Question

In a class test, marks scored by students are given in the following frequency distribution marks- 0-6, 6-12, 12-18, 18-24, 24-30 number of students- 1,4,9,3,3 find the mean and median of the data.

Answer 1

$\sf \bold{\underline{Question:-}}$

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In a class test, marks scored by students are given in the following frequency distribution marks -

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$\begin{tabular}{|c|c|c|c|c|c|c|c|}\cline{1-6} \tt Marks & \tt 0-6 & \tt 6-12 & \tt 12-18 & \tt 18-24 & \tt 24-30 \\\cline{1-8}\tt No. of Students &\tt 1& \tt 4& \tt 9 & \tt 3 & \tt 3\\\cline{1-6}\end{tabular}$

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Find the mean and median of the data.

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$\sf \bold{\underline{AnswEr:-}}$

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$\begin{lgathered}\begin{tabular}{| c | c | c | c | c |}\cline{1-5} \bf{class interval} & \bf{Frequency (fi)} & \bf{(xi)} & \bf{fixi} & \bf{C.f.}\\ \cline{1-5}\sf{0 - 6} & \sf{1} & \sf{3} & \sf{3} & \sf{1}\\ \cline{1-5}\sf{6 - 12} & \sf{4} & \sf{9} & \sf{36} & \sf{5}\\ \cline{1-5}\sf{12 - 18} & \sf{9} & \sf{15} & \sf{135} & \sf{14}\\ \cline{1-5}\sf{18 - 24} & \sf{3} & \sf{21} & \sf{63} & \sf{17}\\ \cline{1-5}\sf{24 - 30} & \sf{3} & \sf{27} & \sf{81} & \sf{20}\\ \cline{1-5}\sf{Total } & \sf{ \sum\limits\ \sf fi = 20} & {} & \sum\limits\sf\ fixi \sf\ = 318 & \sf{}\\ \cline{1-5} \end{tabular}\end{lgathered}$

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${\underline{\sf{\bigstar\;Now,\;We\;have\;to\; calculate\;median\;:}}}$

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Here,

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Sum of frequency, N = 20

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N is an even number.

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we know that,

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Median observation is,

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$:\implies\sf \dfrac{ \frac{n}{2}^{th} + \frac{n + 1}{2}^{th}}{2}$

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$:\implies\sf \dfrac{10^{th} + 11^{th}}{2}$

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$:\implies\sf \dfrac{14 + 14}{2}$

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$:\implies\sf \cancel{ \dfrac{28}{2}}$

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$:\implies\bf 14$

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It lies between interval of 12 - 18

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Therefore, Median class = 12 - 18

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$\boxed{\begin{minipage}{6cm} \bigstar \:\:\sf Median = l + \sf\dfrac{\frac{N}{2}-C.f.}{f}\times h\\\\Here:\\1)\:l=Lower\:limit\:of\:median\:class=12\\2)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=5\\3)\:f= frequency\:of\:median\:class=9\\4)\:h= Class\:interval =18 - 12 = 6\end{minipage}}$

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$\star\;{\boxed{\sf{\purple{Median = l + \dfrac{ \frac{N}{2} - C.f}{f} \times h}}}}$

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

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$:\implies\sf Median = 12 + \dfrac{ \frac{20}{2} - 5}{9} \times 6$

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$:\implies\sf Median = 12 + \dfrac{ 10 - 5}{9} \times 6$

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$:\implies\sf Median = 12 + \dfrac{5}{ \cancel{9}} \times \cancel{6}$

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$:\implies\sf Median = 12 + \dfrac{5}{3} \times 2$

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$:\implies\sf Median = 12 + \dfrac{10}{3}$

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$:\implies\sf Median = 12 + 3.33$

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$:\implies{\underline{\boxed{\sf{\pink{Median = 15.33}}}}}\;\bigstar$

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${\underline{\sf{\bigstar\;Now,\;We\;have\;to\; calculate\;mean\;:}}}$

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$\star\;{\boxed{\sf{\purple{Mean(\bar x) = \dfrac{ \sum\;fixi}{\sum\:fi}}}}}$

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

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$:\implies\sf Mean(\bar x) = \dfrac{318}{20}$

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$:\implies{\underline{\boxed{\sf{\pink{Mean(\bar x) = 15.9}}}}}\;\bigstar$

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$\therefore$ Mean and Median of the given data is 15.9 and 15.33 respectively.