let x_1, x_2, ...,x_n be n observations and X be their arithmetic… - Vidyadip

MCQ

let $x_1, x_2, ...,x_n$ be n observations and $\overline{\text{X}}$ be their arithmetic mean. The standard deviation is given by:

A
$\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
B
$\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
✓
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
D
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$

✓

Answer

Correct option: C.

$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$

It is given that $x_1, x_2, ...,x_n$ be n observations and $\overline{\text{X}}$ be their arithmetic mean.
The standard deviation is given observations is $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}.$
Also,
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}=\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$
Hence, the correct answers are options $(c)$ and $(d).$
Disclaimer: For option $(c)$ to be the only correct answer, option $(d)$ should be different from the given value.

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