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The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & A & 2 A & 3 A & 4 A & 5 A & 6 A \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}$
where $A$ is a positive integer, has a variance of $160 .$ Determine the value of $A$.
Solution
$\begin{array}{|c|c|c|c|} \hline x & f_{i} & f_{1} x_{i} & f x_{i}^{2} \\ \hline A & 2 & 2 A & 2 A^{2} \\ \hline 2 A & 1 & 2 A & 4 A^{2} \\ \hline 3 A & 1 & 3 A & 9 A^{2} \\ \hline 4 A & 1 & 4 A & 16 A^{2} \\ \hline 5 A & 1 & 5 A & 25 A^{2} \\ \hline 6 A & 1 & 6 A & 36 A^{2} \\ \hline \text { Total } & n=7 & \Sigma f_{i}=22 A & \Sigma f_{i}^{2}=92 A^{2} \\ \hline \end{array}$
$\therefore \quad \sigma^{2}=\frac{\Sigma f_{t} x_{1}^{2}}{n}-\left(\frac{\Sigma f_{1} x_{1}}{n}\right)^{2}$
$\Rightarrow \quad 160=\frac{92 A^{2}}{7}-\left(\frac{22 A}{7}\right)^{2} \Rightarrow 160=\frac{92 A^{2}}{7}-\frac{484 A^{2}}{49}$
$\Rightarrow \quad 160=(644-484) \frac{A^{2}}{49} \Rightarrow 160=\frac{160 A^{2}}{49}$
$\Rightarrow \quad A^{2}=49 \quad \therefore \quad A=7$
