S3 pn
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This document discusses measures of central tendency, including the mode, median, quartiles, and percentiles. It provides definitions and formulas for calculating each measure. The mode is defined as the value that occurs most frequently. The median divides the data set into two equal parts. Quartiles divide the data set into four equal parts, with the second quartile being the median. Percentiles divide the data set into 100 equal parts. Several examples are provided to demonstrate calculating the mode, median, quartiles, deciles and percentiles from data sets.
S3 pn
- 1. Session – 6 Measures of Central Tendency 1 Mode It is the value which occurs with the maximum frequency. It is the most typical or common value that receives the height frequency. It represents fashion and often it is used in business. Thus, it corresponds to the values of variable which occurs most frequently. The model class of a frequency distribution is the class with highest frequency. It is denoted by ‘z’. Mode is the value of variable which is repeated the greatest number of times in the series. It is the usual, and not casual, size of item in the series. It lies at the position of greatest density. Ex: If we say modal marks obtained by students in class test is 42, it means that the largest number of student have secured 42 marks. If each observations occurs the same number of times, we can say that there is ‘no mode’. If two observations occur the same number of times, we can say that it is a ‘Bi-modal’. If there are 3 or more observations occurs the same number of times we say that ‘multi-modal’ case. When there is a single observation occurs mot number of times, we can say it is ‘uni-modal’ case. For a grouped data mode can be computed by following equations with usual notations. Mode = h (f f ) m 1 2f f f m 1 2 where, fm = max frequency (modal class frequency) f1 = frequency preceding to modal class. f2 = frequency succeeding to modal class h = class width. or Mode = hf 2 f f 1 2
- 2. 2 Ex: 1. Find the modal for following data. Marks (CI) No. of students (f) 1 – 10 3 11 – 20 16 21 – 30 26 31 – 40 31 Max. frequency 41 – 50 16 51 – 60 8 f = N = 100 We shall identify the modal class being the class of maximum frequency. i.e. 31-40. where, fm = 31 f1 = 26 f2 = 16 h = 10 30 31 2 30.5 Mode (z) = h (f f ) m 1 2f f f m 1 2 Mode = 10 (31 - 26) 2 x 31 26 16 30.5 Mode = 33. Or
- 3. 10 x 16 3 Mode = hf 2 = f f 1 2 (26 16) 30.5 Mode = 34.30 It can be noted that there exists slightly different mode value in the second method. Partition values Median divides in to two equal parts. There are other values also which divides the series partitioned value (PV). Just as one point divides as series in to two equal parts (halves), 3 points divides in to four points (Quartiles) 9 points divides in to 10 points (deciles) and 99 divide in to 100 parts (percentage). The partitioned values are useful to know the exact composition of series. Quartiles A measure, which divides an array, in to four equal parts is known as quartile. Each portion contain equal number of items. The first second and third point are termed as first quartile (Q1). Second quartile (Q2) and third quartile (Qs). The first quartile is also known as lower quartiles as 25% of observation of distribution below it, 75% of observations of the distribution below it and 25% of observation above it. Calculation of quartiles Q1 = size of item N 1 th 4 Q2 = size of item 3 N 1 th 4 Q2 = (median) = C N 2 f h Measures of quartiles The quartile values are located on the principle similar to locating the median value.
