514507502497495506458478463513sample standard deviation

GarethPrice asked Jan 5, 2023

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GarethPrice asked Jan 5, 2023

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23 votes

The sample standard deviation is given by:

$s=\sqrt[]{(1)/(N)\sum^N_(i=1)(x_i-\bar{x})^2}$

Where:

$\begin{gathered} \bar{x}=(514+507+502+497+495+506+458+478+463+513)/(10) \\ \bar{x}=493.3 \end{gathered}$

Therefore:

$\begin{gathered} s=\sqrt[]{(1)/(10)\sum ^(10)_(i=1)(x_i-493.3)^2} \\ s\approx20.155 \end{gathered}$

Erf answered Jan 8, 2023

by Erf

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