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, 25^{th} 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, 25^{th} 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, 25^{th} 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 = 30^{th} value.
Now, 30^{th} value occurs in the cumulative frequency 31, whose corresponding diameter of heart is 132.
So, the median = 132.
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