Let x_i (1 i10) be ten observations of a random variable X.

Question

Let $x_i (1\le i\le10)$ be ten observations of a random variable X. If $\displaystyle\sum_{i=1}^{10}(x_i-p)=3$ and $\displaystyle\sum_{i=1}^{10}(x_i-p)^2=9$ where $0
e p\in\mathbb{R}$, then the standard deviation of these observations is:

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.]

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

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

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

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

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

Solutions · 1