Question

do it fasttt...... ...​

Answer 1

$\large{\underline{\underline{\red{\sf{\hookrightarrow Given:- }}}}}$

A frequency distribution table is given to us.

$\large{\underline{\underline{\red{\sf{\hookrightarrow To\:Find:-}}}}}$

The median of the given data.

$\large{\underline{\underline{\red{\sf{\hookrightarrow Solution:- }}}}}$

Now , given table to us is :

$\begin{tabular}{|c|c|} \cline{1-2}Marks Obtained & No. of students \\ \cline{1-2} Below 10 & 6 \\ \cline{1-2} Below 20 & 15 \\ \cline{1-2} Below 30 & 29 \\ \cline{1-2} Below 40 & 41 \\ \cline{1-2} Below 50 & 60 \\ \cline{1-2} Below 60 & 70 \\ \cline{1-4} \end{tabular}$

Now , we may prepare second table which is :

$\boxed{\begin{tabular}{|c|c|c|}\cline{1-3} Class Interval & Frequency(f_i) & \underset{Frequency}{Cummulative} \\ \cline{1-3} 0-10 &6&6 \\ \cline{1-3} 10-20 &9&15 \\ \cline{1-3} 20-30 &14&29 \\ \cline{1-3} 30-40 &12&41 \\ \cline{1-3} 40-50 &19&60 \\ \cline{1-3} 50-60 &10&70 \\ \cline{1-3} & N=\Sigma f_i = 70 & \\ \cline{1-4}\end{tabular}}$

$\tt{Now }$

$\tt:\implies N = 70$

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

The cumulative frequency just greater than 35 is 41 and the corresponding classes 30 - 40 .

So , the median class is 30 - 40 .

$\blue{\bf Hence\:here ,}$

$\textsc{ l = 30}$

$\textsc {h = 10}$

$\textsc {f = 12}$

$\textsc {cf = 29}$

$\textsc {N / 2 = 35}$

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

$\tt:\implies M_e=30+\Bigg\{10\times\dfrac{35-29}{12}\Bigg\}$

$\tt:\implies M_e=30+\bigg(10\times\dfrac{\cancel{6}}{\cancel{12}}\bigg)$

$\tt:\implies M_e=30+5$

$\underline{\boxed{\red{\tt{\longmapsto\:\: Median\:\:=\:\:35\:\:}}}}$

$\purple{\boxed{\pink{\bf{\dag Hence\:the\:median\:is\:35.}}}}$

Answer 2

Answer:

A frequency distribution table is given to us.

\large{\underline{\underline{\red{\sf{\hookrightarrow To\:Find:-}}}}}

↪ToFind:−

The median of the given data.

\large{\underline{\underline{\red{\sf{\hookrightarrow Solution:- }}}}}

↪Solution:−

Now , given table to us is :

$Now</p><p></p><p>\tt:\implies N = 70:⟹N=70</p><p></p><p>\tt:\implies \dfrac{N}{2}=70:⟹2N=70</p><p></p><p>The cumulative frequency just greater than 35 is 41 and the corresponding classes 30 - 40 .</p><p></p><p>So , the median class is 30 - 40 .</p><p></p><p>\blue{\bf Hence\:here ,}Hencehere,</p><p></p><p>\textsc{ l = 30}\textscl=30</p><p></p><p>\textsc {h = 10}\textsch=10</p><p></p><p>\textsc {f = 12}\textscf=12</p><p></p><p>\textsc {cf = 29}\textsccf=29</p><p></p><p>\textsc {N / 2 = 35}\textscN/2=35</p><p></p><p>\tt :\implies Median (M_e)=l+\Bigg\{h\times\dfrac{\dfrac{N}{2}-cf}{f}\Bigg\}:⟹Median(Me)=l+{h×f2N−cf}</p><p></p><p>\tt:\implies M_e=30+\Bigg\{10\times\dfrac{35-29}{12}\Bigg\}:⟹Me=30+{10×1235−29}</p><p></p><p>\tt:\implies M_e=30+\bigg(10\times\dfrac{\cancel{6}}{\cancel{12}}\bigg):⟹Me=30+(10×126)</p><p></p><p>\tt:\implies M_e=30+5:⟹Me=30+5</p><p></p><p>\underline{\boxed{\red{\tt{\longmapsto\:\: Median\:\:=\:\:35\:\:}}}}⟼Median=35</p><p></p><p>\purple{\boxed{\pink{\bf{\dag Hence\:the\:median\:is\:35.}}}}†Hencethemedianis35.</p><p></p><p>$