Arithmetic mean of 10 observations is 60 and sum of squares of deviations from 50 is 5000. What is the standard deviation of the observations?

Concept:

The standard deviation of N observations is given by: \(\sigma = \sqrt {\frac{1}{N} \times \;\mathop \sum \limits_{i = 1}^N {{\left( {{x_i} - \mu } \right)}^2}}\) where μ is the arithmetic mean.

Calculation:

Given: μ = 60, N =10 and \(\mathop \sum \limits_{i = 1}^{10} {\left( {{x_i} - 50} \right)^2} = 5000\)

As we know, \(\mu = \frac{{\mathop \sum \nolimits_{i = 1}^{10} {x_i}}}{{10\;}} = 60 \Rightarrow \;\mathop \sum \limits_{i = 1}^{10} {x_i} = 600\) 

\( \Rightarrow \mathop \sum \limits_{i = 1}^{10} {\left( {{x_i} - 50} \right)^2} = \;\mathop \sum \limits_{i = 1}^{10} x_i^2 - 100\;\mathop \sum \limits_{i = 1}^{10} {x_i} + 25000 = 5000$\) -----(1)

By substituting the value of \(\mathop \sum \limits_{i = 1}^{10} {x_i} = 600\) in equation (1), we get

\(\Rightarrow \;\mathop \sum \limits_{i = 1}^{10} x_i^2 = 40000\)       -----(2)

As we know that, the standard deviation of N observations is given by: \(\sigma = \sqrt {\frac{1}{N} \times \;\mathop \sum \limits_{i = 1}^N {{\left( {{x_i} - \mu } \right)}^2}}\) where μ is the arithmetic mean.

\(\Rightarrow {\sigma ^2} = \frac{1}{N}\; \times \;\mathop \sum \limits_{i = 1}^N {\left( {{x_i} - \mu } \right)^2}\;\)

\(\Rightarrow {\sigma ^2} = \frac{1}{N} \times \left( {\mathop \sum \limits_{i = 1}^{N\;} x_i^2 + {\mu ^2} \times \;\mathop \sum \limits_{i = 1}^N 1 - 2\mu \times \;\mathop \sum \limits_{i = 1}^N {x_i}} \right)\;\)    ----(3)

By substituting the values of μ = 60, N =10, \(\mathop \sum \limits_{i = 1}^{10} x_i^2\;and\;\mathop \sum \limits_{i = 1}^{10} {x_i}\) in equation (3), we get

\( \Rightarrow {\sigma ^2} = 400\)

⇒ σ = 20