Question

If  \(\sum_{i=1}^{18}(x_i\ -\ 8)\ =\ 3\) and  \(\sum_{i=1}^{18}(x_i\ -\ 8)^2\ =\ 45\)  then standard deviation of x₁, x₂, x₃ ...x₁₈ is:
1. \(\frac{3}{2}\)
2. √2 
3. \(\frac{5}{2}\)
4. √5 
5.

Answer 1

Correct Answer - Option 2 : √2 

Formula used:

Standard deviation:

The standard deviation of the observation set \(\rm \{x_i,i=1,2,3,\cdots\}\) is given as follows:

\(\rm σ=√{\frac{\sum\left(x_i-μ\right)^2}{N}}\) = \(\rm √{\left(\frac{\sum x_i^2}{n}\right)-\left(\frac{\sum x_i}{n}\right)^2}\)

Calculation:

Let, 

x_i - 8 = d

Given that,

\(\sum_{i=1}^{18}d\ =\ 3\)     ----(1)

\(\sum_{i=1}^{18}d^2\ =\ 45\)    ----(2)

We know that, standard deviation 

σ = \(\rm √{\left(\frac{\sum x_i^2}{n}\right)-\left(\frac{\sum x_i}{n}\right)^2}\)

⇒ σ = \(\rm √{\left(\frac{45}{18}\right)-\left(\frac{3}{18}\right)^2}\)

⇒ σ = \(√{\frac{36}{18}}\) 

⇒ σ = √2