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If the mean of the discrete distribution 8,9 , $6,5, x, 4,6,5$ is 6 , then its standard deviation (nearest to two decimal places) is
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Verified Answer
The correct answer is:
1.58
We have,
$\begin{aligned}
& \text { Mean }=6 \\
& \therefore \quad \frac{8+9+6+5+x+4+6+5}{8}=6 \\
& \Rightarrow 43+x=48 \Rightarrow x=5
\end{aligned}$
So, data are $8,9,6,5,5,4,6,5$
$\begin{array}{llrl}\therefore & & x_i=8,9,6,5,5,4,6,5 \\ \Rightarrow & x_i{ }^2 & =64,81,36,25,25,16,36,25\end{array}$
$\begin{aligned} \therefore \quad \Sigma x_i^2 & =64+81+36+25+25+16+36+25 \\ & =308\end{aligned}$
$\begin{aligned} \therefore \quad \operatorname{SD}(x) & =\sqrt{\frac{1}{N} \Sigma x_i^2-(\bar{x})^2} \\ & =\sqrt{\frac{1}{8} \times 308-(6)^2}=\sqrt{38.5-36} \\ & =\sqrt{2.5}=1.58\end{aligned}$
$\begin{aligned}
& \text { Mean }=6 \\
& \therefore \quad \frac{8+9+6+5+x+4+6+5}{8}=6 \\
& \Rightarrow 43+x=48 \Rightarrow x=5
\end{aligned}$
So, data are $8,9,6,5,5,4,6,5$
$\begin{array}{llrl}\therefore & & x_i=8,9,6,5,5,4,6,5 \\ \Rightarrow & x_i{ }^2 & =64,81,36,25,25,16,36,25\end{array}$
$\begin{aligned} \therefore \quad \Sigma x_i^2 & =64+81+36+25+25+16+36+25 \\ & =308\end{aligned}$
$\begin{aligned} \therefore \quad \operatorname{SD}(x) & =\sqrt{\frac{1}{N} \Sigma x_i^2-(\bar{x})^2} \\ & =\sqrt{\frac{1}{8} \times 308-(6)^2}=\sqrt{38.5-36} \\ & =\sqrt{2.5}=1.58\end{aligned}$
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