- 4. Following table shows procedure of locating quartiles. Individual and Discrete Measure senses Continuous series N 1 th 2 N 1 th n item 3 th n item 4 Q1 item 4 item n th 4 Q2 item 4 2 th 4 Q3 N 1 item 4 3 th 4 Ex - 1: From the following marks find Q1, Median and Q3 marks 23, 48, 34, 68, 15, 36, 24, 54, 65, 75, 92, 10, 70, 61, 20, 47, 83, 19, 77 Let us arrange the data in array form. Sl. No. x 1. 10 2. 15 3. 19 4. 20 5. 23 Q1 6. 24 7. 34 8. 36 9. 47 10. 48 Q2 11. 54 12. 61 13. 65 14. 68 15. 70 Q3 16. 75 17. 77 18. 83 19. 92
- 5. 1 Here, n = 19 items 5 1 th a. Q1 = n 1 item 4 Q1 = 19 1 4 1 Q1 = x 20 4 Q1 = 5th item Q1 = 23 2 th b. Q2 = n 1 item 4 2 Q2 = x 20 4 10th item Q2 = 48 3 th c. Q3 = n 1 item 4 3 Q3 = x 20 4 = 15th item Q3 = 70 Ex - 2: Locate the median and quartile from the following data. Size of shoes 4 4.5 5 5.5 6 6.5 7 7.5 8 Frequencies 20 36 44 50 80 30 30 16 14 X f cf 4 20 20 4.5 36 56 5 44 100 Q1 5.5 50 150 6 80 230 Q2 6.5 30 260 Q3 7 30 290 7.5 16 306 8 14 320 N = f = 320
- 6. N C h 3 N C and Q3 = 6 1 th Q1 = n 1 item 4 1 Q1 = 321 4 Q1 = 80.25th item Just above 80.25, the cf is 100. Against 100 cf, value is 5. Q1 = 5 1 th Q2 = n 1 item 2 1 Q2 = x 321 2 160.5th item Just above 160.5, the cf is 230. Against 230 cf value is 6. Q2 = 6 3 th Q3 = n 1 item 4 3 Q3 = x 4 321 = 240.75th item Just above 240.75, the cf is 260. Against 260 cf value is 6.5. Q3 = 6.5 Ex - 3: Compute the quartiles from the following data. CI 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 5 8 7 12 28 20 10 10 (f) 1 h First quartile (Q1) = 4 f 4 f h N and (Q2) = Median = C and 2 f
- 7. CI f cf 0-10 5 5 10-20 8 13 20-30 7 20 30-40 12 32 Q1 40-50 28 60 Q2 50-60 20 80 Q3 60-70 10 90 70-80 10 100 N = f = 100 7 a. First locate Q1 for ¼ N ¼ N = 25 = 30 h = 10 f = 12 c = 20 h 1 N C (Q1) = 4 f 30 30 = 30 2 10 Q1 = 30 25 20 12 Q1 = 34.16 b. Locate Q2 (Median) 40 40 Q2 corresponds to N/2 = 50, 40 2 h N C Q2 = 2 f 10 Q2 = 40 50 32 28 Q2 = 46.42
- 8. 50 50 Q3 corresponds to ¾ N = 75, 50 N C D2 & so,on P26 = and so, on 8 2 h 3 N C Q3 = 4 f 10 Q3 = 50 75 60 20 Q3 = 57.5 Deciles The deciles divide the arrayed set of variates into ten portions of equal frequency and they are some times used to characterize the data for some specific purpose. In this process, we get nine decile values. The fifth decile is nothing but a median value. We can calculate other deciles by following the procedure which is used in computing the quartiles. Formula to compute deciles. D1 N C , 1 10 h f 2 20 h f Percentiles Percentile value divides the distribution into 100 parts of equal frequency. In this process, we get ninety-nine percentile values. The 25th, 50th and 75th percentiles are nothing but quartile first, median and third quartile values respectively. Formula to compute percentiles is given below: 25 h P25 = N C , 100 f N C 26 100 h f Ex: Find the decile 7 and 60th percentile for the given data of patients visited to hospital on a particular day. CI f Cf 10-20 1 1 20-30 3 4 30-40 11 15 40-50 21 36 50-60 43 79 P60 60-70 32 111 D70 70-80 9 120 f = N = 120
- 9. 9 h 7 a. D7 = N C , 10 f 60 60 60 2 7 N 84 10 h = 10, f = 32 c = 79 10 D7 = 60 84 79 32 7th Decile = D7 = 61.562 b. 60th percentile h 60 N C P60 = 100 f 50 50 50 2 h = 10 f = 43 c = 36 60 N 72 100 10 P60 = 50 72 36 43 10 P60 = 50 72 36 43 P60 = 58.37 SOME NUMERICAL EXAMPLES 1. Show that following distribution is symmetrical about the average. Also shows that median is the mid-way between lower and upper quartiles. X 2 3 4 5 6 7 8 9 10 Frequency 2 9 29 57 80 57 29 9 2 To show the given distribution is symmetrical, Mean, Median and Mode must be same.
