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  • Calculate the mean deviation about the mean for the following frequency distribution:Class interval 0-4 Frequency 4

Calculate the mean deviation about the mean for the following frequency distribution:

\begin{array}{|l|l|l|l|l|l|} \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}


 

Answers (1)

A frequency distribution table is given, and we have to find the mean deviation about the mean

 Let us make a table from the given data and fill out the other columns after calculation

\begin{array}{|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} \\ \hline 0-4 & 2 & 4 & 8 \\ \hline 4-8 & 6 & 6 & 36 \\ \hline 8-12 & 10 & 8 & 80 \\ \hline 12-16 & 14 & 5 & 70 \\ \hline 16-20 & 18 & 2 & 36 \\ \hline & \text { Total } & \mathrm{N}=25 & =230 \\ \hline \end{array}

\\ mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{230}{25}=9.2 \\\\ \\ \text{~~ The above column can be re written as} \\\\

 

\begin{array}{|l|l|l|l|l|l|} \hline \begin{array}{l} \text { Class } \\ \text { Interval } \end{array} & \begin{array}{l} \text { Mid value } \\ \left(x_{i}\right) \end{array} & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} & \begin{array}{l} d_{i}=\mid x_{i}- \\ \bar{x} \mid \end{array} & f_{i} d_{i} \\ \hline 0-4 & 2 & 4 & 8 & 7.2 & 28.8 \\ \hline 4-8 & 6 & 6 & 36 & 3.2 & 19.2 \\ \hline 8-12 & 10 & 8 & 80 & 0.8 & 6.4 \\ \hline 12-16 & 14 & 5 & 70 & 1.8 & 24.4 \\ \hline 16-20 & 18 & 2 & 36 & 8.8 & 17.6 \\ \hline & \text { Total } & \mathrm{N}=25 & =230 & & =96.0 \\ \hline \end{array}

\\ mean\: \: Deviation=\frac{ \Sigma f_{i}d_{i}}{ \Sigma f_{i}}=\frac{96}{25}=3.84 \\\\

 

 

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