Question

$\sf Find\;median\;of\:the\:given\:data.$

$\boxed {\begin{array}{c|c|c|c|c|c|c}\bf {Class\:interval} &\sf 20-30 & \sf 30-40 &\sf 40-50 &\sf 50-60 & 60-70 &\sf 70-80 \\&&&&&&\\ \bf Frequency &\sf 10 &\sf 6 &\sf 8 &\sf 12 &\sf 5 &\sf 9 \\&&&&&&\\ \end {array}}$

$\underline{\sf Note:-}$

$\bull$Construct a frequency table with your answer

$\bull$ Spammed answers will be deleted .​

Answer 1

Explanation :

Table :

$\boxed{\begin{array}{c|c|c}\bf \: Class \: interval&\bf \: Frequency&\bf Cumulative \: frequency \\\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad}{}\\\sf20 - 30&\sf10&\sf10\\\sf30 - 40&\sf6&\sf16\\\sf40 - 50&\sf8&\sf24\\\sf50 - 60&\sf12&\sf36\\\sf60 - 70 &\sf5&\sf41\\\sf \: 70 - 80&9&50 \\ \dfrac{\qquad\qquad}{}& \dfrac{\qquad\qquad}{ \sum \sf \: f = 50 = n}& \dfrac{\qquad\qquad}{} \end{array}}$

Median :

We know that,

$\sf \: M = l + \left( \cfrac{ \cfrac{n}{2} - cf}{f} \right) \times h$

Where,

• L is Lower limit of Median Class

• n is Total frequency

• c.f is the cumulative frequency of the class preceding the Median Class

• f is the frequency of the Median class

$\longrightarrow \sf \: M = 50 + \left( \cfrac{ \cfrac{50}{2} - 24 }{12}\right) \times 12 \\ \\ \\ \longrightarrow \sf M = 50 + \left( \dfrac{25 - 24}{12} \right) \times 12 \\ \\ \\ \longrightarrow \sf M = 50 + \left( \dfrac{1}{ \not{12}} \right) \times \not{ 12} \\ \\ \\ \longrightarrow \sf M =50 +1 \\ \\ \\ \longrightarrow \red{\sf M =51}$

Median is 51.

Answer 2

Step-by-step explanation:

Table:

$\begin{gathered}\boxed{\begin{array}{|c|c|c|}\boldsymbol \green{class \: interval}&\: \boldsymbol \green{frequency}&\boldsymbol{ \green{Cumulative \: frequency}} \\\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad}{}\\\tt \red{20 - 30}&\tt \blue{10}&\tt \pink{0 + 10 = 10}\\\tt \red{30 - 40}&\tt \blue6&\tt \pink{10 + 6 = 16}\\\tt \red{40 - 50}&\tt \blue8&\tt \pink{16 + 8 = 24}\\\tt \red{50 - 60}&\tt \blue{12}&\tt \pink{24 + 12 = 36}\\\tt \red{60 - 70 }&\tt \blue5&\tt \pink{36 + 5 = 41}\\\tt \red{ 70 - 80}& \tt \blue9& \tt \pink{41 + 9 = 50 }\\ \dfrac{\qquad\qquad}{}& \dfrac{\qquad\qquad} \purple{ \boldsymbol{\sum \: f = 50 = n}}& \dfrac{\qquad\qquad}{} \end{array}}\end{gathered}$

Formula Used:

$\purple {\huge {\bull{ \boxed{Median \: = \: l + \left( \dfrac{ \frac{n}{2} - cf }{f} \right ) \times h}}}}$

• l = lower limit of median class
• n = number of observation
• cf = cumulative frequency of class preceding the median class
• f = frequency of median class
• h = class size

Solution:

$: \implies \gray{M \: = 50 + \left ( \dfrac{ \dfrac{50}{2} - 24 }{12} \right) 12}$

$: \implies \gray{M \: =50 + \left( \dfrac{25 - 24}{12} \right)12}$

$: \implies \gray{M \: =50 + \left( \dfrac{1}{ \cancel{12}} \right ) \cancel{12}}$

$: \implies \gray{M \: =50 + 1}$

$: \implies \orange{M \: =51}$

•°• Median of the given data is 51 !!

#Make it easy !!