MCQ
let x1, x2, ...,xn 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
    $\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
  • C
    $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
  • D
    $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$

Answer

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

Solution:

It is given that x1, x2, ...,xn 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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