Continuity And Differentiability

Question

If $\sum_{i = 1}^{9} (x_{i} - 5) = 9 a n d \sum_{i = 1}^{9} (x_{i} - 5)^{2} = 45$ then the
standard deviation of the 9 items x₁, x₂, ...., x₉ is

3

9

4

2

Solution

Standard deviation of x_i - 5 is

$σ = \sqrt{\frac{\sum_{i = 1}^{9} (x_{i} - 5)^{2}}{9} - {(\frac{\sum_{i = 1}^{9} (x_{i} - 5)}{9})}^{2}} \Rightarrow σ = \sqrt{5 - 1} = 2$

As, standard deviation remains constant if observations are added/subtracted by a fixed quantity.

so σ of x_i is 2