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  • Determine the mean and standard deviation for the following distribution: Marks 2 Frequency 1

Determine the mean and standard deviation for the following distribution:

\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Marks } & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \text { Frequency } & 1 & 6 & 6 & 8 & 8 & 2 & 2 & 3 & 0 & 2 & 1 & 0 & 0 & 0 & 1 \\ \hline \end{array}

 

Answers (1)

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

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


\begin{array}{|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} \\ \hline 2 & 1 & 2 \\ \hline 3 & 6 & 18 \\ \hline 4 & 6 & 24 \\ \hline 5 & 8 & 40 \\ \hline 6 & 8 & 48 \\ \hline 7 & 2 & 14 \\ \hline 8 & 2 & 16 \\ \hline 9 & 3 & 27 \\ \hline 10 & 0 & 0 \\ \hline 11 & 2 & 22 \\ \hline 12 & 1 & 12 \\ \hline 13 & 0 & 0 \\ \hline 14 & 0 & 0 \\ \hline 15 & 0 & 0 \\ \hline 16 & 1 & 16 \\ \hline \text { Total } & \mathrm{N}=40 & =239 \\ \hline \end{array}

\\ mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{ \Sigma f_{i}}=\frac{239}{40}=5.975\approx6 \\\\

\begin{array}{|l|l|l|l|l|l|} \hline \text { Marks }\left(x_{i}\right) & \begin{array}{l} \text { Frequency } \\ \left(f_{i}\right) \end{array} & f_{i} x_{i} & d_{i}=x_{i}-\bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 2 & 1 & 2 & -4 & -4 & 16 \\ \hline 3 & 6 & 18 & -3 & -18 & 54 \\ \hline 4 & 6 & 24 & -2 & -12 & 24 \\ \hline 5 & 8 & 40 & -1 & -8 & 8 \\ \hline 6 & 8 & 48 & 0 & 0 & 0 \\ \hline 7 & 2 & 14 & 1 & 2 & 2 \\ \hline 8 & 2 & 16 & 2 & 4 & 8 \\ \hline 9 & 3 & 27 & 3 & 9 & 27 \\ \hline 10 & 0 & 0 & 4 & 0 & 0 \\ \hline 11 & 2 & 22 & 5 & 10 & 50 \\ \hline 12 & 1 & 12 & 6 & 6 & 36 \\ \hline 13 & 0 & 0 & 7 & 0 & 0 \\ \hline 14 & 0 & 0 & 8 & 0 & 0 \\ \hline 15 & 0 & 0 & 9 & 0 & 0 \\ \hline 16 & 1 & 16 & 10 & 10 & 100 \\ \hline \text { Total } & \mathrm{N}=40 & =239 & & =-1 & =325 \\ \hline \end{array}

And we know that standard deviation is 

\\ \sigma = \sqrt {\frac{ \Sigma f_{i}d_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}d_{i}}{n} \right) ^{2}}= \sqrt {\frac{325}{40}- \left( \frac{-1}{40} \right) ^{2}}=\sqrt {8.125- \left( 0.025 \right) ^{2}} =2.8721\\\\

 

 

 

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