The median is defined as the positional value in the set of data arranged in order. The median divides the total number of observations into two equal parts. Therefore, there are 50% items more than the median value.
Steps to be remembered
a. Arrange the data in ascending order of magnitude.
b. Construct the cumulative frequency distribution table.
c. Find the median class by using
Median class = (\(\frac{N}{2}\))^{th} class
d. Then find exact median value by using
\( M_d = L + \frac{\frac{N}{2}- c.f.}{f} \times h \)
Where,
L = lower limit of the class
c.f. = cumulative frequency of preceding class
f = frequency of class
h = width of class-interval
\( M_d = L + \frac{\frac{N}{2}- c.f.}{f} \times h \)
Where,
L = lower limit of the class
c.f. = cumulative frequency of preceding class
f = frequency of class
h = width of class-interval
Solution:
\begin{align*}Position \: of \: median &= \frac{N + 1^{th} }{2}term \\ &= \frac{7+1^{th}}{2}term\\ &= 4^{th} term \\ 4^{th}\: represent \: 70 . & \therefore median &= 70 _{Ans}\end{align*}
Solution:
\(Mean (\overline{X}) = 8 \\ Number \: of \: term (N)= 6 \)
\begin{align*} Mean ( \overline {X}) &= \frac{\sum X }{N}\\ or, 8 &= \frac{4 +6+9+11+10+k}{6} \\ or, 48 &= 40 + k \\ or, k &=48 -40 \\ \therefore k &= 8 \end{align*}
The given term write in ascending order: 4, 6, 8, 9, 10, 11
\begin{align*} Position \: of \: median &= \frac{N+1^{th}}{2}term\\&=\frac{6 + 1^{th}}{2}term\\ &= 3.5^{th} term \\ \: \\ Median &= \frac{8+9}{2}\\ &= \frac{17}{2}\\ &= 8.5 _{Ans} \end{align*}
Solution:
First eight multiple of 5 are 5, 10, 15, 20, 25, 30, 35, 40 which are in ascending order. Where N = 8
\begin{align*} Position \: of \: Median &= \frac{N+1^{th}}{2}term \\ &= \frac{8+1^{th}}{2}\\ &= 4.5^{th} term \\ \: \\ Median &= \frac{4^{th}term + 5^{th}term}{2}\\ &= \frac{20+25}{2} \\ &= \frac{45}{2}\\ &= 22.5 _{Ans.} \end{align*}
In which class interval median marks lines?
Marks | 0 - 20 | 20 - 40 | 40 - 60 | 60 - 80 | 80 - 100 |
No. of students | 10 | 16 | 20 | 30 | 10 |
Solution:
Marks | No. of students 9 (f) | c.f. |
0-20 | 10 | 10 |
20-40 | 16 | 26 |
40-60 | 20 | 46 |
60-80 | 30 | 76 |
80-100 | 10 | 86 |
N = 86 |
\begin{align*} Position\:of\:median\:class &= \frac{N^{th}}{2}term\\ &= \frac{86^{th}}{2}term \\ &= 43^{th} \: term \end{align*}
The median class lies in 40 - 60 \(_{Ans}\)
Find the median class from the following data.
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 5 | 7 | 4 | 3 | 11 |
Solution:
Marks | Frequency (f) | c.f. |
0-10 | 5 | 5 |
10-20 | 7 | 12 |
20-30 | 4 | 16 |
30-40 | 3 | 19 |
40-50 | 11 | 30 |
N =30 |
\begin{align*}Position \: of \: median &= \frac{N^{th}}{2}term\\ &= \frac{30^{th}}{2}term\\ &= 15^{th} term. \end{align*}
The median class lies in 20 - 30 \(\: _{Ans.}\)
Solution:
N = 10
Median = 15
\begin{align*} Position \: of \: median &= \frac{N + 1^{th}}{2}term \\ &= \frac{10 + 1^{th}}{2}\\ &= 5.5^{th} term \\ \: \\ Median &= \frac{(5^{th} + 6^{th }) term}{2} \\ 15 &= \frac{2x+2x+x}{2}\\ or, 30 &= 4x + 2 \\ 30 - 2 &= 4x\\ or, x &= \frac{28}{4}\\ \therefore x &= 7 \: \: _{Ans} \end{align*}
Solution:
From the given data we came to know N= 9
Median = 5
\begin{align*} Position \: of \: median &= \frac{N + 1^{th}}{2}term \\ &= \frac{9 + 1^{th}}{2}term \\ &= 5^{th} term \\ \: \\ 5^{th } term \: & represent \: 2x+1 \\ \therefore Median &= 2x + 1 \\ or, 5 &= 2x + 1 \\ or, 2x &= 5 - 1 \\ or, x &= \frac{4}{2}\\ \therefore x &= 2 \end{align*}
Solution:
The even numbers between 60 and 70 are: 62, 64, 66, 68 where N = 4
\(Position \: of \: median = \frac{N + 1^{th}}{2}item = \frac{4 + 1^{th}}{2}item = 2.56^{th} \: item \)
\begin{align*} Median &= \frac{64+66}{2}\\ &= \frac{130}{2}\\ &= 65 \: \: _{Ans} \end{align*}
Solution:
From graph paper sum of the frequency (N) = 30
\begin{align*} Median \: class &= \frac{N^{th}}{2}item \\ &= \frac{30^{th}}{2}item\\ &= 15^{th}\: item \\ \text{The corresponding } & value \: of \: 15^{th} \: item \: is \: 10-15. \end{align*}
Solution:
From the graph,
Total number of students (N) = 60
\begin{align*} Position \: of \: Media \: class &= \frac{N^{th}}{2}term \\ &= \frac{60^{th}}{2}term \\ &= 30^{th} \: term \\ \therefore Median \: class &= 20 - 30 \: _{ans} \end{align*}
