Central tendency

1,791 views

Published on

0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
1,791
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
43
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide

Central tendency

  1. 1. Measures of Central Tendency
  2. 2. Concept of Central Tendency• A measure of central tendency is a typical value around which other figures congregate - Simpson & Kalfa OR An average is a single value which is used to represent all of the values in the series.
  3. 3. Measures of Central Tendency Mean (mathematical Median (positional Mode (positional average) average) average)Arithmetic Mean Geometric mean Harmonic mean Simple Arithmetic Mean Weighted Arithmetic Mean Mean of Composite Group
  4. 4. Basics• Mean Average• Median Mid positional value• Mode Most frequently occurring value 5
  5. 5. Arithmetic Mean Ungrouped (Raw) Data Sum of Observation sx= Number of Observation s ∑ xi = n
  6. 6. EXAMPLETable 4.1 : Equity Holdings of 20 Indian Billionaires ( Rs. in Millions) 2717 2796 3098 3144 3527 3534 3862 4186 4310 4506 4745 4784 4923 5034 5071 5424 5561 6505 6707 6874
  7. 7. ExampleFor the above data, the A.M. is 2717 + 2796 +…… 4645+….. + 5424 + ….+ 6874x = -------------------------------------------------------------------------- 20 = Rs. 4565.4 Millions
  8. 8. Arithmetic Mean Grouped Data x= ∑f x i i• N= n ∑f i = Total frequency ∑f i=1 i• Here, xi is the mid value of the class interval.
  9. 9. example• Calculate arithmetic Marks No.of mean from the students following frequency 25 2 distribution of marks at 30 3 a test in statistics. 35 4 40 8 45 9 50 4 55 3 60 2
  10. 10. • The details of the monthly salary of 100 employees of a firm are given below: Monthly salary (in Rs.) No. of employees 1000 18 1500 26 2000 31 2500 16 3000 5 5000 4
  11. 11. • In grouped data, the middle value of each group is the representative of the group bz when the data are grouped, the exact frequency with which each value of the variable occurs in the distribution is unknown.• We only know the limits within which a certain number of frequencies occur.• So, we make an assumption that the frequencies within each class are distributed uniformly over the range of class interval.
  12. 12. Example• A company manufactures polythene bags. The bags are evaluated on the basis of their strength by buyers. The strength depends on their bursting pressures. The following data relates to the bursting pressure recorded in a sample of 90 bags. Find the average bursting pressure.
  13. 13. exampleBursting No. of Mid Value of Fixi ( 4 )pressure bags Class Interval Col.(4) = Col.(2) x Col. (1) ( fi ) ( 2 ) ( xi ) ( 3 ) (3) 5-10 10 7.5 75 10-15 15 12.5 187.5 15-20 20 17.5 350 20-25 25 22.5 562.5 25-30 20 27.5 550 Sum Σ fi =90 Σ fixi =1725
  14. 14. values of Σ fi and Σ fixi , in formula x= ∑f x i i ∑f i = 1725/90 = 19.17
  15. 15. EXAMPLE (short cut method)• Calculate the mean of Monthly No. of the following wages(in workers distribution of Rs.) monthly wages of 100-120 10 workers in a factory : 120-140 20 140-160 30 160-180 15 180-200 5
  16. 16. • The following frequency distribution represents Time taken (in frequencies the time taken in seconds) seconds to serve 40-60 6 customers at a fast food 60-80 12 take away. Calculate 80-100 15 the mean time taken by 100-120 12 to serve customers 120-140 10 140-160 5
  17. 17. Weighted Arithmetic Mean• It takes into account the importance of each value to the overall data with the help of the weights.• Frequency i.e. the no. of occurrence indicates the relative importance of a particular data in a group of observations.• Used in case the relative importance of each observation differs or when rates, percentages or ratios are being averaged.
  18. 18. • The weighted AM of the n observations: x= ∑ x wi i ∑ wi• AM is considered to be the best measure of central tendency as its computation is based on each and every observation.
  19. 19. Example• 5 students of a B.Sc. (Hons) Stud midt Proje Atten Fin course are marked by using the ent erm ct dnce al following weighing scheme : – Mid-term = 20% 1 65 70 80 80 – Project = 10% – Attendance = 10% 2 48 58 54 60 – Final Exam = 60% Calculate the average marks in the 3 58 63 65 50 examination. Marks of the students in various 4 58 70 54 60 components are: 5 60 65 70 70
  20. 20. • A professor is interested in ranking the following five students in the order of merit on Stud Atte Hom Assi Midt final the basis of data given below: ent nda ewo gnm erm• Attendance average will count for nce rk ent 20% of a student’s grade; the A 85 89 94 87 90 homework 25%; assignment 35%; B 78 84 88 91 92 midterm examination 10% and C 94 88 93 86 89 final examination 10%. What would be the students ranking. D 82 79 88 84 93 E 95 90 92 82 88
  21. 21. Mean of composite group• If two groups contain respectively, n1 and n2 observations with mean X1 and X2, then the combined mean (X) of the combined group of n1+n2 observations is given by : n1 X 1 + n2 X 2 X 12 = n1 + n2
  22. 22. Example• There are two branches of a company employing 100 and 80 employees respectively. If arithmetic means of the monthly salaries paid by two branches are Rs. 4570 and Rs. 6750 respectively, find the A.M. of the salaries of the employees of the company as a whole.
  23. 23. • A factory has 3 shifts :- Morning, evening and night shift. The morning shift has 200 workers, the evening shift has 150 workers and night shift has 100 workers. The mean wage of the morning shift workers is Rs. 200, the evening shift workers is Rs. 180 and the overall mean of the workers is Rs. 160. Find the mean wage of the night shift workers.
