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Central tendency
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Central tendency

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Explains about mean, median and mode

Explains about mean, median and mode

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Central tendency Central tendency Presentation Transcript

  • Central Tendency Prof Vivek Katare Shantiniketan Business School Nagpur
  • Center: a representative or average value that indicates where the middle of the data set is located Center of all colours BLACK
  •  The most common characteristic to measure is the center the dataset. Often people talk about the AVERAGE. ◦ The average American man is six feet, one inches tall
  •  “Average” Is ambiguous, since several different methods can be used to obtain an average Loosely stated, the average means the center of the distribution or the most typical case Measures of Average are also called the Measures of Central Tendency  Mean  Median  Mode
  •  Measures of Center is the data value(s) at the center or middle of a data set Mean Median Mode ◦ We will consider the definition, calculation (formula), advantages, disadvantages, properties, and uses for each measure of central tendency
  •  Notation  Mean of a set of ◦ ∑ (sigma) denotes the sample values (read sum of a set of values as x-bar) ◦ x is the variable usually used to represent the x1 x2 x3 ......... xn individual data values ◦ n represents the x number of values in a n sample ◦ N represents the x number of values in a x population n
  •  Ex: The number of highway miles per gallon of the 10 worst vehicles is given: 12 15 13 14 15 16 17 16 17 18 12 15 13 14 15 16 17 16 17 18x 10 153 x 15.3 10
  • Example (2):Given the following frequencydistribution of first year studentsof a particular School.Age (Years) 13 14 15 16 17Number ofStudents 2 5 13 7 3
  • Solution:The given distribution belongs to a grouped data and the variable involved isages of first year students. While the number of students represent frequencies. Ages (Years) Number of f .x X Students (f) 13 2 26 14 5 70 15 13 195 16 7 112 17 3 51 Total f 30 fx 454 fx 454 x 15.13 f 30
  • Example (3):The following data shows distancecovered by 100 persons to perform theirroutine jobDistance (Km) : 0-10 10-20 20-30 30-40No. of Persons : 10 20 40 30
  • Solution……….. Distance(Km) No. of Mid f.x Persons Points (f) x 0-10 10 5 50 10-20 20 15 300 20-30 40 25 1000 30-40 30 35 1050 f 100 fx 2400 fx 2400 x 24 f 100
  •  Is the middle value when the raw data values are arranged in order from smallest to largest or vice versa Is used when one must find the center or midpoint of a data set Is used when one must determine whether the data values fall into the upper half or lower half of the distribution Does not have to be an original data value Various notations: MD, Med
  •  Arrange data in order  Arrange data in order from smallest to from smallest to largest largest Find the data value in  Find the mean of the the “exact” middle TWO middle numbers (there is no “exact” middle)Odd Number of Data Even Number of DataValues (n is odd) Values (n is even)
  •  The number of highway miles per gallon of the 10 worst vehicles is given: 12 15 13 14 15 16 17 16 17 18 Find the median.
  •  Solution: Arrange data in the ascending order…We get 12 13 14 15 15 16 16 17 17 18 15 16 31 Med 15.5 2 2
  • Example 2:Find the Median of the followingdistributionX : 1 2 3 4 5 6f : 7 12 17 19 21 24Solution………..
  • X f Cumulative Frequency 1 7 7 2 12 19 3 17 36 4 19 55 5 21 76 6 24 100 N=100 N 1Med = Value of 2 th item 100 1 i.e 50.5 th item , which nearly belongs to c.f. of 55Med 2 Hence Median = 4
  • Example 3:Find the Median and Median Class of thefollowing distributionX : 15-25 25-35 35-45 45-55 55-65 65-75f : 4 11 19 14 0 2Solution………..
  • Class f Cumulative Freq. interval 15-25 4 4 25-35 11 15 35-45 19 34 45-55 14 48 55-65 0 48 65-75 2 50 N=50 N 50 Median Number = Value of 2 th element i.e 25th element, 2and this value lies in cumulative freq. (34) for the class interval (35-45).So the median class is (35-45)
  • Hence N cMedian Med Lm ( 2 ).i fWhereLm = 35c = 15f =19i =10 put these values in above formula
  • 25 15Med 35 [ ].10 19 100 35 35 5.26 40.26 19
  • Example 4:Find the Median for following dataValue : 0-4 5-9 10-19 20-29 30-39 40-49 50-59 60-69Freq. : 328 350 720 664 598 524 378 244Solution………..
