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

Hint:
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 \[Var\left( z \right) = {p^2}Var\left( x \right)\].