A PPT FOR DETERMININING MEDIAN

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Presentation transcript:

A PPT FOR DETERMININING MEDIAN CLASS- X SUBJECT-MATHS PRESENTED BY – MAYADHAR PANDA TGT(MATHS) K.V.NO.1 AFS KALAIKUNDA

DEFINITION OF MEDIAN Median is a measure of central tendency which gives the value of the middle – most observation in the data . For finding the median of ungrouped data ,we first arrange the data values of the observations in ascending order and count the total number of observations .

MEDIAN OF UNGROUPED DATA STEPS 1. Arrange data values in ascending order . Count the total number of data values ‘ n’. If n is odd , then median is (n+1)/2 th observation . If n is even ,then median is the average of n/2 th and (n/2 +1 )th observations .

EVALUATION OF MEDIAN OF UNGROUPED DATA Marks obtained by 100 students out of 50 marks in a test are given in the following table: Frequency table after arranging marks in ascending order is made for calculating the median .

FREQUENCY DISTRIBUTION

FINDING MEDIAN In this problem number of data values=n=100, which is even .So the median will be the average of the n/2 th and (n/2 + 1)th observations i.e.,the 50th and 51st observations. To find the 50th and 51st observations we proceed as follows :

TABLE FOR FINDING CUMULATIVE FREQUENCY MARKS OBTAINED NUMBER OF STUDENTS 20 6 UP TO 25 6+20=26 UP TO 28 26+24=50 UP TO 29 50+28=78 UP TO 33 78+15=93 UP TO 38 93+4=97 UP TO 42 97+2=99 UP TO 43 99+1=100 Data given in the right column of the above table are called cumulative frequencies.

TABLE SHOWING CUMULATIVE FREQUENCY MARKS OBTAINED NUMBER OF STUDENTS CUMULATIVE FREQUENCY 20 6 25 20+6=26 28 24 26+24=50 29 50+28=78 33 15 78+15=93 38 4 93+4=97 42 2 97+2=99 43 1 99+1=100 From the above table we find that 50th observation is 28 and 51st observation is 29.Now median = (28+29)/2 = 28.5 .This median mark conveys that about 50 percent students obtained marks less than 28.5 and another 50 percent students obtained marks more than 28.5 .

MEDIAN OF GROUPED DATA Consider a grouped frequency distribution of marks obtained ,out of 100,by 53 students ,in a certain examination ,as follows: Marks obtained Number of students 0 – 10 5 10 – 20 3 20 – 30 4 30 – 40 40 – 50 50 – 60 60 – 70 7 70 – 80 9 80 – 90 90 - 100 8

CUMULATIVE FREQUENCY DISTRIBUTION OF LESS THAN TYPE MARKS NO. OF STUDENTS MARKS OBTAINED CUMULATIVE FREQUENCY 0 -10 5 LESS THAN 10 10 – 20 3 LESS THAN 20 5+3= 8 20 – 30 4 LESS THAN 30 8+4=12 30 – 40 LESS THAN 40 12+3 =15 40 – 50 LESS THAN 50 15+3=18 50 – 60 LESS THAN 60 18+4=22 60 – 70 7 LESS THAN 70 22+7=29 70 – 80 9 LESS THAN 80 29+9=38 80 – 90 LESS THAN 90 38+7=45 90 - 100 8 LESS THAN 100 45+8 = 53 Cumulative frequencies obtained above are called cumulative frequencies of the less than type.

FINDING MEDIAN CLASS Locate the class whose cumulative frequency is greater than and nearest to half of the total number of observations . In the above table the total number of observations = n= 53 n/2= 53/2 = 26.5 The class 60 – 70 has the cumulative frequency 29 which is greater than 26.5 and nearest to it. Therefore,60 – 70 is the median class .

FORMULA OF MEDIAN Median of ungrouped data = l + h/f(n/2 – cf) Where l = lower limit of median class n= total number of observations cf = cumulative frequency of class preceding the median class h = class size (assuming class size to be equal) f = frequency of the median class

CALCULATION OF MEDIAN From the above table n/2 = 26.5 , l= 60,cf = 22 f = 7 , h = 10 Median = l + h/f(n/2 – cf) = 60 + 10/7 (26.5 – 22) = 60 + 45 / 7 = 66.4 So about half the students have scored marks less than 66.4 and the other half have scored marks more than 66.4 .

HOMEWORK Collect the marks of Mathematics scored by students of your class in SA1 and prepare a less than type frequency distribution table using class size of 5 .Find the median of this distribution . THANK YOU