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MEDIAN

MEDIAN. Final Presentation Bs-Cs-II Section-A. Group Members. A li H ider F aran H ashmi I rfan R asheed M ansoor K halid M uhib U llah M . R amzan. Work Distribution. Mansoor Khalid (Introduction AND ungrouped data ) Irfan Rasheed (grouped data from continuous series)

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MEDIAN

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  1. MEDIAN Final Presentation Bs-Cs-II Section-A

  2. Group Members • Ali Hider • FaranHashmi • Irfan Rasheed • Mansoor Khalid • MuhibUllah • M. Ramzan

  3. Work Distribution • Mansoor Khalid (Introduction AND ungrouped data ) • IrfanRasheed (grouped data from continuous series) • M. Ramzan(grouped data from discrete Series) • Muhibullah(quartiles ,deciles ,percentiles from ungrouped data and discrete series ) • Ali haider(quartiles ,deciles ,percentiles from grouped data and continuous series ) • FaranHashmi(Merits & Demerits)

  4. MEDIAN • DEFINITION • MEDIAN is the middle item of a series when it is arranged in ascendingor descending order of magnitude. • Example: Find the Median of 12, 3 and 5 Put them in order: 3, 5, 12 The middle number is 5, so the median is 5.

  5. Properties Of MEDIAN • MEDIAN divides the series in two equal halves in such a way that in one half; the values are less than MEDIAN and in the other half the values more than MEDIAN.

  6. How to Find The MEDIAN….??? • To find the MEDIAN, • First arrange the values in an array, either from the smallest to the largest or from the largest to the smallest. • Second, locate the middle values; that is, the number of values above the median is same as the number of values below the median.

  7. Methods • UNGROUPED DATA • Odd Number • Even Number • GROUPED DATA • Continuous Series • Discrete Series

  8. UNGROUPED DATA • Suppose that the variable X has n values and they r arranged according to magnitude, so that, X1≤X2≤X3≤…….≤Xn

  9. For Odd Numbers • If the number of values i.e. n is an odd number, the median is calculated by the formula given below. Med= the value of(n+1/2)th observation • For example, if 15 items are arranged in order the 8th item will be the median.

  10. Example 1 Find the median of the following items . 5,7,9,12,8,10,7,15,21 Solution: Arranged in ascending order. 5,7,7,8,9,10,12,15,21

  11. Solution

  12. Solution Median= the value of(n+1/2)th observation Here the value of n=9 Median=The value of (9+1/2)th item Median=the value of 5th item Median=9

  13. Even Numbers • If the number of values i.e. n is an even number, the median is obtained by finding the Arithmetic mean of the two middle values i.e. Median=1/2(the value of n/2 th item + the value of n+2/2 th item)

  14. Example • The following are the marks obtained in a paper of statistics in an examination by 12 students. Find the median. 36,56,41,46,54,57,55,51,52,44,37,59 Solution Marks are arranged in ascending order of magnitude 36,37,41,44,46,51,52,54,55,56,57,59

  15. solution • Here n=12(even number) • Median=1/2[the value of n/2th item+the value of n+2/2 th item] • Median=1/2[the value of 12/2th item+the value of 12+2/2 th item] Median=1/2[the value of 6th item+the value of 7 th item] • Median=1/2[51+52] • Median=51.2

  16. Grouped data • For Continuous Series: • When frequency distribution is available in a continuous series, the median is the value of n/2th item from either end. To find the median from frequency distribution, we form cummulative frequency. The cummulative frequency of each class is the sum of the frequency of the class and the frequencies of the previous classes.

  17. For Continuous Series • Median is obtained by formula. Median =L + h / f ( n / 2 - C) • Where: • L=Lower class boundary of median class. • N=Number of items in the data(i.e. total frequency). • f=Frequency of median class. • h=Size of class interval of median class. • C=Cummulative frequency of the class preceding in the median class or sum of frequencies of all classes lower than the median.

  18. Group data Question: 1 In this question, Class find = n/2= 75/2 = 37.5 Lower boundaries “L” = 64.5 Height or difference “h ” =20 Frequency “f ” =22 Cumulative frequency “c.f” =24 • Median =L+ h / f ( n / 2 - C) • Median =64.5+20/22(75/2-24) • Median =64.5+0.90(13.5) • Median =76.65

  19. Question: 2 In this question, Class find = n/2 = 48/2 = 24 Lower boundaries “ l ” = 15 Height or difference “ h ” = 5 Frequency “ f ” = 13 Cumulative frequency “c.f” = 14 • Median= l + h / f ( n / 2 - C) • Median = 15 + 5/13 ( 48/2 - 14) • Median = 15 + (0.38)(10) • Median = 18.86

  20. Question: 3 In this question, Class find = n/2 = 95/2 = 47.5 Lower boundaries “L” = 49.5 Height or difference “h ” = 5 Frequency “f ” =29 Cumulative frequency “c.f” =45 • Median= L+ h / f ( n / 2 - C) • Median = 49.5+5/29(95/2 - 45) • Median = 49.5+(0.17)(2.5) • Median = 49.92

  21. For discrete series • Median-Discrete Series • For calculating Median-Discrete Series, the steps are as follows: • At first arrange the data in ascending or descending order of magnitude. • Then find out the cumulative frequencies. • After that apply the formula median= size of (N + 1)/2 • Now look at the cumulative frequency column and find the total which is either equal to (N + 1)/2 or next higher to that and determine the value of the variable corresponding to it. That gives the value of the median.

