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  • The weights of coffee in 70 jars are shown in the following table: Weight 200-201 Frequency 13 Determine variance and standard deviation of the above distribution.

The weights of coffee in 70 jars are shown in the following table:

\begin{array}{|l|l|} \hline \text { Weight (in grams) } & \text { Frequency } \\ \hline 200-201 & 13 \\ \hline 201-202 & 27 \\ \hline 202-203 & 18 \\ \hline 203-204 & 10 \\ \hline 204-205 & 1 \\ \hline 205-206 & 1 \\ \hline \end{array}

Determine variance and standard deviation of the above distribution.

Answers (1)

A frequency distribution table for the weights of coffee in 70 jars is given and we have to find the variance and standard deviation of the distribution.

 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 { Weight (in } \\ \text { grams) } \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 200-201 & 200.5 & 13 & 2606.5 \\ \hline 201-202 & 201.5 & 27 & 5440.5 \\ \hline 202-203 & 202.5 & 18 & 3645 \\ \hline 203-204 & 203.5 & 10 & 2035 \\ \hline 204-205 & 204.5 & 1 & 204.5 \\ \hline 205-206 & 205.5 & 1 & 205.5 \\ \hline & \text { total } & \mathrm{N}=70 & =14137 \\ \hline & & & \\ \hline \end{array}

\\ mean~\overline{X~}=\frac{ \Sigma f_{i}x_{i}}{N}=\frac{14137}{70}=201.9 \\\\

\begin{array}{|l|l|l|l|l|l|} \hline \begin{array}{l} \text { Weight (in } \\ \text { grams) } \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} & d_{i}=x_{i}-\bar{x} & f_{i} d_{i} & f_{i} d_{i}^{2} \\ \hline 200-201 & 200.5 & 13 & -1.4 & -18.2 & 25.48 \\ \hline 201-202 & 201.5 & 27 & -0.4 & -10.8 & 4.32 \\ \hline 202-203 & 202.5 & 18 & 0.6 & 10.8 & 6.48 \\ \hline 203-204 & 203.5 & 10 & 1.6 & 16 & 25.6 \\ \hline 204-205 & 204.5 & 1 & 2.6 & 2.6 & 6.76 \\ \hline 205-206 & 205.5 & 1 & 3.6 & 3.6 & 12.96 \\ \hline & \text { total } & \mathrm{N}=70 & & =4 & =81.6 \\ \hline & & & & & \\ \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{81.6}{70}- \left( \frac{4}{70} \right) ^{2}}=\sqrt {1.17- \left( 0.057 \right) ^{2}}= \sqrt {1.17}=1.08~~ \\\\

Variance = 1.17

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infoexpert21

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