How can you find standard deviation from a probability distribution?

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How can you find standard deviation from a probability distribution?

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Answer 1 · 2018-02-10T19:10:10

$Standard deviation = \sqrt{E (X^{2}) - {(E (X))}^{2}}$ $"Standard deviation" = sqrt(E(X^2) - (E(X))^2)$

Explanation:

In a PDF, $f (x)$ $f(x)$ , the expected mean is given by $E (X)$ $E(X)$

Where $E (X) = \int_{- \infty}^{\infty} x \cdot f (x) d x$ $E(X) = int_(-oo) ^(oo) x *f(x) dx$

The variance is given by $V a r (x) = E (X^{2}) - {(E (X))}^{2}$ $Var(x) = E(X^2) - ( E(X) )^2$

Where $E (g (X)) = \int_{- \infty}^{\infty} g (x) \cdot f (x) d x$ $E(g(X) ) = int_(-oo) ^(oo) g(x) * f(x) dx$

We know

$Standard Deviation = \sqrt{Variance}$ $"Standard Deviation" = sqrt( "Variance " )$

$\Rightarrow Standard deviation = \sqrt{E (X^{2}) - {(E (X))}^{2}}$ $=> "Standard deviation" = sqrt(E(X^2) - (E(X))^2)$

Or...

$\Rightarrow Standard deviation = \sqrt{\int_{- \infty}^{\infty} x^{2} \cdot f (x) d x - {(\int_{- \infty}^{\infty} x \cdot f (x) d x)}^{2}}$ $=> "Standard deviation" =sqrt( int_(-oo) ^(oo) x^2 *f(x) dx -( int_(-oo) ^(oo) x *f(x) dx)^2$

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How can you find standard deviation from a probability distribution?

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