Question

solve this............​

Answer 1

Answer:

arranging in ascending order=

6 15 29 41 60 70

cancelling from both the sides equally

therefore, we get 29 and 41

to find the median of 29 and 41 divide it by 2

29+41/2

=70/2

therefore,median =35

Step-by-step explanation:

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Answer 2

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

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 .

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

• $\textsc{ l = 30}$
• $\textsc {h = 10}$
• $\textsc {f = 12}$
• $\textsc {cf = 29}$
• $\textsc {N / 2 = 35}$

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

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

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

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

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

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