A PPT FOR DETERMININING MEDIAN

CLASS- X SUBJECT-MATHSPRESENTED 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 .

Marks obtained 20 29 28 33 42 38 43 25 Number of students 6 28 24 15 2 4 1 20

FREQUENCY DISTRIBUTION

Marks obtained 20 29 28 33 42 38 43 25 Number of students 6 28 24 15 2 4 1 20

MARKS OBTAINED NUMBER OF STUDENTS (FREQUENCY)

20 6 25 20 28 24 29 28 33 15 38 4 42 2 43 1

Total 100

FINDING MEDIAN

In this problem number of data values=n=100,which is even .So the median will be the averageof 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 6

25 20 20+6=26

28 24 26+24=50

29 28 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 3

40 – 50 3

50 – 60 4

60 – 70 7

70 – 80 9

80 – 90 7

90 - 100 8

CUMULATIVE FREQUENCY DISTRIBUTION OF LESS THAN TYPE

MARKS NO. OF STUDENTS MARKS OBTAINED CUMULATIVE FREQUENCY

0 -10 5 LESS THAN 10 5

10 – 20 3 LESS THAN 20 5+3= 8

20 – 30 4 LESS THAN 30 8+4=12

30 – 40 3 LESS THAN 40 12+3 =15

40 – 50 3 LESS THAN 50 15+3=18

50 – 60 4 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 7 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