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

Hint: 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.