FINDING MEDIAN OF GROUPED DATA WORKSHEET

Problem 1 :

Find the median for marks of 50 students.

 Marks20273443586589 Number of students246111287

Problem 2 :

Find the median for the following data:

 x12345678 f911568137

Problem 3 :

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

 Height (in cm)156155154153152151150 Number of students846101237

Problem 4 :

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

 Diameter of heart(in mm)130131132133134135 Number ofpatients791512611

Problem 1 :

Find the median for marks of 50 students.

 Marks20273443586589 Number of students246111287

Solution :

 Marks20273443586589 Number of students246111287 Cumulativefrequency22 + 4  = 66 + 6  =  1211 + 12  =  2323 + 12  =  3535 + 8  =  4343 + 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:

 x12345678 f911568137

Solution :

 Marks12345678 Number of students911568137 Cumulativefrequency99 + 11  =  2020 + 5  =  2525 + 6  =  3131 + 8  =  3939 + 1  =  4040 + 3  =  4343 + 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)156155154153152151150 Number of students846101237

Find the meadian.

Solution :

 Marks156155154153152151150 Number of students846101237 Cumulativefrequency88 + 4  =  1212 + 6  =  1818 + 10  =  2828 + 12  =  4040 + 3  =  4343 + 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)130131132133134135 Number ofpatients791512611

Find the meadian.

Solution :

 Marks130131132133134135 Number of students791512611 Cumulativefrequency77 + 9  =  1616 + 15  =  3131 + 12  =  4343 + 6  =  4949 + 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.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

1. Honors Algebra 2 Problems with Solutions (Part - 1)

Aug 09, 24 08:39 PM

Honors Algebra 2 Problems with Solutions (Part - 1)

Read More

2. Honors Algebra 2 Problems with Solutions (Part - 2)

Aug 09, 24 08:36 PM

Honors Algebra 2 Problems with Solutions (Part - 2)

Read More

3. SAT Math Resources (Videos, Concepts, Worksheets and More)

Aug 09, 24 06:15 AM

SAT Math Resources (Videos, Concepts, Worksheets and More)

Read More