Question

If $\sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}=9$ and $\sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-5 \right)}^{2}}}=45$, then the standard deviation of the 9 items ${{x}_{1}},{{x}_{2}},......,{{x}_{9}}$ is: -
(a) 2
(b) 3
(c) 9
(d) 4

Answer 1

Apply the formula for standard deviation given as: - $\sigma =\sqrt{\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}$, where ‘n’ is the number of terms, ‘$\sigma$’ is the notation of standard deviation and ‘$\overline{x}$’ is the average of given terms ${{x}_{1}},{{x}_{2}},{{x}_{3}},......,{{x}_{9}}$. To find $\overline{x}$, use the formula, $\overline{x}=\dfrac{\sum\limits_{i=1}^{9}{{{x}_{i}}}}{9}$. Use the given relation, $\sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}=9$ to get the value of $\overline{x}$ and use the relation $\sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-5 \right)}^{2}}}=45$ to get the value of $\sigma$.

Complete step by step answer:
We have been provided with two relations: -
$\Rightarrow \sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}=9$ - (i)
$\Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-5 \right)}^{2}}}=45$ - (ii)
Now, let us consider relation (i). We have,
$\Rightarrow \sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}=9$
The above expression is written as,
$\Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}}-\sum\limits_{i=1}^{9}{5}=9$ - (iii)
We know that if ‘k’ is a constant, then.
$\Rightarrow \sum\limits_{i=1}^{n}{k}=nk$
Applying this identity in equation (iii), we get,

& \Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}}-5\times 9=9 \\\ & \Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}}=9\times 5+9 \\\ & \Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}}=54 \\\ \end{aligned}$$ Dividing both sides by 9, we get, $$\Rightarrow \dfrac{\sum\limits_{i=1}^{9}{{{x}_{i}}}}{9}=\dfrac{54}{9}=6$$ Since, $$\dfrac{\sum\limits_{i=1}^{9}{{{x}_{i}}}}{9}=\overline{x}$$, $$\Rightarrow \overline{x}=6$$ So, the mean of the given items $${{x}_{1}},{{x}_{2}},{{x}_{3}},......,{{x}_{9}}$$ is 6. Now, we know that standard deviation is given by the formula: - $$\Rightarrow \sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}}$$, where $$\sigma $$ = standard deviation. $$\Rightarrow \sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{9}}$$ Substituting the value of $$\overline{x}$$, we get, $$\Rightarrow \sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}}{9}}$$ - (iv) Now, to find the value of the above equation, we have to consider equation (ii). Therefore, we have, $$\Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-5 \right)}^{2}}}=45$$ This can be written as: - $$\Rightarrow \sum\limits_{i=1}^{9}{{{\left[ \left( {{x}_{i}}-6 \right)+1 \right]}^{2}}}=45$$ Applying the identity, $${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$$, we get, $$\begin{aligned} & \Rightarrow \sum\limits_{i=1}^{9}{\left[ {{\left( {{x}_{i}}-6 \right)}^{2}}+{{1}^{2}}+2\left( {{x}_{i}}-6 \right) \right]}=45 \\\ & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+\sum\limits_{i=1}^{9}{{{1}^{2}}}+2\times \sum\limits_{i=1}^{9}{\left( {{x}_{i}}-6 \right)}=45 \\\ & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+\sum\limits_{i=1}^{9}{1}+2\times \sum\limits_{i=1}^{9}{\left[ \left( {{x}_{i}}-5 \right)-1 \right]}=45 \\\ & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+\sum\limits_{i=1}^{9}{1}+2\sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}-2\sum\limits_{i=1}^{9}{1}=45 \\\ \end{aligned}$$ Substituting the value of $$\sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}$$ in the above relation, we get, $$\begin{aligned} & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+1\times 9+2\times 9-2\times 9=45 \\\ & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+9=45 \\\ & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}=36 \\\ \end{aligned}$$ Substituting the above value in equation (iv), we get, $$\Rightarrow \sigma =\sqrt{\dfrac{36}{9}}$$ $$\Rightarrow \sigma =\sqrt{4}$$ $$\Rightarrow \sigma =2$$ **So, the correct answer is “Option a”.** **Note:** One may note that we have written $$\left( {{x}_{i}}-5 \right)$$ as $$\left( {{x}_{i}}-6+1 \right)$$. This is because in the expression of standard deviation we were required to find the value of $$\sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}$$. This was only possible when we used the above transformation. You must remember that, never try to break the term $$\sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-5 \right)}^{2}}}$$ into 9 terms by removing the summation sign. This will lead us to some difficult calculations.