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Question:
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If the variance of the frequency distribution
$\begin{array}{|c|c|c|c|c|c|c|}
\hline x & c & 2 c & 3 c & 4 c & 5 c & 6 c \\
\hline f & 2 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}$
is 160, then the value of $c \in N$ is
Options:
$\begin{array}{|c|c|c|c|c|c|c|}
\hline x & c & 2 c & 3 c & 4 c & 5 c & 6 c \\
\hline f & 2 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}$
is 160, then the value of $c \in N$ is
Solution:
1967 Upvotes
Verified Answer
The correct answer is:
7
$\begin{array}{|c|c|c|c|c|c|c|}
\hline \mathrm{x} & \mathrm{C} & 2 \mathrm{C} & 3 \mathrm{C} & 4 \mathrm{C} & 5 \mathrm{C} & 6 \mathrm{C} \\
\hline \mathrm{f} & 2 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}$
$\bar{x}=\frac{(2+2+3+4+5+6) C}{7}=\frac{22 C}{7}$
$\begin{aligned} & \operatorname{Var}(x)=\frac{\mathrm{c}^2\left(2+2^2+3^2+4^2+5^2+6^2\right)}{7} \\ & -\left(\frac{22 \mathrm{c}}{7}\right)^2 \\ & =\frac{92 \mathrm{c}^2}{7}-\mathrm{c}^2 \times \frac{484}{49} \\ & =\frac{(644-484) \mathrm{c}^2}{49}=\frac{160 \mathrm{c}^2}{49} \\ & 160=\frac{160 \times \mathrm{c}^2}{49} \Rightarrow \mathrm{c}=7\end{aligned}$
\hline \mathrm{x} & \mathrm{C} & 2 \mathrm{C} & 3 \mathrm{C} & 4 \mathrm{C} & 5 \mathrm{C} & 6 \mathrm{C} \\
\hline \mathrm{f} & 2 & 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}$
$\bar{x}=\frac{(2+2+3+4+5+6) C}{7}=\frac{22 C}{7}$
$\begin{aligned} & \operatorname{Var}(x)=\frac{\mathrm{c}^2\left(2+2^2+3^2+4^2+5^2+6^2\right)}{7} \\ & -\left(\frac{22 \mathrm{c}}{7}\right)^2 \\ & =\frac{92 \mathrm{c}^2}{7}-\mathrm{c}^2 \times \frac{484}{49} \\ & =\frac{(644-484) \mathrm{c}^2}{49}=\frac{160 \mathrm{c}^2}{49} \\ & 160=\frac{160 \times \mathrm{c}^2}{49} \Rightarrow \mathrm{c}=7\end{aligned}$
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