Question

If the standard deviation of ${x_1}$, ${x_2}$, ${x_3}$, ${x_4}$, …., ${x_n}$ is $3.5$, then the standard deviation of $\left( { - 2{x_1} - 3} \right)$ , $\left( { - 2{x_2} - 3} \right)$ , $\left( { - 2{x_3} - 3} \right)$, and $\left( { - 2{x_n} - 3} \right)$ is ?

Answer 1

Hint: In the given question, we are provided with some information about a series where n represents the number of terms and \[{x_i}'s\] represent the observations. So, the standard deviation of all the \[{x_i}'s\] is given to us in the question itself. We have to find the standard deviation of the series obtained by making the changes in the original series as suggested in the question.

Complete step by step solution:
The standard deviation is a measure of the amount of variation or dispersion of a set of values.
So, we have a series of observations: ${x_1}$, ${x_2}$, ${x_3}$, ${x_4}$, …., ${x_n}$.
The standard deviation of the observations ${x_1}$, ${x_2}$, ${x_3}$, ${x_4}$, …., ${x_n}$ is $3.5$.
We know, standard deviation is, $\sigma = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}} $
Where, $n = $total observations
${x_i} = $each term
$\mu = $mean of observations
So, the given series of observations is, ${x_1}$, ${x_2}$, ${x_3}$, ${x_4}$, …., ${x_n}$.
Now, on multiplying $ - 2$ to each of the observations, the series becomes, $ - 2{x_1}$, $ - 2{x_2}$, $ - 2{x_3}$, $ - 2{x_4}$, …., $ - 2{x_n}$.
Therefore, the mean of the new series becomes, ${\mu _1} = \dfrac{{ - 2{x_1} - 2{x_2} - 2{x_3} - ..... - 2{x_n}}}{n}$
$ \Rightarrow {\mu _1} = \dfrac{{ - 2\left( {{x_1} + {x_2} + {x_3} + .... + {x_n}} \right)}}{n}$
[Taking $ - 2$ common from the numerator]
$ \Rightarrow {\mu _1} = - 2\mu $
[$\mu = $older mean]
Now, using this new mean in the formula of standard deviation, for new series, we get,
${\sigma _1} = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( { - 2{x_i} - \left( { - 2\mu } \right)} \right)}^2}} }}{n}} $
Taking, $ - 2$ common from numerator, inside the root, we get,
$ \Rightarrow {\sigma _1} = \sqrt {\dfrac{{{{( - 2)}^2}\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}} $
Now, taking $ - 2$ out of the root, we get,
$ \Rightarrow {\sigma _1} = - 2\sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \mu } \right)}^2}} }}{n}} $
$ \Rightarrow {\sigma _1} = - 2\sigma $
[$\sigma = $older standard deviation]
Now, the series of observations is further changed by subtracting $3$ from the new equation.
That is, $ - 2{x_1} - 3$, $ - 2{x_2} - 3$, $ - 2{x_3} - 3$, $ - 2{x_4} - 3$, …., $ - 2{x_n} - 3$.
Therefore, the mean of the new series becomes, ${\mu _2} = \dfrac{{( - 2{x_1} - 3) + ( - 2{x_2} - 3) + ( - 2{x_3} - 3) + .... + ( - 2{x_n} - 3)}}{n}$
Since, there are $n$ terms, so the number of $ - 3$ will also be $n$.
Therefore, on simplifying, we get,
$ \Rightarrow {\mu _2} = \dfrac{{ - 2\left( {{x_1} + {x_2} + {x_3} + .... + {x_n}} \right) - 3n}}{n}$
Dividing the numerator, by the denominator, gives us,
$ \Rightarrow {\mu _2} = \dfrac{{ - 2\left( {{x_1} + {x_2} + {x_3} + .... + {x_n}} \right)}}{n} - 3$
$ \Rightarrow {\mu _2} = - 2\mu - 3$
[Since, ${\mu _1} = \dfrac{{ - 2\left( {{x_1} + {x_2} + {x_3} + .... + {x_n}} \right)}}{n} = - 2\mu $]
Now, using this new mean in the formula of standard deviation, for new series, we get,
${\sigma _2} = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( {( - 2{x_i} - 3) - \left( { - 2\mu - 3} \right)} \right)}^2}} }}{n}} $
Now, opening the brackets, we get,
${\sigma _2} = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( { - 2{x_i} - 3 + 2\mu + 3} \right)}^2}} }}{n}} $
${\sigma _2} = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( { - 2{x_i} + 2\mu } \right)}^2}} }}{n}} $
This can be written as,
${\sigma _2} = \sqrt {\dfrac{{\sum\limits_{i = 1}^n {{{\left( { - 2{x_i} - ( - 2\mu )} \right)}^2}} }}{n}} $
This is the same as ${\sigma _1}$.
Therefore, ${\sigma _2} = {\sigma _1} = - 2\sigma $
Given, the standard deviation of the older series is $3.5$.
Therefore, the standard deviation of the new series is, ${\sigma _2} = - 2\sigma = - 2.(3.5) = - 7$.
So, we get the standard deviation of $\left( { - 2{x_1} - 3} \right)$ , $\left( { - 2{x_2} - 3} \right)$ , $\left( { - 2{x_3} - 3} \right)$, and $\left( { - 2{x_n} - 3} \right)$ as $ - 7$.
So, the correct answer is “-7”.

Note: The given question revolves around the concepts of statistics. From the above problem we can say that the standard deviation remains constant or unchanged if any constant is added or subtracted from all the observations. But the standard deviation gets multiplied or divided by the constant if all the observations are multiplied or divided by the same constant.