FINDING MEDIAN OF GROUPED DATA WORKSHEET

Problem 1 :

Find the median for marks of 50 students.

Marks

20

27

34

43

58

65

89

Number of students

2

4

6

11

12

8

7

Problem 2 :

Find the median for the following data:

x

1

2

3

4

5

6

7

8

f

9

11

5

6

8

1

3

7

Problem 3 :

The height (in cm) of 50 students in a particular class are given below:

Height (in cm)

156

155

154

153

152

151

150

Number of students

8

4

6

10

12

3

7

Problem 4 :

The hearts of 60 patients were examined through X-ray and the observations obtained are given below:

Diameter of heart

(in mm)

130

131

132

133

134

135

Number of

patients

7

9

15

12

6

11

Problem 1 :

Find the median for marks of 50 students.

Marks

20

27

34

43

58

65

89

Number of students

2

4

6

11

12

8

7

Solution :

Marks


20

27

34

43

58

65

89

Number

of students

2

4

6

11

12

8

7

Cumulative

frequency

2

2 + 4  = 6

6 + 6  =  12

11 + 12  =  23

23 + 12  =  35

35 + 8  =  43

43 + 7  =  50

Here, the total frequency, N = ∑f = 50

N/2  =  50 / 2  =  25

The median is (N/2)th value = 25th value.

Now, 25th value occurs in the cumulative frequency 35, whose corresponding marks is 58.

So, the median = 58.

Problem 2 :

Find the median for the following data:

x

1

2

3

4

5

6

7

8

f

9

11

5

6

8

1

3

7

Solution :

Marks


1

2

3

4

5

6

7

8

Number

of students

9

11

5

6

8

1

3

7

Cumulative

frequency

9

9 + 11  =  20

20 + 5  =  25

25 + 6  =  31

31 + 8  =  39

39 + 1  =  40

40 + 3  =  43

43 + 7  =  50

Here, the total frequency, N = ∑f = 50

N/2  =  50 / 2  =  25

The median is (N/2)th value = 25th value.

Now, 25th value occurs in the cumulative frequency 25, whose corresponding marks is 3.

So, the median = 3.

Problem 3 :

The height (in cm) of 50 students in a particular class are given below:

Height (in cm)

156

155

154

153

152

151

150

Number of students

8

4

6

10

12

3

7

Find the meadian.

Solution :

Marks


156

155

154

153

152

151

150

Number

of students

8

4

6

10

12

3

7

Cumulative

frequency

8

8 + 4  =  12

12 + 6  =  18

18 + 10  =  28

28 + 12  =  40

40 + 3  =  43

43 + 7  =  50

Here, the total frequency, N = ∑f = 50

N/2  =  50 / 2  =  25

The median is (N/2)th value = 25th value.

Now, 25th value occurs in the cumulative frequency 28, whose corresponding marks is 153.

So, the median = 153.

Problem 4 :

The hearts of 60 patients were examined through X-ray and the observations obtained are given below:

Diameter of heart

(in mm)

130

131

132

133

134

135

Number of

patients

7

9

15

12

6

11

Find the meadian.

Solution :

Marks


130

131

132

133

134

135

Number

of students

7

9

15

12

6

11

Cumulative

frequency

7

7 + 9  =  16

16 + 15  =  31

31 + 12  =  43

43 + 6  =  49

49 + 11  =  60

Here, the total frequency, N = ∑f = 60

N/2  =  60 / 2  =  30

The median is (N/2)th value = 30th value.

Now, 30th value occurs in the cumulative frequency 31, whose corresponding diameter of heart is 132.

So, the median = 132.

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Reflection Through Y Axis

    Jul 02, 22 07:41 AM

    Reflection Through Y Axis - Concept - Examples with Step by Step Explanation

    Read More

  2. Reflection Through X Axis

    Jul 02, 22 07:34 AM

    Reflection Through X Axis - Concept - Example with Step by Step Explanation

    Read More

  3. Horizontal and Vertical Translations

    Jul 02, 22 07:23 AM

    Horizontal and Vertical Translations

    Read More