Question

Class | 500-600 | 600-700 | 700-800 | 800-900 | 900-1000|
Frequency| 36. | 32 | 32. | 20. | 30. |

Find the median of given distribution.​

Answer 1

$\begin{gathered}\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6} \tt Class & \tt 500-600 & \tt 600-700 & \tt 700-800 & \tt 800-900 & \tt 900-100 \\\cline{1-6}\tt Frequency &\tt 36 & \tt 32& \tt 32 & \tt 20 & \tt 30 \\\cline{1-6}\end{tabular}\end{gathered}$

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☯ We have to find, Median of given distribution.

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$\boxed{\begin{array}{cccc}\bf Class\: interval&\bf Frequency\: (f)& \bf C.F\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 500 - 600&\sf 36&\sf 36\\\\\sf 600 - 700 &\sf 32&\sf 68\\\\\sf 700-800 &\sf 32&\sf 100\\\\\sf 800 - 900&\sf 20&\sf 120\\\\\sf 900-1000&\sf 30&\sf 150\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\bf & \bf \sum f = 150& \end{array}}$

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$\dag\;{\underline{\frak{Formula\;to\:find\;Median,}}}$

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$\star\;{\boxed{\sf{\pink{l = \dfrac{ \frac{n}{2} - C.F.}{f} \times h}}}}$

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Firstly we have to calculate $\sf \dfrac{n}{2}$, (where N = $\sf \sum F$) = $\sf \dfrac{150}{2} = \bf{75}.$

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So, The value of comulative frequency just greater than or equal to 75 is 100.

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$\dag\;{\underline{\frak{We\;know\;that,}}}$

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$\boxed{\begin{minipage}{6cm} \bigstar \:\:\sf Median = l + \sf\dfrac{\frac{n}{2}-C.f.}{f}\times h\\\\Here: \\1)\:n = \sum f =150\\2)\:l=Lower\:limit\:of\:median\:class=700\\3)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=68\\4)\:f= frequency\:of\:median\:class=32\\5)\:h= Class\:interval =700-800 = 100\end{minipage}}$

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${\underline{\sf{\bigstar\;Putting\;values\;in\;formula\;:}}}$

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$:\implies\sf 700 + \dfrac{ \frac{150}{2} - 68}{32} \times 100$

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$:\implies\sf 700 + \dfrac{ 75 - 68}{32} \times 100$

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$:\implies\sf 700 + \dfrac{7}{32} \times 100$

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$:\implies\sf 700 + 21.872$

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$:\implies{\underline{\boxed{\frak{\purple{721.875}}}}}\;\bigstar$

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$\therefore\;{\underline{\sf{Median\;of\;given\; distribution\;is\; \textbf{721.875}}}}$

Answer 2

Question :

Find the median of given distribution :

$\underbrace{\overbrace{\begin{tabular}{|c|c|}\cline{1-2}Class & Frequency \\\cline{1-2} 500 - 600 & 36 \\ 600 - 700 & 32 \\ 700 - 800 & 32 \\800 - 900 & 20 \\ 900 - 1000 & 30 \\\cline{1-2}\end{tabular}}}}$

Theory :

Steps to find Median of a grouped distribution

1) Find the cumulative frequency ( c.f)

2) Find $\sf\dfrac{N}{2}$ where , $\sf\:N=\sum\:F$

3) See the c.f just greater than N/2 and determine the corresponding class

4) Then , use Formula of median;

${\purple{\boxed{\large{\bold{Median=l+(\dfrac{\frac{N}{2}-F}{f})\times\:h}}}}}$

Here ,l = lower limit of the median class

f=frequency of the median class

h= size of the median class

F= c.f of thr class of the class preceding the median class $\sf\:N=\sum\:F$

Solution :

We have to find the median of the given distribution.

Let's solve the problem :

First , we have to find c.f to compute the median :

$\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency\: (f) &\sf C.F\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 500 - 600&\sf 36&\sf 36\\\\\sf 600 - 700 &\sf 32&\sf 68\\\\\sf 700-800 &\sf 32&\sf 100\\\\\sf 800 - 900&\sf 20&\sf 120\\\\\sf 900-1000&\sf 30&\sf 150\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf & \sf \sum f = 150& \end{array}}$

We have ,

$\sf\sum\:F=N=150$

$\sf\implies\dfrac{N}{2}=75$

The cumulative frequency just greater than N/2 is 100 and the corresponding corresponding class is 700-800 . Therefore , 700-800 is the median class :

Thus , l = 700

h = 100

F = 68

f= 32

$\tt\:Median=l+[\dfrac{\frac{N}{2}-F}{f}]\times\:h$

$\sf=700+\dfrac{75-68}{32}\times100$

$\sf=700+\dfrac{7}{32}\times100$

$\sf=700+0.218\times100$

$\sf=700+21.8$

$\sf=721.8$

Therefore, Median of given distribution is 721.8