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9 3 11 29 4 16 45 5 20 65 6 25 90 7 15 105 8 9 114 9 6 120=N N/2=60. The c.f. just greater than N/2 is 65 and the value of x corresponding to 65 is 5. Thus , median is 5. In case of grouped frequency distribution, median is obtained as follows: Let us consider the grouped frequency distribution: Class intervals frequency cumulative frequency x1 - x2 f1F1 x2 –x3 f2F2 …….….xp –xp+1 fpFp….…..…xn –xn+1 fnFn where Fk=∑. Let the smallest c.f. greater then N/2 is Fp. Then the median classis xp –xp+1. We assume that frequency of a class is uniformly distributed over the class interval. Let the c.f. for the class just above the median class be c . Thus (N/2-c) is the frequency of the interval between the median and the lower limit of the median class . the length of the interval corresponding to the frequency (N/2-c) is I, where f is frequency of the median class, I is the length of the class interval of the median class . Hence the median is L0+I, where L0is lower limit of the median class. Properties of Median Median is a positional average and hence is not influenced by extreme valuesMedian can be calculated even in the case of open end intervalsMedian can be located even if the data is incompleteIt is not a good representative of data if the number of observations is smallIt is not amenable to algebraic treatmentIt is susceptible to sampling fluctuations
10 Quartilesare thsose variate values which divide the total frequency into four equal parts; decilesandpercentiles divide into ten and hundred equal parts respectively. Suppose the values of the variate have been arranged in ascending order of magnitude, then the value of the quartile having the position between the lower extreme and the median , is the first quartile Q1and that between the median and the upper extreme is the third quartile Q3. The median is the second quartile Q2, is the fifth decile D5and the fiftieth percentile P50. For a grouped frequency distribution, the quartiles, deciles and percentiles are given by Qi=l+h, i=1,2,3 Dj= l+h, j=1,…,9Pk= l+h, k=1,…,99where l is the lower limit of the class in which the particular quartile/decile/percentile lies, f is the frequency of the class , h is the width of this class, C is the cumulative frequency upto and including the class preceding the class in which the particular quartile/decile/percentile lies and N is the total frequency. Calculate the three quartiles for the following frequency distribution of the number of marks obtained by 49 students in a class: Marks No. of students Marks No. of students 5-10 5 25-30 5 10-15 6 30-35 4 15-20 15 35-40 2 20-25 10 40-45 2 Cumulative Frequency Table Class Frequency Cumulative Frequency(less than) 5-10 5 5 10-15 6 11 15-20 15 26 20-25 10 36 25-30 5 41 30-35 4 45
11 35-40 2 47 40-45 2 49=N The cumulative frequency immediately greater than N/4=49/4 is 26; hence to find Q1, L=15, h=15-10=5, C=11, f=15. Thus Q1= 15+5= 15.47 marks.