  • Class Class f Cumulative interval Distribution Freq. 0-9 -0.5-9.5 678 678 10-19 9.5-19.5 720 1398 20-29 19.5-29.5 664 2062 30-39 29.5-39.5 598 2660 40-49 39.5-49.5 524 3184 50-59 49.5-59.5 378 3562 60-69 59.5-69.5 244 3806 N Median Number = Value of 2 th element i.e 3806 1903rd element, 2and this value lies in cumulative freq. (2062) for theclass distribution (19.5-29.5).So the median class is(19.5-29.5)
  • Hence N cMedian Med Lm ( 2 ).i fWhereLm = 19.5C = 1398f =664I =10 put these values in above formula
  • 1903 1398Med 19.5 [ ].10 664 505 5050 19.5 [ ].10 19.5 664 664 19.5 7.60 27.1
  • Example 5:Find missing frequency from the followingdata, given that the median marks is 23Marks : 0-10 10-20 20-30 30-40 40-50No. ofStudents : 5 8 ? 6 3
  • Solution……….. Let, the missing frequency is β. Marks (x) No. of students (f) Cumulative Frequency 0-10 5 5 10-20 8 13 20-30 β 13+β 30-40 6 19+β 40-50 3 22+β Given median is 23. hence the median class is (20-30). and N= (22+β)
  • Hence N cMedian Med Lm ( 2 ).i fWhereLm = 20c = 13f =βi =10 put these values in above formula
  • 22 13Med 20 [ 2 ].10 22 2623 20 [ ].5 110 5 13023 20 110 25 13023
  • 23 110 25 130 2 20 10So, the missing frequency is 10
  •  Is the data value(s) that occurs most often in a data set Is not always unique. A data set can have more than one mode, or the mode may not exist for a data set Has no “special” symbol Look for the number(s) that occur the most often in the data set
  • Example 1)Find mode of following data2, 7, 10, 15, 10, 17, 8, 10, 2Solution:Size of element : 2 7 8 10 15 17No. of times it occur : 2 1 1 3 1 1Number 10 is observed three times in abovedata setHence 10 is the mode of above data set.
  • To find the Mode of grouped data we have following formula fm f1 Mode Lm [ ]i 2 fm f1 f2Where• Lm is the lower class limit of the modal class• fm is the frequency of the modal class• f1 is the frequency of the class before the modal class in the frequency table• f2 is the frequency of the class after the modal class in the frequency table• i is the class interval of the modal class
  • Example 2:Find the Mode of following dataWeight (in Kg.) : 30-35 35-40 40-45 45-50 50-55 55-60No. of Students : 5 9 14 22 16 4Solution……….. Prof Vivek Katare Shantiniketan Business School Nagpur
  • Here the class interval (45-50) has the maximum frequency,i.e. 22, therefore, modal class is (45-50).Hence fm f 1 Mode Lm [ ]i 2 fm f 1 f 2We have, • Lm =45 • fm =22 • f1 =14 • f2 =16 • i =5 put these values in above formula 22 14 Mode 45 [ ]5 2(22) 14 16
  • 8Mode 45 [ ]5 44 14 16 8 45 [ ]5 14 20 45 7 45 2.85 47.85
  • Example 3:The following are the marks obtained by students in aclass test. Find the modal marks. Marks Students 32-35 10 36-39 37 40-43 65 44-47 80 48-51 51 52-55 35 56-59 18 60-63 4
  • Solution: Class(Marks) Class frequency Distribution 32-35 31.5-35.5 10 36-39 35.5-39.5 37 40-43 39.5-43.5 65 44-47 43.5-47.5 80 Modal class 48-51 47.5-51.5 51 52-55 51.5-55.5 35 56-59 55.5-59.5 18 60-63 59.5-63.5 4It is clear from the data that the mode lies in the class 43.5-47.5
  • Hence fm f 1Mode Lm [ ]i 2 fm f 1 f 2We have• Lm =43.5• fm =80• f1 =65• f2 =51• i =4 put these values in above formula 80 65 Mode 43.5 [ ]4 2(80) 65 51
  • 15Mode 43.5 [ ]4 160 65 51 15 43.5 [ ]4 44 15 43.5 11 43.5 1.36 44.86
  • Assignment1) Calculate Mean, Median and Mode of the data given below: X 0.5 1.2 0.9 1 0.6 0.8 1.8 0.9 2.2 1.1 2) Calculate Mean, Median and Mode of the information given below:Class 0-4 5-9 10- 15- 20- 25- 30- 35- 40-Interval 14 19 24 29 34 39 44Freq. 8 10 11 13 25 13 10 5 5 3) Find Mean, Median of the following data: Age Belo 20 30 40 50 60 70 80 w 10 No. of 5 25 60 105 180 250 275 320 Persons
  • 4) The mean marks of 100 students were found to be 40. Later on, it was discovered that a score of 53 was misread as 83. Find the correct mean corresponding to the correct score.5) Mean of 200 observations is found to be 50. If at the time of computation, two items are wrongly taken as 40 and 28 instead of 4 and 82. Find the correct mean.6) Calculate Median and Mode for the distribution of the weights of 150 students from the data given below: Class 30-40 40-50 50-60 60-70 70-80 80-90 Interval Freq. 18 37 45 27 15 8
  • 7) Construct the Frequency table for the following data regarding annual profits, in thousands of rupees in 50 firms by taking class intervals as per your convenience. Find Mean, Median and Mode. 28 35 61 29 36 48 57 67 69 50 48 40 47 42 41 37 51 62 63 33 31 31 34 40 38 37 60 51 54 56 37 46 42 38 61 59 58 44 39 57 38 44 45 45 47 38 44 47 47 64 Prof Vivek Katare Shantiniketan Business School Nagpur