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  • Find the mean deviation about the mean of the distribution: Size 20 Frequency 6

Find the mean deviation about the mean of the distribution:

\begin{array}{|l|l|l|l|l|l|} \hline \text { Size } & 20 & 21 & 22 & 23 & 24 \\ \hline \text { Frequency } & 6 & 4 & 5 & 1 & 4 \\ \hline \end{array}

Answers (1)

We have to find the mean deviation about the mean 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 { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} \\ \hline 20 & 6 & 120 \\ \hline 21 & 4 & 84 \\ \hline 22 & 5 & 110 \\ \hline 23 & 1 & 23 \\ \hline 24 & 4 & 96 \\ \hline \text { Total } & 20 & 433 \\ \hline \end{array}

\text{Here, mean }\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{433}{20}=21.65 \\\\ \\ \text{ So the above table can be rewritten as } \\\\

\begin{array}{|l|l|l|l|l|} \hline \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} & d_{i}=\left|x_{i}-\bar{x}\right| & f_{i} d_{i} \\ \hline 20 & 6 & 120 & 1.65 & 9.90 \\ \hline 21 & 4 & 84 & 0.65 & 2.60 \\ \hline 22 & 5 & 110 & 0.35 & 1.75 \\ \hline 23 & 1 & 23 & 1.35 & 1.35 \\ \hline 24 & 4 & 96 & 2.35 & 9.40 \\ \hline \text { Total } & 20 & 433 & 6.35 & 25.00 \\ \hline \end{array}

 

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

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