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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.

### 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%

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60

69

156

31

134

13

56

54

123

45

43

6

66

12

9

79.876

79.375

79.123

80

45

90

35

53

25

70

45

52

45
34
22
47

765

890

110

444

68

90

10

77

60.345

80.432

50.864

79.375

65

55

45

35

102

120

108

125

65

75

55

45

## DISCUSSIONS ABOUT THIS NOTE

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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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