Question

If the standard deviation of a variable $X$ is $\sigma$, then the standard deviation of the variable $\dfrac{{aX + b}}{c}$ is
(a) $a\sigma$
(b) $\dfrac{a}{c}\sigma$
(c) $\left| {\dfrac{a}{c}} \right|\sigma$
(d) $\dfrac{{a\sigma + b}}{c}$

Answer 1

Here, we need to find the standard deviation of the variable $\dfrac{{aX + b}}{c}$. Let $Y = \dfrac{{aX + b}}{c}$. First, we will find the variance of the variable $X$. Then, we will find the mean of the variable $Y$. Using the mean of $Y$, we will find the variance of $Y$. Then, using the variance of $X$, we will simplify the expression for variance of $Y$. Finally, we will use the variance of $Y$ to find the standard deviation of $Y$, and hence, the standard deviation of the variable $\dfrac{{aX + b}}{c}$.
Formula Used: The mean of a variable of a variable $X$ is given by the formula $\dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}}$.
The variance of a variable of a variable $X$ is given by the formula $\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline X } \right)}^2}}$.
The standard deviation of a variable is equal to the square root of the variance of the variable.

Complete step by step solution:
The mean of a variable $X$ is given by the formula $\dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}}$.
Therefore, we get
$\overline X = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}}$
The standard deviation of a variable is equal to the square root of the variance of the variable.
The standard deviation of a variable $X$ is $\sigma$.
The variance of a variable of a variable $X$ is given by the formula $\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline X } \right)}^2}}$.
Therefore, we get
$\Rightarrow {\sigma ^2} = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline X } \right)}^2}}$
Now, let $Y = \dfrac{{aX + b}}{c}$.
Therefore, we get ${y_i} = \dfrac{{a{x_i} + b}}{c}$.
We know that if all the terms of a series are increased, decreased, multiplied, or divided by the same number, the mean is also increased, divided, multiplied, or divided by the same number.
This means that if $z = px + q$, then $\overline z = p\overline x + q$.
Therefore, we get
$\Rightarrow \overline Y = \dfrac{{a\overline X + b}}{c}$
Now, we will find the variance of the variable $Y$.
Using the formula for variance of a variable, we get
$Var\left( Y \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{y_i} - \overline Y } \right)}^2}}$
Substituting ${y_i} = \dfrac{{a{x_i} + b}}{c}$ and $\overline Y = \dfrac{{a\overline X + b}}{c}$ in the expression, we get
$\Rightarrow Var\left( Y \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{a{x_i} + b}}{c} - \dfrac{{a\overline X + b}}{c}} \right)}^2}}$
Splitting the L.C.M., we get
$\Rightarrow Var\left( Y \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{a{x_i}}}{c} + \dfrac{b}{c} - \dfrac{{a\overline X }}{c} - \dfrac{b}{c}} \right)}^2}}$
Subtracting the like terms, we get
$\Rightarrow Var\left( Y \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{a{x_i}}}{c} - \dfrac{{a\overline X }}{c}} \right)}^2}}$
Factoring the terms, we get
$\Rightarrow Var\left( Y \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{a\left( {{x_i} - \overline X } \right)}}{c}} \right)}^2}}$
Simplifying the expression, we get
$\Rightarrow Var\left( Y \right) = \dfrac{1}{n} \times {\left( {\dfrac{a}{c}} \right)^2} \times \sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline X } \right)}^2}}$
Applying the exponent on the base and rewriting, we get
$\Rightarrow Var\left( Y \right) = \dfrac{{{a^2}}}{{{c^2}}}\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline X } \right)}^2}}$
Substituting ${\sigma ^2} = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline X } \right)}^2}}$ in the expression, we get
$\Rightarrow Var\left( Y \right) = \dfrac{{{a^2}}}{{{c^2}}}{\sigma ^2}$
The standard deviation of $Y$ is the square root of $Var\left( Y \right)$.
Therefore, taking the square root of both sides, we get
$\Rightarrow S.D.\left( Y \right) = \sqrt {\dfrac{{{a^2}}}{{{c^2}}}{\sigma ^2}}$
Simplifying the expression, we get
$\Rightarrow S.D.\left( Y \right) = \left| {\dfrac{a}{c}} \right|\sigma$
Substituting $Y = \dfrac{{aX + b}}{c}$ in the equation, we get
$\Rightarrow S.D.\left( {\dfrac{{aX + b}}{c}} \right) = \left| {\dfrac{a}{c}} \right|\sigma$
Therefore, we get the standard deviation of the variable $\dfrac{{aX + b}}{c}$ as $\left| {\dfrac{a}{c}} \right|\sigma$.

Thus, the correct option is option (c).

Note:
We simplified $Var\left( Y \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{a\left( {{x_i} - \overline X } \right)}}{c}} \right)}^2}}$ to $Var\left( Y \right) = \dfrac{1}{n} \times {\left( {\dfrac{a}{c}} \right)^2} \times \sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline X } \right)}^2}}$.
If all the terms of a series are multiplied, or divided by the same number, then the variance of the series is multiplied, or divided by the square of the same number.
This means that if $z = px$, then $Var\left( z \right) = {p^2}Var\left( x \right)$.