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Pair of Linear Equations in Two Variables
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- Relation Between Co-efficient
Arithmetic Progressions
Quadratic Equations
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Triangles
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Trigonometry
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Introduction to Trigonometry
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Statistics and Probability
Probability
Statistics
Coordinate Geometry
Lines (In Two-dimensions)
Mensuration
Areas Related to Circles
- Perimeter and Area of a Circle - A Review
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Surface Areas and Volumes
- Concept of Surface Area, Volume, and Capacity
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- Concept of Surface Area, Volume, and Capacity
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Internal Assessment
Notes
Median is the middle most term of the data. Median means that when the data are arranged, the median is the middle value if the number of values is odd and the mean of the two middle values if the number of values is even. In 9th class we studied that median of ungrouped data is
1)Median for odd number of data= `(n+1)/2`, where n is the total number of data given.
Example: Find the median of 1,2,3,4,5
Median= `(n+1)/2= (5+1)/2= 3`
2)Median for even number of data
= `{(n/2)th + [(n/2)+1]th}/2`
Example: Find the median of 1,2,3,4,5,6
Median for even number of data= `{(n/2)th + [(n/2)+1]th}/2`
= `{(6/2)th + [(6/2)+1]th}/2`
= `{3rd+ 4th}/2`
= `3+4/2`
Median for even number of data= 3.5
But in this concept of class 10th we will study how to find median of grouped data. The formula to find median of grouped data is
Median= `l+ {[(N/2)- cf]/f} xx h`
l= lower limit of the median class
N= ∑fi= sum of the frequencies
cf= cumulative frequency
f= frequency of the median class
h= Class size
Example- Find the median of the following data:
Marks |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
80-90 |
No. of students |
5 |
15 |
25 |
20 |
7 |
8 |
10 |
Solution:
Marks |
f |
cf |
20-30 |
5 |
5 |
30-40 |
15 |
20 |
40-50 |
25 |
45 |
50-60 |
20 |
65 |
60-70 |
7 |
72 |
70-80 |
8 |
80 |
80-90 |
10 |
90 |
|
N=∑fi= 90 |
|
Cf of the last median class should always be equal to N
`N/2= 90/2= 45`
Median class is the class of that cf which is just more than `N/2`
therefore, Median class= 50-60, l=50,
cf is the cumulative frequency preceeding the median class
therefore, cf= 45,
f is the frequency of the median class
therefore, f=20, h=10
Median=` l+ {[(N/2)- cf]/f} xx h`
= `50+ {[45- 45]/20} xx 10`
= `50+ {0/20} xx 10`
= `50+ 0`
Median= 50
-
There is a empirical relationship between the three measures of central tendency :
`3 "Median" = "Mode" + 2 "Mean"`
Video Tutorials
Shaalaa.com | Statistics part 2 (Mean Median Mode)
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Related QuestionsVIEW ALL [21]
Estimate the median for the given data by drawing an ogive:
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 4 | 9 | 15 | 14 | 8 |
If the mean of the following distribution is 3, find the value of p.
x | 1 | 2 | 3 | 5 | p + 4 |
f | 9 | 6 | 9 | 3 | 6 |
In th following table, Σf = 200 and mean = 73. find the missing frequencies f1, and f2.
x | 0 | 50 | 100 | 150 | 200 | 250 |
f | 46 | f1 | f2 | 25 | 10 | 5 |
The marks obtained by 30 students in a class assignment of 5 marks are given below.
Marks | 0 | 1 | 2 | 3 | 4 | 5 |
No. of Students |
1 | 3 | 6 | 10 | 5 | 5 |
Calculate the mean, median and mode of the above distribution