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Question:
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Calculate the mean deviation about the mean for the following frequency distribution:
$$
\begin{array}{|c|c|c|c|c|c|}
\hline \text { Class interval } & 0-4 & 4-8 & 8-12 & 12-16 & 16-20 \\
\hline \text { Frequency } & 4 & 6 & 8 & 5 & 2 \\
\hline
\end{array}
$$
$$
\begin{array}{|c|c|c|c|c|c|}
\hline \text { Class interval } & 0-4 & 4-8 & 8-12 & 12-16 & 16-20 \\
\hline \text { Frequency } & 4 & 6 & 8 & 5 & 2 \\
\hline
\end{array}
$$
Solution:
2103 Upvotes
Verified Answer
$$
\begin{array}{|c|c|c|c|c|c|}
\hline \begin{array}{l}
\text { Class } \\
\text { interval }
\end{array} & \begin{array}{l}
\text { Mid } \\
\text { Value } \\
\left(\mathbf{x}_{\mathrm{i}}\right)
\end{array} & \begin{array}{l}
\text { Frequency } \\
(\boldsymbol{f})
\end{array} & \boldsymbol{f x} & (\boldsymbol{x}-\overline{\boldsymbol{x}}) & \boldsymbol{f} \cdot|\boldsymbol{x}-\overline{\boldsymbol{x}}| \\
\hline 0-4 & 2 & 4 & 8 & 7.2 & 28.8 \\
4-8 & 6 & 6 & 36 & 3.2 & 19.2 \\
8-12 & 10 & 8 & 80 & 0.8 & 6.4 \\
12-16 & 14 & 5 & 70 & 4.8 & 24.4 \\
16-20 & 18 & 2 & 36 & 8.8 & 17.6 \\
\hline & & 25 & 230 & & 96.0 \\
\hline
\end{array}
$$
$$
\therefore \quad \bar{x}=\frac{230}{25}=9.2
$$
Required M.D $=\frac{96}{25}=3.84$
\begin{array}{|c|c|c|c|c|c|}
\hline \begin{array}{l}
\text { Class } \\
\text { interval }
\end{array} & \begin{array}{l}
\text { Mid } \\
\text { Value } \\
\left(\mathbf{x}_{\mathrm{i}}\right)
\end{array} & \begin{array}{l}
\text { Frequency } \\
(\boldsymbol{f})
\end{array} & \boldsymbol{f x} & (\boldsymbol{x}-\overline{\boldsymbol{x}}) & \boldsymbol{f} \cdot|\boldsymbol{x}-\overline{\boldsymbol{x}}| \\
\hline 0-4 & 2 & 4 & 8 & 7.2 & 28.8 \\
4-8 & 6 & 6 & 36 & 3.2 & 19.2 \\
8-12 & 10 & 8 & 80 & 0.8 & 6.4 \\
12-16 & 14 & 5 & 70 & 4.8 & 24.4 \\
16-20 & 18 & 2 & 36 & 8.8 & 17.6 \\
\hline & & 25 & 230 & & 96.0 \\
\hline
\end{array}
$$
$$
\therefore \quad \bar{x}=\frac{230}{25}=9.2
$$
Required M.D $=\frac{96}{25}=3.84$
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