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Note on Median

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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. 

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Calculation of median in continuous series

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

.

Very Short Questions

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*}

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 \(_{Ans}\)

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*}

0%
  • Find the median of:

    64, 60, 70, 72, 68, 80, 85, 56

    60


    79


    156


    69


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    16, 13, 10, 14, 11, 12, 15

    13


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    50, 60, (frac {3x + 5}2), 80, 90

    54


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    45


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  • 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


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    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.

    45


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    53


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    45
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    34
    22
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    444


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    890


    110


  • x - 1, 2x + 1, x + 5 and 3x + 1 are in ascending order. If its median is 18, find the value of x.

    90


    10


    77


    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.

    60.345


    50.864


    79.375


    80.432


  • 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.

    55


    65


    45


    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.

    108


    102


    120


    125


  • 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


    45


    75


    55


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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?


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L n/2


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