First slide

Introduction to statistics

Question

If the S.D. of a variate $X$ is $σ,$ then the S.D. of a $a X + b$ is

Moderate

A

$| a | σ$
B

$σ$
C

$a σ$
D

$a σ + b$

Solution

Let $x_{1}, x_{2}, \dots, x_{n}$ be $n$ values of $X$ . Then,
$σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{X})}^{2}$ ….(i)
The variable $a X + b$ takes values $a x_{1} + b, a x_{2} + b, \dots, a x_{n} + b$
with mean $a \bar{X} + b$ .
$∴$ Var $(a X + b) = \frac{1}{n} \sum_{i = 1}^{n} {\{(a x_{i} + b) - (a \bar{X} + b)\}}^{2}$
$= a^{2} \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{X})}^{2}$
$\Rightarrow (S.D. of a X + b) = \sqrt{a^{2} \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{X})}^{2}} = | a | σ$

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