 ## Median

Subject: Compulsory Mathematics

#### Overview

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.

##### Median

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 ##### Things to remember

$M_d = L + \frac{\frac{N}{2}- c.f.}{f} \times h$

Where,
L = lower limit of the class
c.f. = cumulative frequency of preceding class
f = frequency of class
h = width of class-interval

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Median ##### Finding mean, median, and mode | Descriptive statistics | Probability and Statistics | Khan Academy ##### Statistics Quartiles Median Type 1

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*}