Proving That Two Different Variance Formulas Are Equal To One Another

Prove that:

\frac { \Sigma { \left( x-\overline { x }  \right)  }^{ 2 } }{ n } =\frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n }  \right)  }^{ 2 }

Well you first identify these truths:

x=\left\{ a,\quad b,\quad c \right\} \\ \\ \therefore \quad \overline { x } =\frac { \Sigma x }{ n } =\frac { a+b+c }{ n } ,\quad \\ \\ \therefore \quad n=3\\ \\ \therefore \quad n\overline { x } =a+b+c\\ \\ \therefore \quad \frac { \Sigma { x }^{ 2 } }{ n } =\frac { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }{ n } .\\ \\ \therefore \quad \Sigma { x }^{ 2 }={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }

And next:

variance proof

The standard deviation formula looks like this:

\sigma =\sqrt { \frac { \Sigma { \left( x-\overline { x }  \right)  }^{ 2 } }{ n }  } =\sqrt { \frac { \Sigma { x }^{ 2 } }{ n } -{ \left( \frac { \Sigma x }{ n }  \right)  }^{ 2 } }

Other posts you may be interested in...