Question

If 𝜎 is the standard deviation of a random variable X, then, the standard deviation of the random variable \[\left( {\dfrac{{aX + b}}{c}} \right)\], where a, b ϵ R is
a) \[a\sum + b\]
b) \[\left| {\dfrac{a}{c}} \right|\sigma \]
c) \[\left| a \right|\sum + b\]
d) \[{a^2}\sum \]

Answer 1

Hint: Variance is the average of the squared deviations from the mean value. To calculate the variance, subtract the mean from each number and then square the results to find the squared differences and then find the average of those squared differences. The standard deviation can be calculated by taking the square root of variance.

Formula Used:
\[{\sigma _Y}^2 = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{({Y_i} - \overline Y )}^2}} \]

Complete step by step Solution:
Given: Standard deviation of a variable X is \[\sigma \]i.e. \[{\sigma _X} = \sigma \].
Let \[Y = \left( {\dfrac{{aX + b}}{c}} \right)\]
The first step is to find the mean value of Y.
Therefore \[\overline Y = \overline {\dfrac{{aX + b}}{c}} \] --- eq (1)
In eq (1), \[\overline Y \]represents the mean value of Y. X is a random variable whereas a, b, and c are constants.
Therefore, \[\overline Y = \dfrac{a}{c}\overline X + \dfrac{b}{c}\]
\[\overline Y = \dfrac{{a\overline X + b}}{c}\] --eq (2)
Now calculate the variance of Y
\[{\sigma _Y}^2 = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{({Y_i} - \overline Y )}^2}} \] -- eq (3)
\[{Y_i} = \dfrac{{a{X_i} + b}}{c}\] -- eq (4)
Substitute eq (2) and eq (4) in eq (3).
\[{\sigma _Y}^2 = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{a{X_i} + b}}{c} - \dfrac{{a\overline X + b}}{c}} \right)}^2}} \]
Simplify the above equation to get,
\[{\sigma _Y}^2 = \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{a{X_i} - a\overline X }}{c}} \right)}^2}} \]
Take \[\dfrac{{{a^2}}}{{{c^2}}}\] as a common term.
\[{\sigma _Y}^2 = \dfrac{{{a^2}}}{{{c^2}}}\,\,\dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \overline X } \right)}^2}} \]-- eq (5)
Since the term \[\dfrac{1}{{n\,}}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \overline X } \right)}^2}} \]can be written as Variance of X i.e. \[{\sigma _X}^2\], eq (5) becomes:
\[{\sigma _Y}^2 = \dfrac{{{a^2}}}{{{c^2}}}\,\,{\sigma _X}^2\]-- eq (6)
From the definition, we know that the standard deviation is the square root of Variance. Therefore, taking the square root of eq (6).
\[{\sigma _Y} = \left| {\dfrac{a}{c}} \right|\,\,{\sigma _X}\] - eq (7)
\[{\sigma _X}\]= Standard deviation of Variable X
\[{\sigma _X} = \sigma \] [Given: 𝜎 is the standard deviation of a random variable X.] Put \[{\sigma _X} = \sigma \] in eq (7)
\[{\sigma _Y} = \left| {\dfrac{a}{c}} \right|\,\,\sigma \]

Hence, the correct option is 2.

Note: To solve the statistics questions related to Variance and standard deviation, one must be thorough with the theoretical concepts. Also keep in mind the formulas related to mean value, variance, and standard deviation. Once a variance is calculated, the standard deviation can be easily known by taking the square root of the variance.