Mathematics › Statistics › Measures of Dispersion
The mean deviation about the mean for the following data \(\begin{array}{llllll} \hline \text {Marks…
The mean deviation about the mean for the following data
\(\begin{array}{llllll}
\hline \text {Marks obtained } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\
\hline \text {Number of Boys } 6 & 8 & 10 & 4 & 2 \\
\hline
\end{array}\)
is
9.33 5.6 8.33 9.6 Solution
Calculation of mean deviation about mean
\(\begin{array}{cccccc}
\hline \begin{array}{c}
\text { Marks } \\
\text { obtained }
\end{array} & \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} \boldsymbol{x}_{\boldsymbol{i}} & \left|\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{2 1}\right| & \boldsymbol{f}_{\boldsymbol{i}}\left|\boldsymbol{x}_{\boldsymbol{i}}-\mathbf{2 1}\right| \\
\hline 0-10 & 5 & 6 & 30 & 16 & 96 \\
\hline 10-20 & 15 & 8 & 120 & 6 & 48 \\
\hline 20-30 & 25 & 10 & 250 & 4 & 40 \\
\hline 30-40 & 35 & 4 & 140 & 14 & 56 \\
\hline 40-50 & 45 & 2 & 90 & 24 & 48 \\
\hline & \begin{array}{c}
N=\Sigma f_i \\
=30
\end{array} & \begin{array}{c}
\Sigma f_i x_i \\
=630
\end{array} & & \begin{array}{c}
\Sigma f_i \mid x_i-21 \\
=288
\end{array} \\
\hline
\end{array}\)
Mean
\(\begin{gathered}
\qquad \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{630}{30}=21 \\
\therefore \text { Mean deviation }=\frac{1}{N} \Sigma f_i\left|x_i-21\right| \\
=\frac{288}{30}=9.6
\end{gathered}\)
Asked in: AP EAMCET 2019 (22 Apr Shift 1)
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