Question

If $\,\sum\limits_{i=1}^{n}{(x_i-5)}=9$ and $\,\sum\limits_{i=1}^{n}{{{(x_i-5)}^{2}}}=45$ , then the standard deviation of the 9 times $\,{{x}_{1}},{{x}_{2}},\,......{{x}_{9}}$ is
A. $9$
B. $4$
C. $3$
D. $2$

Answer 1

In the given question, it is written that standard deviation of the 9 times $\,{{x}_{1}},{{x}_{2}},\,......{{x}_{9}}$ that means values of $\,\,n=9$ . To find out the standard deviation, that is it can be denoted as $\,\sigma$ . For that we have to use the formula for Variance ${{V}_{ar}}(x)=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{d}_{i}}^{2}}-{{\left( \dfrac{1}{n}\sum\limits_{i=1}^{n}{{{d}_{i}}} \right)}^{2}}$ where $\,{{d}_{i}}=\,\,$ derivation $\,\,;{{d}_{i}}=\,\,({{x}_{i}}-A)$ .

Complete step by step answer:
According to the given data in the question $\,\sum\limits_{i=1}^{n}{(x_i-5)}=9$ and $\,\sum\limits_{i=1}^{n}{{{(x_i-5)}^{2}}}=45$ .
We have to find the standard deviation which is denoted as $\,\sigma$ so, to find the standard deviation
We need to find the variance which is denoted as ${{V}_{ar}}(x)$ ,
Formula for variance that is ${{V}_{ar}}(x)$ ,
${{V}_{ar}}(x)=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{d}_{i}}^{2}}-{{\left( \dfrac{1}{n}\sum\limits_{i=1}^{n}{{{d}_{i}}} \right)}^{2}}---(1)$
Where $\,{{d}_{i}}=\,\,$ derivation $\,\,;{{d}_{i}}=\,\,({{x}_{i}}-A)$ .
But according to question $\,\,{{d}_{i}}=\,\,({{x}_{i}}-5)----(2)$
By substituting the value of equation $(2)$ in equation $(1)$
And also substitute the value of $n=9$ in equation $(1)$
${{V}_{ar}}(x)=\dfrac{1}{9}\sum\limits_{i=1}^{n}{{{({{x}_{i}}-5)}^{2}}}-{{\left( \dfrac{1}{9}\sum\limits_{i=1}^{n}{({{x}_{i}}-5)} \right)}^{2}}---(3)$
By substituting the values of $\,\sum\limits_{i=1}^{n}{(x_i-5)}=9$ and $\,\sum\limits_{i=1}^{n}{{{(x_i-5)}^{2}}}=45$ in equation $(3)$
${{V}_{ar}}(x)=\left( \dfrac{1}{9}\times 45 \right)-{{\left( \dfrac{1}{9}\times 9 \right)}^{2}}$
By further simplifying we get:
${{V}_{ar}}(x)=\left( \dfrac{45}{9} \right)-{{\left( \dfrac{9}{9} \right)}^{2}}$
${{V}_{ar}}(x)=\left( 5 \right)-{{\left( 1 \right)}^{2}}$
By solving this, we get:
${{V}_{ar}}(x)=5-1$
${{V}_{ar}}(x)=4----(4)$
To get the value of standard deviation we have to use the formula which shows the relation between standard deviation that is $\sigma$ and variance that is ${{V}_{ar}}(x)$ .
$\sigma =\sqrt{{{V}_{ar}}(x)}----(5)$
Substitute the value of equation $(4)$ and equation $(5)$ .
$\sigma =\sqrt{4}$
$\sigma =2$

So, the correct answer is “Option D”.

Note: In this type of question they have written indirectly that standard deviation of the 9 times $\,{{x}_{1}},{{x}_{2}},\,......{{x}_{9}}$ . That is $\,n=9$ . Remember that to find the standard deviation $\,\sigma$ we need to find the variance ${{V}_{ar}}(x)$ . By using formulas and proper steps to solve the similar type of problem.