Question

A data consists of n observations: \({x_1},{x_2}, \ldots .,{x_{\rm{n}}}.{\rm{\;If\;}}\mathop \sum \limits_{i = 1}^{\rm{n}} {\left( {{x_i} + 1} \right)^2} = 9{\rm{n\;and\;}}\mathop \sum \limits_{i = 1}^{\rm{n}} {\left( {{x_i} - 1} \right)^2} = 5{\rm{n}}\), then the standard deviation of this data is:
1. 2
2. √5
3. 5
4. √7

Answer 1

Correct Answer - Option 2 : √5

Variance is given by,

\({σ ^2} = \frac{1}{n}\mathop \sum \limits_{i = 1}^n x_i^2 - {\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {x_i}} \right)^2}\)

\({σ ^2} = \frac{1}{n}A - \frac{1}{{{n^2}}}{B^2}\)     ----(1)

Here, \(A = \mathop \sum \limits_{i = 1}^n x_i^2{\rm{\;and\;}}B = \mathop \sum \limits_{i = 1}^n {x_i}\) 

\(\because \underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left( {{x}_{i}}+1 \right)}^{2}}=9n\)

⇒ A + n + 2B = 9n

⇒ A + 2B = 8n     ----(2)

\(\because ~\underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left( {{x}_{i}}-1 \right)}^{2}}=5n\)

⇒ A + n – 2B = 5n

⇒ A – 2B = 4n     ----(3)

Solving (2) and (3), we get

⇒ 2A = 12n

\(\Rightarrow A=\frac{12n}{2}=6n\)

Put A = 6n in equation (3)

⇒ 6n – 2B = 4n

⇒ – 2B = 4n – 6n = – 2n

⇒ B = n

Substitute A = 6n, B = n in equation (1), we get

\(\Rightarrow {{σ }^{2}}=\frac{1}{n}\times 6n-\frac{1}{{{n}^{2}}}\times {{n}^{2}}\)

= 6 – 1 = 5 

∴ Standard deviation, σ  = √5