 40+ x 10 = = P 80 = 47.5 SUMMARY: Central Tendency:- It may be defined as e tendency of a given set of observation to cluster around a single central or middle value. And is single value known as Measure of Central Tendency or location or average. Different Measures of Central tendency:- 1) Arimetic Mean (AM) ) Median (Me) 3) Mode (Mo/Z) 4) Geometric Mean (GM) 5) Harmonic Mean (HM) Criteria for an Ideal Measure of central tendency:- 1) It should be properly and unambiguously defined. ) It should be easy to comprehend. 3) It should be simple to compute. 4) It should based on all observations. 5) It should have certain desirable maematical properties. 6) It should be least affected by e presence of extreme observations. Arimetic Mean (AM): It is defined as e sum of observations to e number of observations. Σx X = i N Simple (Ungrouped) frequency distribution:- Σfixi Σfixi X = = Σx N 36 i For Grouped Frequency distribution:- Σfidi X = A + x c N Where, X = AM N = Total No. of observation A = Assumed mean x d i = i A c c = Class Wid Properties of AM:- 1) If all e observations are equal e AM is at number itself. ) The algebric sum of deviation from AM is zero. i.e. (x i X) = 0 and ƒ i (x i X) = 0 3) The two groups wi n 1 and n observations and X 1 and X AM, en eir combined mean n1x + nx X = n + n 1 Median:- It is defined as e middle most value when e observations are arranged in ascending or descending order. For Discrete Variable:- Median can be found out by inspection. For Simple (Ungrouped) Frequency distribution:- Median can be found out by finding e N observation. For Grouped Frequency distribution:- Median can be find out by following formula N f1 Micro economics = L 1 + f Where, l 1 N f 1 f m c = Lower limit of median class = Total Frequency = c.f. of pre median class = frequency of median class = leng or wid of median class Partition Values:- It may be defined as values dividing a given set of observations into number of equal parts. m

37 Quartiles:- These are e values which divides e given set of observation into 4 equal parts. So, ere are 3 Quartiles Q 1, Q and Q 3. Q p = (n + 1) x 4 p Deciles:- These are e values which divides e given set of observations into 10 equal parts. So, ere are 9 deciles. i.e. D 1, D, D 3... D 9 Dp = (n + 1) x 10 P Percentiles or Centiles:- These are e values which divides e given set of observations into 100 equal parts. So, ere are 99 percentiles. i.e. P 1, P, P 3...P 99. P PP = (n+ 1) x 100 Where, n = Total observations Mode :- It is defined as e value at occurs e maximum number of times. Depending upon e observation values of modes may one or more or none. For unclassified Data:- Mode can be find out by inspection. G = ( x 1, x 1, x 3... x n ) 1/n For unclassified data f1 f f3 fn 1/ n = ( ) x... For frequency 1,x,x3,... xn distribution Properties:- 1) Logarim G for a set of observations is e AM of e logarim of e observations. ) If all e observations are equal (say K) en eir GM is also K. 3) GM of e product of variable is e product of eir GM s if z = xy en (GM) of z = (GM) of x X (GM) of y. 4) GM of e ratio of variables is e ratio of e GM s of e variable. i.e. If z = x/y en, (GM) ofx (GM) of z = (GM) ofy Harmonic Mean (HM):- It is defined as e reciprocal of e AM of e reciprocal of e observations. n H = 1 Σ[ ] x Frequency distribution:- N H = fi Σ[ ] x i i For Frequency (Ungrouped) distribution:- Mode is e observation having maximum frequency. For Frequency (grouped) Distribution:- Mode can be calculated as, f Mode = l f1 x c f0 f1 f Where, l 1 = LCB of modal class ƒ 0 = frequency of modal class ƒ 1 = frequency of pre modal class ƒ = frequency of post modal class c = Class leng. Mean Mode = 3(Mean Median) Geometric Mean:- It is defined as e n root of e product of e observation. 37 Properties:- 1) If all e observations are equal (say k) en e HM is also k. ) If ere are groups wi n 1 and n observations and H 1 and H as respective HM s en e combined HM is given by n1 + n H = n1 n + H H 3) Relation between AM, GM and HM. For any set of +ve observations AM GM HM Properties of GM and HM:- 1) Bo possess some maematical properties. ) Rigidly defined 3) Based on all observations 4) Difficult to comprehend 5) Difficult to Compute

