JEE Math Practice Question
If $$\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9$$ and $$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$$, then the standard deviation of the 9 items $${x_1},{x_2},.......,{x_9}$$ is
- A.3
- B.9
- C.4
- D.2
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