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  • The frequency distribution: xA f2 where A is a positive integer, has a variance of 160. Determine the value of A.

The frequency distribution:

\begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{x} & \mathrm{A} & 2 \mathrm{A} & 3 \mathrm{A} & 4 \mathrm{A} & 5 \mathrm{A} & 6 \mathrm{A} \\ \hline \mathrm{f} & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}

where A is a positive integer, has a variance of 160. Determine the value of A.

 

 

Answers (1)

A frequency distribution table is given where variance =160. We have to find the value of A, where A is a positive number. Let us make a table from the given data and fill out the other columns after calculation

\begin{array}{|l|l|l|c|} \hline \text { Size }\left(x_{i}\right) & \text { Frequency }\left(f_{i}\right) & f_{i} x_{i} & f_{i} x_{i}^{2} \\ \hline \text { A } & 2 & 2 \mathrm{A} & 2 A^{2} \\ \hline 2 \mathrm{A} & 1 & 2 \mathrm{A} & 4 A^{2} \\ \hline 3 \mathrm{A} & 1 & 3 \mathrm{A} & 9 A^{2} \\ \hline 4 \mathrm{A} & 1 & 4 \mathrm{A} & 16 A^{2} \\ \hline 5 \mathrm{A} & 1 & 5 \mathrm{A} & 25 \mathrm{A}^{2} \\ \hline 6 \mathrm{A} & 1 & 6 \mathrm{A} & 36 A^{2} \\ \hline \text { Total } & \mathrm{N}=7 & 22 \mathrm{A} & 92 \mathrm{A}^{2} \\ \hline \end{array}

 

And we know that variance is \\\\ ~ \sigma ^{2}= \frac{ \Sigma f_{i}x_{i}^{2}}{n}- \left( \frac{ \Sigma f_{i}x_{i}}{n} \right) ^{2}~ \\\\


 \\ \text{ Substituting the values from the above table and also given that variance}=160 \\ we~get~ 160=\frac{92A^{2}}{7}- \left( \frac{22A}{7} \right) ^{2}= \left( \frac{92A^{2}}{7}-\frac{484A^{2}}{49} \right) =\frac{7\ast92A^{2}-484A^{2}}{49}=\frac{160A^{2}}{49}~ \\\ ~~~ 160=\frac{160A^{2}}{49}~ \\\\ \\ ~~A^{2}=49 \\\\ \\ A=7 \\\\ \\ \text{ Hence, the value of A is 7} \\\\

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