38 Select e correct alternative out of e given ones: 1) Measures of central tendency are known as (A) Difference (B) Averages (C) Bo (D) None of ese ) Measures of central tendency for a given set of observations measures (A) The scatter ness of e observation (B) The central location of e observations (C) Bo (A) and (B) (D) None of ese 3) The average has relevance for (A) Homogeneous population (B) Heterogeneous population (C) Bo (A) and (B) (D) None of ese 4) A measure of central tendency tries to estimate e (A) Central value (B) Lowe value (C) Upper value (D) None of ese 5) The number of measure of central tendency is (A) Two (B) Three (C) Four (D) five 6) A. M. is never less e G.M. (A) True (B) False (C) Bo (A) and (B) (D) None of ese 7) While computing e A.M. from a grouped frequency distribution, we assume at (A) The classes are of equal leng (B) The classes have equal frequency (C) All e values of a class are equal to e mid-value of at class (D) None of ese 8) If e class interval is open-end, en it is difficult to find (A) Frequency (B) A.M. (C) Bo (A) and (B) (D) None of ese 9) The words mean or average only refer to (A) A.M (B) G.M (C) H.M (D) None of ese 10) Which of e following statements is wrong? (A) Mean in rigidly defined (B) Mean is not affected due to sampling fluctuations. (C) Mean has some maematical properties. (D) All ese 11) When all values occur wi equal frequency, ere is No. (A) Mode (B) Mean (C) Median (D) None of ese 1) Weighted A.M is related to (A) G.M (B) Frequency (C) H.M (D) None of ese 13) The most commonly used measure of central tendency is (A) A.M (B) Median (C) Mode (D) Bo G.M and H.M 14) Weighted averages are considered when (A) The data are not classified. (B) The data are put in e form of grouped frequency distribution (C) All e observations are not of equal importance. (D) Bo (A) and (C) 15) The average discovers (A) Uniformity in variability. (B) Variability in uniformity of distribution (C) Bo (A) and (B) (D) None of ese 16) Each different values is considered only once for (A) Simple average (B) Weight average (C) Bo (A) and (B) (D) None of ese 17) Which of e following statements is true? (A) Usually means is e best measure of central tendency. (B) Usually median is e best measure of central tendency. (C) Usually mode is e best measure of central tendency. (D) Normally, G.M is e best measure of central tendency. 18) When a frequency distribution is given, e frequencies of values are emselves treated as weights. (A) True (B) False (C) Bo (A) and (B) 38

39 (D) None of ese 19) The sum of e squares of deviations of a set of observations has e smallest value, when e deviations are taken from eir (A) A.M (B) H.M (C) G.M (D) None of ese 0) For a given set of observations A.M is greater an G.M (A) True (B) False (C) Bo (A) and (B) (D) None of ese 1) Frequencies are generally used as (A) Range (B) Weights (C) Mean (D) None of ese ) The values of all s are taken into consideration in e calculation of (A) Median (B) Mean (C) Mode (D) None of ese 3) The word average used in simple average and weighted average generally refers to (A) Median (B) Mode (C) A.M or G.M or H.M (D) None of ese 4) The total of a set of observations is equal to e product of eir numbers and e (A) A.M (B) G.M (C) A.M (D) None of ese 5) Simple average is sometimes called (A) Weighted average (B) Unweighted average (C) Relative average (D) None of ese 6) The algebraic sum of deviations of observations from eir A.M is (A) (B) -1 (C) 1 (D) 0. 7) G.M is less an H.M (A) True (B) False (C) Bo (D) None of ese 8) Which of e following measure of e central tendency is difficult to compute? (A) Mean (B) Median (C) Mode (D) G.M 9) You are given e population of India for e courses of 1981 and you are to find e population of India at e middle of e period by averaging ese population figures, assuming a consistent rate of increase of population. What is e suitable of average in is case (A) A.M (B) G.M (C) H.M (D) None of ese 30) Which measure(s) of central tendency is(are) considered for finding e average rates? (A) A.m (B) G.m (C) H.m (D) Bo (B) and (C) 31) Calculation of G.M is more difficult an (A) A.M (B) H.m (C) Median (D) None of ese 3) When a firm registers bo profits and losses, which of e following measure of central tendency cannot be considered? (A) A.m (B) G.m (C) Median (D) Mode. 33) G.M is defined only wi (A) All observation have e same sign and none is zero (B) All observations have e different sign and none is zero (C) All observations have e same sign and one is zero (D) All observation have e different sign and one is zero 34) More laborious numerical calculations involves in G.M an A.M (A) True (B) False (C) Bo (D) None of ese 35) G.M is useful in construction of index number (A) True (B) False 39

40 (C) Bo (D) None of ese 36) In... e quantities are in ratios. (A) A.M (B) G.M (C) H.M (D) None of ese 37)... is not mush affected by fluctuations of sampling (A) A.M (B) G.M (C) H.M (D) None of ese 38)... is e most stable of all e measures of central tendency (A) G.M (B) H.M (C) S.M (D) None of ese 39)... of a set of observations is defined to be eir sum, divided by e number of observations. (A) H.M (B) G.M (C) A.M (D) None of ese 40)... is used when variability has also to be calculated. (A) A.M (B) G.M (C) H.M (D) None of ese 41) Logarim of G.M is e... of e different values. (A) Weighted mean (B) Simple mean (C) Bo (D) None of ese 4) Mean is of... types. (A) 3 (B) 4 (C) 8 (D) 5. 43)... average is obtained on dividing e total of a set of observations by eir number (A) Simple (B) Weighted (C) Bo (D) None of ese 44)... is useful in averaging ratios, rates and percentages. (A) A.M (B) G.M (C) H.M (D) None of ese 45) G.M of a set on n observations is e... root not eir product. (A) (n/) (B) (n+1) (C) n (D) (n-1). 46)... and... cannot be calculated if any observation is zero. (A) G.M & A.M (B) H.M & A.M (C) H.M & G.M (D) None of ese 47)... has a limited use (A) A.M (B) G.M (C) H.M (D) None of ese 48)... is e reciprocal of e A.M of reciprocal of observations. (A) H.M (B) G.M (C) Bo (D) None of ese 49)... is used when e sum of deviations from e average should be least. (A) Mean (B) Mode (C) Median (D) None of ese 50) Half of e numb3er in an ordered set have values less an e... values greater an e... (A) Mean, median (B) Median, median (C) Mode, mean (D) None of ese 51) In e case of continuous frequency distribution e size of e... indicates class interval in which e median lies. (A) (n-1)/ (B) (n+1)/ (C) (n/) (D) None of ese 5)... is a good substitute to a weighted average. (A) A.M (B) G.M (C) H.M (D) None of ese 40