If the standard deviation of the data 1, 3, 5,7, ... , 2017 is σ then σ is (where . denotes the greatest interger function)

Solution:

Let $\bar{X}$ be the mean of the given data. Then,

$\bar{X} = \frac{1 + 3 + 5 + 7 + \dots + 2017}{109} = \frac{(109)^{2}}{109} = 109$

$\begin{array}{ll} ∴ Variance (σ^{2}) = \frac{1}{1009} (1^{2} + 3^{2} + \dots + 2017^{2}) - (1009)^{2} \\ \Rightarrow σ^{2} = 339360 \\ ∴ σ = 582.54 \end{array}$

Hence, $[σ] = 582$

If the standard deviation of the data 1, 3, 5,7, ... , 2017 is $σ$ then $[σ]$ is (where $[.]$ denotes the greatest interger function)

Solution:

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