Question

A data consists of n observations: \[{{x}_{1}},{{x}_{2}},....,{{x}_{n}}\]. If \[\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}+1 \right)}^{2}}}=9n\] and \[\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-1 \right)}^{2}}}=5n\], then the standard deviation of this data is: -
(a) 5
(b) \[\sqrt{5}\]
(c) \[\sqrt{7}\]
(d) 2

Answer 1

Hint: Apply the identity: - \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] to break the terms of \[{{\left( {{x}_{i}}+1 \right)}^{2}}\] and assume the summation of this as equation (i). Now, apply the identity, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] to break the terms of \[{{\left( {{x}_{i}}-1 \right)}^{2}}\] and assume the summation of this as equation (ii). Subtract equation (ii) from equation (i) and find the value of \[\overline{x}\], which is the mean of the given ‘n’ observation. Now, apply the formula: - \[\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}}\], where ‘\[\sigma \]’ is the standard deviation, to find the value of standard deviation.

Complete step-by-step solution:
We have been provided with two relations: - \[\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}+1 \right)}^{2}}}=9n\] and \[\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-1 \right)}^{2}}}=5n\]. We have to find standard deviation.
Consider the first relation: -
\[\Rightarrow \] \[\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}+1 \right)}^{2}}}=9n\]
Applying the identity: - \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], we get,
\[\Rightarrow \sum\limits_{i=1}^{n}{x_{i}^{2}+{{1}^{2}}+2{{x}_{i}}}=9n\]
\[\Rightarrow \sum\limits_{i=1}^{n}{\left( x_{i}^{2}+1+2{{x}_{i}} \right)}=9n\] - (i)
Now, let us consider the second relation: -
\[\Rightarrow \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-1 \right)}^{2}}}=5n\]
Applying the identity: - \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we get,
\[\Rightarrow \sum\limits_{i=1}^{n}{\left( x_{i}^{2}+{{1}^{2}}-2{{x}_{i}} \right)}=5n\]
\[\Rightarrow \sum\limits_{i=1}^{n}{\left( x_{i}^{2}+{{1}^{2}}-2{{x}_{i}} \right)}=5n\] - (ii)
Substituting equation (i) and (ii), we get,
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{n}{\left( x_{i}^{2}+1+2{{x}_{i}} \right)}-\sum\limits_{i=1}^{n}{\left( x_{i}^{2}+1-2{{x}_{i}} \right)}=9n-5n \\
 & \Rightarrow \sum\limits_{i=1}^{n}{\left[ \left( x_{i}^{2}+1+2{{x}_{i}} \right)-\left( x_{i}^{2}+1-2{{x}_{i}} \right) \right]}=9n-5 \\
 & \Rightarrow \sum\limits_{i=1}^{n}{\left[ \left( x_{i}^{2}+1+2{{x}_{i}}-x_{i}^{2}-1+2{{x}_{i}} \right) \right]=4n} \\
\end{align}\]
Cancelling the like terms, we get,
\[\begin{align}
  & \Rightarrow \sum\limits_{i=1}^{n}{4{{x}_{i}}}=4n \\
 & \Rightarrow 4\sum\limits_{i=1}^{n}{{{x}_{i}}}=4n \\
 & \Rightarrow \sum\limits_{i=1}^{n}{{{x}_{i}}}=n \\
 & \Rightarrow \dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}=1 \\
\end{align}\]
Since, mean of ‘n’ observation is \[\overline{x}=\sum\limits_{i=1}^{n}{\dfrac{{{x}_{i}}}{n}}\]. Therefore, the above relation becomes: -
\[\Rightarrow \overline{x}=1\] - (iii)
We know that standard deviation of ‘n’ observations are given by: - \[\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}}\], where ‘\[\sigma \]’ is the standard deviation.
\[\Rightarrow \sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}}\]
Now, substituting \[\overline{x}=1\], from equation (iii), we get,
\[\Rightarrow \sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n}}\]
Substituting the value of \[\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-1 \right)}^{2}}}\] from equation (ii), we get,
\[\begin{align}
  & \Rightarrow \sigma =\sqrt{\dfrac{5n}{n}} \\
 & \Rightarrow \sigma =\sqrt{5} \\
\end{align}\]
Hence, option (b) is the correct answer.

Note: One may note that we have subtracted equation (ii) from equation (i). This is because we had to find the mean of the ‘n’ observations. If we add the two equations then the term \[{{x}_{i}}\] will get canceled and the term \[x_{i}^{2}\] will be left. But we need the term \[{{x}_{i}}\] for mean value. So, the addition was not preferred. This value of mean was necessary for the calculation of standard deviation.