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The mean deviation from the median for the following distribution (corrected to two decimals) is…
The mean deviation from the median for the following distribution (corrected to two decimals) is
\begin{array}{lllllllll}
\hline \boldsymbol{x}_{\boldsymbol{i}} & 6 & 9 & 3 & 12 & 15 & 13 & 21 & 22 \\
\hline \boldsymbol{f}_{\boldsymbol{i}} & 4 & 5 & 3 & 2 & 5 & 4 & 4 & 3 \\
\hline
\end{array}
13.42 5.40 4.97 11.25 Solution
\begin{array}{c|c|c|c|c}
\hline \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \begin{array}{c}\text { Cumulative } \\
\text { frequency }\end{array} & \left(\boldsymbol{d}_{\boldsymbol{i}}\right)=\left|\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{1 5}\right| & \boldsymbol{f}_{\boldsymbol{i}}\left|\boldsymbol{d}_{\boldsymbol{i}}\right| \\
\hline 6 & 4 & 4 & 9 & 36 \\
\hline 9 & 5 & 9 & 6 & 30 \\
\hline 3 & 3 & 12 & 12 & 36 \\
\hline 12 & 2 & 14 & 3 & 6 \\
\hline 15 & 5 & 19 & 0 & 0 \\
\hline 13 & 4 & 23 & 2 & 8 \\
\hline 21 & 4 & 27 & 6 & 24 \\
\hline 22 & 3 & 30 & 7 & 21 \\
\hline & \begin{array}{c}N=\boldsymbol{\Sigma} f_i \\
=30\end{array} & & & \begin{array}{c}\boldsymbol{\Sigma} f d_i \mid \\
=161\end{array} \\
\hline
\end{array}
Clearly,
$\begin{aligned}
N & =30 \\
\Rightarrow\frac{N}{2} & =15
\end{aligned}$
The cumulative frequency just greater than $\frac{N}{2}$ is 19 and corresponding value of $x$ is 15 .
Therefore, median $=15$.
Clearly, $\Sigma f_i\left|x_i-15\right|=\Sigma f_i d_i=161$ and $N=30$
$\therefore$ Mean deviation $=\frac{161}{30}=5 \cdot 40($ approx $)$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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