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  • Find the mean deviation about the median of the following distribution:Marks Obtained 10 No of students2

Find the mean deviation about the median of the following distribution:


\begin{array}{|l|l|l|l|l|l|} \hline \text { Marks obtained } & 10 & 11 & 12 & 14 & 15 \\ \hline \text { No. of students } & 2 & 3 & 8 & 3 & 4 \\ \hline \end{array}

Answers (1)

The data distribution is given in the question.

We have to find the mean deviation about the median of the distribution in this question.

 Let us make a table of the given data and fill up the other columns after calculations

\begin{array}{|l|l|l|} \hline \text { Marks obtained }\left(x_{i}\right) & \text { Number of students }\left(f_{i}\right) & \text { Cumulative frequency } \\ \hline 10 & 2 & 2 \\ \hline 11 & 3 & 5 \\ \hline 12 & 8 & 13 \\ \hline 14 & 3 & 16 \\ \hline 15 & 4 & 20 \\ \hline \text { Total } & 20 & \\ \hline \end{array}

\\ \text{Now here N=20 which is even}.\\\\ Here, median \( M= \frac{1}{2} \left[ \left( \frac{N}{2} \right) ^{th}observation+ \left( \frac{N}{2}+1 \right) ^{th}~observation \right] ~ \\ \\ ~ M=\frac{1}{2} \left[ \left( \frac{20}{2} \right) ^{th} observation+ \left( \frac{20}{2}+1 \right) ^{th}~observation~ \right] ~~ \\\\ \\ ~ M=\frac{1}{2}~ \left[ 10^{th}observation+11^{th}observation~ \right] ~~ \\\\ \\ \text{ Both these observations lie in cumulative frequency 13, for which corresponding observation is 12 } \\\\ \\ M=\frac{1}{2} \left[ 12+12 \right] =12 \\\\ \\ \text{~ So the above updated table is as shown below } \\\\

\begin{array}{|l|l|l|l|l|} \hline \begin{array}{l} \text { Marks obtained } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Number of } \\ \text { students }\left(f_{i}\right) \end{array} & \begin{array}{l} \text { Cumulative } \\ \text { frequency } \end{array} & d_{i}=\left|x_{i}-M\right| & f_{i} d_{i} \\ \hline 10 & 2 & 2 & 2 & 4 \\ \hline 11 & 3 & 5 & 1 & 3 \\ \hline 12 & 8 & 13 & 0 & 0 \\ \hline 14 & 3 & 16 & 2 & 6 \\ \hline 15 & 4 & 20 & 3 & 12 \\ \hline \text { Total } & 20 & & 6.35 & 25 \\ \hline \end{array}

 Hence, mean deviation becomes=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{25}{20}=1.25~ \\\\

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