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Question
Show that the two formulae for the standard deviation of ungrouped data.
`sigma = sqrt((x_i - barx)^2/n)` and `sigma`' = `sqrt((x^2_i)/n - barx^2)` are equivalent.
Solution
We have `(x_i - barx)^2 = (x_i^2 - 2barx x_i + barx^2)`
= `x_i^2 + -2barx x_i + barx^2`
= `x_i^2 - 2barx x_i + (barx)^2` 1
= `x_i^2 - 2 barx (nbarx) + n barx^2`
= `x_i^2 - nbarx^2`
Dividing both sides by n and taking their square root
We get `sigma = sigma`'.
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