- 10. To show median is mid-way between the lower and upper quartile i.e., Q2 – Q1 10 = Q3 – Q2. Mid-point x Class interval CI f d = (x – 6) fd cf Cum. freq. 2 1.5 – 2.5 2 -4 -8 2 3 2.5 – 3.5 9 -3 -27 11 4 3.5 – 4.5 29 -2 -58 40 5 4.5 – 5.5 57 -1 -57 97 Q1 class 6 5.5 – 6.5 80 0 0 177 Q2 class 7 6.5 – 7.5 57 1 57 234 Q3 class 8 7.5 – 8.5 29 2 58 263 9 8.5 – 9.5 9 3 27 272 10 8.5 – 10.5 2 4 8 274 N=274 fd = 0 Let A = 6 Mean = h fd N A 1x 0 6 = 6 274 Mean = 6. Median h N C Q2 = 2 f 137 N 274 2 2 C = 97 1 Q2 = 5. 137 97 80 Q2 = 5.5 + 0.5 Median = Q2 = 6.
- 11. Modal class 5.5 – 6.5 11 Mode Mode = h f f m 1 2f f f m 1 2 Mode = 1 80 57 2 x 80 57 57 5.5 Mode = 6. Since, Mean = Mode = Median. The given distribution is symmetrical. Q1 calculation h 4 N C Q1 = 2 f 1 Q1 = 6.5 68.5 40 57 Q1 = 7. Now, Q2 – Q1 = Q3 – Q2 i.e. 6 – 5 = 7 – 5 2 = 2 2. Find the mean for the set of observations given below. 6, 7, 5, 4 6 8 7 8 4 5 N xi x i 1 n 30 = 6 5 3. Find the mean for the following data. CI f xi fx 0-10 3 5.5 16.5 11-20 16 15.5 248 21-30 26 23.5 683 31-40 31 35.5 1180.5 41-50 16 45.5 728 51-60 8 55.5 444 N = f = 100 3300
- 12. 12 3200 100 fx x N x 32 4. Find the mean profit of the organisation for the given data below: Profit CI f xi fx 100-200 10 150 1500 200-300 18 250 4500 300-400 20 350 7000 400-500 26 450 11700 500-600 30 550 16500 600-700 28 650 18200 700-800 18 750 13500 N = f = 150 72900 x1 = 100 200 2 x1 = 300 2 x1 = 150 fx N x = 72900 150 x 486 Step Deviation Method x = a + hd d = x a h fd N x a h a = Arbitrary constant h = class width
- 13. Profit CI f xi d fd 100-200 10 150 -3 -30 200-300 18 250 -2 -36 300-400 20 350 -1 -20 400-500 26 450 0 0 500-600 30 550 +1 30 600-700 28 650 +2 56 700-800 18 750 +3 34 N = f = 150 fm = 54 2430 x 4 2590 x 28 2870 x 31 3390 x 16 4730 x 3 5160 x 2 13 fd N x a h 54 150 x 450 100 x 486 5. In an office there are 84 employees and there salaries are given below. Salary 2430 2590 2870 3390 4720 5160 Employees 4 28 31 16 3 2 1. Find the mean salary of the employees 2. What is the total salary of the employees? fx N x = 84 fx N x 249930 84 x Rs. 2975.36 1. x 2975.36 2. Total salary = 2,49,930 (Rs.)
- 14. 6. The average marks secured by 36 students was 52 but it was discovered that on item 64 was misread as 46. Find the correct me of the marks. 14 fx N x fx 56 52 fx = 52 x 36 = 1872 fx = fx - incorrect + correct correct = 1872 – 46 64 = 1890 x fx correct N x 1890 36 x 52.5 7. The mean of 100 items is 46, later it was discovered that an item 16 was misread as 61 and another item 43 was misread as 34 and also found that the total number of items are 90 not 100 find the correct mean value. fx N x fx 100 46 fx = 4600 fx = fx - incorrect + correct = 4600– 61 - 34 + 16 + 43 = 4564 x fx correct N x 4564 90 = 50.71
- 15. 8. Calculate the mean for the following data. 15 Value Frequency < 10 4 < 20 10 < 30 15 < 40 25 < 50 30 CI f ‘m’ mid point fm 0-10 4 5 20 10-20 10 15 150 20-30 15 25 375 30-40 25 35 875 40-50 30 45 1350 f = 84 fx 2770 fm N x 2770 84 x 32.97 9. For a given frequency table, find out the missing data. The average accident are 1.46. No. of accidents Frequency 0 46 1 ? 2 ? 3 25 4 10 5 5
- 16. 16 No. of accidents (x) Frequency (f) fx 0 46 0 1 ? f1 2 ? 2f1 3 25 75 4 10 40 5 5 25 N = 200 fx = 140 + f1 + 2f2 1.46 = 140 f 2f 1 2 200 292 = 140 + f1 + 2f2 f1 + 2f2 = 152 ----(1) w.k.t. N = f 200 = 86 + f1 + f2 f1 + f2 = 114 ----(2) f1 + 2f2 = 152 ----(1) f1 + f2 = 114 ----(2) (1) – (2) --------------------------------- f2 = 38 --------------------------------- f2 = 38 f1 + f2 = 114 f1 + 114 – 38 f1 = 76
- 17. 10. Find out the missing values of the variate for the following data with mean is 17 31.87. xi F 12 8 20 16 27 48 33 90 ? 30 54 8 N = 200 xi f fx 12 8 96 20 16 320 27 48 1296 33 90 2970 x 30 30x 54 8 432 N = 200 fx = 5114 + 30x x 31.87 fx N x fx 200 31.87 fx = 6374 ----(1) fx = 5114 + 30x ----(2) (1) = (2) 6374 = 5114 + 30x 6374 - 5114 = 30x 30x = 1260 x = 42.