Solution:
From the graph,
Total number of students (N) = 60
\begin{align*} Position \: of \: median \: class &= \frac{N^{th}}{2}term \\ &= \frac{100^{th}}{2} term \\ &= 50^{th} term \\ \therefore Median \: class &= 20 - 30 \end{align*}
Solution:
From the graph,
Total number of students (N) = 35
\begin{align*} Position \: of \: median \: class &= \frac{N^{th}}{4}term\\ &= \frac{35^{th}}{2}term \\ &= 17.5^{th} \: term \end{align*}
From the figure
Class | Frequency (f) | cf |
5 - 10 | 5 | 5 |
10 - 15 | 5 | 10 |
15 - 20 | 5 | 15 |
20 - 25 | 4 | 19 |
25 - 30 | 6 | 25 |
30 - 35 | 5 | 30 |
35 - 40 | 5 | 35 |
N = 35 |
Median class = 20 - 25 = 20 \(_{Ans}\)
Solution:
From graph, the number of men (N) = 60
\begin{align*}\text{Position of median class}&=\frac{N^{th}}{2}term \\ &= \frac{60^{th}}{2}term\\ &= 30^{th} \\ median \: class &= 30 - 40 \: _{Ans} \end{align*}
Solution:
From graph,
Number of term (N) = 60
\begin{align*} Position \: of\:Q_1\: class &= \frac{N^{th}}{4}term \\ &= \frac{60^{th}}{4}term \\ &=15^{th} \:term\\ Q_1 \: class &= 10-20 \: _{Ans} \\ \: \\ Position \: of \: Q_3\:class&= \frac{3N}{4}term\\ &= \frac{3\times60^{th}}{4} \\ &= 45^{th} \: term \\ Q_3 \: class &= 40-50 \:_{Ans} \end{align*}
Solution:
From graph, Number of students (N) = 20
\begin{align*} Position \: of \: Q_1 &= \frac{N^{th}}{4}term \\ &= \frac{20^{th}}{4}term \\ &= 5^{th} \: term \\ class\:of\:Q_1 &= 8 - 12 \: _{Ans} \end{align*}
Solution:
From graph paper number of the student (N) = 120
\begin{align*}Position\:of\:the \: Q_1\: class &= \frac{N^{th}}{4}term \\ &= \frac{120^{th}}{4}term \\ From\:graph\:the\:corres&ponding \: \text{class of } 30^{th}\:term \:is\: 20 - 40 \\ \therefore First \: quartile \: class &= 20-40 \end{align*}
Find the median of:
64, 60, 70, 72, 68, 80, 85, 56
79
60
69
156
Find the median of:
16, 13, 10, 14, 11, 12, 15
56
31
13
134
If the given data is in ascending order and the median is 70, find the value of x.
50, 60, (frac {3x + 5}2), 80, 90
54
45
43
123
x + 1, 2x - 1, x + 7 and 3x + 4 are in ascending order. If the median is 12, find the value of x.
6
9
12
66
The median class of a continuous data is 70 - 80, its corresponding frequency is 16 and the sum of frequencies of the data is 60. If the total preceding term of 70 - 80 is 15, find the median of the data.
79.123
79.375
79.876
80
The median of a continuous data is 30 - 40, its corresponding frequency is 14 and the sum of frequencies of data is 60. If the total preceding term of 30 - 40 is 23, find the median of the data.
35
90
45
53
Find the median of:
51, 53, 49, 51, 53, 50, 54, 56
52
70
45
25
Find the median of:
47, 61, 13, 34, 56, 22, 8
If the given data is in ascending order and the median is 140, find the value of x.
100, 120, (frac {5x + 10}4), 160, 180
890
110
444
765
x - 1, 2x + 1, x + 5 and 3x + 1 are in ascending order. If its median is 18, find the value of x.
90
77
10
68
The median class of a continuous data is 70-80, its corresponding frequency is 16 and the sum of frequencies of the data is 60.If the total preceding term of 70-80 is 15, find the median of the data.
80.432
50.864
79.375
60.345
If median class is 30-40, cumulative frequency of pre-median class is 8, the frequency of median class if 14 and total number of data is 30 then find the value of median.
65
45
55
35
If lower of median class interval (L) =100, number of frequency (N) =34, frequency of median class interval (f) =10, cumulative frequency of the class preceding the median class (c.f) =15 and class interval of the median class(i) =10.Find the median.
102
120
125
108
If lower of median class interval (L) =40, number of frequency (N) =60, frequency of median class interval (f) =10, cumulative frequency of the class preceding the median class (c.f) =25 and class interval of the median class(i) =10.Find the median.
65
75
55
45
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raabi gautam
how to find frequency from the cumalative frequency table to find the median class or first quartile class or third quartile class?
Mar 10, 2017
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L n/2
Feb 05, 2017
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