  22. For discrete series • Illustration:From the following data find the value of median. • Income ($) 5000 5500 6800 8000 8500 7800 • No. of persons 24 26 16 20 6 30

  23. Solution:   • Calculation of median • Income arranged in ascending order No. of persons c.f • 5,000 24 24 • 5,500 26 50 • 6,800 20 70 • 7800 30 100 • 8000 16 116 • 8500 6 122

  24. Median = size of {n + (1/2)} th item • Median = 122 + (1/2) • Median = 61.5th item.Size of 61.5th item = $ 6,800, hence the median income is $ 6,800.

  25. Quartiles, Deciles and Percentiles • Median divides the set of observations into two equal parts so that the number of observations less than median is equal to the number of observations greater than median. Similarly we can divide the set of observations into a fixed number of equal parts. These are called partitions. The values that divide the setof observations into different partitions are called partition values or Quantiles • Some of the important partition values are Quartiles, Deciles and Percentiles.

  26. Quartiles • Quartiles • There are three Quartiles,Q1,Q2,Q3 • which divide the set of observations into four equal parts. They are named as first quartileQ1, second quartileQ2and third quartileQ3 • Of themQ2 is the median andQ1andQ3are also called lower and upper quartiles respectively.

  27. Deciles • The values which divide an arrayed set of data into ten equal parts are called deciles and are denoted by D1,D2,…….D9 respectively. • Percentiles • Percentiles are the values in an arranged distribution which divide the distribution into hundred equal parts. Percentiles are denoted by P1,P2,…….P25,….P50,…..P75,….P99.

  28. Quartiles, Deciles, Percentiles, from un group data and discrete series • Quartiles: The quartiles are calculated by the following • Q1=The value of (n+1/4)th item • Q2=The value of 2(n+1/4)th item=Median • Q3=The value of 3(n+1/4)th item

  29. Deciles • Deciles are calculated by the following D1=The value of (n+1/10)th item • D2=The value of 2(n+1/10)th item=Median • D3=The value of 3(n+1/10)th item • D5=The value of 5(n+1/10)th item=Median • D9=The value of 9(n+1/10)th item

  30. Percentiles • Percentiles are calculated by the following P1=The value of (n+1/100)th item • P2=The value of 2(n+1/100)th item=Median • P25=The value of 25(n+1/100)th item=Q1 • P50=The value of 50(n+1/100)th item=Median • P75=The value of 75(n+1/100)th item=Q3 • P99=The value of 99(n+1/100)th item

  31. Cocclusion • From above discussion we find the following relation • Median=Q2=D5=P50 • Q1=P25, Q3=P75

  32. Question • Percentiles for the ungrouped data • Example: find • The 1st quartiles • The 7th deciles • The 75thpercentiles. • For the following height data collected from students  91, 89, 88, 87, 89, 91, 87, 92, 90, 98, 95, 97, 96, 100, 101, 96, 98, 99, 98, 100, 102, 99, 101, 105, 103, 107, 105, 106, 107, 112. • Solution: • The ordered observations of the data are 87, 87, 88, 89, 89, 90, 91, 91, 92, 95, 96, 96, 97, 98, 98, 98, 99, 99, 100, 100, 101, 101, 102, 103, 105, 105, 106, 107, 107, 112. • N=30

  33. Quartiles Q1=?solution Q1=The value of (n+1/4)th item Q1=The value of (30+1/4)th item Q1=The value of (7.8)th item Q1=The value of 7th item+0.8(8th item-7th item) Q1=91+0.8(91-91) 7th term=91 8Th term=91 Q1=91+0.8(0) Q1=91+0 Q1=91

  34. Deciles D7=?solution • D7=The value of 7(n+1/10)th item • D7=The value of 7(30+1/10)th item • D7=The value of 7(3.1)th item • D7=The value of 21.7th item • D7=The value of 21th item+0.7(8th item-7th item) • D7=101+0.8(91-91) 7th term=91 8Th term=91 • D7=101+0.8(0) 21th item=101 • D7=101+0 • D7=101

  35. Percentiles P75=?solution • P75=The value of 75(n+1/100)th item • P75=The value of 75(30+1/100)th item • P75=The value of 75(0.31)th item • P75=The value of 23.3th item • P75=The value of 23th item+0.3(4th item-3th item) • P75=102+0.3(89-88) 23th term=102 4Th term=89 • P75=102+0.3(1) 3th item=88 • P75=102+0.3 • P75=102.3

  36. Quartiles In Continuous Series In the case of continuous series, we find the cumulative frequency first and then use the interpolation formula.Q1 = L+H/F(n/4-C) Q2 = L+H/F(2n/4-C) = median Q3 = L+H/F(3n/4-C) Where, L = lower limit of the Q1, Q2 and Q3 classes respectively.C = cumulative frequency of the class just preceding the corresponding classes.F = frequency of the Q1, Q2 and Q3 classes respectively andH= class size of the corresponding classes.