- 18. 11. The average rainfall of a city from Monday to Saturday is 0.3 inches. Due to heavy rainfall Sunday the average rainfall for the week increased to 0.5 inches. What is the rainfall on Sunday? Given: Mon – Sat = 0.3” Sun = 0.5” fx1 = 1.8 fx2 = 3.5 18 fx x 1 N fx 0.3 1 6 fx x 2 N fx 0.5 2 7 Rainfall on Sunday = fx2 – fx1 = 3.5 – 1.8 = 1.7” 12. The average salary of male employees in a firm was Rs. 520 and that of females Rs. 420 the mean of salary of all the employees as a whole is Rs. 500. Find the percentage of male and female employees. Given: x 520 1 x 420 2 x 500 n1 = Male persons. n2 = Female persons. n x n x 1 1 2 2 n n 1 2 x n x 520 n x 420 1 2 n n 1 2 500 520n 420n 1 2 n n 1 2 500 500n1 + 500n2 = 520n1 – 420n2 80n2 = 20n1 n1 = 4n2 Let n1 + n2 = 100 4n2 + n2 = 100 5n2 = 100 n2 = 20% Female n1 = 80% Male 20% and 80% are male and females in the firm.
- 19. 13. The A-M of two observations is 25 and there GM is 15. Find the HM. 19 Given: AM = 25 a b 2 x a b 2 x a b 2 25 a + b = 50 GM = 15 GM = 2 ab GM = ab 15 = ab (15)2 = ( ab )2 ab = 225 HM = ? HM = 1 b 1 a 2 HM = 2ab a b HM = 2 x 225 50 HM = 9 a + b = 50 ab = 225 a = 225 b HM = 9 14. The GM is 60 an HM is 28.24. Find AM for two observations. AM GM HM a b 2 x 254 95b 2 x = 127.475 60 = ab 602 = ab ab = 3600 28.24 = 2ab a b a + b = 2ab 28.4 = 2 x 3600 28.4 a + b = 254.95
- 20. 15. Calculate the missing frequency from the data if the median is 50. CI f cf 10-20 2 2 20-30 8 10 30-40 6 16 40-50 ? f1 16+f1 50-60 15 31+f1 median class 60-70 10 41+f1 20 f = 41 + f1 h N C Q = 2 f (16 f ) 2 10 50 = 50 + N 15 (16 f ) 2 10 1 50 – 50 = N 15 1 (16 f ) 2 10 0 = N 15 (16 f ) 2 1 0 = N 10 1 (16 f ) 2 0 = N 1 N (16 f ) 1 2 16 + f1 = ½ (41 + f1) 2 (16 + f1) = 41 + f1 32 + 2f1 = 41 + f1 f1 = 9
- 21. SOURCES AND REFERENCES 1. Statistics for Management, Richard I Levin, PHI / 2000. 2. Statistics, RSN Pillai and Bagavathi, S. Chands, Delhi. 3. An Introduction to Statistical Method, C.B. Gupta, & Vijaya Gupta, Vikasa 21 Publications, 23e/2006. 4. Business Statistics, C.M. Chikkodi and Salya Prasad, Himalaya Publications, 2000. 5. Statistics, D.C. Sancheti and Kappor, Sultan Chand and Sons, New Delhi, 2004. 6. Fundamentals of Statistics, D.N. Elhance and Veena and Aggarwal, KITAB Publications, Kolkata, 2003. 7. Business Statistics, Dr. J.S. Chandan, Prof. Jagit Singh and Kanna, Vikas Publications, 2006.