If _i=1^9(x_i-5)=9 and _i=1^9(x_i-5)^2=45, then the standard deviation of the 9 items x_1,…

Question

If $\displaystyle\sum_{i=1}^9(x_i-5)=9$ and $\displaystyle\sum_{i=1}^9(x_i-5)^2=45$, then the standard deviation of the 9 items $x_1,x_2,\ldots,x_9$ is:

Accepted Answer

IMPORTANT POINT :-

When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged.

so let ${x_i} - 5 = {y_i}$

So, new equation is $$\sum\limits_{i = 1}^9 {{y_i}}  = 9$$

and $$\sum\limits_{i = 1}^9 {y_i^2}  = 45$$

As, we know. Standard Deviation (S.D)

$$ = \sqrt {{{\sum\limits_{i = 1}^9 {y_i^2} } \over 9} - {{\left( {{{\sum\limits_{i = 1}^9 {yi} } \over 9}} \right)}^2}} $$

$$ = \sqrt {{{45} \over 9} - {{\left( {{9 \over 9}} \right)}^2}} $$

$$ = \sqrt {5 - 1} $$

$=$ $$\sqrt 4 $$

$=2$

Solutions · 1