"Let xi (1 $\le$ i $\le$ 10) be ten observations of a random variable X. If $\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$ and $\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$ where 0 $\ne$ p $\in$ R, then the standard deviation of these observations is :"

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Accepted Answer

"Standard deviation = $\sqrt {Variance}$

$= \sqrt {{{\sum {x_1^2} } \over n} - {{(\overline x )}^2}}$

$$ = \sqrt {{{\sum\limits_{i = 1}^{10} {{{({x_i} - p)}^2}} } \over {10}} - {{\left( {{{\sum\limits_{i = 1}^{10} {({x_i} - p)} } \over {10}}} \right)}^2}} $$

[ Standard deviation is free from shifting of origin.]

$= \sqrt {{9 \over {10}} - {{\left( {{3 \over {10}}} \right)}^2}}$

$= \sqrt {{9 \over {10}} - {9 \over {100}}}$

$= \sqrt {{{90 - 9} \over {100}}}$

$= \sqrt {{{81} \over {100}}}$

$= {9 \over {10}}$"

Let x_i (1 $\le$ i $\le$ 10) be ten observations of a random variable X. If
$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$ and $\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$
where 0 $\ne$ p $\in$ R, then the standard deviation of these observations is :

Solution

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Let xi (1 $\le$ i $\le$ 10) be ten observations of a random variable X. If $\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$ and $\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$ where 0 $\ne$ p $\in$ R, then the standard deviation of these observations is :

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More Statistics questions

Drill 25 more like these. Every day. Free.

Let x_i (1 $\le$ i $\le$ 10) be ten observations of a random variable X. If
$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$ and $\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$
where 0 $\ne$ p $\in$ R, then the standard deviation of these observations is :