Notes on Median | Grade 10 > Compulsory Mathematics > Statistics | KULLABS.COM

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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*}

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69

60

156

79

13

134

56

31

45

43

123

54

6

9

12

66

79.876

79.375

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79.123

53

45

90

35

70

25

45

52

22
34
47
45

890

444

110

765

90

77

68

10

80.432

60.345

50.864

79.375

35

65

55

45

120

102

125

108

55

65

45

75