Question

Find the median of the following data :​

Answer 1

$\underline{\underline{\blue{\textbf{Step - By - Step - Explanation : -}}}}$

To Find :

• we need to find median of the given data

Solution :

Classes ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Frequency

|10 - 20⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀10⠀⠀⠀⠀⠀ |

|20 - 30⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀12⠀⠀⠀⠀⠀ |

|30 - 40⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀8⠀⠀⠀⠀⠀⠀|

|40 - 50⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀16⠀⠀⠀⠀⠀ |

|50 - 60⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀14⠀⠀⠀⠀⠀ |

|60 - 70⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀10⠀⠀⠀⠀⠀ |

Arranging frequency in ascending order

Frequency : 8, 10, 10, 12, 14, 16

Numbers of observations = 6

Number of observations is even then Formula for finding median of data is $\green{\textbf{(n/2)}}$

where n denotes number of observations

Median of Data :-

= (n/2)th term

= 6/2th term

= 3rd term

So,

Third term is 10

$\large\dag \large { \red{\underline{\bf{Hence }}}}$

$\boxed{\underline{\blue{\textrm{Median of data is 10}}}}$

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Answer 2

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

A frequency distribution table is given to us .

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

The median of the following data.

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } FormulA\:UseD:-}}}}}$

$\orange{\underline{\boxed{\purple{\tt{\dag Median(M_e)=l+\Bigg\{ h\times\dfrac{\dfrac{N}{2}-cf}{f}\Bigg\} }}}}}$

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } SolutioN:-}}}}}$

$\tt Given\:table\:to\:us\:is:-$

$\boxed{\begin{tabular}{|c|c|c|c|c|c|c|} \cline{1-7} \bf Classes & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 \\ \cline{1-7} \bf Frequency & 10&12&8&16&14&10 \\ \cline{1-7}\end{tabular}}$

Now we may prepare Cummulative Frequency Table , as :

$\boxed{\begin{tabular}{|c|c|c|} \cline{1-3} \bf Class Interval &\bf Frequency & \bf Cummulative Frequency \\ \cline{1-3} 10-30 & 10&10\\ \cline{1-3} 20-30 &12&22\\\cline{1-3} 30-40 &8&30\\\cline{1-3} 40-50 &16&46\\\cline{1-3} 50-60 &14&60\\\cline{1-3} 60-70 &10&70 \\ \cline{1-3} & N=\Sigma f_i =70 & \\ \cline{1-3} \end{tabular}}$

[ If LaTeX doesn't work on app refer to attachment:]

$\tt Now\:here,$

$\tt:\implies N=70$

$\tt:\implies\dfrac{N}{2}=\dfrac{70}{2}$

$\underline{\boxed{\red{\tt{\longmapsto\:\:\dfrac{N}{2}\:\:=\:\:35}}}}$

Now here Cummulative Frequency just greater than 35 is 46 and the corresponding class is 40-50.

$\green{\tt Thus\: median\:class\:is\:40-50.}$

$\tt Now\:here$

• l ( Lower limit of class ) = 40.
• h ( width of median class ) = 10.
• cf ( c.f of preceding class ) = 30.
• N/2 = 35.
• f ( frequency of median class)=16

$\tt Hence\: median\:can\:be\: calculated\:as:-$

$\tt:\implies Median(M_e)=l+\Bigg\{ h\times\dfrac{\dfrac{N}{2}-cf}{f}\Bigg\}$

$\tt:\implies M_e = 40+\Bigg\{ 10\times \dfrac{35-30}{16}\bigg\}$

$\tt:\implies M_e=40+\bigg\{\cancel{10}\times\dfrac{5}{\cancel{16}}\bigg\}$.

$\tt:\implies M_e=40+\dfrac{25}{8}$

$\tt:\implies M_e=\dfrac{320+25}{8}$

$\tt:\implies M_e=\dfrac{345}{8}$

$\underline{\boxed{\red{\tt{\longmapsto\:\:Median\:\:=\:\:43.125}}}}$

$\orange{\boxed{\green{\bf \dag Hence\: required\: Median\:is\:43.125.}}}$