  37. DECILES IN CONTINUOUS SERIES In the case of continuous series, we find the cumulative frequency first and then use the interpolation formula.D1 = L+H/F(n/10-C) D2 = L+H/F(2n/10-C) . . D5 = L+H/F(3n/10-C) = median . . D9= L+H/F(9n/10-C) Where, L = lower limit of the D1, D2,D5andD9, classes respectively.C = cumulative frequency of the class just preceding the corresponding classes.F = frequency of the D1, D2,D5andD9  classes respectively andH= class size of the corresponding classes.

  38. PERCENTILES IN CONTINUOUS SERIES In the case of continuous series, we find the cumulative frequency first and then use the interpolation formula.P1 = L+H/F(n/100-C)P2 = L+H/F(2n/100-C) . . P25 = L+H/F(25n/100-C)= Q1 . P50= L+H/F(50n/100-C) median . P75= L+H/F(75n/100-C)=Q3 . P99= L+H/F(99n/100-C) Where, L = lower limit of the D1, D2,D5andD9, classes respectively.C = cumulative frequency of the class just preceding the corresponding classes.F = frequency of the D1, D2,D5andD9  classes respectively andH= class size of the corresponding classes.

  39. Question Find the Quartiles ,deciles , percentiles of the following data:-

  40. solution

  41. solution • FIND Q1=???? Q1 = L+H/F(n/4-C) L=10,H=5,F=20 ,n=102 ,n/4,102/4=25.5,C=25 Q1 = 10+5/20(102/4-25) Q1 = 10+5/20(25.5-25) Q1=10+(5*.5/20) Q1=10+0.125 Q1=10.125

  42. solution • FIND Q3=???? Q3 = L+H/F(n/4-C) L=20,H=5,F=18,n=102 ,3n/4,3*102/4=76.5,C=70 Q3= 20+5/18(3*102/4-70) Q3 = 20+5/18(76.5-70) Q3=20+(0.27)(6.5) Q3=20+1.8 = 21.8

  43. solution • FIND D7=???? D7= L+H/F(7n/10-C) L=20,H=5,F=18,n=102 ,7n/10,7*102/10=71.4,C=70 D7= 20+5/18(71.4-70) D7 = 20+5/18(1.4) D7=20+(0.39) D7=20.39

  44. solution • FIND P75=???? P75= L+H/F(75n/100-C) L=20,H=5,F=18,n=102 ,7n/10,75*102/100=76.5,C=70 P75= 20+5/18(76.5-70) P75= 20+5/18(6.5) P75=20+(1.81) P75=21.81

  45. Question#2 • The height of 100 college students measured the nearest inch is given below. • Height in Inches: 63-64-65-66-67-68-69-70-71 • No. of Students: 4---6--10-20-30-13-12--3--2 • Calculate the Median Q1, Q3, D7and P65????

  46. Solution The values are given as mid points. The data related to heights, we therefore, treat the data as continuous. We find the class boundaries as shown below. N=100

  47. Solution • Median =L+h/f[n/2-c] • Median=66.5+1/30(50-40) • Median=66.5+0.33 • Median=66.83 • n/4=25 • Q1=L/f[n/4-c] • Q1=65.5+1/20(25-20) • Q1=66.5+0.25 • Q1=65.75

  48. Solution • 3n/4=75 • Q3=L+h/f[3n/4-c] • Q3=67.5+1/30(75-70) • Q3=67.5+0.38 • Q3=67.88 • 7n/10=7*100/10=70 • D7=L+h/f[7n/10-c] • D7=66.5+1/30(70-40) • D7=66.5+30/30 • D7=66.5+1 • D7=67.5

  49. Solution • 65n/100=65*100/100=65 • P65=l+h/f[65n/100-c] • P65=70.5+1/2(65*100/100-98) • P65=70.5+.5(65-98) • P65=80(-33) • P65=

  50. MERITS AND DEMERITS MERITS 1) It is easy to compute and understand.2) It is well defined an ideal average should be.3) It can also be computed in case of frequency distribution with open ended classes.4) It is not affected by extreme values and also interdependent of range or dispersion of the data. 5) It can be determined graphically.6) It is proper average for qualitative data where items are not measured but are scored.

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