  24. 24. Properties of A.M.• If a constant amount is added or subtracted from each value in the series, mean is also added or subtracted by the same constant amount. E.g. Consider the values 3,5,9,15,16 A.M. = 9.6 If 2 is added to each value, then A.M. = 11.6 = 9.6 + 2. Thus, mean is also added by 2.• Sum of the deviations of a set of observations say x1, x2, , xn from their mean is equal to zero. A.M. is dependent on both change in origin and scale. The sum of the squares of the deviations of a set of observation from any number say A is least when A is X.
  25. 25. Merits and demerits of Arithmetic Mean Advantages Disadvantages(i) Easy to understand and (i ) Unduly influenced by extreme calculate values(ii) Makes use of full data (ii) Cannot be(iii) Based upon all the calculated from the data with observations. open-end class.e.g. below 10 or above 90 (iii) It cannot be obtained if a single observation is missing. (iv) It cannot be used if we are dealing with qualitative characteristics which cannot be measured quantitatively; intelligence, honesty, beauty
  26. 26. Harmonic Mean The harmonic mean (H.M.) is defined as the reciprocalof the arithmetic mean of the reciprocals of theobservations. For example, if x1 and x2 are two observations, then thearithmetic means of their reciprocals viz 1/x1 and 1/ x2 is {(1 / x1) + (1 / x2)} / 2 = (x2 + x1) / 2 x1 x2 The reciprocal of this arithmetic mean is 2 x1 x2 / (x2 +x1). This is called the harmonic mean. Thus the harmonic mean of two observations x1 and x2is 2 x1 x2 -----------------
  27. 27. • In general, for the set of n observations X1,X2……..Xn, HM is given by : n HM = 1 ∑x i• And for the same set of observations with frequencies f1,f2……..fn, HM is given by: n HM = fi ∑x i
  28. 28. • HM gives the largest weight to the smallest item and the smallest weight of the largest item• If each observation is divided by a constant, K then HM is also divided by the same constant.• If each observation is multiplied by a constant, K then HM is also multiplied by the same constant.• It is used in averaging speed, price of articles.
  29. 29. • If time varies w.r.t. a fixed distance then HM determines the average speed.• If distance varies w.r.t. a fixed time then AM determines the average speed.• EXAMPLE : If a man moves along the sides of a square with speed v1, v2, v3, v4 km/hr, the average speed for the whole journey = 4 (1/v1)+(1/v2)+(1/v3)+(1/v4)
  30. 30. EXAMPLE• In a certain factory a unit of work is completed by A in 4 min, by B in 5 min, by C in 6 min, by D in 10 min, and by E in 12 minutes. – What is the average no. of units of work completed per minute?
  31. 31. Example• The profit earned by 19 Profit No. of companies is given (lakhs) companies below: 20-25 4 calculate the HM of 25-30 7 profit earned. 30-35 4 35-40 4
  32. 32. Geometric MeanNeither mean, median or mode is the appropriate average incalculating the average % rate of change over time. For this G.M. isused.The Geometric Mean ( G. M.) of a series of observations with x 1, x2,x3, ……..,xn is defined as the nth root of the product of thesevalues . MathematicallyG.M. = { ( x1 )( x2 )( x3 )…………….(xn ) } (1/ n )It may be noted that the G.M. cannot be defined if any value of x iszero as the whole product of various values becomes zero.
  33. 33. • When the no. of observation is three or more then to simplify the calculations logarithms are used. log G.M. = log X1 + log X2 + ……+ log Xn N G.M. = antilog (log X1 + log X2 + ……+ log Xn) NFor grouped data, G.M. = antilog (f1log X1 + f2log X2 + ……+ fnlog Xn) N
  34. 34. Geometric mean• GM is often used to calculate the rate of change of population growth.• GM is also useful in averaging ratios, rates and percentages.
  35. 35. EXAMPLE• A machinery is assumed to depreciate 44% in value in first year, 15% in second year and 10% in next three years, each percentage being calculated on diminishing value. What is the average % of depreciation for the entire period?• Compared to the previous year the overhead expenses went up by 32% in 2002; they increased by 40% in the next year and by 50% in the following year. Calculate the average rate of increase in the overhead expenses over the three years.
  36. 36. Example• The annual rate of growth for a factory for 5 years is 7%,8%,4%,6%,10%respectively.What is the average rate of growth per annum for this period.• The price of the commodity increased by 8% from 1993 to 1994,12%from 1994 to 1995 and 76% from 1995 to 1996.the average price increase from 1993 to 1996 is quoted as 28.64% and not 32%.Explain and verify the result. 37
  37. 37. Combined G.M. of Two Sets of Data If G1 & G2 are the Geometric means of two setsof observations of sizes n1 and n2, then thecombined Geometric mean, say G, of thecombined series is given by : n1 log G1 + n2 log G2log G = ------------------------------- n1 + n2
  38. 38. Example• The GM of two series of sizes 10 and 12 are 12.5 and 10 respectively. Find the combined GM of the 22 observations.
  39. 39. Combined G.M. of Two Sets of Data 10 log 12.5 + 12log 10log G = ------------------------------- 10 + 12 22.9691 = ------------ = 1.04405 22Therefore, G = antilog 1.04405 = xThus the combined average rate of growth for the period of 22years is x%.
  40. 40. Relationship Among A.M. G.M. and H.M.The relationships among the magnitudes of the threetypes of Means calculated from the same data are asfollows: (i) H.M. ≤ G.M. ≤ A.M. i.e. the arithmetic mean is greater than or equalto the geometric which is greater than or equal to theharmonic mean.( ii ) G.M. = A.M * H .M .i.e. geometric mean is the square root of the product ofarithmetic mean and harmonic mean.( iii) H.M. = ( G.M.) 2 / A .M.

×