Discussion Forum

1)

If $\sum_\left(i=1\right)^9\left(x_{i}-5\right)=9$ and $\sum_\left(i=1\right)^9\left(x_{i}-5\right)^{2}=45$

then the standard deviation of the 9 items x₁, x₂,........ x₉  is 

Answer:

Option C

Explanation:

key idea  Standard deviation is remain unchanged, if observations are added or substracted by a fixed number.

  We have,

$\sum_\left(i=1\right)^9\left(x_{i}-5\right)=9$ and $\sum_\left(i=1\right)^9\left(x_{i}-5\right)^{2}=45$

$SD=\sqrt{\frac{\sum_\left(i=1\right)^9\left(x_{1}-5\right)^{2}}{9}-\left(\frac{\sum_\left(i=1\right)^9\left(x_{1}-5\right)}{9}\right)^{2}}$

$\Rightarrow$   SD= $\sqrt{\frac{45}{9}-( \frac{9}{9})^{2}}$

$\Rightarrow$    SD= $\sqrt{5-1}=\sqrt{4}=2$