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

Hint: 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,
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}}-5\times 9=9 \\
 & \Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}}=9\times 5+9 \\
 & \Rightarrow \sum\limits_{i=1}^{9}{{{x}_{i}}}=54 \\
\end{align}\]
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,
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{9}{\left[ {{\left( {{x}_{i}}-6 \right)}^{2}}+{{1}^{2}}+2\left( {{x}_{i}}-6 \right) \right]}=45 \\
 & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+\sum\limits_{i=1}^{9}{{{1}^{2}}}+2\times \sum\limits_{i=1}^{9}{\left( {{x}_{i}}-6 \right)}=45 \\
 & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+\sum\limits_{i=1}^{9}{1}+2\times \sum\limits_{i=1}^{9}{\left[ \left( {{x}_{i}}-5 \right)-1 \right]}=45 \\
 & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+\sum\limits_{i=1}^{9}{1}+2\sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}-2\sum\limits_{i=1}^{9}{1}=45 \\
\end{align}\]
Substituting the value of \[\sum\limits_{i=1}^{9}{\left( {{x}_{i}}-5 \right)}\] in the above relation, we get,
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+1\times 9+2\times 9-2\times 9=45 \\
 & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}+9=45 \\
 & \Rightarrow \sum\limits_{i=1}^{9}{{{\left( {{x}_{i}}-6 \right)}^{2}}}=36 \\
\end{align}\]
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.